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Theorem 2.1.11. Suppose that \( X \) with density \( f \) and \( Y \) with distribution function \( G \) are independent. Then \( X + Y \) has density\n\n\[ h\left( x\right) = \int f\left( {x - y}\right) {dG}\left( y\right) \]\n\nWhen \( Y \) has density \( g \), the last formula can be written as\n\n\[ h\left( x\right... | Proof. From Theorem 2.1.10, the definition of density function, and Fubini's theorem (Theorem 1.7.2), which is justified since everything is nonnegative, we get\n\n\[ P\left( {X + Y \leq z}\right) = \int F\left( {z - y}\right) {dG}\left( y\right) = \iint {\int }_{-\infty }^{z}f\left( {x - y}\right) {dxdG}\left( y\right... | Yes |
Theorem 2.1.12. If \( X = \operatorname{gamma}\left( {\alpha ,\lambda }\right) \) and \( Y = \operatorname{gamma}\left( {\beta ,\lambda }\right) \) are independent then \( X + Y \) is gamma \( \left( {\alpha + \beta ,\lambda }\right) \) . Consequently if \( {X}_{1},\ldots {X}_{n} \) are independent exponential \( \left... | Proof. Writing \( {f}_{X + Y}\left( z\right) \) for the density function of \( X + Y \) and using Theorem 2.1.11\n\n\[ \n{f}_{X + Y}\left( x\right) = {\int }_{0}^{x}\frac{{\lambda }^{\alpha }{\left( x - y\right) }^{\alpha - 1}}{\Gamma \left( \alpha \right) }{e}^{-\lambda \left( {x - y}\right) }\frac{{\lambda }^{\beta }... | Yes |
Theorem 2.1.13. If \( X = \operatorname{normal}\left( {\mu, a}\right) \) and \( Y = \operatorname{normal}\left( {\nu, b}\right) \) are independent then \( X + Y = \operatorname{normal}\left( {\mu + \nu, a + b}\right) . | Proof. It is enough to prove the result for \( \mu = \nu = 0 \) . Suppose \( {Y}_{1} = \operatorname{normal}\left( {0, a}\right) \) and \( {Y}_{2} = \operatorname{normal}\left( {0, b}\right) \) . Then Theorem 2.1.11 implies\n\n\[ \n{f}_{{Y}_{1} + {Y}_{2}}\left( z\right) = \frac{1}{{2\pi }\sqrt{ab}}\int {e}^{-{x}^{2}/{2... | Yes |
Theorem 2.1.15. If \( S \) is a Borel subset of a complete separable metric space \( M \), and \( \mathcal{S} \) is the collection of Borel subsets of \( S \), then \( \left( {S,\mathcal{S}}\right) \) is nice. | Proof. We begin with the special case \( S = \lbrack 0,1{)}^{\mathbf{N}} \) with metric\n\n\[ \rho \left( {x, y}\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\left| {{x}_{n} - {y}_{n}}\right| /{2}^{n} \]\n\nIf \( x = \left( {{x}^{1},{x}^{2},{x}^{3},\ldots }\right) \), expand each component in binary \( {x}^{j} = .... | No |
Theorem 2.2.1. Let \( {X}_{1},\ldots ,{X}_{n} \) have \( E\left( {X}_{i}^{2}\right) < \infty \) and be uncorrelated. Then\n\n\[ \operatorname{var}\left( {{X}_{1} + \cdots + {X}_{n}}\right) = \operatorname{var}\left( {X}_{1}\right) + \cdots + \operatorname{var}\left( {X}_{n}\right) \]\n\nwhere \( \operatorname{var}\left... | Proof. Let \( {\mu }_{i} = E{X}_{i} \) and \( {S}_{n} = \mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} \) . Since \( E{S}_{n} = \mathop{\sum }\limits_{{i = 1}}^{n}{\mu }_{i} \), using the definition of the variance, writing the square of the sum as the product of two copies of the sum, and then expanding, we have\n\n\[ \op... | Yes |
