id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-2600 | Additive Function is Strongly Additive | Let $\SS$ be an algebra of sets.
Let $f: \SS \to \overline \R$ be an additive function on $\SS$.
Then $f$ is also strongly additive.
That is:
:$\forall A, B \in \SS: \map f {A \cup B} + \map f {A \cap B} = \map f A + \map f B$ | From Set Difference and Intersection form Partition:
:$A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$
:$B$ is the union of the two disjoint sets $B \setminus A$ and $A \cap B$.
So, by the definition of additive function:
:$\map f A = \map f {A \setminus B} + \map f {A \cap B}$
:$\map f B = \map... | Let $\SS$ be an [[Definition:Algebra of Sets|algebra of sets]].
Let $f: \SS \to \overline \R$ be an [[Definition:Additive Function (Measure Theory)|additive function]] on $\SS$.
Then $f$ is also [[Definition:Strongly Additive Function|strongly additive]].
That is:
:$\forall A, B \in \SS: \map f {A \cup B} + \map f... | From [[Set Difference and Intersection form Partition]]:
:$A$ is the [[Definition:Set Union|union]] of the two [[Definition:Disjoint Sets|disjoint sets]] $A \setminus B$ and $A \cap B$
:$B$ is the [[Definition:Set Union|union]] of the two [[Definition:Disjoint Sets|disjoint sets]] $B \setminus A$ and $A \cap B$.
So,... | Additive Function is Strongly Additive/Proof 1 | https://proofwiki.org/wiki/Additive_Function_is_Strongly_Additive | https://proofwiki.org/wiki/Additive_Function_is_Strongly_Additive/Proof_1 | [
"Additive Function is Strongly Additive",
"Set Systems",
"Additive Functions"
] | [
"Definition:Algebra of Sets",
"Definition:Additive Function (Measure Theory)",
"Definition:Strongly Additive Function"
] | [
"Set Difference and Intersection form Partition",
"Definition:Set Union",
"Definition:Disjoint Sets",
"Definition:Set Union",
"Definition:Disjoint Sets",
"Definition:Additive Function (Measure Theory)",
"Set Difference is Disjoint with Reverse",
"Sum of Additive Function Values is Well-Defined",
"Se... |
proofwiki-2601 | Additive Function is Strongly Additive | Let $\SS$ be an algebra of sets.
Let $f: \SS \to \overline \R$ be an additive function on $\SS$.
Then $f$ is also strongly additive.
That is:
:$\forall A, B \in \SS: \map f {A \cup B} + \map f {A \cap B} = \map f A + \map f B$ | Recall that $\sqcup$ denotes the disjoint union.
First, since:
{{begin-eqn}}
{{eqn | l = A \cup B
| r = A \cup \paren {B \setminus A}
| c = Set Difference Union Second Set is Union
}}
{{eqn | r = A \sqcup \paren {B \setminus A}
| c = in view of {{Defof|Set Difference}}
}}
{{end-eqn}}
we have:
:$\paren... | Let $\SS$ be an [[Definition:Algebra of Sets|algebra of sets]].
Let $f: \SS \to \overline \R$ be an [[Definition:Additive Function (Measure Theory)|additive function]] on $\SS$.
Then $f$ is also [[Definition:Strongly Additive Function|strongly additive]].
That is:
:$\forall A, B \in \SS: \map f {A \cup B} + \map f... | Recall that $\sqcup$ denotes the [[Definition:Disjoint Union (Set Theory)|disjoint union]].
First, since:
{{begin-eqn}}
{{eqn | l = A \cup B
| r = A \cup \paren {B \setminus A}
| c = [[Set Difference Union Second Set is Union]]
}}
{{eqn | r = A \sqcup \paren {B \setminus A}
| c = in view of {{Defof|... | Additive Function is Strongly Additive/Proof 2 | https://proofwiki.org/wiki/Additive_Function_is_Strongly_Additive | https://proofwiki.org/wiki/Additive_Function_is_Strongly_Additive/Proof_2 | [
"Additive Function is Strongly Additive",
"Set Systems",
"Additive Functions"
] | [
"Definition:Algebra of Sets",
"Definition:Additive Function (Measure Theory)",
"Definition:Strongly Additive Function"
] | [
"Definition:Disjoint Union (Set Theory)",
"Set Difference Union Second Set is Union",
"Set Difference Union Intersection",
"Set Difference and Intersection are Disjoint",
"Set Difference Union Second Set is Union",
"Definition:Extended Real Addition"
] |
proofwiki-2602 | Set Difference is Disjoint with Reverse | :$\paren {S \setminus T} \cap \paren {T \setminus S} = \O$ | We assume that $S, T \subseteq \mathbb U$ where $\mathbb U$ is the universal set.
Then we can use the definition of Set Difference as Intersection with Complement.
{{begin-eqn}}
{{eqn | o =
| r = \paren {S \setminus T} \cap \paren {T \setminus S}
| c =
}}
{{eqn | r = \paren {S \cap \map \complement T} \cap... | :$\paren {S \setminus T} \cap \paren {T \setminus S} = \O$ | We assume that $S, T \subseteq \mathbb U$ where $\mathbb U$ is the [[Definition:Universal Set|universal set]].
Then we can use the definition of [[Set Difference as Intersection with Complement]].
{{begin-eqn}}
{{eqn | o =
| r = \paren {S \setminus T} \cap \paren {T \setminus S}
| c =
}}
{{eqn | r = \par... | Set Difference is Disjoint with Reverse | https://proofwiki.org/wiki/Set_Difference_is_Disjoint_with_Reverse | https://proofwiki.org/wiki/Set_Difference_is_Disjoint_with_Reverse | [
"Set Difference",
"Set Intersection"
] | [] | [
"Definition:Universal Set",
"Set Difference as Intersection with Complement",
"Set Difference as Intersection with Complement",
"Intersection is Associative",
"Intersection is Commutative",
"Intersection with Complement",
"Empty Set Disjoint with Itself",
"Category:Set Difference",
"Category:Set Int... |
proofwiki-2603 | Measure is Monotone | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is monotone, that is:
:$\forall E, F \in \Sigma: E \subseteq F \implies \map \mu E \le \map \mu F$ | A direct corollary of Non-Negative Additive Function is Monotone.
{{qed}} | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Then $\mu$ is [[Definition:Monotone (Measure Theory)|monotone]], that is:
:$\forall E, F \in \Sigma: E \subseteq F \implies \map \mu E \le \map \mu F$ | A direct [[Definition:Corollary|corollary]] of [[Non-Negative Additive Function is Monotone]].
{{qed}} | Measure is Monotone | https://proofwiki.org/wiki/Measure_is_Monotone | https://proofwiki.org/wiki/Measure_is_Monotone | [
"Measures",
"Measure is Monotone"
] | [
"Definition:Measure Space",
"Definition:Monotone (Measure Theory)"
] | [
"Definition:Corollary",
"Non-Negative Additive Function is Monotone"
] |
proofwiki-2604 | Measure is Monotone | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is monotone, that is:
:$\forall E, F \in \Sigma: E \subseteq F \implies \map \mu E \le \map \mu F$ | From Set Difference Union Second Set is Union:
:$A \cup B = \paren {B \setminus A} \cup A$
From Set Difference Intersection with Second Set is Empty Set:
:$\paren {B \setminus A} \cap A = \O$
From the Addition Law of Probability:
:$\map \Pr {A \cup B} = \map \Pr {B \setminus A} + \map \Pr A$
From Union with Superset is... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Then $\mu$ is [[Definition:Monotone (Measure Theory)|monotone]], that is:
:$\forall E, F \in \Sigma: E \subseteq F \implies \map \mu E \le \map \mu F$ | From [[Set Difference Union Second Set is Union]]:
:$A \cup B = \paren {B \setminus A} \cup A$
From [[Set Difference Intersection with Second Set is Empty Set]]:
:$\paren {B \setminus A} \cap A = \O$
From the [[Addition Law of Probability]]:
:$\map \Pr {A \cup B} = \map \Pr {B \setminus A} + \map \Pr A$
From [[Uni... | Probability Measure is Monotone/Proof 1 | https://proofwiki.org/wiki/Measure_is_Monotone | https://proofwiki.org/wiki/Probability_Measure_is_Monotone/Proof_1 | [
"Measures",
"Measure is Monotone"
] | [
"Definition:Measure Space",
"Definition:Monotone (Measure Theory)"
] | [
"Set Difference Union Second Set is Union",
"Set Difference Intersection with Second Set is Empty Set",
"Addition Law of Probability",
"Union with Superset is Superset",
"Definition:Probability Measure"
] |
proofwiki-2605 | Measure is Monotone | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is monotone, that is:
:$\forall E, F \in \Sigma: E \subseteq F \implies \map \mu E \le \map \mu F$ | As by definition a probability measure is a measure, we can directly use the result Measure is Monotone.
{{qed}} | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Then $\mu$ is [[Definition:Monotone (Measure Theory)|monotone]], that is:
:$\forall E, F \in \Sigma: E \subseteq F \implies \map \mu E \le \map \mu F$ | As by definition a [[Definition:Probability Measure|probability measure]] is a [[Definition:Measure (Measure Theory)|measure]], we can directly use the result [[Measure is Monotone]].
{{qed}} | Probability Measure is Monotone/Proof 2 | https://proofwiki.org/wiki/Measure_is_Monotone | https://proofwiki.org/wiki/Probability_Measure_is_Monotone/Proof_2 | [
"Measures",
"Measure is Monotone"
] | [
"Definition:Measure Space",
"Definition:Monotone (Measure Theory)"
] | [
"Definition:Probability Measure",
"Definition:Measure (Measure Theory)",
"Measure is Monotone"
] |
proofwiki-2606 | Countably Additive Function also Finitely Additive | Let $\AA$ be a $\sigma$-algebra.
Let $\overline \R$ denote the extended set of real numbers.
Let $f: \AA \to \overline \R$ be a countably additive function.
Then $f$ is a finitely additive function. | We have that $f$ is defined as countably additive {{iff}}:
:$\ds \map f {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map f {A_i}$
where $\sequence {A_i}$ is any sequence of pairwise disjoint elements of $\AA$.
We need to show that:
:$\ds \forall n \in \N: \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \m... | Let $\AA$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]].
Let $\overline \R$ denote the [[Definition:Extended Real Number Line|extended set of real numbers]].
Let $f: \AA \to \overline \R$ be a [[Definition:Countably Additive Function|countably additive function]].
Then $f$ is a [[Definition:Finitely Additive F... | We have that $f$ is defined as [[Definition:Countably Additive Function|countably additive]] {{iff}}:
:$\ds \map f {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map f {A_i}$
where $\sequence {A_i}$ is any [[Definition:Sequence|sequence]] of [[Definition:Pairwise Disjoint|pairwise disjoint]] elements of $\... | Countably Additive Function also Finitely Additive | https://proofwiki.org/wiki/Countably_Additive_Function_also_Finitely_Additive | https://proofwiki.org/wiki/Countably_Additive_Function_also_Finitely_Additive | [
"Set Systems",
"Measure Theory"
] | [
"Definition:Sigma-Algebra",
"Definition:Extended Real Number Line",
"Definition:Countably Additive Function",
"Definition:Additive Function (Measure Theory)"
] | [
"Definition:Countably Additive Function",
"Definition:Sequence",
"Definition:Pairwise Disjoint",
"Definition:Constant Mapping",
"Countably Additive Function Dichotomy by Empty Set",
"Definition:Natural Numbers",
"Definition:Sequence",
"Definition:Pairwise Disjoint",
"Definition:Element",
"Category... |
proofwiki-2607 | Countably Additive Function of Null Set | Let $\AA$ be a $\sigma$-algebra.
Let $\overline \R$ denote the extended set of real numbers.
Let $f: \AA \to \overline \R$ be a function be a countably additive function.
Suppose that there exists at least one $A \in \AA$ where $\map f A$ is a finite number.
Then:
: $\map f \O = 0$ | Suppose that $A \in \AA$ such that $\map f A$ is a finite number.
So, let $\map f A = x$.
Consider the sequence $\sequence {S_i} \subseteq \AA$ defined as:
:$\forall i \in \N: S_i = \begin{cases} A & : i = 0 \\ \O & : i > 0 \end{cases}$
Then:
:$\ds \bigcup_{i \mathop \ge 0} S_i = A$
Hence:
{{begin-eqn}}
{{eqn | l = ... | Let $\AA$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]].
Let $\overline \R$ denote the [[Definition:Extended Real Number Line|extended set of real numbers]].
Let $f: \AA \to \overline \R$ be a [[Definition:Mapping|function]] be a [[Definition:Countably Additive Function|countably additive function]].
Suppose t... | Suppose that $A \in \AA$ such that $\map f A$ is a [[Definition:Finite Extended Real Number|finite number]].
So, let $\map f A = x$.
Consider the [[Definition:Sequence|sequence]] $\sequence {S_i} \subseteq \AA$ defined as:
:$\forall i \in \N: S_i = \begin{cases} A & : i = 0 \\ \O & : i > 0 \end{cases}$
Then:
:$\... | Countably Additive Function of Null Set | https://proofwiki.org/wiki/Countably_Additive_Function_of_Null_Set | https://proofwiki.org/wiki/Countably_Additive_Function_of_Null_Set | [
"Set Systems",
"Countably Additive Functions"
] | [
"Definition:Sigma-Algebra",
"Definition:Extended Real Number Line",
"Definition:Mapping",
"Definition:Countably Additive Function",
"Definition:Finite Extended Real Number"
] | [
"Definition:Finite Extended Real Number",
"Definition:Sequence"
] |
proofwiki-2608 | Non-Negative Additive Function is Monotone | Let $\SS$ be an algebra of sets.
Let $f: \SS \to \overline \R$ be an additive function, that is:
:$\forall A, B \in \SS: A \cap B = \O \implies \map f {A \cup B} = \map f A + \map f B$
If $\forall A \in \SS: \map f A \ge 0$, then $f$ is monotone, that is:
:$A \subseteq B \implies \map f A \le \map f B$ | Let $A \subseteq B$.
Then:
{{begin-eqn}}
{{eqn | l = B
| r = \paren {B \setminus A} \cup \paren {A \cap B}
| c = Set Difference Union Intersection
}}
{{eqn | r = \paren {B \setminus A} \cup A
| c = Intersection with Subset is Subset
}}
{{eqn | l = \O
| r = \paren {B \setminus A} \cap A
| c... | Let $\SS$ be an [[Definition:Algebra of Sets|algebra of sets]].
Let $f: \SS \to \overline \R$ be an [[Definition:Additive Function (Measure Theory)|additive function]], that is:
:$\forall A, B \in \SS: A \cap B = \O \implies \map f {A \cup B} = \map f A + \map f B$
If $\forall A \in \SS: \map f A \ge 0$, then $f$ is... | Let $A \subseteq B$.
Then:
{{begin-eqn}}
{{eqn | l = B
| r = \paren {B \setminus A} \cup \paren {A \cap B}
| c = [[Set Difference Union Intersection]]
}}
{{eqn | r = \paren {B \setminus A} \cup A
| c = [[Intersection with Subset is Subset]]
}}
{{eqn | l = \O
| r = \paren {B \setminus A} \cap A... | Non-Negative Additive Function is Monotone | https://proofwiki.org/wiki/Non-Negative_Additive_Function_is_Monotone | https://proofwiki.org/wiki/Non-Negative_Additive_Function_is_Monotone | [
"Set Systems"
] | [
"Definition:Algebra of Sets",
"Definition:Additive Function (Measure Theory)",
"Definition:Monotone (Measure Theory)"
] | [
"Set Difference Union Intersection",
"Intersection with Subset is Subset",
"Set Difference Intersection with Second Set is Empty Set",
"Category:Set Systems"
] |
proofwiki-2609 | Power Set is Sigma-Algebra | The power set of a set is a sigma-algebra. | Let $S$ be a set, and let $\powerset S$ be its power set.
We have that a power set is an algebra of sets, and so:
:$(1): \quad \forall A, B \in \powerset S: A \cup B \in \powerset S$
:$(2): \quad \relcomp S A \in \powerset S$
Let $\sequence {A_i}$ be a countably infinite sequence of sets in $\powerset S$.
Then from Pow... | The [[Definition:Power Set|power set]] of a [[Definition:Set|set]] is a [[Definition:Sigma-Algebra|sigma-algebra]]. | Let $S$ be a set, and let $\powerset S$ be its [[Definition:Power Set|power set]].
We have that a [[Power Set is Algebra of Sets|power set is an algebra of sets]], and so:
:$(1): \quad \forall A, B \in \powerset S: A \cup B \in \powerset S$
:$(2): \quad \relcomp S A \in \powerset S$
Let $\sequence {A_i}$ be a [[Defi... | Power Set is Sigma-Algebra | https://proofwiki.org/wiki/Power_Set_is_Sigma-Algebra | https://proofwiki.org/wiki/Power_Set_is_Sigma-Algebra | [
"Sigma-Algebras",
"Examples of Sigma-Algebras"
] | [
"Definition:Power Set",
"Definition:Set",
"Definition:Sigma-Algebra"
] | [
"Definition:Power Set",
"Power Set is Algebra of Sets",
"Definition:Countable Set",
"Definition:Sequence",
"Definition:Set",
"Power Set is Closed under Countable Unions",
"Definition:Sigma-Algebra"
] |
proofwiki-2610 | Probability Measure on Equiprobable Outcomes | Let $\struct {\Omega, \Sigma, \Pr}$ be an equiprobability space.
Let $\card \Omega = n$.
Then:
:$\forall \omega \in \Omega: \map \Pr \omega = \dfrac 1 n$
:$\forall A \subseteq \Omega: \map \Pr A = \dfrac {\card A} n$. | By definition, $\map \Pr {\omega_i} = \map \Pr {\omega_j}$ for all $\omega_i, \omega_j \in \Omega$.
So let $\map \Pr {\omega_i} = p$.
Also, by definition of probability measure, we have:
:$\map \Pr \Omega = 1$
We have that $\set {\omega_i} \cap \set {\omega_j} = \O$ when $i \ne j$ and so, by definition of definition of... | Let $\struct {\Omega, \Sigma, \Pr}$ be an [[Definition:Equiprobability Space|equiprobability space]].
Let $\card \Omega = n$.
Then:
:$\forall \omega \in \Omega: \map \Pr \omega = \dfrac 1 n$
:$\forall A \subseteq \Omega: \map \Pr A = \dfrac {\card A} n$. | By definition, $\map \Pr {\omega_i} = \map \Pr {\omega_j}$ for all $\omega_i, \omega_j \in \Omega$.
So let $\map \Pr {\omega_i} = p$.
Also, by [[Definition:Probability Measure|definition of probability measure]], we have:
:$\map \Pr \Omega = 1$
We have that $\set {\omega_i} \cap \set {\omega_j} = \O$ when $i \ne j... | Probability Measure on Equiprobable Outcomes | https://proofwiki.org/wiki/Probability_Measure_on_Equiprobable_Outcomes | https://proofwiki.org/wiki/Probability_Measure_on_Equiprobable_Outcomes | [
"Probability Theory"
] | [
"Definition:Equiprobability Space"
] | [
"Definition:Probability Measure",
"Definition:Probability Measure",
"Definition:Cardinality",
"Definition:Iverson's Convention",
"Category:Probability Theory"
] |
proofwiki-2611 | Conditional Probability Defines Probability Space | Let $\struct {\Omega, \Sigma, \Pr}$ be a measure space.
Let $B \in \Sigma$ such that $\map \Pr B > 0$.