Lemma 2.2.2. If \( p > 0 \) and \( E{\left| {Z}_{n}\right| }^{p} \rightarrow 0 \) then \( {Z}_{n} \rightarrow 0 \) in probability. | Proof. Chebyshev’s inequality, Theorem 1.6.4, with \( \varphi \left( x\right) = {x}^{p} \) and \( X = \left| {Z}_{n}\right| \) implies that if \( \epsilon > 0 \) then \( P\left( {\left| {Z}_{n}\right| \geq \epsilon }\right) \leq {\epsilon }^{-p}E{\left| {Z}_{n}\right| }^{p} \rightarrow 0 \) . | Yes |
Theorem 2.2.3. \( {L}^{2} \) weak law. Let \( {X}_{1},{X}_{2},\ldots \) be uncorrelated random variables with \( E{X}_{i} = \mu \) and \( \operatorname{var}\left( {X}_{i}\right) \leq C < \infty \) . If \( {S}_{n} = {X}_{1} + \ldots + {X}_{n} \) then as \( n \rightarrow \infty \) , \( {S}_{n}/n \rightarrow \mu \) in \( ... | Proof. To prove \( {L}^{2} \) convergence, observe that \( E\left( {{S}_{n}/n}\right) = \mu \), so\n\n\[ E{\left( {S}_{n}/n - \mu \right) }^{2} = \operatorname{var}\left( {{S}_{n}/n}\right) = \frac{1}{{n}^{2}}\left( {\operatorname{var}\left( {X}_{1}\right) + \cdots + \operatorname{var}\left( {X}_{n}\right) }\right) \le... | Yes |
Example 2.2.1. Polynomial approximation. Let \( f \) be a continuous function on \( \left\lbrack {0,1}\right\rbrack \), and let\n\n\[ \n{f}_{n}\left( x\right) = \mathop{\sum }\limits_{{m = 0}}^{n}\left( \begin{matrix} n \\ m \end{matrix}\right) {x}^{m}{\left( 1 - x\right) }^{n - m}f\left( {m/n}\right) \;\text{ where }\... | Proof. First observe that if \( {S}_{n} \) is the sum of \( n \) independent random variables with \( P\left( {{X}_{i} = 1}\right) = p \) and \( P\left( {{X}_{i} = 0}\right) = 1 - p \) then \( E{X}_{i} = p \), var \( \left( {X}_{i}\right) = p\left( {1 - p}\right) \) and\n\n\[ \nP\left( {{S}_{n} = m}\right) = \left( \be... | Yes |
A high-dimensional cube is almost the boundary of a ball. Let \( {X}_{1},{X}_{2},\ldots \) be independent and uniformly distributed on \( \left( {-1,1}\right) \) . Let \( {Y}_{i} = {X}_{i}^{2} \) , which are independent since they are functions of independent random variables. \( E{Y}_{i} = 1/3 \) and \( \operatorname{... | \[ \left( {{X}_{1}^{2} + \ldots + {X}_{n}^{2}}\right) /n \rightarrow 1/3\;\text{ in probability as }n \rightarrow \infty \] Let \( {A}_{n,\epsilon } = \left\{ {x \in {\mathbf{R}}^{n} : \left( {1 - \epsilon }\right) \sqrt{n/3} < \left| x\right| < \left( {1 + \epsilon }\right) \sqrt{n/3}}\right\} \) where \( \left| x\rig... | Yes |
Theorem 2.2.4. Let \( {\mu }_{n} = E{S}_{n},{\sigma }_{n}^{2} = \operatorname{var}\left( {S}_{n}\right) \) . If \( {\sigma }_{n}^{2}/{b}_{n}^{2} \rightarrow 0 \) then\n\n\[ \frac{{S}_{n} - {\mu }_{n}}{{b}_{n}} \rightarrow 0\;\text{ in probability } \] | Proof. Our assumptions imply \( E{\left( \left( {S}_{n} - {\mu }_{n}\right) /{b}_{n}\right) }^{2} = {b}_{n}^{-2}\operatorname{var}\left( {S}_{n}\right) \rightarrow 0 \), so the desired conclusion follows from Lemma 2.2.2. | Yes |
Coupon collector’s problem. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. uniform on \( \{ 1,2,\ldots, n\} \) . To motivate the name, think of collecting baseball cards (or coupons). Suppose that the \( i \) th item we collect is chosen at random from the set of possibilities and is independent of the previous choices. Le... | It is easy to see that \( {\tau }_{1}^{n} = 1 \) . To make later formulas work out nicely, we will set \( {\tau }_{0}^{n} = 0 \) . For \( 1 \leq k \leq n,{X}_{n, k} \equiv {\tau }_{k}^{n} - {\tau }_{k - 1}^{n} \) represents the time to get a choice different from our first \( k - 1 \), so \( {X}_{n, k} \) has a geometr... | Yes |
Example 2.2.4. Random permutations. Let \( {\Omega }_{n} \) consist of the \( n \) ! permutations (i.e., one-to-one mappings from \( \{ 1,\ldots, n\} \) onto \( \{ 1,\ldots, n\} \) ) and make this into a probability space by assuming all the permutations are equally likely. This application of the weak law concerns the... | Lemma 2.2.5. \( {X}_{n, | No |
Lemma 2.2.5. \( {X}_{n,1},\ldots ,{X}_{n, n} \) are independent and \( P\left( {{X}_{n, j} = 1}\right) = \frac{1}{n - j + 1} \) . | Proof. To prove this, it is useful to generate the permutation in a special way. Let \( {i}_{1} = 1 \) . Pick \( {j}_{1} \) at random from \( \{ 1,\ldots, n\} \) and let \( \pi \left( {i}_{1}\right) = {j}_{1} \) . If \( {j}_{1} \neq 1 \), let \( {i}_{2} = {j}_{1} \) . If \( {j}_{1} = 1 \), let \( {i}_{2} = 2 \) . In ei... | Yes |
Example 2.2.5. An occupancy problem. Suppose we put \( r \) balls at random in \( n \) boxes, i.e., all \( {n}^{r} \) assignments of balls to boxes have equal probability. Let \( {A}_{i} \) be the event that the \( i \) th box is empty and \( {N}_{n} = \) the number of empty boxes. It is easy to see that\n\n\[ P\left( ... | A little calculus (take logarithms) shows that if \( r/n \rightarrow c, E{N}_{n}/n \rightarrow {e}^{-c} \) . (For a proof, see Lemma 3.1.1.) To compute the variance of \( {N}_{n} \), we observe that\n\n\[ E{N}_{n}^{2} = E{\left( \mathop{\sum }\limits_{{m = 1}}^{n}{1}_{{A}_{m}}\right) }^{2} = \mathop{\sum }\limits_{{1 \... | Yes |
Theorem 2.2.6. Weak law for triangular arrays. For each \( n \) let \( {X}_{n, k},1 \leq k \leq n \) , be independent. Let \( {b}_{n} > 0 \) with \( {b}_{n} \rightarrow \infty \), and let \( {\bar{X}}_{n, k} = {X}_{n, k}{1}_{\left( \left| {X}_{n, k}\right| \leq {b}_{n}\right) } \) . Suppose that as \( n \rightarrow \in... | Proof. Let \( {\bar{S}}_{n} = {\bar{X}}_{n,1} + \cdots + {\bar{X}}_{n, n} \) . Clearly,\n\n\[ P\left( {\left| \frac{{S}_{n} - {a}_{n}}{{b}_{n}}\right| > \epsilon }\right) \leq P\left( {{S}_{n} \neq {\bar{S}}_{n}}\right) + P\left( {\left| \frac{{\bar{S}}_{n} - {a}_{n}}{{b}_{n}}\right| > \epsilon }\right) \]\n\nTo estima... | Yes |
Theorem 2.2.7. Weak law of large numbers. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with\n\n\[ \n{xP}\left( {\left| {X}_{i}\right| > x}\right) \rightarrow 0\;\text{ as }x \rightarrow \infty \n\]\n\nLet \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) and let \( {\mu }_{n} = E\left( {{X}_{1}{1}_{\left( \left| {X}_{1}\right| ... | Proof. We will apply Theorem 2.2.6 with \( {X}_{n, k} = {X}_{k} \) and \( {b}_{n} = n \) . To check (i), we note\n\n\[ \n\mathop{\sum }\limits_{{k = 1}}^{n}P\left( {\left| {X}_{n, k}\right| > n}\right) = {nP}\left( {\left| {X}_{i}\right| > n}\right) \rightarrow 0 \n\]\n\nby assumption. To check (ii), we need to show \(... | No |