Let $Q: \Sigma \to \R$ be the real-valued function defined as:
:$\map Q A = \condprob A B$
where:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} }{\map \Pr B}$
is the conditional probability of $A$ given $B$.
Then $\st... | It is to be shown that $Q$ is a probability measure on $\struct {\Omega, \Sigma}$.
As $\Pr$ is a measure, we have that:
:$\forall A \in \Omega: \map Q A \ge 0$
Also, we have that:
{{begin-eqn}}
{{eqn | l = \map Q \Omega
| r = \condprob \Omega B
| c =
}}
{{eqn | r = \frac {\map \Pr {\Omega \cap B} } {\map \P... | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Measure Space|measure space]].
Let $B \in \Sigma$ such that $\map \Pr B > 0$.
Let $Q: \Sigma \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
:$\map Q A = \condprob A B$
where:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} }{... | It is to be shown that $Q$ is a [[Definition:Probability Measure|probability measure]] on $\struct {\Omega, \Sigma}$.
As $\Pr$ is a [[Definition:Measure (Measure Theory)|measure]], we have that:
:$\forall A \in \Omega: \map Q A \ge 0$
Also, we have that:
{{begin-eqn}}
{{eqn | l = \map Q \Omega
| r = \condpro... | Conditional Probability Defines Probability Space | https://proofwiki.org/wiki/Conditional_Probability_Defines_Probability_Space | https://proofwiki.org/wiki/Conditional_Probability_Defines_Probability_Space | [
"Conditional Probabilities",
"Probability Theory"
] | [
"Definition:Measure Space",
"Definition:Real-Valued Function",
"Definition:Conditional Probability",
"Definition:Probability Space"
] | [
"Definition:Probability Measure",
"Definition:Measure (Measure Theory)",
"Intersection with Universal Set",
"Definition:Disjoint Sets",
"Definition:Event",
"Intersection Distributes over Union",
"Definition:Measure (Measure Theory)"
] |
proofwiki-2612 | Bayes' Theorem | Let $\Pr$ be a probability measure on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\condprob A B$ denote the conditional probability of $A$ given $B$.
Let $\map \Pr A > 0$ and $\map \Pr B > 0$.
Then:
:$\condprob B A = \dfrac {\condprob A B \, \map \Pr B} {\map \Pr A}$ | From the definition of conditional probabilities, we have:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
:$\condprob B A = \dfrac {\map \Pr {A \cap B} } {\map \Pr A}$
After some algebra:
:$\condprob A B \, \map \Pr B = \map \Pr {A \cap B} = \condprob B A \, \map \Pr A$
Dividing both sides by $\map \Pr A... | Let $\Pr$ be a [[Definition:Probability Measure|probability measure]] on a [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
Let $\condprob A B$ denote the [[Definition:Conditional Probability|conditional probability of $A$ given $B$]].
Let $\map \Pr A > 0$ and $\map \Pr B > 0$.
Th... | From the definition of [[Definition:Conditional Probability|conditional probabilities]], we have:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
:$\condprob B A = \dfrac {\map \Pr {A \cap B} } {\map \Pr A}$
After some algebra:
:$\condprob A B \, \map \Pr B = \map \Pr {A \cap B} = \condprob B A \, \m... | Bayes' Theorem | https://proofwiki.org/wiki/Bayes'_Theorem | https://proofwiki.org/wiki/Bayes'_Theorem | [
"Bayes' Theorem",
"Bayesian Decision Theory",
"Probability Theory"
] | [
"Definition:Probability Measure",
"Definition:Probability Space",
"Definition:Conditional Probability"
] | [
"Definition:Conditional Probability"
] |
proofwiki-2613 | Pairwise Independence does not imply Independence | Just because all the events in a family of events in a probability space are pairwise independent, it does not mean that the family is independent. | Consider throwing a fair four-sided die.
This gives us an event space $\Omega = \set {1, 2, 3, 4}$, with each $\omega \in \Omega$ equally likely to occur:
:$\forall \omega \in \Omega: \map \Pr \omega = \dfrac 1 4$
Consider the set of events:
:$\SS = \set {A, B, C}$
where:
:$A = \set {1, 2}, B = \set {1, 3}, C = \set {1... | Just because all the [[Definition:Event|events]] in a [[Definition:Indexed Family of Sets|family]] of [[Definition:Event|events]] in a [[Definition:Probability Space|probability space]] are [[Definition:Pairwise Independent|pairwise independent]], it does not mean that the [[Definition:Indexed Family of Sets|family]] i... | Consider throwing a fair four-sided [[Definition:Die|die]].
This gives us an [[Definition:Event Space|event space]] $\Omega = \set {1, 2, 3, 4}$, with each $\omega \in \Omega$ equally likely to occur:
:$\forall \omega \in \Omega: \map \Pr \omega = \dfrac 1 4$
Consider the set of [[Definition:Event|events]]:
:$\SS = ... | Pairwise Independence does not imply Independence | https://proofwiki.org/wiki/Pairwise_Independence_does_not_imply_Independence | https://proofwiki.org/wiki/Pairwise_Independence_does_not_imply_Independence | [
"Independent Events"
] | [
"Definition:Event",
"Definition:Indexing Set/Family of Sets",
"Definition:Event",
"Definition:Probability Space",
"Definition:Independent Events/Pairwise Independent",
"Definition:Indexing Set/Family of Sets",
"Definition:Independent Events/General Definition"
] | [
"Definition:Die",
"Definition:Event Space",
"Definition:Event",
"Definition:Independent Events/Pairwise Independent",
"Definition:Independent Events/Pairwise Independent",
"Definition:Independent Events/General Definition"
] |
proofwiki-2614 | Event Independence is Symmetric | Let $A$ and $B$ be events in a probability space.
Let $A$ be independent of $B$.
Then $B$ is independent of $A$.
That is, '''is independent of''' is a symmetric relation. | We assume throughout that $\map \Pr A > 0$ and $\map \Pr B > 0$.
Let $A$ be independent of $B$.
Then by definition:
:$\condprob A B = \map \Pr A$
From the definition of conditional probabilities, we have:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
and also:
:$\condprob B A = \dfrac {\map \Pr {A \cap ... | Let $A$ and $B$ be [[Definition:Event|events]] in a [[Definition:Probability Space|probability space]].
Let $A$ be [[Definition:Independent Events|independent]] of $B$.
Then $B$ is [[Definition:Independent Events|independent]] of $A$.
That is, '''is [[Definition:Independent Events|independent]] of''' is a [[Definiti... | We assume throughout that $\map \Pr A > 0$ and $\map \Pr B > 0$.
Let $A$ be [[Definition:Independent Events|independent]] of $B$.
Then by definition:
:$\condprob A B = \map \Pr A$
From the definition of [[Definition:Conditional Probability|conditional probabilities]], we have:
:$\condprob A B = \dfrac {\map \Pr {A \... | Event Independence is Symmetric | https://proofwiki.org/wiki/Event_Independence_is_Symmetric | https://proofwiki.org/wiki/Event_Independence_is_Symmetric | [
"Independent Events"
] | [
"Definition:Event",
"Definition:Probability Space",
"Definition:Independent Events",
"Definition:Independent Events",
"Definition:Independent Events",
"Definition:Symmetry (Relation)"
] | [
"Definition:Independent Events",
"Definition:Conditional Probability",
"Definition:Independent Events"
] |
proofwiki-2615 | Disjoint Independent Events means One is Void | Let $A$ and $B$ be events in a probability space.
Suppose $A$ and $B$ are:
:disjoint
:independent.
Then either $\map \Pr A = 0$ or $\map \Pr B = 0$.
That is, if two events are disjoint and independent, at least one of them can't happen. | For $A$ and $B$ to be independent:
:$\map \Pr {A \cap B} = \map \Pr A \, \map \Pr B$
For $A$ and $B$ to be disjoint:
:$\map \Pr {A \cap B} = 0$
So:
:$\map \Pr A \, \map \Pr B = 0$
Hence the result.
{{qed}} | Let $A$ and $B$ be [[Definition:Event|events]] in a [[Definition:Probability Space|probability space]].
Suppose $A$ and $B$ are:
:[[Definition:Disjoint Events|disjoint]]
:[[Definition:Independent Events|independent]].
Then either $\map \Pr A = 0$ or $\map \Pr B = 0$.
That is, if two events are [[Definition:Disjoint... | For $A$ and $B$ to be [[Definition:Independent Events|independent]]:
:$\map \Pr {A \cap B} = \map \Pr A \, \map \Pr B$
For $A$ and $B$ to be [[Definition:Disjoint Events|disjoint]]:
:$\map \Pr {A \cap B} = 0$
So:
:$\map \Pr A \, \map \Pr B = 0$
Hence the result.
{{qed}} | Disjoint Independent Events means One is Void | https://proofwiki.org/wiki/Disjoint_Independent_Events_means_One_is_Void | https://proofwiki.org/wiki/Disjoint_Independent_Events_means_One_is_Void | [
"Probability Theory"
] | [
"Definition:Event",
"Definition:Probability Space",
"Definition:Disjoint Events",
"Definition:Independent Events",
"Definition:Disjoint Events",
"Definition:Independent Events",
"Definition:Can't Happen/Probability Theory"
] | [
"Definition:Independent Events",
"Definition:Disjoint Events"
] |
proofwiki-2616 | Set Difference and Intersection form Partition | Let $S$ and $T$ be sets such that:
:$S \setminus T \ne \O$
:$S \cap T \ne \O$
where $S \setminus T$ denotes set difference and $S \cap T$ denotes set intersection.
Then $S \setminus T$ and $S \cap T$ form a partition of $S$. | From Set Difference and Intersection are Disjoint:
:$S \setminus T$ and $S \cap T$ are disjoint.
Next from Set Difference Union Intersection:
:$S = \paren {S \setminus T} \cup \paren {S \cap T}$
Thus by definition $S \setminus T$ and $S \cap T$ form a partition of $S$.
{{qed}}
Category:Set Difference
Category:Set Inter... | Let $S$ and $T$ be [[Definition:Set|sets]] such that:
:$S \setminus T \ne \O$
:$S \cap T \ne \O$
where $S \setminus T$ denotes [[Definition:Set Difference|set difference]] and $S \cap T$ denotes [[Definition:Set Intersection|set intersection]].
Then $S \setminus T$ and $S \cap T$ form a [[Definition:Set Partition|par... | From [[Set Difference and Intersection are Disjoint]]:
:$S \setminus T$ and $S \cap T$ are [[Definition:Disjoint Sets|disjoint]].
Next from [[Set Difference Union Intersection]]:
:$S = \paren {S \setminus T} \cup \paren {S \cap T}$
Thus by definition $S \setminus T$ and $S \cap T$ form a [[Definition:Set Partition|... | Set Difference and Intersection form Partition | https://proofwiki.org/wiki/Set_Difference_and_Intersection_form_Partition | https://proofwiki.org/wiki/Set_Difference_and_Intersection_form_Partition | [
"Set Difference",
"Set Intersection",
"Set Partitions",
"Set Difference and Intersection form Partition"
] | [
"Definition:Set",
"Definition:Set Difference",
"Definition:Set Intersection",
"Definition:Set Partition"
] | [
"Set Difference and Intersection are Disjoint",
"Definition:Disjoint Sets",
"Set Difference Union Intersection",
"Definition:Set Partition",
"Category:Set Difference",
"Category:Set Intersection",
"Category:Set Partitions",
"Category:Set Difference and Intersection form Partition"
] |
proofwiki-2617 | Independent Events are Independent of Complement | Let $A$ and $B$ be events in a probability space $\struct {\Omega, \Sigma, \Pr}$.
Then $A$ and $B$ are independent {{iff}} $A$ and $\Omega \setminus B$ are independent. | For $A$ and $B$ to be independent:
:$\map \Pr {A \cap B} = \map \Pr A \map \Pr B$
We need to show that:
:$\map \Pr {A \cap \paren {\Omega \setminus B} } = \map \Pr A \map \Pr {\Omega \setminus B}$
First note that $\Omega \setminus B \equiv \relcomp \Omega B$ where $\complement_\Omega$ denotes the relative complement.
F... | Let $A$ and $B$ be [[Definition:Event|events]] in a [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
Then $A$ and $B$ are [[Definition:Independent Events|independent]] {{iff}} $A$ and $\Omega \setminus B$ are [[Definition:Independent Events|independent]]. | For $A$ and $B$ to be [[Definition:Independent Events|independent]]:
:$\map \Pr {A \cap B} = \map \Pr A \map \Pr B$
We need to show that:
:$\map \Pr {A \cap \paren {\Omega \setminus B} } = \map \Pr A \map \Pr {\Omega \setminus B}$
First note that $\Omega \setminus B \equiv \relcomp \Omega B$ where $\complement_\Omeg... | Independent Events are Independent of Complement | https://proofwiki.org/wiki/Independent_Events_are_Independent_of_Complement | https://proofwiki.org/wiki/Independent_Events_are_Independent_of_Complement | [
"Independent Events"
] | [
"Definition:Event",
"Definition:Probability Space",
"Definition:Independent Events",
"Definition:Independent Events"
] | [
"Definition:Independent Events",
"Definition:Relative Complement",
"Set Difference as Intersection with Relative Complement",
"Set Difference and Intersection form Partition",
"Axiom:Kolmogorov Axioms",
"Definition:Independent Events",
"Elementary Properties of Probability Measure",
"Definition:Indepe... |
proofwiki-2618 | Probability of Independent Events Not Happening | Let $\EE = \struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $A_1, A_2, \ldots, A_m \in \Sigma$ be independent events in the event space of $\EE$.
Then the probability of none of $A_1$ to $A_m$ occurring is:
:$\ds \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$ | Let $A_1, A_2, \ldots, A_m \in \Sigma$ be independent events.
From Independent Events are Independent of Complement, we have that $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m \in \Sigma$ are also independent.
From the definition of occurrence, if $A$ does not happen then $\Omega \setminus A... | Let $\EE = \struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $A_1, A_2, \ldots, A_m \in \Sigma$ be [[Definition:Independent Events|independent events]] in the [[Definition:Event Space|event space]] of $\EE$.
Then the [[Definition:Probability|probability]] of none of $A_1$ to... | Let $A_1, A_2, \ldots, A_m \in \Sigma$ be [[Definition:Independent Events|independent events]].
From [[Independent Events are Independent of Complement]], we have that $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m \in \Sigma$ are also [[Definition:Independent Events|independent]].
From th... | Probability of Independent Events Not Happening | https://proofwiki.org/wiki/Probability_of_Independent_Events_Not_Happening | https://proofwiki.org/wiki/Probability_of_Independent_Events_Not_Happening | [
"Independent Events"
] | [
"Definition:Probability Space",
"Definition:Independent Events",
"Definition:Event Space",
"Definition:Probability",
"Definition:Event/Occurrence"
] | [
"Definition:Independent Events",
"Independent Events are Independent of Complement",
"Definition:Independent Events",
"Definition:Event/Occurrence",
"Elementary Properties of Probability Measure",
"Definition:Probability",
"Definition:Event/Occurrence"
] |
proofwiki-2619 | Probability of Occurrence of At Least One Independent Event | Let $\EE = \struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $A_1, A_2, \ldots, A_m \in \Sigma$ be independent events in the event space of $\EE$.
Then the probability of at least one of $A_1$ to $A_m$ occurring is:
:$\ds 1 - \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$ | Follows as a direct result of Probability of Independent Events Not Happening.
Let $B$ be the event "None of $A_1$ to $A_m$ happen".
From Probability of Independent Events Not Happening:
:$\ds \map \Pr B = \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$
Then $\Omega \setminus B$ is the event "''Not'' none of $A_1... | Let $\EE = \struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $A_1, A_2, \ldots, A_m \in \Sigma$ be [[Definition:Independent Events|independent events]] in the [[Definition:Event Space|event space]] of $\EE$.
Then the [[Definition:Probability|probability]] of at least one of ... | Follows as a direct result of [[Probability of Independent Events Not Happening]].
Let $B$ be the event "None of $A_1$ to $A_m$ happen".
From [[Probability of Independent Events Not Happening]]:
:$\ds \map \Pr B = \prod_{i \mathop = 1}^m \paren {1 - \map \Pr {A_i} }$
Then $\Omega \setminus B$ is the event "''Not'' ... | Probability of Occurrence of At Least One Independent Event | https://proofwiki.org/wiki/Probability_of_Occurrence_of_At_Least_One_Independent_Event | https://proofwiki.org/wiki/Probability_of_Occurrence_of_At_Least_One_Independent_Event | [
"Probability Theory"
] | [
"Definition:Probability Space",
"Definition:Independent Events",
"Definition:Event Space",
"Definition:Probability",
"Definition:Event/Occurrence",
"Definition:Event",
"Definition:Event Space",
"Definition:Probability Space",
"Definition:Independent Events",
"Definition:Probability",
"Definition... | [
"Probability of Independent Events Not Happening",
"Probability of Independent Events Not Happening",
"Elementary Properties of Probability Measure"
] |
proofwiki-2620 | Total Probability Theorem | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\set {B_1, B_2, \ldots}$ be a partition of $\Omega$ such that $\forall i: \map \Pr {B_i} > 0$.
Then:
:$\ds \forall A \in \Sigma: \map \Pr A = \sum_i \condprob A {B_i} \map \Pr {B_i}$ | {{begin-eqn}}
{{eqn | l = \map \Pr A
| r = \map \Pr {A \cap \paren {\bigcup_i B_i} }
| c = Intersection with Subset is Subset: $\ds \bigcup_i B_i = \Omega$ and $A \subseteq \Omega$
}}
{{eqn | r = \map \Pr {\bigcup_i \paren {A \cap B_i} }
| c = Intersection Distributes over Union
}}
{{eqn | r = \sum_i ... | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $\set {B_1, B_2, \ldots}$ be a [[Definition:Partition (Probability Theory)|partition of $\Omega$]] such that $\forall i: \map \Pr {B_i} > 0$.
Then:
:$\ds \forall A \in \Sigma: \map \Pr A = \sum_i \condprob A {B_i} \map \... | {{begin-eqn}}
{{eqn | l = \map \Pr A
| r = \map \Pr {A \cap \paren {\bigcup_i B_i} }
| c = [[Intersection with Subset is Subset]]: $\ds \bigcup_i B_i = \Omega$ and $A \subseteq \Omega$
}}
{{eqn | r = \map \Pr {\bigcup_i \paren {A \cap B_i} }
| c = [[Intersection Distributes over Union]]
}}
{{eqn | r =... | Total Probability Theorem | https://proofwiki.org/wiki/Total_Probability_Theorem | https://proofwiki.org/wiki/Total_Probability_Theorem | [
"Probability Theory",
"Named Theorems",
"Total Probability Theorem"
] | [
"Definition:Probability Space",
"Definition:Partition (Probability Theory)"
] | [
"Intersection with Subset is Subset",
"Intersection Distributes over Union",
"Definition:Disjoint Events"
] |
proofwiki-2621 | Boole's Inequality | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $A_1, A_2, \ldots, A_n$ be events in $\Sigma$.
Then:
:$\ds \map \Pr {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map \Pr {A_i}$ | A direct consequence of the facts that:
:a Probability Measure is Subadditive
:the result Finite Union of Sets in Subadditive Function which gives:
::$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map f {A_i}$
:for a subadditive function $f$.
{{qed}} | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $A_1, A_2, \ldots, A_n$ be [[Definition:Event|events]] in $\Sigma$.