Lemma 2.2.8. If \( Y \geq 0 \) and \( p > 0 \) then \( E\left( {Y}^{p}\right) = {\int }_{0}^{\infty }p{y}^{p - 1}P\left( {Y > y}\right) {dy} \) . | Proof. Using the definition of expected value, Fubini's theorem (for nonnegative random variables), and then calculating the resulting integrals gives\n\n\[ \n{\int }_{0}^{\infty }p{y}^{p - 1}P\left( {Y > y}\right) {dy} = {\int }_{0}^{\infty }{\int }_{\Omega }p{y}^{p - 1}{1}_{\left( Y > y\right) }{dPdy} \n\]\n\n\[ \n= ... | Yes |
Theorem 2.2.9. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E\left| {X}_{i}\right| < \infty \) . Let \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) and let \( \mu = E{X}_{1} \) . Then \( {S}_{n}/n \rightarrow \mu \) in probability. | Proof. Two applications of the dominated convergence theorem imply\n\n\[ \n{xP}\left( {\left| {X}_{1}\right| > x}\right) \leq E\left( {\left| {X}_{1}\right| {1}_{\left( \left| {X}_{1}\right| > x\right) }}\right) \rightarrow 0\;\text{ as }x \rightarrow \infty \]\n\n\[ \n{\mu }_{n} = E\left( {{X}_{1}{1}_{\left( \left| {X... | Yes |
For an example where the weak law does not hold, suppose \( {X}_{1},{X}_{2},\ldots \) are independent and have a Cauchy distribution: | As \( x \rightarrow \infty \) ,\n\n\[ P\left( {\left| {X}_{1}\right| > x}\right) = 2{\int }_{x}^{\infty }\frac{dt}{\pi \left( {1 + {t}^{2}}\right) } \sim \frac{2}{\pi }{\int }_{x}^{\infty }{t}^{-2}{dt} = \frac{2}{\pi }{x}^{-1} \]\n\nFrom the necessity of the condition above, we can conclude that there is no sequence of... | No |
Theorem 2.3.1. Borel-Cantelli lemma. If \( \mathop{\sum }\limits_{{n = 1}}^{\infty }P\left( {A}_{n}\right) < \infty \) then\n\n\[ P\left( {{A}_{n}\text{ i.o. }}\right) = 0. \] | Proof. Let \( N = \mathop{\sum }\limits_{k}{1}_{{A}_{k}} \) be the number of events that occur. Fubini’s theorem implies \( {EN} = \mathop{\sum }\limits_{k}P\left( {A}_{k}\right) < \infty \), so we must have \( N < \infty \) a.s. | Yes |
Theorem 2.3.2. \( {X}_{n} \rightarrow X \) in probability if and only if for every subsequence \( {X}_{n\left( m\right) } \) there is a further subsequence \( {X}_{n\left( {m}_{k}\right) } \) that converges almost surely to \( X \) . | Proof. Let \( {\epsilon }_{k} \) be a sequence of positive numbers that \( \downarrow 0 \) . For each \( k \), there is an \( n\left( {m}_{k}\right) > n\left( {m}_{k - 1}\right) \) so that \( P\left( {\left| {{X}_{n\left( {m}_{k}\right) } - X}\right| > {\epsilon }_{k}}\right) \leq {2}^{-k} \) . Since\n\n\[ \mathop{\sum... | Yes |
Theorem 2.3.3. Let \( {y}_{n} \) be a sequence of elements of a topological space. If every subsequence \( {y}_{n\left( m\right) } \) has a further subsequence \( {y}_{n\left( {m}_{k}\right) } \) that converges to \( y \) then \( {y}_{n} \rightarrow y \) . | Proof. If \( {y}_{n} \nrightarrow y \) then there is an open set \( G \) containing \( y \) and a subsequence \( {y}_{n\left( m\right) } \) with \( {y}_{n\left( m\right) } \notin G \) for all \( m \), but clearly no subsequence of \( {y}_{n\left( m\right) } \) converges to \( y \) . | Yes |
Theorem 2.3.4. If \( f \) is continuous and \( {X}_{n} \rightarrow X \) in probability then \( f\left( {X}_{n}\right) \rightarrow f\left( X\right) \) in probability. If, in addition, \( f \) is bounded then \( {Ef}\left( {X}_{n}\right) \rightarrow {Ef}\left( X\right) \) . | Proof. If \( {X}_{n\left( m\right) } \) is a subsequence then Theorem 2.3.2 implies there is a further subsequence \( {X}_{n\left( {m}_{k}\right) } \rightarrow X \) almost surely. Since \( f \) is continuous, Exercise 1.3.3 implies \( f\left( {X}_{n\left( {m}_{k}\right) }\right) \rightarrow f\left( X\right) \) almost s... | No |
Theorem 2.3.5. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E{X}_{i} = \mu \) and \( E{X}_{i}^{4} < \infty \) . If \( {S}_{n} = \) \( {X}_{1} + \cdots + {X}_{n} \) then \( {S}_{n}/n \rightarrow \mu \) a.s. | Proof. By letting \( {X}_{i}^{\prime } = {X}_{i} - \mu \), we can suppose without loss of generality that \( \mu = 0 \) . Now\n\n\[ E{S}_{n}^{4} = E{\left( \mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i}\right) }^{4} = E\mathop{\sum }\limits_{{1 \leq i, j, k,\ell \leq n}}{X}_{i}{X}_{j}{X}_{k}{X}_{\ell } \]\n\nTerms in the s... | Yes |
Theorem 2.3.6. The second Borel-Cantelli lemma. If the events \( {A}_{n} \) are independent then \( \sum P\left( {A}_{n}\right) = \infty \) implies \( P\left( {A}_{n}\right. \) i.o. \( ) = 1 \) . | Proof. Let \( M < N < \infty \) . Independence and \( 1 - x \leq {e}^{-x} \) imply\n\n\[ P\left( {{ \cap }_{n = M}^{N}{A}_{n}^{c}}\right) = \mathop{\prod }\limits_{{n = M}}^{N}\left( {1 - P\left( {A}_{n}\right) }\right) \leq \mathop{\prod }\limits_{{n = M}}^{N}\exp \left( {-P\left( {A}_{n}\right) }\right) \]\n\n\[ = \e... | Yes |
Theorem 2.3.7. If \( {X}_{1},{X}_{2},\ldots \) are i.i.d. with \( E\left| {X}_{i}\right| = \infty \), then \( P\left( {\left| {X}_{n}\right| \geq n\text{i.o.}}\right) = 1 \) . So if \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) then \( P\left( {\lim {S}_{n}/n}\right. \) exists \( \left. { \in \left( {-\infty ,\infty }\rig... | Proof. From Lemma 2.2.8, we get\n\n\[ E\left| {X}_{1}\right| = {\int }_{0}^{\infty }P\left( {\left| {X}_{1}\right| > x}\right) {dx} \leq \mathop{\sum }\limits_{{n = 0}}^{\infty }P\left( {\left| {X}_{1}\right| > n}\right) \]\n\nSince \( E\left| {X}_{1}\right| = \infty \) and \( {X}_{1},{X}_{2},\ldots \) are i.i.d., it f... | Yes |
Theorem 2.3.8. If \( {A}_{1},{A}_{2},\ldots \) are pairwise independent and \( \mathop{\sum }\limits_{{n = 1}}^{\infty }P\left( {A}_{n}\right) = \infty \) then as \( n \rightarrow \infty \)\n\[ \mathop{\sum }\limits_{{m = 1}}^{n}{1}_{{A}_{m}}/\mathop{\sum }\limits_{{m = 1}}^{n}P\left( {A}_{m}\right) \rightarrow 1\;\tex... | Proof. Let \( {X}_{m} = {1}_{{A}_{m}} \) and let \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) . Since the \( {A}_{m} \) are pairwise independent, the \( {X}_{m} \) are uncorrelated and hence Theorem 2.2.1 implies\n\n\[ \operatorname{var}\left( {S}_{n}\right) = \operatorname{var}\left( {X}_{1}\right) + \cdots + \operatorn... | Yes |