Then:
:$\ds \map \Pr {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map \Pr {A_i}$ | A direct consequence of the facts that:
:a [[Probability Measure is Subadditive]]
:the result [[Finite Union of Sets in Subadditive Function]] which gives:
::$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map f {A_i}$
:for a [[Definition:Subadditive Function (Measure Theory)|subadditive funct... | Boole's Inequality | https://proofwiki.org/wiki/Boole's_Inequality | https://proofwiki.org/wiki/Boole's_Inequality | [
"Boole's Inequality",
"Unions of Events"
] | [
"Definition:Probability Space",
"Definition:Event"
] | [
"Probability Measure is Subadditive",
"Finite Union of Sets in Subadditive Function",
"Definition:Subadditive Function (Measure Theory)"
] |
proofwiki-2622 | Additive Nowhere Negative Function is Subadditive | Let $\AA$ be an algebra of sets.
Let $f: \AA \to \overline \R$ be an additive function such that:
:$\forall A \in \AA: \map f A \ge 0$
Then $f$ is subadditive. | If $f$ is additive then by Additive Function is Strongly Additive:
:$\forall A, B \in \AA: \map f {A \cup B} = \map f A + \map f B - \map f {A \cap B}$
As $\map f {A \cap B} \ge 0$, the result follows by definition of subadditive:
:$\forall A, B \in \AA: \map f {A \cup B} \le \map f A + \map f B$
{{qed}}
Category:Set S... | Let $\AA$ be an [[Definition:Algebra of Sets|algebra of sets]].
Let $f: \AA \to \overline \R$ be an [[Definition:Additive Function (Measure Theory)|additive function]] such that:
:$\forall A \in \AA: \map f A \ge 0$
Then $f$ is [[Definition:Subadditive Function (Measure Theory)|subadditive]]. | If $f$ is [[Definition:Additive Function (Measure Theory)|additive]] then by [[Additive Function is Strongly Additive]]:
:$\forall A, B \in \AA: \map f {A \cup B} = \map f A + \map f B - \map f {A \cap B}$
As $\map f {A \cap B} \ge 0$, the result follows by definition of [[Definition:Subadditive Function (Measure The... | Additive Nowhere Negative Function is Subadditive | https://proofwiki.org/wiki/Additive_Nowhere_Negative_Function_is_Subadditive | https://proofwiki.org/wiki/Additive_Nowhere_Negative_Function_is_Subadditive | [
"Set Systems"
] | [
"Definition:Algebra of Sets",
"Definition:Additive Function (Measure Theory)",
"Definition:Subadditive Function (Measure Theory)"
] | [
"Definition:Additive Function (Measure Theory)",
"Additive Function is Strongly Additive",
"Definition:Subadditive Function (Measure Theory)",
"Category:Set Systems"
] |
proofwiki-2623 | Probability Measure is Subadditive | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Then $\Pr$ is a subadditive function. | By definition, a probability measure is a measure.
The result follows from Measure is Subadditive
{{qed}}
Category:Probability Measures
dfid6kx4vypxcmoju6wzculb0u571x3 | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Then $\Pr$ is a [[Definition:Subadditive Function (Measure Theory)|subadditive function]]. | By definition, a [[Definition:Probability Measure|probability measure]] is a [[Definition:Measure (Measure Theory)|measure]].
The result follows from [[Measure is Subadditive]]
{{qed}}
[[Category:Probability Measures]]
dfid6kx4vypxcmoju6wzculb0u571x3 | Probability Measure is Subadditive | https://proofwiki.org/wiki/Probability_Measure_is_Subadditive | https://proofwiki.org/wiki/Probability_Measure_is_Subadditive | [
"Probability Theory",
"Measure Theory",
"Probability Measures",
"Probability Measures"
] | [
"Definition:Probability Space",
"Definition:Subadditive Function (Measure Theory)"
] | [
"Definition:Probability Measure",
"Definition:Measure (Measure Theory)",
"Measure is Subadditive",
"Category:Probability Measures"
] |
proofwiki-2624 | Inclusion-Exclusion Principle | Let $\SS$ be an algebra of sets.
Let $A_1, A_2, \ldots, A_n$ be finite sets.
Let $f: \SS \to \R$ be an additive function.
Then:
{{begin-eqn}}
{{eqn | l = \map f {\bigcup_{i \mathop = 1}^n A_i}
| r = \sum_{i \mathop = 1}^n \map f {A_i}
| c =
}}
{{eqn | o =
| ro= -
| r = \sum_{1 \mathop \le i \m... | Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
{{begin-eqn}}
{{eqn | l = \map f {\bigcup_{i \mathop = 1}^n A_i}
| r = \sum_{i \mathop = 1}^n \map f {A_i}
| c =
}}
{{eqn | o =
| ro= -
| r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {A_i \cap A_j}
... | Let $\SS$ be an [[Definition:Algebra of Sets|algebra of sets]].
Let $A_1, A_2, \ldots, A_n$ be [[Definition:Finite|finite sets]].
Let $f: \SS \to \R$ be an [[Definition:Additive Function (Measure Theory)|additive function]].
Then:
{{begin-eqn}}
{{eqn | l = \map f {\bigcup_{i \mathop = 1}^n A_i}
| r = \sum_{i... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
{{begin-eqn}}
{{eqn | l = \map f {\bigcup_{i \mathop = 1}^n A_i}
| r = \sum_{i \mathop = 1}^n \map f {A_i}
| c =
}}
{{eqn | o =
| ro= -
| r = \sum... | Inclusion-Exclusion Principle | https://proofwiki.org/wiki/Inclusion-Exclusion_Principle | https://proofwiki.org/wiki/Inclusion-Exclusion_Principle | [
"Inclusion-Exclusion Principle",
"Set Union",
"Set Systems",
"Probability Theory",
"Combinatorics",
"Named Theorems"
] | [
"Definition:Algebra of Sets",
"Definition:Finite",
"Definition:Additive Function (Measure Theory)"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-2625 | Existence of Probability Space and Discrete Random Variable | Let $I$ be an arbitrary countable indexing set.
Let $S = \set {s_i: i \in I} \subset \R$ be a countable set of real numbers.
Let $\set {\pi_i: i \in I} \subset \R$ be a countable set of real numbers which satisfies:
:$\ds \forall i \in I: \pi_i \ge 0, \sum_{i \mathop \in I} \pi_i = 1$
Then there exists:
:a probability ... | Take $\Omega = S$ and $\Sigma = \powerset S$ (the power set of $S$).
Then let:
:$\ds \map \Pr A = \sum_{i: s_i \mathop \in A} \pi_i$
for all $A \in \Sigma$.
Then we can define $X: \Omega \to \R$ by:
:$\forall \omega \in \Omega: \map X \omega = \omega$
This suits the conditions of the assertion well enough.
{{qed}} | Let $I$ be an arbitrary [[Definition:Countable Set|countable]] [[Definition:Indexing Set|indexing set]].
Let $S = \set {s_i: i \in I} \subset \R$ be a [[Definition:Countable Set|countable set]] of [[Definition:Real Number|real numbers]].
Let $\set {\pi_i: i \in I} \subset \R$ be a [[Definition:Countable Set|countabl... | Take $\Omega = S$ and $\Sigma = \powerset S$ (the [[Definition:Power Set|power set]] of $S$).
Then let:
:$\ds \map \Pr A = \sum_{i: s_i \mathop \in A} \pi_i$
for all $A \in \Sigma$.
Then we can define $X: \Omega \to \R$ by:
:$\forall \omega \in \Omega: \map X \omega = \omega$
This suits the conditions of the asserti... | Existence of Probability Space and Discrete Random Variable | https://proofwiki.org/wiki/Existence_of_Probability_Space_and_Discrete_Random_Variable | https://proofwiki.org/wiki/Existence_of_Probability_Space_and_Discrete_Random_Variable | [
"Probability Theory"
] | [
"Definition:Countable Set",
"Definition:Indexing Set",
"Definition:Countable Set",
"Definition:Real Number",
"Definition:Countable Set",
"Definition:Real Number",
"Definition:Probability Space",
"Definition:Random Variable/Discrete",
"Definition:Probability Mass Function"
] | [
"Definition:Power Set"
] |
proofwiki-2626 | Countable Function on Power Set of Sample Space is Discrete Random Variable | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space such that $\Sigma$ is the power set of $\Omega$.
Let $f: \Omega \to \R$ be a function such that $\Img f$ is countable.
Then $f$ is a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$. | By definition, $\map {f^{-1} } x \subseteq \Omega$.
But then $\map {f^{-1} } x \in \powerset \Omega$.
Hence the result.
{{qed}} | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]] such that $\Sigma$ is the [[Definition:Power Set|power set]] of $\Omega$.
Let $f: \Omega \to \R$ be a [[Definition:Function|function]] such that $\Img f$ is [[Definition:Countable|countable]].
Then $f$ is a [[Definition:Discr... | By definition, $\map {f^{-1} } x \subseteq \Omega$.
But then $\map {f^{-1} } x \in \powerset \Omega$.
Hence the result.
{{qed}} | Countable Function on Power Set of Sample Space is Discrete Random Variable | https://proofwiki.org/wiki/Countable_Function_on_Power_Set_of_Sample_Space_is_Discrete_Random_Variable | https://proofwiki.org/wiki/Countable_Function_on_Power_Set_of_Sample_Space_is_Discrete_Random_Variable | [
"Probability Theory"
] | [
"Definition:Probability Space",
"Definition:Power Set",
"Definition:Function",
"Definition:Countable Set",
"Definition:Random Variable/Discrete"
] | [] |
proofwiki-2627 | Characteristic Function on Event is Discrete Random Variable | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $E \in \Sigma$ be any event of $\struct {\Omega, \Sigma, \Pr}$.
Let $\chi_E: \Omega \to \set {0, 1}$ be the characteristic function of $E$.
Then $\chi_E$ is a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$. | By definition of characteristic function:
:$\forall \omega \in \Omega: \chi_E = \begin{cases}
1 & : \omega \in E \\
0 & : \omega \notin E \\
\end{cases}$
Then clearly:
:$\forall x \in \R: \map { {\chi_E}^{-1} } x = \begin{cases}
E & : x = 1 \\
\Omega \setminus E & : x = 0 \\
\O & : x \notin \set {0, 1}
\end{cases}$
So ... | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $E \in \Sigma$ be any [[Definition:Event|event]] of $\struct {\Omega, \Sigma, \Pr}$.
Let $\chi_E: \Omega \to \set {0, 1}$ be the [[Definition:Characteristic Function of Set|characteristic function]] of $E$.
Then $\chi_E... | By definition of [[Definition:Characteristic Function of Set|characteristic function]]:
:$\forall \omega \in \Omega: \chi_E = \begin{cases}
1 & : \omega \in E \\
0 & : \omega \notin E \\
\end{cases}$
Then clearly:
:$\forall x \in \R: \map { {\chi_E}^{-1} } x = \begin{cases}
E & : x = 1 \\
\Omega \setminus E & : x = 0... | Characteristic Function on Event is Discrete Random Variable | https://proofwiki.org/wiki/Characteristic_Function_on_Event_is_Discrete_Random_Variable | https://proofwiki.org/wiki/Characteristic_Function_on_Event_is_Discrete_Random_Variable | [
"Probability Theory"
] | [
"Definition:Probability Space",
"Definition:Event",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Random Variable/Discrete"
] | [
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Preimage/Mapping/Element"
] |
proofwiki-2628 | Poisson Distribution Gives Rise to Probability Mass Function | Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the poisson distribution with parameter $\lambda$ (where $\lambda > 0$).
Then $X$ gives rise to a probability mass function. | By definition:
:$\Img X = \N$
:$\map \Pr {X = k} = \dfrac 1 {k!} \lambda^k e^{-\lambda}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
| r = \sum_{k \mathop \in \N} \frac 1 {k!} \lambda^k e^{-\lambda}
| c =
}}
{{eqn | r = e^{-\lambda} \sum_{k \mathop \in \N} \frac 1 {k!} \lambda^k
| c =
}}
{{eqn |... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] on a [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the [[Definition:Poisson Distribution|poisson distribution with parameter $\lambda$]] (where $\lambda > 0$).
Then $X$ gives rise to a [[... | By [[Definition:Poisson Distribution|definition]]:
:$\Img X = \N$
:$\map \Pr {X = k} = \dfrac 1 {k!} \lambda^k e^{-\lambda}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
| r = \sum_{k \mathop \in \N} \frac 1 {k!} \lambda^k e^{-\lambda}
| c =
}}
{{eqn | r = e^{-\lambda} \sum_{k \mathop \in \N} \frac 1... | Poisson Distribution Gives Rise to Probability Mass Function | https://proofwiki.org/wiki/Poisson_Distribution_Gives_Rise_to_Probability_Mass_Function | https://proofwiki.org/wiki/Poisson_Distribution_Gives_Rise_to_Probability_Mass_Function | [
"Poisson Distribution",
"Probability Mass Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Space",
"Definition:Poisson Distribution",
"Definition:Probability Mass Function"
] | [
"Definition:Poisson Distribution",
"Power Series Expansion for Exponential Function"
] |
proofwiki-2629 | Geometric Distribution Gives Rise to Probability Mass Function | Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the geometric distribution with parameter $p$ (where $0 < p < 1$).
Then $X$ gives rise to a probability mass function. | By definition:
:$\map \Omega X = \N = \set {0, 1, 2, \ldots}$
:$\map \Pr {X = k} = p^k \paren {1 - p}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
| r = \sum_{k \mathop \ge 0} p^k \paren {1 - p}
| c = {{Defof|Geometric Distribution}}
}}
{{eqn | r = \paren {1 - p} \sum_{k \mathop \ge 0} p^k
| c =
... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] on a [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the [[Definition:Geometric Distribution|geometric distribution with parameter $p$]] (where $0 < p < 1$).
Then $X$ gives rise to a [[Defin... | By [[Definition:Geometric Distribution|definition]]:
:$\map \Omega X = \N = \set {0, 1, 2, \ldots}$
:$\map \Pr {X = k} = p^k \paren {1 - p}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
| r = \sum_{k \mathop \ge 0} p^k \paren {1 - p}
| c = {{Defof|Geometric Distribution}}
}}
{{eqn | r = \paren {1 - p... | Geometric Distribution Gives Rise to Probability Mass Function | https://proofwiki.org/wiki/Geometric_Distribution_Gives_Rise_to_Probability_Mass_Function | https://proofwiki.org/wiki/Geometric_Distribution_Gives_Rise_to_Probability_Mass_Function | [
"Geometric Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Space",
"Definition:Geometric Distribution",
"Definition:Probability Mass Function"
] | [
"Definition:Geometric Distribution",
"Sum of Infinite Geometric Sequence",
"Category:Geometric Distribution"
] |
proofwiki-2630 | Negative Binomial Distribution Gives Rise to Probability Mass Function/Type 1 | Let $X$ have the Type $1$ negative binomial distribution with parameters $n$ and $p$ ($0 < p < 1$).
Then $X$ gives rise to a probability mass function. | By definition:
:$\Img X = \set {n, n + 1, n + 2, \ldots}$
:$\map \Pr {X = k} = \dbinom {k - 1} {n - 1} p^n \paren {1 - p}^{k - n}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
| r = \sum_{k \mathop \ge n} \binom {k - 1} {n - 1} p^n \paren {1 - p}^{k - n}
| c =
}}
{{eqn | r = p^n \sum_{j \mathop \ge 0} \... | Let $X$ have the [[Definition:Negative Binomial Distribution (Type 1)|Type $1$ negative binomial distribution with parameters $n$ and $p$]] ($0 < p < 1$).
Then $X$ gives rise to a [[Definition:Probability Mass Function|probability mass function]]. | By [[Definition:Negative Binomial Distribution (Type 1)|definition]]:
:$\Img X = \set {n, n + 1, n + 2, \ldots}$
:$\map \Pr {X = k} = \dbinom {k - 1} {n - 1} p^n \paren {1 - p}^{k - n}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
| r = \sum_{k \mathop \ge n} \binom {k - 1} {n - 1} p^n \paren {1 - p}^{k - ... | Negative Binomial Distribution Gives Rise to Probability Mass Function/Type 1 | https://proofwiki.org/wiki/Negative_Binomial_Distribution_Gives_Rise_to_Probability_Mass_Function/Type_1 | https://proofwiki.org/wiki/Negative_Binomial_Distribution_Gives_Rise_to_Probability_Mass_Function/Type_1 | [
"Negative Binomial Distribution Gives Rise to Probability Mass Function",
"Negative Binomial Distribution (Type 1)",
"Probability Mass Functions"
] | [
"Definition:Negative Binomial Distribution/Type 1",
"Definition:Probability Mass Function"
] | [
"Definition:Negative Binomial Distribution/Type 1",
"Negated Upper Index of Binomial Coefficient",
"Binomial Theorem"
] |
proofwiki-2631 | Coin-Tossing Modeled by Bernoulli Trial | The game of coin-tossing can be modeled as a Bernoulli trial.
This applies whether the coin is fair or biased. | When a coin is tossed, it can land either with the head side up, or the tail side up.
Not taking into account tricks and teases in which the coin is given due consideration for landing on its edge, these are the only two possible outcomes.
Let the probability that it lands heads be $p$.
Let $\EE$ be the experiment of t... | The game of [[Definition:Coin-Tossing|coin-tossing]] can be modeled as a [[Definition:Bernoulli Trial|Bernoulli trial]].
This applies whether the [[Definition:Coin|coin]] is [[Definition:Fair Coin|fair]] or [[Definition:Biased Coin|biased]]. | When a [[Definition:Coin|coin]] is [[Definition:Coin-Tossing|tossed]], it can land either with the [[Definition:Head of Coin|head]] [[Definition:Side of Coin|side]] up, or the [[Definition:Tail of Coin|tail]] [[Definition:Side of Coin|side]] up.
Not taking into account tricks and teases in which the [[Definition:Coin|... | Coin-Tossing Modeled by Bernoulli Trial | https://proofwiki.org/wiki/Coin-Tossing_Modeled_by_Bernoulli_Trial | https://proofwiki.org/wiki/Coin-Tossing_Modeled_by_Bernoulli_Trial | [
"Coin-Tossing",
"Bernoulli Trials"
] | [
"Definition:Coin/Coin-Tossing",
"Definition:Bernoulli Trial",
"Definition:Coin",
"Definition:Fair/Coin",
"Definition:Coin/Biased"
] | [
"Definition:Coin",
"Definition:Coin/Coin-Tossing",
"Definition:Coin/Head",
"Definition:Coin/Side",
"Definition:Coin/Tail",
"Definition:Coin/Side",
"Definition:Coin",
"Definition:Elementary Event",
"Definition:Probability",
"Definition:Coin/Head",
"Definition:Experiment",
"Definition:Coin/Coin-... |
proofwiki-2632 | Binomial Experiment has Binomial Distribution | Let $\sequence {X_i}$ be a finite binomial experiment of length $n$ such that each of the $X_i$ in the sequence is a Bernoulli trial with parameter $p$.
Then the number of successes in $\sequence {X_i}$ is modelled by a binomial distribution with parameters $n$ and $p$. | Consider the sample space $\Omega$ of all sequences $\sequence {X_i}$ of length $n$.
The $i$th entry of any such sequence is the result of the $i$th trial.
We have that $\Omega$ is finite.
Let us take the event space $\Sigma$ to be the power set of $\Omega$.
As the elements of $\Omega$ are independent, by definition of... | Let $\sequence {X_i}$ be a [[Definition:Finite|finite]] [[Definition:Binomial Experiment|binomial experiment]] of length $n$ such that each of the $X_i$ in the [[Definition:Sequence|sequence]] is a [[Definition:Bernoulli Trial|Bernoulli trial with parameter $p$]].