Claim. The \( {A}_{k} \) are independent with \( P\left( {A}_{k}\right) = 1/k \) . | To prove this, we start by observing that since \( F \) is continuous \( P\left( {{X}_{j} = {X}_{k}}\right) = 0 \) for any \( j \neq k \) (see Exercise 2.1.8), so we can let \( {Y}_{1}^{n} > {Y}_{2}^{n} > \cdots > {Y}_{n}^{n} \) be the random variables \( {X}_{1},\ldots ,{X}_{n} \) put into decreasing order and define ... | No |
Example 2.3.3. Head runs. Let \( {X}_{n}, n \in \mathbf{Z} \), be i.i.d. with \( P\left( {{X}_{n} = 1}\right) = P\left( {{X}_{n} = }\right. \) \( - 1) = 1/2 \) . Let \( {\ell }_{n} = \max \left\{ {m : {X}_{n - m + 1} = \ldots = {X}_{n} = 1}\right\} \) be the length of the run of +1’s at time \( n \), and let \( {L}_{n}... | To prove (2.3.1), we begin by observing\n\n\[ \nP\left( {{\ell }_{n} \geq \left( {1 + \epsilon }\right) {\log }_{2}n}\right) \leq {n}^{-\left( {1 + \epsilon }\right) } \n\]\n\nfor any \( \epsilon > 0 \), so it follows from the Borel-Cantelli lemma that \( {\ell }_{n} \leq \left( {1 + \epsilon }\right) {\log }_{2}n \) f... | Yes |
Theorem 2.4.1. Strong law of large numbers. Let \( {X}_{1},{X}_{2},\ldots \) be pairwise independent identically distributed random variables with \( E\left| {X}_{i}\right| < \infty \) . Let \( E{X}_{i} = \mu \) and \( {S}_{n} = {X}_{1} + \ldots + {X}_{n} \) . Then \( {S}_{n}/n \rightarrow \mu \) a.s. as \( n \rightarr... | Proof. As in the proof of the weak law of large numbers, we begin by truncating. | No |
Lemma 2.4.2. Let \( {Y}_{k} = {X}_{k}{1}_{\left( \left| {X}_{k}\right| \leq k\right) } \) and \( {T}_{n} = {Y}_{1} + \cdots + {Y}_{n} \) . It is sufficient to prove that \( {T}_{n}/n \rightarrow \mu \) a.s. | Proof. \( \mathop{\sum }\limits_{{k = 1}}^{\infty }P\left( {\left| {X}_{k}\right| > k}\right) \leq {\int }_{0}^{\infty }P\left( {\left| {X}_{1}\right| > t}\right) {dt} = E\left| {X}_{1}\right| < \infty \) so \( P\left( {{X}_{k} \neq {Y}_{k}}\right. \) i.o. \( ) = 0 \) . This shows that \( \left| {{S}_{n}\left( \omega \... | Yes |
Lemma 2.4.3. \( \mathop{\sum }\limits_{{k = 1}}^{\infty }\operatorname{var}\left( {Y}_{k}\right) /{k}^{2} \leq {4E}\left| {X}_{1}\right| < \infty \) . | Proof. To bound the sum, we observe\n\n\[\n\operatorname{var}\left( {Y}_{k}\right) \leq E\left( {Y}_{k}^{2}\right) = {\int }_{0}^{\infty }{2yP}\left( {\left| {Y}_{k}\right| > y}\right) {dy} \leq {\int }_{0}^{k}{2yP}\left( {\left| {X}_{1}\right| > y}\right) {dy}\n\]\n\nso using Fubini’s theorem (since everything is \( \... | No |
Lemma 2.4.4. If \( y \geq 0 \) then \( {2y}\mathop{\sum }\limits_{{k > y}}{k}^{-2} \leq 4 \) . | Proof. We begin with the observation that if \( m \geq 2 \) then\n\n\[ \mathop{\sum }\limits_{{k \geq m}}{k}^{-2} \leq {\int }_{m - 1}^{\infty }{x}^{-2}{dx} = {\left( m - 1\right) }^{-1} \]\n\nWhen \( y \geq 1 \), the sum starts with \( k = \left\lbrack y\right\rbrack + 1 \geq 2 \), so\n\n\[ {2y}\mathop{\sum }\limits_{... | Yes |