Then the number of [[Definition:Success|successes]] i... | Consider the [[Definition:Sample Space|sample space]] $\Omega$ of all [[Definition:Sequence|sequences]] $\sequence {X_i}$ of length $n$.
The $i$th entry of any such sequence is the result of the $i$th [[Definition:Bernoulli Trial|trial]].
We have that $\Omega$ is [[Definition:Finite Set|finite]].
Let us take the [[D... | Binomial Experiment has Binomial Distribution | https://proofwiki.org/wiki/Binomial_Experiment_has_Binomial_Distribution | https://proofwiki.org/wiki/Binomial_Experiment_has_Binomial_Distribution | [
"Binomial Experiments",
"Binomial Distribution"
] | [
"Definition:Finite",
"Definition:Binomial Experiment",
"Definition:Sequence",
"Definition:Bernoulli Trial",
"Definition:Bernoulli Distribution",
"Definition:Binomial Distribution"
] | [
"Definition:Sample Space",
"Definition:Sequence",
"Definition:Bernoulli Trial",
"Definition:Finite Set",
"Definition:Event Space",
"Definition:Power Set",
"Definition:Independent Events",
"Definition:Binomial Experiment",
"Definition:Bernoulli Distribution",
"Definition:Random Variable/Discrete",
... |
proofwiki-2633 | Binomial Distribution Approximated by Poisson Distribution | Let $X$ be a discrete random variable which has the binomial distribution with parameters $n$ and $p$.
Then for $\lambda = n p$, $X$ can be approximated by a Poisson distribution with parameter $\lambda$:
:$\ds \lim_{n \mathop \to \infty} \binom n k p^k \paren {1 - p}^{n - k} = \frac {\lambda^k} {k!} e^{-\lambda}$ | Let $X$ be as described.
Let $k \in \Z_{\ge 0}$ be fixed.
We write $p = \dfrac \lambda n$ and suppose that $n$ is large.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \binom n k p^k \paren {1 - p}^{n - k}
| r = \lim_{n \mathop \to \infty} \binom n k \paren {\frac \lambda n}^k \paren {1 - \frac \lamb... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] which has the [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]].
Then for $\lambda = n p$, $X$ can be approximated by a [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]]:
... | Let $X$ be as described.
Let $k \in \Z_{\ge 0}$ be fixed.
We write $p = \dfrac \lambda n$ and suppose that $n$ is large.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \binom n k p^k \paren {1 - p}^{n - k}
| r = \lim_{n \mathop \to \infty} \binom n k \paren {\frac \lambda n}^k \paren {1 - \frac ... | Binomial Distribution Approximated by Poisson Distribution | https://proofwiki.org/wiki/Binomial_Distribution_Approximated_by_Poisson_Distribution | https://proofwiki.org/wiki/Binomial_Distribution_Approximated_by_Poisson_Distribution | [
"Binomial Distribution",
"Poisson Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Poisson Distribution"
] | [
"Limit to Infinity of Binomial Coefficient over Power",
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"Combination Theorem for Limits of Functions/Product Rule"
] |
proofwiki-2634 | Bernoulli Process as Geometric Distribution | Let $\sequence {X_i}$ be a binomial experiment with parameter $p$.
Let $\EE$ be the experiment which consists of performing the Bernoulli trial $X_i$ until a failure occurs, and then stop.
Let $k$ be the number of successes before a failure is encountered.
Then $k$ is modelled by a geometric distribution with parameter... | Follows directly from the definition of geometric distribution.
Let $X$ be the discrete random variable defined as the number of successes before a failure is encountered.
Thus the last trial (and the last trial only) will be a failure, and the others will be successes.
The probability that $k$ successes are followed b... | Let $\sequence {X_i}$ be a [[Definition:Binomial Experiment|binomial experiment with parameter $p$]].
Let $\EE$ be the experiment which consists of performing the [[Definition:Bernoulli Trial|Bernoulli trial]] $X_i$ until a [[Definition:Failure|failure]] occurs, and then stop.
Let $k$ be the number of [[Definition:Su... | Follows directly from the definition of [[Definition:Geometric Distribution|geometric distribution]].
Let $X$ be the [[Definition:Discrete Random Variable|discrete random variable]] defined as the number of [[Definition:Success|successes]] before a [[Definition:Failure|failure]] is encountered.
Thus the last trial (a... | Bernoulli Process as Geometric Distribution | https://proofwiki.org/wiki/Bernoulli_Process_as_Geometric_Distribution | https://proofwiki.org/wiki/Bernoulli_Process_as_Geometric_Distribution | [
"Bernoulli Distribution",
"Geometric Distribution"
] | [
"Definition:Binomial Experiment",
"Definition:Bernoulli Trial",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Distribution",
"Definition:Geometric Distribution/Formulation 1"
] | [
"Definition:Geometric Distribution",
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Distribution"
] |
proofwiki-2635 | Function of Discrete Random Variable | Let $X$ be a discrete random variable on the probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $g: \R \to \R$ be any real function.
Then $Y = g \sqbrk X$, defined as:
:$\forall \omega \in \Omega: \map Y \omega = g \sqbrk {\map X \omega}$
is also a discrete random variable. | As $\Img X$ is countable, then so is $\Img {g \sqbrk X}$.
Now consider $g^{-1} \sqbrk Y$.
We have that:
:$\forall x \in \R: \map {X^{-1} } x \in \Sigma$
We also have that:
:$\ds \forall y \in \R: \map {g^{-1} } y = \bigcup_{x: \map g x = y} \set x$
But $\Sigma$ is a sigma-algebra and therefore closed for unions.
Thus:
... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] on the [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
Let $g: \R \to \R$ be any [[Definition:Real Function|real function]].
Then $Y = g \sqbrk X$, defined as:
:$\forall \omega \in \Omega: \map Y \omega =... | As $\Img X$ is [[Definition:Countable|countable]], then so is $\Img {g \sqbrk X}$.
Now consider $g^{-1} \sqbrk Y$.
We have that:
:$\forall x \in \R: \map {X^{-1} } x \in \Sigma$
We also have that:
:$\ds \forall y \in \R: \map {g^{-1} } y = \bigcup_{x: \map g x = y} \set x$
But $\Sigma$ is a [[Definition:Sigma-Alge... | Function of Discrete Random Variable | https://proofwiki.org/wiki/Function_of_Discrete_Random_Variable | https://proofwiki.org/wiki/Function_of_Discrete_Random_Variable | [
"Probability Theory"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Space",
"Definition:Real Function",
"Definition:Random Variable/Discrete"
] | [
"Definition:Countable Set",
"Definition:Sigma-Algebra",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Set Union"
] |
proofwiki-2636 | Expectation of Function of Discrete Random Variable | Let $X$ be a discrete random variable.
Let $\expect X$ be the expectation of $X$.
Let $g: \R \to \R$ be a real function.
Then:
:$\ds \expect {g \sqbrk X} = \sum_{x \mathop \in \Omega_X} \map g x \, \map \Pr {X = x}$
whenever the sum is absolutely convergent. | Let $\Omega_X = \Img X = I$.
Let $Y = g \sqbrk X$.
Thus:
:$\Omega_Y = \Img Y = g \sqbrk I$
So:
{{begin-eqn}}
{{eqn | l = \expect Y
| r = \sum_{y \mathop \in g \sqbrk I} y \, \map \Pr {Y = y}
| c =
}}
{{eqn | r = \sum_{y \mathop \in g \sqbrk I} y \sum_{ {x \mathop \in I} \atop {\map g x \mathop = y} } \map ... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]].
Let $\expect X$ be the [[Definition:Expectation|expectation]] of $X$.
Let $g: \R \to \R$ be a [[Definition:Real Function|real function]].
Then:
:$\ds \expect {g \sqbrk X} = \sum_{x \mathop \in \Omega_X} \map g x \, \map \Pr {X = x}$
when... | Let $\Omega_X = \Img X = I$.
Let $Y = g \sqbrk X$.
Thus:
:$\Omega_Y = \Img Y = g \sqbrk I$
So:
{{begin-eqn}}
{{eqn | l = \expect Y
| r = \sum_{y \mathop \in g \sqbrk I} y \, \map \Pr {Y = y}
| c =
}}
{{eqn | r = \sum_{y \mathop \in g \sqbrk I} y \sum_{ {x \mathop \in I} \atop {\map g x \mathop = y} } \m... | Expectation of Function of Discrete Random Variable | https://proofwiki.org/wiki/Expectation_of_Function_of_Discrete_Random_Variable | https://proofwiki.org/wiki/Expectation_of_Function_of_Discrete_Random_Variable | [
"Expectation"
] | [
"Definition:Random Variable/Discrete",
"Definition:Expectation",
"Definition:Real Function",
"Definition:Absolutely Convergent Series"
] | [
"Probability Mass Function of Function of Discrete Random Variable",
"Definition:Expectation",
"Definition:Absolutely Convergent Series"
] |
proofwiki-2637 | Probability Mass Function of Function of Discrete Random Variable | Let $X$ be a discrete random variable.
Let $Y = \map g X$, where $g: \R \to \R$ is a real function.
Then the probability mass function of $Y$ is given by:
:$\ds \map {p_Y} y = \sum_{x \mathop \in \inv g y} \map \Pr {X = x}$ | By Function of Discrete Random Variable we have that $Y$ is itself a discrete random variable.
Thus:
{{begin-eqn}}
{{eqn | l = \map {p_Y} y
| r = \map \Pr {Y = y}
| c =
}}
{{eqn | r = \map \Pr {\map g X = y}
| c =
}}
{{eqn | r = \map \Pr {X \in \inv g y}
| c =
}}
{{eqn | r = \sum_{x \mathop \... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]].
Let $Y = \map g X$, where $g: \R \to \R$ is a [[Definition:Real Function|real function]].
Then the [[Definition:Probability Mass Function|probability mass function]] of $Y$ is given by:
:$\ds \map {p_Y} y = \sum_{x \mathop \in \inv g y} \... | By [[Function of Discrete Random Variable]] we have that $Y$ is itself a [[Definition:Discrete Random Variable|discrete random variable]].
Thus:
{{begin-eqn}}
{{eqn | l = \map {p_Y} y
| r = \map \Pr {Y = y}
| c =
}}
{{eqn | r = \map \Pr {\map g X = y}
| c =
}}
{{eqn | r = \map \Pr {X \in \inv g y}... | Probability Mass Function of Function of Discrete Random Variable | https://proofwiki.org/wiki/Probability_Mass_Function_of_Function_of_Discrete_Random_Variable | https://proofwiki.org/wiki/Probability_Mass_Function_of_Function_of_Discrete_Random_Variable | [
"Probability Mass Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Real Function",
"Definition:Probability Mass Function"
] | [
"Function of Discrete Random Variable",
"Definition:Random Variable/Discrete"
] |
proofwiki-2638 | Variance as Expectation of Square minus Square of Expectation | Let $X$ be a random variable.
Then the variance of $X$ can be expressed as:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
That is, it is the expectation of the square of $X$ minus the square of the expectation of $X$. | === Discrete Random Variable ===
{{:Variance as Expectation of Square minus Square of Expectation/Discrete}} | Let $X$ be a [[Definition:Random Variable|random variable]].
Then the [[Definition:Variance|variance]] of $X$ can be expressed as:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
That is, it is the [[Definition:Expectation|expectation]] of the [[Definition:Squaring|square]] of $X$ minus the [[Definition:Squaring|sq... | === [[Variance as Expectation of Square minus Square of Expectation/Discrete|Discrete Random Variable]] ===
{{:Variance as Expectation of Square minus Square of Expectation/Discrete}} | Variance as Expectation of Square minus Square of Expectation | https://proofwiki.org/wiki/Variance_as_Expectation_of_Square_minus_Square_of_Expectation | https://proofwiki.org/wiki/Variance_as_Expectation_of_Square_minus_Square_of_Expectation | [
"Variance",
"Expectation",
"Variance as Expectation of Square minus Square of Expectation"
] | [
"Definition:Random Variable",
"Definition:Variance",
"Definition:Expectation",
"Definition:Square/Function",
"Definition:Square/Function",
"Definition:Expectation"
] | [
"Variance as Expectation of Square minus Square of Expectation/Discrete"
] |
proofwiki-2639 | Derivative of Geometric Sequence | Let $x \in \R: \size x < 1$.
Then:
:$\ds \sum_{n \mathop \ge 1} n x^{n - 1} = \frac 1 {\paren {1 - x}^2}$ | We have from Power Rule for Derivatives that:
:$\ds \frac \d {\d x} \sum_{n \mathop \ge 1} x^n = \sum_{n \mathop \ge 1} n x^{n - 1}$
But from Sum of Infinite Geometric Sequence: Corollary:
:$\ds \sum_{n \mathop \ge 1} x^n = \frac x {1 - x}$
The result follows by Power Rule for Derivatives and the Chain Rule for Derivat... | Let $x \in \R: \size x < 1$.
Then:
:$\ds \sum_{n \mathop \ge 1} n x^{n - 1} = \frac 1 {\paren {1 - x}^2}$ | We have from [[Power Rule for Derivatives]] that:
:$\ds \frac \d {\d x} \sum_{n \mathop \ge 1} x^n = \sum_{n \mathop \ge 1} n x^{n - 1}$
But from [[Sum of Infinite Geometric Sequence/Corollary 1|Sum of Infinite Geometric Sequence: Corollary]]:
:$\ds \sum_{n \mathop \ge 1} x^n = \frac x {1 - x}$
The result follows by ... | Derivative of Geometric Sequence | https://proofwiki.org/wiki/Derivative_of_Geometric_Sequence | https://proofwiki.org/wiki/Derivative_of_Geometric_Sequence | [
"Analysis",
"Differential Calculus",
"Geometric Sequences"
] | [] | [
"Power Rule for Derivatives",
"Sum of Infinite Geometric Sequence/Corollary 1",
"Power Rule for Derivatives",
"Derivative of Composite Function"
] |
proofwiki-2640 | Expectation of Shifted Geometric Distribution | Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac 1 p$ | From the definition of expectation:
:$\expect X = \ds \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
By definition of shifted geometric distribution:
:$\expect X = \ds \sum_{k \mathop \in \Omega_X} k p \paren {1 - p}^{k - 1}$
Let $q = 1 - p$:
{{begin-eqn}}
{{eqn | l = \expect X
| r = p \sum_{k \mathop \ge 0} k... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Shifted Geometric Distribution|shifted geometric distribution with parameter $p$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = \dfrac 1 p$ | From the definition of [[Definition:Expectation|expectation]]:
:$\expect X = \ds \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
By definition of [[Definition:Shifted Geometric Distribution|shifted geometric distribution]]:
:$\expect X = \ds \sum_{k \mathop \in \Omega_X} k p \paren {1 - p}^{k - 1}$
Let $q = 1 - p... | Expectation of Shifted Geometric Distribution/Proof 1 | https://proofwiki.org/wiki/Expectation_of_Shifted_Geometric_Distribution | https://proofwiki.org/wiki/Expectation_of_Shifted_Geometric_Distribution/Proof_1 | [
"Expectation of Shifted Geometric Distribution",
"Expectation",
"Geometric Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution/Shifted",
"Definition:Expectation"
] | [
"Definition:Expectation",
"Definition:Geometric Distribution/Shifted",
"Derivative of Geometric Sequence"
] |
proofwiki-2641 | Expectation of Shifted Geometric Distribution | Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac 1 p$ | From the Probability Generating Function of Shifted Geometric Distribution, we have:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r = \map {\frac \d ... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Shifted Geometric Distribution|shifted geometric distribution with parameter $p$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = \dfrac 1 p$ | From the [[Probability Generating Function of Shifted Geometric Distribution]], we have:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
where $q = 1 - p$.
From [[Expectation of Discrete Random Variable from PGF]], we have:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r =... | Expectation of Shifted Geometric Distribution/Proof 2 | https://proofwiki.org/wiki/Expectation_of_Shifted_Geometric_Distribution | https://proofwiki.org/wiki/Expectation_of_Shifted_Geometric_Distribution/Proof_2 | [
"Expectation of Shifted Geometric Distribution",
"Expectation",
"Geometric Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution/Shifted",
"Definition:Expectation"
] | [
"Probability Generating Function of Shifted Geometric Distribution",
"Expectation of Discrete Random Variable from PGF",
"Derivatives of PGF of Shifted Geometric Distribution"
] |
proofwiki-2642 | Expectation of Bernoulli Distribution | Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.
Then the expectation of $X$ is given by:
:$\expect X = p$ | From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$
By definition of Bernoulli distribution:
:$\expect X = 1 \times p + 0 \times \paren {1 - p}$
Hence the result.
{{qed}} | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = p$ | From the definition of [[Definition:Expectation|expectation]]:
:$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$
By definition of [[Definition:Bernoulli Distribution|Bernoulli distribution]]:
:$\expect X = 1 \times p + 0 \times \paren {1 - p}$
Hence the result.
{{qed}} | Expectation of Bernoulli Distribution/Proof 1 | https://proofwiki.org/wiki/Expectation_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Expectation_of_Bernoulli_Distribution/Proof_1 | [
"Expectation",
"Bernoulli Distribution",
"Expectation of Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Expectation"
] | [
"Definition:Expectation",
"Definition:Bernoulli Distribution"
] |
proofwiki-2643 | Expectation of Bernoulli Distribution | Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.
Then the expectation of $X$ is given by:
:$\expect X = p$ | Follows directly from Expectation of Binomial Distribution, putting $n = 1$.
{{qed}} | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = p$ | Follows directly from [[Expectation of Binomial Distribution]], putting $n = 1$.
{{qed}} | Expectation of Bernoulli Distribution/Proof 2 | https://proofwiki.org/wiki/Expectation_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Expectation_of_Bernoulli_Distribution/Proof_2 | [
"Expectation",
"Bernoulli Distribution",
"Expectation of Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Expectation"
] | [
"Expectation of Binomial Distribution"
] |
proofwiki-2644 | Expectation of Bernoulli Distribution | Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.
Then the expectation of $X$ is given by:
:$\expect X = p$ | From the Probability Generating Function of Bernoulli Distribution, we have:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
:$\expect X = \map { {\Pi_X}'} 1$
From Derivatives of PGF of Bernoulli Distribution:
:$\map { {\Pi_X}'} s = p$
Hence the result.
{{q... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = p$ | From the [[Probability Generating Function of Bernoulli Distribution]], we have:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$.
From [[Expectation of Discrete Random Variable from PGF]], we have:
:$\expect X = \map { {\Pi_X}'} 1$
From [[Derivatives of PGF of Bernoulli Distribution]]:
:$\map { {\Pi_X}'} s = p$
H... | Expectation of Bernoulli Distribution/Proof 3 | https://proofwiki.org/wiki/Expectation_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Expectation_of_Bernoulli_Distribution/Proof_3 | [
"Expectation",
"Bernoulli Distribution",
"Expectation of Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Expectation"
] | [
"Probability Generating Function of Bernoulli Distribution",
"Expectation of Discrete Random Variable from PGF",
"Derivatives of PGF of Bernoulli Distribution"
] |
proofwiki-2645 | Expectation of Bernoulli Distribution | Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.
Then the expectation of $X$ is given by:
:$\expect X = p$ | From Moment Generating Function of Bernoulli Distribution, the moment generating function of $X$, $M_X$, is given by:
:$\map {M_X} t = q + p e^t$
where $q = 1 - p$.