Theorem 2.4.5. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E{X}_{i}^{ + } = \infty \) and \( E{X}_{i}^{ - } < \infty \) . If \( {S}_{n} = \) \( {X}_{1} + \cdots + {X}_{n} \) then \( {S}_{n}/n \rightarrow \infty \) a.s. | Proof. Let \( M > 0 \) and \( {X}_{i}^{M} = {X}_{i} \land M \) . The \( {X}_{i}^{M} \) are i.i.d. with \( E\left| {X}_{i}^{M}\right| < \infty \), so if \( {S}_{n}^{M} = {X}_{1}^{M} + \cdots + {X}_{n}^{M} \) then Theorem 2.4.1 implies \( {S}_{n}^{M}/n \rightarrow E{X}_{i}^{M} \) . Since \( {X}_{i} \geq {X}_{i}^{M} \) , ... | Yes |
Theorem 2.4.6. If \( E{X}_{1} = \mu \leq \infty \) then as \( t \rightarrow \infty \) , \[ {N}_{t}/t \rightarrow 1/\mu \text{ a.s. }\;\left( {1/\infty = 0}\right) . | Proof. By Theorems 2.4.1 and 2.4.5, \( {T}_{n}/n \rightarrow \mu \) a.s. From the definition of \( {N}_{t} \), it follows that \( T\left( {N}_{t}\right) \leq t < T\left( {{N}_{t} + 1}\right) \), so dividing through by \( {N}_{t} \) gives \[ \frac{T\left( {N}_{t}\right) }{{N}_{t}} \leq \frac{t}{{N}_{t}} \leq \frac{T\lef... | Yes |
Theorem 2.4.7. The Glivenko-Cantelli theorem. As \( n \rightarrow \infty \) ,\n\n\[ \mathop{\sup }\limits_{x}\left| {{F}_{n}\left( x\right) - F\left( x\right) }\right| \rightarrow 0\;\text{ a.s. } \] | Proof. Fix \( x \) and let \( {Y}_{n} = {1}_{\left( {X}_{n} \leq x\right) } \) . Since the \( {Y}_{n} \) are i.i.d. with \( E{Y}_{n} = P\left( {{X}_{n} \leq x}\right) = \) \( F\left( x\right) \), the strong law of large numbers implies that \( {F}_{n}\left( x\right) = {n}^{-1}\mathop{\sum }\limits_{{m = 1}}^{n}{Y}_{m} ... | Yes |
Example 2.4.3. Shannon’s theorem. Let \( {X}_{1},{X}_{2},\ldots \in \{ 1,\ldots, r\} \) be independent with \( P\left( {{X}_{i} = k}\right) = p\left( k\right) > 0 \) for \( 1 \leq k \leq r \) . Here we are thinking of \( 1,\ldots, r \) as the letters of an alphabet, and \( {X}_{1},{X}_{2},\ldots \) are the successive l... | \[ - {n}^{-1}\log {\pi }_{n}\left( \omega \right) \rightarrow H \equiv - \mathop{\sum }\limits_{{k = 1}}^{r}p\left( k\right) \log p\left( k\right) \text{ a.s. } \] The constant \( H \) is called the entropy of the source and is a measure of how random it is. The last result is the asymptotic equipartition property: If ... | Yes |
If \( {B}_{n} \in \mathcal{R} \) then \( \left\{ {{X}_{n} \in {B}_{n}\text{i.o.}}\right\} \in \mathcal{T} \) . | If we let \( {X}_{n} = {1}_{{A}_{n}} \) and \( {B}_{n} = \{ 1\} \), this example becomes \( \left\{ {A}_{n}\right. \) i.o. \( \} \) . | No |
Let \( {S}_{n} = {X}_{1} + \ldots + {X}_{n} \). It is easy to check that | \( \left\{ {\mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{n}\text{ exists }}\right\} \in \mathcal{T} \)\n\n\( \left\{ {\lim \mathop{\sup }\limits_{{n \rightarrow \infty }}{S}_{n} > 0}\right\} \notin \mathcal{T}, \)\n\n\( \left\{ {\lim \mathop{\sup }\limits_{{n \rightarrow \infty }}{S}_{n}/{c}_{n} > x}\right\} \in... | No |
Theorem 2.5.1. Kolmogorov’s 0-1 law. If \( {X}_{1},{X}_{2},\ldots \) are independent and \( A \in \mathcal{T} \) then \( P\left( A\right) = 0 \) or 1 . | Proof. We will show that \( A \) is independent of itself, that is, \( P\left( {A \cap A}\right) = P\left( A\right) P\left( A\right) \) , so \( P\left( A\right) = P{\left( A\right) }^{2} \), and hence \( P\left( A\right) = 0 \) or 1 . We will sneak up on this conclusion in two steps:\n\n(a) \( A \in \sigma \left( {{X}_... | Yes |