By Moment in terms of Moment Generating Function:
:$\expect X = \map {M_X'} 0$
We have:
{{begin-eqn}}
{{eqn | l = \map {M_X'} t
| r = \frac \d {\d t} \... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = p$ | From [[Moment Generating Function of Bernoulli Distribution]], the [[Definition:Moment Generating Function|moment generating function]] of $X$, $M_X$, is given by:
:$\map {M_X} t = q + p e^t$
where $q = 1 - p$.
By [[Moment in terms of Moment Generating Function]]:
:$\expect X = \map {M_X'} 0$
We have:
{{begin-... | Expectation of Bernoulli Distribution/Proof 4 | https://proofwiki.org/wiki/Expectation_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Expectation_of_Bernoulli_Distribution/Proof_4 | [
"Expectation",
"Bernoulli Distribution",
"Expectation of Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Expectation"
] | [
"Moment Generating Function of Bernoulli Distribution",
"Definition:Moment Generating Function",
"Moment in terms of Moment Generating Function",
"Derivative of Constant",
"Derivative of Exponential Function",
"Exponential of Zero"
] |
proofwiki-2646 | Expectation of Binomial Distribution | Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.
Then the expectation of $X$ is given by:
:$\expect X = n p$ | From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
Thus:
{{begin-eqn}}
{{eqn | l = \expect X
| r = \sum_{k \mathop = 0}^n k \binom n k p^k q^{n - k}
| c = {{Defof|Binomial Distribution}}, with $p + q = 1$
}}
{{eqn | r = \sum_{k \mathop = 1}^n k \binom n k... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]] for some $n \in \N$ and $0 \le p \le 1$.
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = n p$ | From the definition of [[Definition:Expectation|expectation]]:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
Thus:
{{begin-eqn}}
{{eqn | l = \expect X
| r = \sum_{k \mathop = 0}^n k \binom n k p^k q^{n - k}
| c = {{Defof|Binomial Distribution}}, with $p + q = 1$
}}
{{eqn | r = \sum... | Expectation of Binomial Distribution/Proof 1 | https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution | https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution/Proof_1 | [
"Expectation of Binomial Distribution",
"Expectation",
"Binomial Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Expectation"
] | [
"Definition:Expectation",
"Factors of Binomial Coefficient",
"Binomial Theorem"
] |
proofwiki-2647 | Expectation of Binomial Distribution | Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.
Then the expectation of $X$ is given by:
:$\expect X = n p$ | From Binomial Experiment has Binomial Distribution, we see that $X$ as defined here is a sum of discrete random variables $Y_i$ that model the Bernoulli distribution:
:$\ds X = \sum_{i \mathop = 1}^n Y_i$
Each of the Bernoulli trials is independent of each other, by definition of a binomial experiment.
It follows that:... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]] for some $n \in \N$ and $0 \le p \le 1$.
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = n p$ | From [[Binomial Experiment has Binomial Distribution]], we see that $X$ as defined here is a sum of [[Definition:Discrete Random Variable|discrete random variables]] $Y_i$ that model the [[Definition:Bernoulli Distribution|Bernoulli distribution]]:
:$\ds X = \sum_{i \mathop = 1}^n Y_i$
Each of the [[Definition:Bernou... | Expectation of Binomial Distribution/Proof 2 | https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution | https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution/Proof_2 | [
"Expectation of Binomial Distribution",
"Expectation",
"Binomial Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Expectation"
] | [
"Binomial Experiment has Binomial Distribution",
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Trial",
"Definition:Independent Events",
"Definition:Binomial Experiment",
"Sum of Expectations of Independent Trials",
"Expectation of Bernoulli Distribut... |
proofwiki-2648 | Expectation of Binomial Distribution | Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.
Then the expectation of $X$ is given by:
:$\expect X = n p$ | From the Probability Generating Function of Binomial Distribution, we have:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r = \map {\frac \d {\d s} }... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]] for some $n \in \N$ and $0 \le p \le 1$.
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = n p$ | From the [[Probability Generating Function of Binomial Distribution]], we have:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$.
From [[Expectation of Discrete Random Variable from PGF]], we have:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r = \map {\fra... | Expectation of Binomial Distribution/Proof 3 | https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution | https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution/Proof_3 | [
"Expectation of Binomial Distribution",
"Expectation",
"Binomial Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Expectation"
] | [
"Probability Generating Function of Binomial Distribution",
"Expectation of Discrete Random Variable from PGF",
"Derivatives of PGF of Binomial Distribution"
] |
proofwiki-2649 | Expectation of Binomial Distribution | Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.
Then the expectation of $X$ is given by:
:$\expect X = n p$ | From Moment Generating Function of Binomial Distribution, the moment generating function of $X$, $M_X$, is given by:
:$\ds \map {M_X} t = \paren {1 - p + p e^t}^n$
By Moment in terms of Moment Generating Function:
:$\ds \expect X = \map {M_X'} 0$
We have:
{{begin-eqn}}
{{eqn | l = \map {M_X'} t
| r = \frac \d {\d t}... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]] for some $n \in \N$ and $0 \le p \le 1$.
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = n p$ | From [[Moment Generating Function of Binomial Distribution]], the [[Definition:Moment Generating Function|moment generating function]] of $X$, $M_X$, is given by:
:$\ds \map {M_X} t = \paren {1 - p + p e^t}^n$
By [[Moment in terms of Moment Generating Function]]:
:$\ds \expect X = \map {M_X'} 0$
We have:
{{begin-... | Expectation of Binomial Distribution/Proof 4 | https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution | https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution/Proof_4 | [
"Expectation of Binomial Distribution",
"Expectation",
"Binomial Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Expectation"
] | [
"Moment Generating Function of Binomial Distribution",
"Definition:Moment Generating Function",
"Moment in terms of Moment Generating Function",
"Derivative of Composite Function",
"Derivative of Exponential Function",
"Power Rule for Derivatives",
"Exponential of Zero"
] |
proofwiki-2650 | Expectation of Poisson Distribution | Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the expectation of $X$ is given by:
:$\expect X = \lambda$ | From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$
By definition of Poisson distribution:
:$\ds \expect X = \sum_{k \mathop \ge 0} k \frac 1 {k!} \lambda^k e^{-\lambda}$
Then:
{{begin-eqn}}
{{eqn | l = \expect X
| r = \lambda e^{-\lambda} \sum_{k \mathop \ge 1} \... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = \lambda$ | From the definition of [[Definition:Expectation|expectation]]:
:$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$
By definition of [[Definition:Poisson Distribution|Poisson distribution]]:
:$\ds \expect X = \sum_{k \mathop \ge 0} k \frac 1 {k!} \lambda^k e^{-\lambda}$
Then:
{{begin-eqn}}
{{eqn | l = ... | Expectation of Poisson Distribution/Proof 1 | https://proofwiki.org/wiki/Expectation_of_Poisson_Distribution | https://proofwiki.org/wiki/Expectation_of_Poisson_Distribution/Proof_1 | [
"Expectation of Poisson Distribution",
"Expectation",
"Poisson Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Poisson Distribution",
"Definition:Expectation"
] | [
"Definition:Expectation",
"Definition:Poisson Distribution",
"Power Series Expansion for Exponential Function"
] |
proofwiki-2651 | Expectation of Poisson Distribution | Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the expectation of $X$ is given by:
:$\expect X = \lambda$ | From Probability Generating Function of Poisson Distribution:
:$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$
From Expectation of Discrete Random Variable from PGF:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r = \frac \d {\d s} e^{-\lambda \paren {1 - s} }
| c =
}}... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = \lambda$ | From [[Probability Generating Function of Poisson Distribution]]:
:$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$
From [[Expectation of Discrete Random Variable from PGF]]:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r = \frac \d {\d s} e^{-\lambda \paren {1 - s} }
... | Expectation of Poisson Distribution/Proof 2 | https://proofwiki.org/wiki/Expectation_of_Poisson_Distribution | https://proofwiki.org/wiki/Expectation_of_Poisson_Distribution/Proof_2 | [
"Expectation of Poisson Distribution",
"Expectation",
"Poisson Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Poisson Distribution",
"Definition:Expectation"
] | [
"Probability Generating Function of Poisson Distribution",
"Expectation of Discrete Random Variable from PGF",
"Derivatives of PGF of Poisson Distribution",
"Exponential of Zero"
] |
proofwiki-2652 | Expectation of Poisson Distribution | Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the expectation of $X$ is given by:
:$\expect X = \lambda$ | From Moment Generating Function of Poisson Distribution, the moment generating function of $X$, $M_X$, is given by:
:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
By Moment in terms of Moment Generating Function:
:$\expect X = \map {M_X'} 0$
We have:
{{begin-eqn}}
{{eqn | l = \map {M_X'} t
| r = \map {\frac \d {\d... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = \lambda$ | From [[Moment Generating Function of Poisson Distribution]], the [[Definition:Moment Generating Function|moment generating function]] of $X$, $M_X$, is given by:
:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
By [[Moment in terms of Moment Generating Function]]:
:$\expect X = \map {M_X'} 0$
We have:
{{begin-eqn... | Expectation of Poisson Distribution/Proof 3 | https://proofwiki.org/wiki/Expectation_of_Poisson_Distribution | https://proofwiki.org/wiki/Expectation_of_Poisson_Distribution/Proof_3 | [
"Expectation of Poisson Distribution",
"Expectation",
"Poisson Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Poisson Distribution",
"Definition:Expectation"
] | [
"Moment Generating Function of Poisson Distribution",
"Definition:Moment Generating Function",
"Moment in terms of Moment Generating Function",
"Derivative of Composite Function",
"Derivative of Exponential Function",
"Exponential of Zero"
] |
proofwiki-2653 | Variance of Bernoulli Distribution | Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
:$X \sim \Bernoulli p$
Then the variance of $X$ is given by:
:$\var X = p \paren {1 - p}$ | From the definition of variance:
:$\var X = \expect {\paren {X - \expect X}^2}$
From the Expectation of Bernoulli Distribution, we have $\expect X = p$.
Then by definition of Bernoulli distribution:
{{begin-eqn}}
{{eqn | l = \expect {\paren {X - \expect X}^2}
| r = \paren {1 - p}^2 \times p + \paren {0 - p}^2 \ti... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]]:
:$X \sim \Bernoulli p$
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = p \paren {1 - p}$ | From the definition of [[Definition:Variance of Discrete Random Variable|variance]]:
:$\var X = \expect {\paren {X - \expect X}^2}$
From the [[Expectation of Bernoulli Distribution]], we have $\expect X = p$.
Then by definition of [[Definition:Bernoulli Distribution|Bernoulli distribution]]:
{{begin-eqn}}
{{eqn | l ... | Variance of Bernoulli Distribution/Proof 1 | https://proofwiki.org/wiki/Variance_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Variance_of_Bernoulli_Distribution/Proof_1 | [
"Variance",
"Bernoulli Distribution",
"Variance of Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Variance/Discrete"
] | [
"Definition:Variance/Discrete",
"Expectation of Bernoulli Distribution",
"Definition:Bernoulli Distribution"
] |
proofwiki-2654 | Variance of Bernoulli Distribution | Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
:$X \sim \Bernoulli p$
Then the variance of $X$ is given by:
:$\var X = p \paren {1 - p}$ | From Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \, \map \Pr {X = x}$
So:
{{begin-eqn}}
{{eqn | l = \expect {X^2}
| r = 1^2 \times p +... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]]:
:$X \sim \Bernoulli p$
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = p \paren {1 - p}$ | From [[Variance as Expectation of Square minus Square of Expectation]]:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From [[Expectation of Function of Discrete Random Variable]]:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \, \map \Pr {X = x}$
So:
{{begin-eqn}}
{{eqn | l = \expect {X^2}
| r = 1^... | Variance of Bernoulli Distribution/Proof 2 | https://proofwiki.org/wiki/Variance_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Variance_of_Bernoulli_Distribution/Proof_2 | [
"Variance",
"Bernoulli Distribution",
"Variance of Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Variance/Discrete"
] | [
"Variance as Expectation of Square minus Square of Expectation",
"Expectation of Function of Discrete Random Variable",
"Expectation of Bernoulli Distribution"
] |
proofwiki-2655 | Variance of Bernoulli Distribution | Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
:$X \sim \Bernoulli p$
Then the variance of $X$ is given by:
:$\var X = p \paren {1 - p}$ | We can simply use the Variance of Binomial Distribution, putting $n = 1$.
{{qed}} | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]]:
:$X \sim \Bernoulli p$
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = p \paren {1 - p}$ | We can simply use the [[Variance of Binomial Distribution]], putting $n = 1$.
{{qed}} | Variance of Bernoulli Distribution/Proof 3 | https://proofwiki.org/wiki/Variance_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Variance_of_Bernoulli_Distribution/Proof_3 | [
"Variance",
"Bernoulli Distribution",
"Variance of Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Variance/Discrete"
] | [
"Variance of Binomial Distribution"
] |
proofwiki-2656 | Variance of Bernoulli Distribution | Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
:$X \sim \Bernoulli p$
Then the variance of $X$ is given by:
:$\var X = p \paren {1 - p}$ | From Variance of Discrete Random Variable from PGF, we have:
:$\var X = \map { {\Pi_X}''} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Bernoulli Distribution, we have:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$.
From Expectation of Bernoulli Distribu... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]]:
:$X \sim \Bernoulli p$
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = p \paren {1 - p}$ | From [[Variance of Discrete Random Variable from PGF]], we have:
:$\var X = \map { {\Pi_X}''} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the [[Definition:Expectation|expectation]] of $X$.
From the [[Probability Generating Function of Bernoulli Distribution]], we have:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$.
... | Variance of Bernoulli Distribution/Proof 4 | https://proofwiki.org/wiki/Variance_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Variance_of_Bernoulli_Distribution/Proof_4 | [
"Variance",
"Bernoulli Distribution",
"Variance of Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Variance/Discrete"
] | [
"Variance of Discrete Random Variable from PGF",
"Definition:Expectation",
"Probability Generating Function of Bernoulli Distribution",
"Expectation of Bernoulli Distribution",
"Derivatives of PGF of Bernoulli Distribution"
] |
proofwiki-2657 | Variance of Bernoulli Distribution | Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
:$X \sim \Bernoulli p$
Then the variance of $X$ is given by:
:$\var X = p \paren {1 - p}$ | From Moment Generating Function of Bernoulli Distribution, the moment generating function $M_X$ of $X$ is given by:
:$\map {M_X} t = q + p e^t$
From Variance as Expectation of Square minus Square of Expectation, we have:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Moment in terms of Moment Generating Functi... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]]:
:$X \sim \Bernoulli p$
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = p \paren {1 - p}$ | From [[Moment Generating Function of Bernoulli Distribution]], the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by:
:$\map {M_X} t = q + p e^t$
From [[Variance as Expectation of Square minus Square of Expectation]], we have:
:$\var X = \expect {X^2} - \paren {\expect X... | Variance of Bernoulli Distribution/Proof 5 | https://proofwiki.org/wiki/Variance_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Variance_of_Bernoulli_Distribution/Proof_5 | [
"Variance",
"Bernoulli Distribution",
"Variance of Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Variance/Discrete"
] | [
"Moment Generating Function of Bernoulli Distribution",
"Definition:Moment Generating Function",
"Variance as Expectation of Square minus Square of Expectation",
"Moment in terms of Moment Generating Function",
"Derivative of Constant",
"Derivative of Exponential Function",
"Derivative of Exponential Fu... |
proofwiki-2658 | Variance of Poisson Distribution | Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the variance of $X$ is given by:
:$\var X = \lambda$ | From the definition of Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \, \map \Pr {X = x}$
So:
{{begin-eqn}}
{{eqn | l = \expect {X^2}
... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]].
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = \lambda$ | From the definition of [[Variance as Expectation of Square minus Square of Expectation]]:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From [[Expectation of Function of Discrete Random Variable]]:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \, \map \Pr {X = x}$
So:
{{begin-eqn}}
{{eqn | l = \expect ... | Variance of Poisson Distribution/Proof 1 | https://proofwiki.org/wiki/Variance_of_Poisson_Distribution | https://proofwiki.org/wiki/Variance_of_Poisson_Distribution/Proof_1 | [
"Variance of Poisson Distribution",
"Variance",
"Poisson Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Poisson Distribution",
"Definition:Variance/Discrete"
] | [
"Variance as Expectation of Square minus Square of Expectation",
"Expectation of Function of Discrete Random Variable",
"Power Series Expansion for Exponential Function",
"Expectation of Poisson Distribution"
] |
proofwiki-2659 | Variance of Poisson Distribution | Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the variance of $X$ is given by:
:$\var X = \lambda$ | From Variance of Discrete Random Variable from PGF, we have:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Poisson Distribution, we have:
:$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$
From Expectation of Poisson Distribution... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]].
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = \lambda$ | From [[Variance of Discrete Random Variable from PGF]], we have:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the [[Definition:Expectation|expectation]] of $X$.
From the [[Probability Generating Function of Poisson Distribution]], we have:
:$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$
F... | Variance of Poisson Distribution/Proof 2 | https://proofwiki.org/wiki/Variance_of_Poisson_Distribution | https://proofwiki.org/wiki/Variance_of_Poisson_Distribution/Proof_2 | [
"Variance of Poisson Distribution",
"Variance",
"Poisson Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Poisson Distribution",
"Definition:Variance/Discrete"
] | [
"Variance of Discrete Random Variable from PGF",
"Definition:Expectation",
"Probability Generating Function of Poisson Distribution",
"Expectation of Poisson Distribution",
"Derivatives of PGF of Poisson Distribution"
] |
proofwiki-2660 | Variance of Poisson Distribution | Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the variance of $X$ is given by:
:$\var X = \lambda$ | From Moment Generating Function of Poisson Distribution, the moment generating function of $X$, $M_X$, is given by:
:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
From Variance as Expectation of Square minus Square of Expectation, we have:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Moment in terms of Mome... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]].
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = \lambda$ | From [[Moment Generating Function of Poisson Distribution]], the [[Definition:Moment Generating Function|moment generating function]] of $X$, $M_X$, is given by:
:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
From [[Variance as Expectation of Square minus Square of Expectation]], we have:
:$\var X = \expect {X^2}... | Variance of Poisson Distribution/Proof 3 | https://proofwiki.org/wiki/Variance_of_Poisson_Distribution | https://proofwiki.org/wiki/Variance_of_Poisson_Distribution/Proof_3 | [
"Variance of Poisson Distribution",
"Variance",
"Poisson Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Poisson Distribution",
"Definition:Variance/Discrete"
] | [
"Moment Generating Function of Poisson Distribution",
"Definition:Moment Generating Function",
"Variance as Expectation of Square minus Square of Expectation",
"Moment in terms of Moment Generating Function",
"Expectation of Poisson Distribution",
"Derivative of Composite Function",
"Power Rule for Deri... |
proofwiki-2661 | Variance of Binomial Distribution | Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$.
Then the variance of $X$ is given by:
:$\var X = n p \paren {1 - p}$ | From the definition of Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \Pr \paren {X = x}$
To simplify the algebra a bit, let $q = 1 - p$, so $p... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = n p \paren {1 - p}$ | From the definition of [[Variance as Expectation of Square minus Square of Expectation]]:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From [[Expectation of Function of Discrete Random Variable]]:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \Pr \paren {X = x}$
To simplify the algebra a bit, let $q = 1 ... | Variance of Binomial Distribution/Proof 1 | https://proofwiki.org/wiki/Variance_of_Binomial_Distribution | https://proofwiki.org/wiki/Variance_of_Binomial_Distribution/Proof_1 | [
"Variance of Binomial Distribution",
"Variance",
"Binomial Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Variance/Discrete"
] | [
"Variance as Expectation of Square minus Square of Expectation",
"Expectation of Function of Discrete Random Variable",
"Factors of Binomial Coefficient",
"Factors of Binomial Coefficient",
"Binomial Theorem",
"Expectation of Binomial Distribution"
] |
proofwiki-2662 | Variance of Binomial Distribution | Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$.
Then the variance of $X$ is given by:
:$\var X = n p \paren {1 - p}$ | From Variance of Discrete Random Variable from PGF:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Binomial Distribution:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$.
From Expectation of Binomial Distribution:
:$\mu... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = n p \paren {1 - p}$ | From [[Variance of Discrete Random Variable from PGF]]:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the [[Definition:Expectation|expectation]] of $X$.
From the [[Probability Generating Function of Binomial Distribution]]:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$.
From [[Ex... | Variance of Binomial Distribution/Proof 2 | https://proofwiki.org/wiki/Variance_of_Binomial_Distribution | https://proofwiki.org/wiki/Variance_of_Binomial_Distribution/Proof_2 | [
"Variance of Binomial Distribution",
"Variance",
"Binomial Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Variance/Discrete"
] | [
"Variance of Discrete Random Variable from PGF",
"Definition:Expectation",
"Probability Generating Function of Binomial Distribution",
"Expectation of Binomial Distribution",
"Derivatives of PGF of Binomial Distribution"
] |
proofwiki-2663 | Variance of Binomial Distribution | Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$.
Then the variance of $X$ is given by:
:$\var X = n p \paren {1 - p}$ | From Binomial Experiment has Binomial Distribution, we see that $X$ as defined here is the sum of the discrete random variables that model the Bernoulli distribution.
Each of the Bernoulli trials is independent of each other.
Hence we can use Sum of Variances of Independent Trials.
The Variance of Bernoulli Distributio... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = n p \paren {1 - p}$ | From [[Binomial Experiment has Binomial Distribution]], we see that $X$ as defined here is the sum of the [[Definition:Discrete Random Variable|discrete random variables]] that model the [[Definition:Bernoulli Distribution|Bernoulli distribution]].
Each of the [[Definition:Bernoulli Trial|Bernoulli trials]] is [[Defin... | Variance of Binomial Distribution/Proof 3 | https://proofwiki.org/wiki/Variance_of_Binomial_Distribution | https://proofwiki.org/wiki/Variance_of_Binomial_Distribution/Proof_3 | [
"Variance of Binomial Distribution",
"Variance",
"Binomial Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Variance/Discrete"
] | [
"Binomial Experiment has Binomial Distribution",
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Trial",
"Definition:Independent Events",
"Sum of Variances of Independent Trials",
"Variance of Bernoulli Distribution"
] |
proofwiki-2664 | Sum of Expectations of Independent Trials | Let $\EE_1, \EE_2, \ldots, \EE_n$ be a sequence of experiments whose outcomes are independent of each other.
Let $X_1, X_2, \ldots, X_n$ be discrete random variables on $\EE_1, \EE_2, \ldots, \EE_n$ respectively.
Let $\expect {X_j}$ denote the expectation of $X_j$ for $j \in \set {1, 2, \ldots, n}$.
Then we have, whene... | The proof proceeds by induction on the number of terms $n$ in the sum.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \expect {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \expect {X_j}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\ds \expect {\sum_{j \mathop = 1}^1 X_j} = \s... | Let $\EE_1, \EE_2, \ldots, \EE_n$ be a sequence of [[Definition:Experiment|experiments]] whose [[Definition:Outcome|outcomes]] are [[Definition:Independent Events|independent]] of each other.
Let $X_1, X_2, \ldots, X_n$ be [[Definition:Discrete Random Variable|discrete random variables]] on $\EE_1, \EE_2, \ldots, \EE_... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on the number of terms $n$ in the [[Definition:Summation|sum]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \expect {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \expect {X_j}$
=== Basi... | Sum of Expectations of Independent Trials/Proof 1 | https://proofwiki.org/wiki/Sum_of_Expectations_of_Independent_Trials | https://proofwiki.org/wiki/Sum_of_Expectations_of_Independent_Trials/Proof_1 | [
"Expectation",
"Sum of Expectations of Independent Trials"
] | [
"Definition:Experiment",
"Definition:Elementary Event",
"Definition:Independent Events",
"Definition:Random Variable/Discrete",
"Definition:Expectation",
"Definition:Summation",
"Definition:Expectation",
"Definition:Expectation",
"Definition:Summation"
] | [
"Principle of Mathematical Induction",
"Definition:Summation",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Definition:Independent Events",
"Sum of Expectations of Independent Trials/Proof 1",
"Principle of Mathemati... |
proofwiki-2665 | Sum of Expectations of Independent Trials | Let $\EE_1, \EE_2, \ldots, \EE_n$ be a sequence of experiments whose outcomes are independent of each other.
Let $X_1, X_2, \ldots, X_n$ be discrete random variables on $\EE_1, \EE_2, \ldots, \EE_n$ respectively.
Let $\expect {X_j}$ denote the expectation of $X_j$ for $j \in \set {1, 2, \ldots, n}$.
Then we have, whene... | The proof proceeds by induction on the number of terms $n$ in the sum.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \expect {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \expect {X_j}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\ds \expect {\sum_{j \mathop = 1}^1 X_j} = \s... | Let $\EE_1, \EE_2, \ldots, \EE_n$ be a sequence of [[Definition:Experiment|experiments]] whose [[Definition:Outcome|outcomes]] are [[Definition:Independent Events|independent]] of each other.
Let $X_1, X_2, \ldots, X_n$ be [[Definition:Discrete Random Variable|discrete random variables]] on $\EE_1, \EE_2, \ldots, \EE_... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on the number of terms $n$ in the [[Definition:Summation|sum]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \expect {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \expect {X_j}$
=== Basi... | Sum of Expectations of Independent Trials/Proof 2 | https://proofwiki.org/wiki/Sum_of_Expectations_of_Independent_Trials | https://proofwiki.org/wiki/Sum_of_Expectations_of_Independent_Trials/Proof_2 | [
"Expectation",
"Sum of Expectations of Independent Trials"
] | [
"Definition:Experiment",
"Definition:Elementary Event",
"Definition:Independent Events",
"Definition:Random Variable/Discrete",
"Definition:Expectation",
"Definition:Summation",
"Definition:Expectation",
"Definition:Expectation",
"Definition:Summation"
] | [
"Principle of Mathematical Induction",
"Definition:Summation",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Expectation is Linear",
"Sum of Expectations of Independent Trials/Proof 2",
"Principle of Mathematical Indu... |
proofwiki-2666 | Sum of Variances of Independent Trials | Let $\EE_1, \EE_2, \ldots, \EE_n$ be a sequence of experiments whose outcomes are independent of each other.
Let $X_1, X_2, \ldots, X_n$ be discrete random variables on $\EE_1, \EE_2, \ldots, \EE_n$ respectively.
Let $\var {X_j}$ be the variance of $X_j$ for $j \in \set {1, 2, \ldots, n}$.
Then:
:$\ds \var {\sum_{j \ma... | {{begin-eqn}}
{{eqn | l = \var {\sum_{j \mathop = 1}^n X_j}
| r = \expect {\paren {\sum_{j \mathop = 1}^n X_j}^2} - \expect {\sum_{j \mathop = 1}^n X_j}^2
| c = Variance as Expectation of Square minus Square of Expectation: Discrete
}}
{{eqn | r = \expect {\sum_{0 \mathop < i \mathop < j \mathop \le n} 2 X_... | Let $\EE_1, \EE_2, \ldots, \EE_n$ be a sequence of [[Definition:Experiment|experiments]] whose [[Definition:Outcome|outcomes]] are [[Definition:Independent Events|independent]] of each other.
Let $X_1, X_2, \ldots, X_n$ be [[Definition:Discrete Random Variable|discrete random variables]] on $\EE_1, \EE_2, \ldots, \EE_... | {{begin-eqn}}
{{eqn | l = \var {\sum_{j \mathop = 1}^n X_j}
| r = \expect {\paren {\sum_{j \mathop = 1}^n X_j}^2} - \expect {\sum_{j \mathop = 1}^n X_j}^2
| c = [[Variance as Expectation of Square minus Square of Expectation/Discrete|Variance as Expectation of Square minus Square of Expectation: Discrete]]
... | Sum of Variances of Independent Trials | https://proofwiki.org/wiki/Sum_of_Variances_of_Independent_Trials | https://proofwiki.org/wiki/Sum_of_Variances_of_Independent_Trials | [
"Variance"
] | [
"Definition:Experiment",
"Definition:Elementary Event",
"Definition:Independent Events",
"Definition:Random Variable/Discrete",
"Definition:Variance/Discrete",
"Definition:Variance/Discrete",
"Definition:Variance/Discrete"
] | [
"Variance as Expectation of Square minus Square of Expectation/Discrete",
"Expectation is Linear/Discrete",
"Variance as Expectation of Square minus Square of Expectation/Discrete",
"Category:Variance"
] |
proofwiki-2667 | Variance of Shifted Geometric Distribution | Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
Then the variance of $X$ is given by:
:$\var X = \dfrac {1 - p} {p^2}$ | From the definition of Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \map \Pr {X = x}$
To simplify the algebra a bit, let $q = 1 - p$, so $p +... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Shifted Geometric Distribution|shifted geometric distribution with parameter $p$]].
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = \dfrac {1 - p} {p^2}$ | From the definition of [[Variance as Expectation of Square minus Square of Expectation]]:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From [[Expectation of Function of Discrete Random Variable]]:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \map \Pr {X = x}$
To simplify the algebra a bit, let $q = 1 - ... | Variance of Shifted Geometric Distribution/Proof 1 | https://proofwiki.org/wiki/Variance_of_Shifted_Geometric_Distribution | https://proofwiki.org/wiki/Variance_of_Shifted_Geometric_Distribution/Proof_1 | [
"Variance of Shifted Geometric Distribution",
"Variance",
"Geometric Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution/Shifted",
"Definition:Variance/Discrete"
] | [
"Variance as Expectation of Square minus Square of Expectation",
"Expectation of Function of Discrete Random Variable",
"Expectation of Shifted Geometric Distribution",
"Derivative of Geometric Sequence/Corollary",
"Expectation of Shifted Geometric Distribution"
] |
proofwiki-2668 | Variance of Shifted Geometric Distribution | Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
Then the variance of $X$ is given by:
:$\var X = \dfrac {1 - p} {p^2}$ | From Variance of Discrete Random Variable from PGF, we have:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Shifted Geometric Distribution, we have:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
where $q = 1 - p$.
From Expectation ... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Shifted Geometric Distribution|shifted geometric distribution with parameter $p$]].
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = \dfrac {1 - p} {p^2}$ | From [[Variance of Discrete Random Variable from PGF]], we have:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the [[Definition:Expectation|expectation]] of $X$.
From the [[Probability Generating Function of Shifted Geometric Distribution]], we have:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$... | Variance of Shifted Geometric Distribution/Proof 2 | https://proofwiki.org/wiki/Variance_of_Shifted_Geometric_Distribution | https://proofwiki.org/wiki/Variance_of_Shifted_Geometric_Distribution/Proof_2 | [
"Variance of Shifted Geometric Distribution",
"Variance",
"Geometric Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution/Shifted",
"Definition:Variance/Discrete"
] | [
"Variance of Discrete Random Variable from PGF",
"Definition:Expectation",
"Probability Generating Function of Shifted Geometric Distribution",
"Expectation of Shifted Geometric Distribution",
"Derivatives of PGF of Shifted Geometric Distribution"
] |
proofwiki-2669 | Expectation of Geometric Distribution | Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
=== Formulation 1 ===
{{:Expectation of Geometric Distribution/Formulation 1}}
=== Formulation 2 ===
{{:Expectation of Geometric Distribution/Formulation 2}} | From the Probability Generating Function of Geometric Distribution:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r = \map {\frac \d {\d s} } {\frac q {1 - p s} }
... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Geometric Distribution|geometric distribution with parameter $p$]] for some $0 < p < 1$.
=== [[Expectation of Geometric Distribution/Formulation 1|Formulation 1]] ===
{{:Expectation of Geometric Distribution/Formulatio... | From the [[Probability Generating Function of Geometric Distribution]]:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From [[Expectation of Discrete Random Variable from PGF]]:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r = \map {\frac \d {\d s} } {\fra... | Expectation of Geometric Distribution/Formulation 1/Proof 2 | https://proofwiki.org/wiki/Expectation_of_Geometric_Distribution | https://proofwiki.org/wiki/Expectation_of_Geometric_Distribution/Formulation_1/Proof_2 | [
"Expectation",
"Geometric Distribution",
"Expectation of Geometric Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution",
"Expectation of Geometric Distribution/Formulation 1",
"Expectation of Geometric Distribution/Formulation 2"
] | [
"Probability Generating Function of Geometric Distribution",
"Expectation of Discrete Random Variable from PGF",
"Derivatives of PGF of Geometric Distribution"
] |
proofwiki-2670 | Expectation of Geometric Distribution | Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
=== Formulation 1 ===
{{:Expectation of Geometric Distribution/Formulation 1}}
=== Formulation 2 ===
{{:Expectation of Geometric Distribution/Formulation 2}} | From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
Then
{{begin-eqn}}
{{eqn | l = \expect X
| r = \sum_{k \mathop \in \N} k p^k \paren {1 - p}
| c = {{Defof|Geometric Distribution}}
}}
{{eqn | r = \sum_{k \mathop \ge 1} k p^k \paren {1 - p}
| c = as... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Geometric Distribution|geometric distribution with parameter $p$]] for some $0 < p < 1$.
=== [[Expectation of Geometric Distribution/Formulation 1|Formulation 1]] ===
{{:Expectation of Geometric Distribution/Formulatio... | From the definition of [[Definition:Expectation|expectation]]:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
Then
{{begin-eqn}}
{{eqn | l = \expect X
| r = \sum_{k \mathop \in \N} k p^k \paren {1 - p}
| c = {{Defof|Geometric Distribution}}
}}
{{eqn | r = \sum_{k \mathop \ge 1} k p^k ... | Expectation of Geometric Distribution/Formulation 1/Proof 3 | https://proofwiki.org/wiki/Expectation_of_Geometric_Distribution | https://proofwiki.org/wiki/Expectation_of_Geometric_Distribution/Formulation_1/Proof_3 | [
"Expectation",
"Geometric Distribution",
"Expectation of Geometric Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution",
"Expectation of Geometric Distribution/Formulation 1",
"Expectation of Geometric Distribution/Formulation 2"
] | [
"Definition:Expectation",
"Real Multiplication Distributes over Addition",
"Ratio Test",
"Absolutely Convergent Series is Convergent/Real Numbers",
"Convergent Series can be Added Term by Term",
"Translation of Index Variable of Summation",
"Sum of Infinite Geometric Sequence"
] |
proofwiki-2671 | Expectation of Geometric Distribution | Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
=== Formulation 1 ===
{{:Expectation of Geometric Distribution/Formulation 1}}
=== Formulation 2 ===
{{:Expectation of Geometric Distribution/Formulation 2}} | From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
By definition of geometric distribution:
:$\ds \expect X = \sum_{k \mathop \in \Omega_X} k p^k \paren {1 - p}$
Let $q = 1 - p$:
{{begin-eqn}}
{{eqn | l = \expect X
| r = q \sum_{k \mathop \ge 0} k p^k
| c... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Geometric Distribution|geometric distribution with parameter $p$]] for some $0 < p < 1$.
=== [[Expectation of Geometric Distribution/Formulation 1|Formulation 1]] ===
{{:Expectation of Geometric Distribution/Formulatio... | From the definition of [[Definition:Expectation|expectation]]:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
By definition of [[Definition:Geometric Distribution|geometric distribution]]:
:$\ds \expect X = \sum_{k \mathop \in \Omega_X} k p^k \paren {1 - p}$
Let $q = 1 - p$:
{{begin-eqn}}
{{eq... | Expectation of Geometric Distribution/Proof 1 | https://proofwiki.org/wiki/Expectation_of_Geometric_Distribution | https://proofwiki.org/wiki/Expectation_of_Geometric_Distribution/Proof_1 | [
"Expectation",
"Geometric Distribution",
"Expectation of Geometric Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution",
"Expectation of Geometric Distribution/Formulation 1",
"Expectation of Geometric Distribution/Formulation 2"
] | [
"Definition:Expectation",
"Definition:Geometric Distribution",
"Derivative of Geometric Sequence"
] |
proofwiki-2672 | Expectation of Geometric Distribution | Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
=== Formulation 1 ===
{{:Expectation of Geometric Distribution/Formulation 1}}
=== Formulation 2 ===
{{:Expectation of Geometric Distribution/Formulation 2}} | From the Probability Generating Function of Geometric Distribution:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r = \map {\frac \d {\d s} } {\frac q {1 - p s} }
... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Geometric Distribution|geometric distribution with parameter $p$]] for some $0 < p < 1$.
=== [[Expectation of Geometric Distribution/Formulation 1|Formulation 1]] ===
{{:Expectation of Geometric Distribution/Formulatio... | From the [[Probability Generating Function of Geometric Distribution]]:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From [[Expectation of Discrete Random Variable from PGF]]:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
| r = \map {\frac \d {\d s} } {\fra... | Expectation of Geometric Distribution/Proof 2 | https://proofwiki.org/wiki/Expectation_of_Geometric_Distribution | https://proofwiki.org/wiki/Expectation_of_Geometric_Distribution/Proof_2 | [
"Expectation",
"Geometric Distribution",
"Expectation of Geometric Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution",
"Expectation of Geometric Distribution/Formulation 1",
"Expectation of Geometric Distribution/Formulation 2"
] | [
"Probability Generating Function of Geometric Distribution",
"Expectation of Discrete Random Variable from PGF",
"Derivatives of PGF of Geometric Distribution"
] |
proofwiki-2673 | Variance of Geometric Distribution | Let $p \in \R$ be a real number such that $0 < p < 1$.
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.
=== Formulation 1 ===
{{:Variance of Geometric Distribution/Formulation 1}}
=== Formulation 2 ===
{{:Variance of Geometric Distribution/Formulation 2}} | From Variance of Discrete Random Variable from PGF, we have:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \map E x$ is the expectation of $X$.
From the Probability Generating Function of Geometric Distribution, we have:
:$\map {\Pi_X} s = \dfrac q {1 - ps}$
where $q = 1 - p$.
From Expectation of Geometric D... | Let $p \in \R$ be a [[Definition:Real Number|real number]] such that $0 < p < 1$.
Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Geometric Distribution|geometric distribution with parameter $p$]].
=== [[Variance of Geometric Distribution/Formulation 1|Formulation ... | From [[Variance of Discrete Random Variable from PGF]], we have:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \map E x$ is the [[Definition:Expectation|expectation]] of $X$.
From the [[Probability Generating Function of Geometric Distribution]], we have:
:$\map {\Pi_X} s = \dfrac q {1 - ps}$
where $q = 1 ... | Variance of Geometric Distribution/Formulation 1/Proof 2 | https://proofwiki.org/wiki/Variance_of_Geometric_Distribution | https://proofwiki.org/wiki/Variance_of_Geometric_Distribution/Formulation_1/Proof_2 | [
"Variance",
"Geometric Distribution",
"Variance of Geometric Distribution"
] | [
"Definition:Real Number",
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution",
"Variance of Geometric Distribution/Formulation 1",
"Variance of Geometric Distribution/Formulation 2"
] | [
"Variance of Discrete Random Variable from PGF",
"Definition:Expectation",
"Probability Generating Function of Geometric Distribution",
"Expectation of Geometric Distribution",
"Derivatives of PGF of Geometric Distribution"
] |
proofwiki-2674 | Variance of Geometric Distribution | Let $p \in \R$ be a real number such that $0 < p < 1$.
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.
=== Formulation 1 ===
{{:Variance of Geometric Distribution/Formulation 1}}
=== Formulation 2 ===
{{:Variance of Geometric Distribution/Formulation 2}} | By Moment Generating Function of Geometric Distribution, the moment generating function of $X$ is given by:
:$\map {M_X} t = \dfrac p {1 - \paren {1 - p} e^t}$
for $t < -\map \ln {1 - p}$, and is undefined otherwise.
From Variance as Expectation of Square minus Square of Expectation:
:$\ds \var X = \expect {X^2} - \p... | Let $p \in \R$ be a [[Definition:Real Number|real number]] such that $0 < p < 1$.
Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Geometric Distribution|geometric distribution with parameter $p$]].
=== [[Variance of Geometric Distribution/Formulation 1|Formulation ... | By [[Moment Generating Function of Geometric Distribution]], the [[Definition:Moment Generating Function|moment generating function]] of $X$ is given by:
:$\map {M_X} t = \dfrac p {1 - \paren {1 - p} e^t}$
for $t < -\map \ln {1 - p}$, and is undefined otherwise.
From [[Variance as Expectation of Square minus Squar... | Variance of Geometric Distribution/Formulation 2/Proof 2 | https://proofwiki.org/wiki/Variance_of_Geometric_Distribution | https://proofwiki.org/wiki/Variance_of_Geometric_Distribution/Formulation_2/Proof_2 | [
"Variance",
"Geometric Distribution",
"Variance of Geometric Distribution"
] | [
"Definition:Real Number",
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution",
"Variance of Geometric Distribution/Formulation 1",
"Variance of Geometric Distribution/Formulation 2"
] | [
"Moment Generating Function of Geometric Distribution",
"Definition:Moment Generating Function",
"Variance as Expectation of Square minus Square of Expectation",
"Expectation of Geometric Distribution/Formulation 2",
"Moment Generating Function of Geometric Distribution/Formulation 2/Examples/Second Moment"... |
proofwiki-2675 | Variance of Geometric Distribution | Let $p \in \R$ be a real number such that $0 < p < 1$.
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.
=== Formulation 1 ===
{{:Variance of Geometric Distribution/Formulation 1}}
=== Formulation 2 ===
{{:Variance of Geometric Distribution/Formulation 2}} | From the definition of Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$
To simplify the algebra a bit, let $q = 1 - p$, so $p... | Let $p \in \R$ be a [[Definition:Real Number|real number]] such that $0 < p < 1$.
Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Geometric Distribution|geometric distribution with parameter $p$]].
=== [[Variance of Geometric Distribution/Formulation 1|Formulation ... | From the definition of [[Variance as Expectation of Square minus Square of Expectation]]:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From [[Expectation of Function of Discrete Random Variable]]:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$
To simplify the algebra a bit, let $q = 1 ... | Variance of Geometric Distribution/Proof 1 | https://proofwiki.org/wiki/Variance_of_Geometric_Distribution | https://proofwiki.org/wiki/Variance_of_Geometric_Distribution/Proof_1 | [
"Variance",
"Geometric Distribution",
"Variance of Geometric Distribution"
] | [
"Definition:Real Number",
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution",
"Variance of Geometric Distribution/Formulation 1",
"Variance of Geometric Distribution/Formulation 2"
] | [
"Variance as Expectation of Square minus Square of Expectation",
"Expectation of Function of Discrete Random Variable",
"Variance of Shifted Geometric Distribution/Proof 1",
"Expectation of Geometric Distribution"
] |
proofwiki-2676 | Variance of Geometric Distribution | Let $p \in \R$ be a real number such that $0 < p < 1$.
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.
=== Formulation 1 ===
{{:Variance of Geometric Distribution/Formulation 1}}
=== Formulation 2 ===
{{:Variance of Geometric Distribution/Formulation 2}} | From Variance of Discrete Random Variable from PGF:
:$\var X = \map {\Pi''_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Geometric Distribution:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From Expectation of Geometric Distribution:
:$\m... | Let $p \in \R$ be a [[Definition:Real Number|real number]] such that $0 < p < 1$.
Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Geometric Distribution|geometric distribution with parameter $p$]].
=== [[Variance of Geometric Distribution/Formulation 1|Formulation ... | From [[Variance of Discrete Random Variable from PGF]]:
:$\var X = \map {\Pi''_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the [[Definition:Expectation|expectation]] of $X$.
From the [[Probability Generating Function of Geometric Distribution]]:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From [[Ex... | Variance of Geometric Distribution/Proof 2 | https://proofwiki.org/wiki/Variance_of_Geometric_Distribution | https://proofwiki.org/wiki/Variance_of_Geometric_Distribution/Proof_2 | [
"Variance",
"Geometric Distribution",
"Variance of Geometric Distribution"
] | [
"Definition:Real Number",
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution",
"Variance of Geometric Distribution/Formulation 1",
"Variance of Geometric Distribution/Formulation 2"
] | [
"Variance of Discrete Random Variable from PGF",
"Definition:Expectation",
"Probability Generating Function of Geometric Distribution",
"Expectation of Geometric Distribution",
"Derivatives of PGF of Geometric Distribution"
] |
proofwiki-2677 | Discrete Random Variable is Random Variable | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.
Then $X$ fulfils the condition:
:$\forall x \in \R: \set {\omega \in \Omega: \map X \omega \le x} \in \Sigma$
That is, $X$ fulfils the condition for it to be a random variable. | Let $X$ be a discrete random variable.
Then by definition:
:$\forall x \in \R: \set {\omega \in \Omega: \map X \omega = x} \in \Sigma$
But see that:
:$\ds \set {\omega \in \Omega: \map X \omega \le x} = \bigcup_{\substack {y \mathop \in \Omega_X \\ y \mathop \le x} } \set {\omega \in \Omega: \map X \omega = y}$
This is... | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] on $\struct {\Omega, \Sigma, \Pr}$.
Then $X$ fulfils the condition:
:$\forall x \in \R: \set {\omega \in \Omega: \map X \omega \le x} \in \Sigma$
... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]].
Then by definition:
:$\forall x \in \R: \set {\omega \in \Omega: \map X \omega = x} \in \Sigma$
But see that:
:$\ds \set {\omega \in \Omega: \map X \omega \le x} = \bigcup_{\substack {y \mathop \in \Omega_X \\ y \mathop \le x} } \set {\om... | Discrete Random Variable is Random Variable | https://proofwiki.org/wiki/Discrete_Random_Variable_is_Random_Variable | https://proofwiki.org/wiki/Discrete_Random_Variable_is_Random_Variable | [
"Probability Theory"
] | [
"Definition:Probability Space",
"Definition:Random Variable/Discrete",
"Definition:Random Variable"
] | [
"Definition:Random Variable/Discrete",
"Definition:Countable Set",
"Definition:Set Union",
"Definition:Event",
"Definition:Sigma-Algebra",
"Category:Probability Theory"
] |
proofwiki-2678 | Expectation of Discrete Uniform Distribution | Let $X$ be a discrete random variable with the discrete uniform distribution with parameter $n$.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac {n + 1} 2$ | From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
Thus:
{{begin-eqn}}
{{eqn | l = \expect X
| r = \sum_{k \mathop = 1}^n k \paren {\frac 1 n}
| c = {{Defof|Discrete Uniform Distribution}}
}}
{{eqn | r = \frac 1 n \sum_{k \mathop = 1}^n k
| c =
}}
... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Discrete Uniform Distribution|discrete uniform distribution with parameter $n$]].
Then the [[Definition:Expectation|expectation]] of $X$ is given by:
:$\expect X = \dfrac {n + 1} 2$ | From the definition of [[Definition:Expectation|expectation]]:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
Thus:
{{begin-eqn}}
{{eqn | l = \expect X
| r = \sum_{k \mathop = 1}^n k \paren {\frac 1 n}
| c = {{Defof|Discrete Uniform Distribution}}
}}
{{eqn | r = \frac 1 n \sum_{k \math... | Expectation of Discrete Uniform Distribution | https://proofwiki.org/wiki/Expectation_of_Discrete_Uniform_Distribution | https://proofwiki.org/wiki/Expectation_of_Discrete_Uniform_Distribution | [
"Expectation",
"Discrete Uniform Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Uniform Distribution/Discrete",
"Definition:Expectation"
] | [
"Definition:Expectation",
"Closed Form for Triangular Numbers",
"Category:Expectation",
"Category:Discrete Uniform Distribution"
] |
proofwiki-2679 | Variance of Discrete Uniform Distribution | Let $X$ be a discrete random variable with the discrete uniform distribution with parameter $n$.
Then the variance of $X$ is given by:
:$\var X = \dfrac {n^2 - 1} {12}$ | From the definition of Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$
So:
{{begin-eqn}}
{{eqn | l = \expect {X^2}
| r... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Discrete Uniform Distribution|discrete uniform distribution with parameter $n$]].
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is given by:
:$\var X = \dfrac {n^2 - 1} {12}$ | From the definition of [[Variance as Expectation of Square minus Square of Expectation]]:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From [[Expectation of Function of Discrete Random Variable]]:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \map \Pr {X = x}$
So:
{{begin-eqn}}
{{eqn | l = \expect {X^... | Variance of Discrete Uniform Distribution | https://proofwiki.org/wiki/Variance_of_Discrete_Uniform_Distribution | https://proofwiki.org/wiki/Variance_of_Discrete_Uniform_Distribution | [
"Variance",
"Discrete Uniform Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Uniform Distribution/Discrete",
"Definition:Variance/Discrete"
] | [
"Variance as Expectation of Square minus Square of Expectation",
"Expectation of Function of Discrete Random Variable",
"Sum of Sequence of Squares",
"Expectation of Discrete Uniform Distribution",
"Category:Variance",
"Category:Discrete Uniform Distribution"
] |
proofwiki-2680 | Probability Generating Function of Degenerate Distribution | Let $X$ be the degenerate distribution:
:$\forall x \in \N: \map {p_X} x = \begin{cases}
1 & : x = k \\
0 & : x \ne k
\end{cases}$
where $k \in \N$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} x = s^k$ | Follows directly from the definition:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
As $\map {p_X} x \ne 0$ for only one value of $x$, all the terms vanish except that one.
Hence the result.
{{qed}}
Category:Probability Generating Functions
Category:Degenerate Distribution
ot9gqsg5ebkqpkpy24m99p8zg4dd... | Let $X$ be the [[Definition:Degenerate Distribution|degenerate distribution]]:
:$\forall x \in \N: \map {p_X} x = \begin{cases}
1 & : x = k \\
0 & : x \ne k
\end{cases}$
where $k \in \N$.
Then the [[Definition:Probability Generating Function|p.g.f.]] of $X$ is:
:$\map {\Pi_X} x = s^k$ | Follows directly from the [[Definition:Probability Generating Function|definition]]:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
As $\map {p_X} x \ne 0$ for only one value of $x$, all the terms vanish except that one.
Hence the result.
{{qed}}
[[Category:Probability Generating Functions]]
[[Cate... | Probability Generating Function of Degenerate Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Degenerate_Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Degenerate_Distribution | [
"Probability Generating Functions",
"Degenerate Distribution"
] | [
"Definition:Degenerate Distribution",
"Definition:Probability Generating Function"
] | [
"Definition:Probability Generating Function",
"Category:Probability Generating Functions",
"Category:Degenerate Distribution"
] |
proofwiki-2681 | Probability Generating Function of Bernoulli Distribution | Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$. | From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the Bernoulli distribution:
:$\map {p_X} x = \begin{cases}
p & : x = a \\
1 - p & : x = b \\
0 & : x \notin \set {a, b} \\
\end{cases}$
So:
{{begin-eqn}}
{{eqn | l=\map {\Pi_X} s
| r=\map {p_X} 0 s... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]].
Then the [[Definition:Probability Generating Function|p.g.f.]] of $X$ is:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$. | From the definition of [[Definition:Probability Generating Function|p.g.f]]:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the [[Definition:Bernoulli Distribution|Bernoulli distribution]]:
:$\map {p_X} x = \begin{cases}
p & : x = a \\
1 - p & : x = b \\
0 & : x \notin \set {a,... | Probability Generating Function of Bernoulli Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Bernoulli_Distribution | [
"Probability Generating Functions",
"Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Probability Generating Function"
] | [
"Definition:Probability Generating Function",
"Definition:Bernoulli Distribution"
] |
proofwiki-2682 | Probability Generating Function of Binomial Distribution | Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$. | From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map {p_X} k s^k$
From the definition of the binomial distribution:
:$\map {p_X} k = \dbinom n k p^k \paren {1 - p}^{n - k}$
So:
{{begin-eqn}}
{{eqn | l = \map {\Pi_X} s
| r = \sum_{k \mathop = 0}^n \binom n k p^k \paren {1 - p}^{n - k} s... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Probability Generating Function|p.g.f.]] of $X$ is:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$. | From the definition of [[Definition:Probability Generating Function|p.g.f]]:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map {p_X} k s^k$
From the definition of the [[Definition:Binomial Distribution|binomial distribution]]:
:$\map {p_X} k = \dbinom n k p^k \paren {1 - p}^{n - k}$
So:
{{begin-eqn}}
{{eqn | l = \m... | Probability Generating Function of Binomial Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Binomial_Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Binomial_Distribution | [
"Binomial Distribution",
"Examples of Probability Generating Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Probability Generating Function"
] | [
"Definition:Probability Generating Function",
"Definition:Binomial Distribution",
"Binomial Theorem"
] |
proofwiki-2683 | Probability Generating Function of Shifted Geometric Distribution | Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
where $q = 1 - p$. | From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the shifted geometric distribution:
:$\forall k \in \N, k \ge 1: \map {p_X} k = p q^{k - 1}$
So:
{{begin-eqn}}
{{eqn | l = \map {\Pi_X} s
| r = \sum_{k \mathop \ge 1} p q^{k - 1} s^k
| c =
}... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Shifted Geometric Distribution|shifted geometric distribution with parameter $p$]].
Then the [[Definition:Probability Generating Function|p.g.f.]] of $X$ is:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
where $q = 1 - p$... | From the definition of [[Definition:Probability Generating Function|p.g.f]]:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the [[Definition:Shifted Geometric Distribution|shifted geometric distribution]]:
:$\forall k \in \N, k \ge 1: \map {p_X} k = p q^{k - 1}$
So:
{{begin-e... | Probability Generating Function of Shifted Geometric Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Shifted_Geometric_Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Shifted_Geometric_Distribution | [
"Probability Generating Functions",
"Geometric Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution/Shifted",
"Definition:Probability Generating Function"
] | [
"Definition:Probability Generating Function",
"Definition:Geometric Distribution/Shifted",
"Sum of Infinite Geometric Sequence"
] |
proofwiki-2684 | Probability Generating Function of Geometric Distribution | Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$. | From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the geometric distribution:
:$\forall k \in \N, k \ge 0: \map {p_X} k = q p^k$
So:
{{begin-eqn}}
{{eqn | l = \map {\Pi_X} s
| r = \sum_{k \mathop \ge 0} q p^k s^k
| c =
}}
{{eqn | r = q \sum... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Geometric Distribution|geometric distribution with parameter $p$]].
Then the [[Definition:Probability Generating Function|p.g.f.]] of $X$ is:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$. | From the definition of [[Definition:Probability Generating Function|p.g.f]]:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the [[Definition:Geometric Distribution|geometric distribution]]:
:$\forall k \in \N, k \ge 0: \map {p_X} k = q p^k$
So:
{{begin-eqn}}
{{eqn | l = \map {... | Probability Generating Function of Geometric Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Geometric_Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Geometric_Distribution | [
"Probability Generating Functions",
"Geometric Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution",
"Definition:Probability Generating Function"
] | [
"Definition:Probability Generating Function",
"Definition:Geometric Distribution",
"Sum of Infinite Geometric Sequence",
"Category:Probability Generating Functions",
"Category:Geometric Distribution"
] |
proofwiki-2685 | Probability Generating Function of Poisson Distribution | Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$ | From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the Poisson distribution:
:$\ds \forall k \in \N, k \ge 0: \map {p_X} k = \frac {e^{-\lambda} \lambda^k} {k!}$
So:
{{begin-eqn}}
{{eqn | l = \map {\Pi_X} s
| r = \sum_{k \mathop \ge 0} \frac {e^{-\... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]].
Then the [[Definition:Probability Generating Function|p.g.f.]] of $X$ is:
:$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$ | From the definition of [[Definition:Probability Generating Function|p.g.f]]:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the [[Definition:Poisson Distribution|Poisson distribution]]:
:$\ds \forall k \in \N, k \ge 0: \map {p_X} k = \frac {e^{-\lambda} \lambda^k} {k!}$
So:
{{... | Probability Generating Function of Poisson Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Poisson_Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Poisson_Distribution | [
"Probability Generating Functions",
"Poisson Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Poisson Distribution",
"Definition:Probability Generating Function"
] | [
"Definition:Probability Generating Function",
"Definition:Poisson Distribution",
"Power Series Expansion for Exponential Function"
] |
proofwiki-2686 | Expectation of Discrete Random Variable from PGF | Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.
Then the expectation of $X$ is the value of the first derivative of $\map {\Pi_X} s$ {{WRT|Differentiation}} $s$ at $s = 1$.
That is:
:$\expect X = \map {\Pi'_X} 1$ | For ease of notation, write $\map p x$ for $\map \Pr {X = x}$.
From the definition of the probability generating function:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map p x s^x = \map p 0 + \map p 1 s + \map p 2 s^2 + \map p 3 s^3 + \cdots$
Differentiate this {{WRT|Differentiation}} $s$:
{{begin-eqn}}
{{eqn | l = ... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] whose [[Definition:Probability Generating Function|probability generating function]] is $\map {\Pi_X} s$.
Then the [[Definition:Expectation|expectation]] of $X$ is the value of the [[Definition:Derivative|first derivative]] of $\map {\Pi_X}... | For ease of notation, write $\map p x$ for $\map \Pr {X = x}$.
From the definition of the [[Definition:Probability Generating Function|probability generating function]]:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map p x s^x = \map p 0 + \map p 1 s + \map p 2 s^2 + \map p 3 s^3 + \cdots$
[[Definition:Differenti... | Expectation of Discrete Random Variable from PGF | https://proofwiki.org/wiki/Expectation_of_Discrete_Random_Variable_from_PGF | https://proofwiki.org/wiki/Expectation_of_Discrete_Random_Variable_from_PGF | [
"Expectation of Discrete Random Variable from PGF",
"Expectation",
"Probability Generating Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Generating Function",
"Definition:Expectation",
"Definition:Derivative"
] | [
"Definition:Probability Generating Function",
"Definition:Differentiation",
"Abel's Theorem",
"Power Rule for Derivatives",
"Definition:Expectation"
] |
proofwiki-2687 | Variance of Discrete Random Variable from PGF | Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.
Then the variance of $X$ can be obtained from the second derivative of $\map {\Pi_X} s$ {{WRT|Differentiation}} $s$ at $x = 1$:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$. | From the definition of the probability generating function:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 2} \map p x s^x$
From Derivatives of Probability Generating Function at One:
:$\ds \map {\Pi' '_X} s = \sum_{x \mathop \ge 2} x \paren {x - 1} \map p x s^{x - 2}$
But it also holds when you include $x = 0$ and $x = 1$... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] whose [[Definition:Probability Generating Function|probability generating function]] is $\map {\Pi_X} s$.
Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ can be obtained from the [[Definition:Second Derivative|s... | From the definition of the [[Definition:Probability Generating Function|probability generating function]]:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 2} \map p x s^x$
From [[Derivatives of Probability Generating Function at One]]:
:$\ds \map {\Pi' '_X} s = \sum_{x \mathop \ge 2} x \paren {x - 1} \map p x s^{x - 2}$
... | Variance of Discrete Random Variable from PGF | https://proofwiki.org/wiki/Variance_of_Discrete_Random_Variable_from_PGF | https://proofwiki.org/wiki/Variance_of_Discrete_Random_Variable_from_PGF | [
"Variance of Discrete Random Variable from PGF",
"Variance",
"Probability Generating Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Generating Function",
"Definition:Variance/Discrete",
"Definition:Derivative/Higher Derivatives/Second Derivative",
"Definition:Expectation"
] | [
"Definition:Probability Generating Function",
"Derivatives of Probability Generating Function at One",
"Definition:Variance/Discrete"
] |
proofwiki-2688 | Nth Derivative of Mth Power | Let $m \in \Z$ be an integer such that $m \ge 0$.
The $n$th derivative of $x^m$ {{WRT|Differentiation}} $x$ is:
:$\dfrac {\d^n} {\d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ \\ 0 & : n > m \end{cases}$
where $m^{\underline n}$ denotes the falling factorial. | Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\dfrac {\d^n} {\d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ \\ 0 & : n > m \end{cases}$ | Let $m \in \Z$ be an [[Definition:Integer|integer]] such that $m \ge 0$.
The [[Definition:Higher Derivative|$n$th derivative]] of $x^m$ {{WRT|Differentiation}} $x$ is:
:$\dfrac {\d^n} {\d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ \\ 0 & : n > m \end{cases}$
where $m^{\underline n}$ denotes the... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\dfrac {\d^n} {\d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ \\ 0 & : n > m \end{cases}$ | Nth Derivative of Mth Power | https://proofwiki.org/wiki/Nth_Derivative_of_Mth_Power | https://proofwiki.org/wiki/Nth_Derivative_of_Mth_Power | [
"Derivatives"
] | [
"Definition:Integer",
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Falling Factorial"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-2689 | Derivative of Complex Power Series | Let $\xi \in \C$ be a complex number.
Let $\sequence {a_n}$ be a sequence in $\C$.
Let $\ds \map f z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a power series in a complex variable $z \in \C$ about $\xi$.
Let $R$ be the radius of convergence of the series defining $\map f z$.
Let $\cmod {z - \xi} < R$.
Th... | Define:
:$\ds \map g z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
From Radius of Convergence of Derivative of Complex Power Series, it follows that $g$ has radius of convergence $R$.
Fix an $\epsilon > 0$ satisfying $\epsilon < R - \cmod {z - \xi}$.
Define:
:$\ds M = \paren {R - \epsilon - \cmod {z ... | Let $\xi \in \C$ be a [[Definition:Complex Number|complex number]].
Let $\sequence {a_n}$ be a [[Definition:Sequence|sequence]] in $\C$.
Let $\ds \map f z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a [[Definition:Complex Power Series|power series]] in a [[Definition:Complex Number|complex]] variable $z ... | Define:
:$\ds \map g z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
From [[Radius of Convergence of Derivative of Complex Power Series]], it follows that $g$ has [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]] $R$.
Fix an $\epsilon > 0$ satisfying $\epsilon < R - \... | Derivative of Complex Power Series/Proof 1 | https://proofwiki.org/wiki/Derivative_of_Complex_Power_Series | https://proofwiki.org/wiki/Derivative_of_Complex_Power_Series/Proof_1 | [
"Derivative of Complex Power Series",
"Complex Differential Calculus",
"Complex Power Series",
"Complex Analysis"
] | [
"Definition:Complex Number",
"Definition:Sequence",
"Definition:Power Series/Complex Domain",
"Definition:Complex Number",
"Definition:Radius of Convergence/Complex Domain",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Derivative/Complex Function"
] | [
"Radius of Convergence of Derivative of Complex Power Series",
"Definition:Radius of Convergence/Complex Domain",
"Binomial Theorem",
"Difference of Absolutely Convergent Series",
"Triangle Inequality/Complex Numbers",
"Binomial Theorem"
] |
proofwiki-2690 | Derivative of Complex Power Series | Let $\xi \in \C$ be a complex number.
Let $\sequence {a_n}$ be a sequence in $\C$.
Let $\ds \map f z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a power series in a complex variable $z \in \C$ about $\xi$.
Let $R$ be the radius of convergence of the series defining $\map f z$.
Let $\cmod {z - \xi} < R$.
Th... | === Lemma ===
{{:Derivative of Complex Power Series/Proof 2/Lemma}}{{qed|lemma}}
Define:
:$\ds \map g z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
Fix an $\epsilon > 0$ satisfying $\epsilon < R - \cmod {z - \xi}$.
Let:
:$\ds M = \sum_{n \mathop = 2}^\infty \dfrac {n \paren {n - 1} } 2 \cmod {a_n} \pa... | Let $\xi \in \C$ be a [[Definition:Complex Number|complex number]].
Let $\sequence {a_n}$ be a [[Definition:Sequence|sequence]] in $\C$.
Let $\ds \map f z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a [[Definition:Complex Power Series|power series]] in a [[Definition:Complex Number|complex]] variable $z ... | === [[Derivative of Complex Power Series/Proof 2/Lemma|Lemma]] ===
{{:Derivative of Complex Power Series/Proof 2/Lemma}}{{qed|lemma}}
Define:
:$\ds \map g z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
Fix an $\epsilon > 0$ satisfying $\epsilon < R - \cmod {z - \xi}$.
Let:
:$\ds M = \sum_{n \mathop... | Derivative of Complex Power Series/Proof 2 | https://proofwiki.org/wiki/Derivative_of_Complex_Power_Series | https://proofwiki.org/wiki/Derivative_of_Complex_Power_Series/Proof_2 | [
"Derivative of Complex Power Series",
"Complex Differential Calculus",
"Complex Power Series",
"Complex Analysis"
] | [
"Definition:Complex Number",
"Definition:Sequence",
"Definition:Power Series/Complex Domain",
"Definition:Complex Number",
"Definition:Radius of Convergence/Complex Domain",
"Definition:Differentiable Mapping/Complex Function",
"Definition:Derivative/Complex Function"
] | [
"Derivative of Complex Power Series/Proof 2/Lemma",
"Nth Root Test",
"Derivative of Complex Power Series/Proof 2/Lemma",
"Triangle Inequality",
"Triangle Inequality",
"Difference of Two Powers",
"Closed Form for Triangular Numbers",
"Triangle Inequality",
"Difference of Two Powers",
"Difference of... |
proofwiki-2691 | Probability Generating Function of Discrete Uniform Distribution | Let $X$ be a discrete random variable with the discrete uniform distribution with parameter $n$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = \dfrac {s \paren {1 - s^n} } {n \paren {1 - s} }$ | From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the discrete uniform distribution:
:$\forall k \in \N, 1 \le k \le n: \map {p_X} k = \dfrac 1 n$
So:
{{begin-eqn}}
{{eqn | l = \map {\Pi_X} s
| r = \sum_{k \mathop = 1}^n \frac 1 n s^k
| c = ... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Discrete Uniform Distribution|discrete uniform distribution with parameter $n$]].
Then the [[Definition:Probability Generating Function|p.g.f.]] of $X$ is:
:$\map {\Pi_X} s = \dfrac {s \paren {1 - s^n} } {n \paren {1 -... | From the definition of [[Definition:Probability Generating Function|p.g.f]]:
:$\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map {p_X} x s^x$
From the definition of the [[Definition:Discrete Uniform Distribution|discrete uniform distribution]]:
:$\forall k \in \N, 1 \le k \le n: \map {p_X} k = \dfrac 1 n$
So:
{{begi... | Probability Generating Function of Discrete Uniform Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Discrete_Uniform_Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_of_Discrete_Uniform_Distribution | [
"Probability Generating Functions",
"Discrete Uniform Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Uniform Distribution/Discrete",
"Definition:Probability Generating Function"
] | [
"Definition:Probability Generating Function",
"Definition:Uniform Distribution/Discrete",
"Sum of Geometric Sequence",
"Category:Probability Generating Functions",
"Category:Discrete Uniform Distribution"
] |
proofwiki-2692 | Derivatives of PGF of Bernoulli Distribution | Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$.
Then the derivatives of the PGF of $X$ {{WRT|Differentiation}} $s$ are:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \begin{cases} p & : k = 1 \\ 0 & : k > 1 \end{cases}$ | The Probability Generating Function of Bernoulli Distribution is:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$.
We have that for a given Bernoulli distribution, $p$ and $q$ are constant.
So, from Derivative of Constant, Sum Rule for Derivatives, Derivative of Identity Function and Derivative of Constant Multiple:
:$\d... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]].
Then the [[Definition:Higher Derivative|derivatives]] of the [[Definition:Probability Generating Function|PGF]] of $X$ {{WRT|Differentiation}} $s$ are:... | The [[Probability Generating Function of Bernoulli Distribution]] is:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$.
We have that for a given Bernoulli distribution, $p$ and $q$ are constant.
So, from [[Derivative of Constant]], [[Sum Rule for Derivatives]], [[Derivative of Identity Function]] and [[Derivative of Co... | Derivatives of PGF of Bernoulli Distribution/Proof 1 | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Bernoulli_Distribution/Proof_1 | [
"Bernoulli Distribution",
"Derivatives of PGFs",
"Derivatives of PGF of Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Probability Generating Function"
] | [
"Probability Generating Function of Bernoulli Distribution",
"Derivative of Constant",
"Sum Rule for Derivatives",
"Derivative of Identity Function",
"Derivative of Constant Multiple",
"Derivative of Constant",
"Derivative of Constant"
] |
proofwiki-2693 | Derivatives of PGF of Bernoulli Distribution | Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$.
Then the derivatives of the PGF of $X$ {{WRT|Differentiation}} $s$ are:
:$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \begin{cases} p & : k = 1 \\ 0 & : k > 1 \end{cases}$ | Follows directly from Derivatives of PGF of Binomial Distribution, setting $n = 1$.
{{qed}} | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]].
Then the [[Definition:Higher Derivative|derivatives]] of the [[Definition:Probability Generating Function|PGF]] of $X$ {{WRT|Differentiation}} $s$ are:... | Follows directly from [[Derivatives of PGF of Binomial Distribution]], setting $n = 1$.
{{qed}} | Derivatives of PGF of Bernoulli Distribution/Proof 2 | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Bernoulli_Distribution | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Bernoulli_Distribution/Proof_2 | [
"Bernoulli Distribution",
"Derivatives of PGFs",
"Derivatives of PGF of Bernoulli Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Probability Generating Function"
] | [
"Derivatives of PGF of Binomial Distribution"
] |
proofwiki-2694 | Derivatives of PGF of Binomial Distribution | Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$.
Then the derivatives of the PGF of $X$ {{WRT|Differentiation}} $s$ are:
:<nowiki>$\dfrac {\d^k} {\d s^k} \map {\Pi_X} s = \begin {cases}
n^{\underline k} p^k \paren {q + p s}^{n-k} & : k \le n \\
0 & : k > n
\end {cases}$<... | The Probability Generating Function of Binomial Distribution is:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$.
From Derivatives of Function of $a x + b$:
:$\map {\dfrac {\d^k} {\d s^k} } {\map f {q + p s} } = p^k \dfrac {\d^k} {\d z^k} \paren {\map f z}$
where $z = q + p s$.
Here we have that $\map f z = z^... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Higher Derivative|derivatives]] of the [[Definition:Probability Generating Function|PGF]] of $X$ {{WRT|Differentiation}} $... | The [[Probability Generating Function of Binomial Distribution]] is:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$.
From [[Derivatives of Function of a x + b|Derivatives of Function of $a x + b$]]:
:$\map {\dfrac {\d^k} {\d s^k} } {\map f {q + p s} } = p^k \dfrac {\d^k} {\d z^k} \paren {\map f z}$
where $z... | Derivatives of PGF of Binomial Distribution | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Binomial_Distribution | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Binomial_Distribution | [
"Binomial Distribution",
"Derivatives of PGFs"
] | [
"Definition:Random Variable/Discrete",
"Definition:Binomial Distribution",
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Probability Generating Function",
"Definition:Falling Factorial"
] | [
"Probability Generating Function of Binomial Distribution",
"Derivatives of Function of a x + b",
"Nth Derivative of Mth Power"
] |
proofwiki-2695 | Higher Derivatives of Exponential Function | Let $\exp x$ be the exponential function.
Then:
:$\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$ | Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$ | Let $\exp x$ be the [[Definition:Exponential Function|exponential function]].
Then:
:$\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$ | Higher Derivatives of Exponential Function | https://proofwiki.org/wiki/Higher_Derivatives_of_Exponential_Function | https://proofwiki.org/wiki/Higher_Derivatives_of_Exponential_Function | [
"Derivatives involving Exponential Function"
] | [
"Definition:Exponential Function"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-2696 | Derivatives of PGF of Poisson Distribution | Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the derivatives of the PGF of $X$ {{WRT|Differentiation}} $s$ are:
:$\dfrac {d^k} {\d s^k} \, \map {\Pi_X} s = \lambda^k e^{- \lambda \paren {1 - s} }$ | The Probability Generating Function of Poisson Distribution is:
{{begin-eqn}}
{{eqn | l = \map {\Pi_X} s
| r = e^{-\lambda \paren {1 - s} }
| c =
}}
{{eqn | r = e^{-\lambda + \lambda s}
| c =
}}
{{eqn | r = e^{-\lambda} e^{\lambda s}
| c = Exponential of Sum
}}
{{end-eqn}}
We have that for a g... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]].
Then the [[Definition:Higher Derivative|derivatives]] of the [[Definition:Probability Generating Function|PGF]] of $X$ {{WRT|Differentiation}} $s$ ar... | The [[Probability Generating Function of Poisson Distribution]] is:
{{begin-eqn}}
{{eqn | l = \map {\Pi_X} s
| r = e^{-\lambda \paren {1 - s} }
| c =
}}
{{eqn | r = e^{-\lambda + \lambda s}
| c =
}}
{{eqn | r = e^{-\lambda} e^{\lambda s}
| c = [[Exponential of Sum]]
}}
{{end-eqn}}
We have t... | Derivatives of PGF of Poisson Distribution | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Poisson_Distribution | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Poisson_Distribution | [
"Poisson Distribution",
"Derivatives of PGFs"
] | [
"Definition:Random Variable/Discrete",
"Definition:Poisson Distribution",
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Probability Generating Function"
] | [
"Probability Generating Function of Poisson Distribution",
"Exponential of Sum",
"Definition:Poisson Distribution",
"Higher Derivatives of Exponential Function",
"Derivative of Constant Multiple",
"Higher Derivatives of Exponential Function",
"Category:Poisson Distribution",
"Category:Derivatives of P... |
proofwiki-2697 | Derivatives of Function of a x + b | Let $f$ be a real function which is differentiable on $\R$.
Let $a, b \in \R$ be constants.
Then:
:$\map {\dfrac {\d^n} {\d x^n} } {\map f {a x + b} } = a^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = a x + b$. | Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\map {\dfrac {\d^n} {\d x^n} } {\map f {a x + b} } = a^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = a x + b$. | Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Differentiable on Interval|differentiable]] on $\R$.
Let $a, b \in \R$ be [[Definition:Constant|constants]].
Then:
:$\map {\dfrac {\d^n} {\d x^n} } {\map f {a x + b} } = a^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = a x + b$. | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\map {\dfrac {\d^n} {\d x^n} } {\map f {a x + b} } = a^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = a x + b$. | Derivatives of Function of a x + b | https://proofwiki.org/wiki/Derivatives_of_Function_of_a_x_+_b | https://proofwiki.org/wiki/Derivatives_of_Function_of_a_x_+_b | [
"Differential Calculus",
"Derivatives"
] | [
"Definition:Real Function",
"Definition:Differentiable Mapping/Real Function/Interval",
"Definition:Constant"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-2698 | Nth Derivative of Reciprocal of Mth Power | Let $m \in \Z$ be an integer such that $m > 0$.
The $n$th derivative of $\dfrac 1 {x^m}$ {{WRT|Differentiation}} $x$ is:
:$\dfrac {\d^n} {\d x^n} \dfrac 1 {x^m} = \dfrac {\paren {-1}^n m^{\overline n}} {x^{m + n}}$
where $m^{\overline n}$ denotes the rising factorial. | Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\dfrac {\d^n} {\d x^n} \dfrac 1 {x^m} = \dfrac {\paren {-1}^n m^{\overline n}} {x^{m + n}}$ | Let $m \in \Z$ be an [[Definition:Integer|integer]] such that $m > 0$.
The [[Definition:Higher Derivative|$n$th derivative]] of $\dfrac 1 {x^m}$ {{WRT|Differentiation}} $x$ is:
:$\dfrac {\d^n} {\d x^n} \dfrac 1 {x^m} = \dfrac {\paren {-1}^n m^{\overline n}} {x^{m + n}}$
where $m^{\overline n}$ denotes the [[Definition... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\dfrac {\d^n} {\d x^n} \dfrac 1 {x^m} = \dfrac {\paren {-1}^n m^{\overline n}} {x^{m + n}}$ | Nth Derivative of Reciprocal of Mth Power | https://proofwiki.org/wiki/Nth_Derivative_of_Reciprocal_of_Mth_Power | https://proofwiki.org/wiki/Nth_Derivative_of_Reciprocal_of_Mth_Power | [
"Derivatives",
"Reciprocals"
] | [
"Definition:Integer",
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Rising Factorial"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition"
] |
proofwiki-2699 | Derivatives of PGF of Geometric Distribution | Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.
Then the derivatives of the PGF of $X$ {{WRT|Differentiation}} $s$ are:
:$\dfrac {\d^n} {\d s^n} \map {\Pi_X} s = \dfrac {q p^n n!} {\paren {1 - p s}^{n + 1} }$
where $q = 1 - p$. | The Probability Generating Function of Geometric Distribution is:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From Derivatives of Function of $a x + b$, we have that:
:$\map {\dfrac {\d^n} {\d s^n} } {\map f {1 - p s} } = \paren {-p}^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$
where $z = 1 - ps$.
Here we... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Geometric Distribution|geometric distribution with parameter $p$]].
Then the [[Definition:Higher Derivative|derivatives]] of the [[Definition:Probability Generating Function|PGF]] of $X$ {{WRT|Differentiation}} $s$ are... | The [[Probability Generating Function of Geometric Distribution]] is:
:$\map {\Pi_X} s = \dfrac q {1 - p s}$
where $q = 1 - p$.
From [[Derivatives of Function of a x + b|Derivatives of Function of $a x + b$]], we have that:
:$\map {\dfrac {\d^n} {\d s^n} } {\map f {1 - p s} } = \paren {-p}^n \map {\dfrac {\d^n} {\d z... | Derivatives of PGF of Geometric Distribution | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Geometric_Distribution | https://proofwiki.org/wiki/Derivatives_of_PGF_of_Geometric_Distribution | [
"Geometric Distribution",
"Derivatives of PGFs"
] | [
"Definition:Random Variable/Discrete",
"Definition:Geometric Distribution",
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Probability Generating Function"
] | [
"Probability Generating Function of Geometric Distribution",
"Derivatives of Function of a x + b",
"Nth Derivative of Reciprocal of Mth Power",
"Definition:Factorial",
"Category:Geometric Distribution",
"Category:Derivatives of PGFs"
] |
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