Theorem 2.5.2. Kolmogorov’s maximal inequality. Suppose \( {X}_{1},\ldots ,{X}_{n} \) are independent with \( E{X}_{i} = 0 \) and \( \operatorname{var}\left( {X}_{i}\right) < \infty \) . If \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) then\n\n\[ P\left( {\mathop{\max }\limits_{{1 \leq k \leq n}}\left| {S}_{k}\right| \geq... | Proof. Let \( {A}_{k} = \left\{ {\left| {S}_{k}\right| \geq x}\right. \) but \( \left. {\left| {S}_{j}\right| < x\text{for}j < k}\right\} \), i.e., we break things down according to the time that \( \left| {S}_{k}\right| \) first exceeds \( x \) . Since the \( {A}_{k} \) are disjoint and \( \left( {{S}_{n} - }\right. \... | Yes |
Theorem 2.5.3. Suppose \( {X}_{1},{X}_{2},\ldots \) are independent and have \( E{X}_{n} = 0 \) . If\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\operatorname{var}\left( {X}_{n}\right) < \infty \]\n\nthen with probability one \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{X}_{n}\left( \omega \right) \) converges. | Proof. Let \( {S}_{N} = \mathop{\sum }\limits_{{n = 1}}^{N}{X}_{n} \) . From Theorem 2.5.2, we get\n\n\[ P\left( {\mathop{\max }\limits_{{M \leq m \leq N}}\left| {{S}_{m} - {S}_{M}}\right| > \epsilon }\right) \leq {\epsilon }^{-2}\operatorname{var}\left( {{S}_{N} - {S}_{M}}\right) = {\epsilon }^{-2}\mathop{\sum }\limit... | Yes |
Theorem 2.5.4. Kolmogorov’s three-series theorem. Let \( {X}_{1},{X}_{2},\ldots \) be independent. Let \( A > 0 \) and let \( {Y}_{i} = {X}_{i}{1}_{\left( \left| {X}_{i}\right| \leq A\right) } \) . In order that \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{X}_{n} \) converges a.s., it is necessary and sufficient that\n... | Proof. We will prove the necessity in Example 3.4.7 as an application of the central limit theorem. To prove the sufficiency, let \( {\mu }_{n} = E{Y}_{n} \) . (iii) and Theorem 2.5.3 imply that \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\left( {{Y}_{n} - {\mu }_{n}}\right) \) converges a.s. Using (ii) now gives that ... | Yes |
Theorem 2.5.5. Kronecker’s lemma. If \( {a}_{n} \uparrow \infty \) and \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{x}_{n}/{a}_{n} \) converges then\n\n\[ {a}_{n}^{-1}\mathop{\sum }\limits_{{m = 1}}^{n}{x}_{m} \rightarrow 0 \] | Proof. Let \( {a}_{0} = 0,{b}_{0} = 0 \), and for \( m \geq 1 \), let \( {b}_{m} = \mathop{\sum }\limits_{{k = 1}}^{m}{x}_{k}/{a}_{k} \) . Then \( {x}_{m} = \) \( {a}_{m}\left( {{b}_{m} - {b}_{m - 1}}\right) \) and so\n\n\[ {a}_{n}^{-1}\mathop{\sum }\limits_{{m = 1}}^{n}{x}_{m} = {a}_{n}^{-1}\left\{ {\mathop{\sum }\lim... | Yes |
Theorem 2.5.6. The strong law of large numbers. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. random variables with \( E\left| {X}_{i}\right| < \infty \) . Let \( E{X}_{i} = \mu \) and \( {S}_{n} = {X}_{1} + \ldots + {X}_{n} \) . Then \( {S}_{n}/n \rightarrow \mu \) a.s. as \( n \rightarrow \infty \) . | Proof. Let \( {Y}_{k} = {X}_{k}{1}_{\left( \left| {X}_{k}\right| \leq k\right) } \) and \( {T}_{n} = {Y}_{1} + \cdots + {Y}_{n} \) . By (a) in the proof of Theorem 2.4.1 it suffices to show that \( {T}_{n}/n \rightarrow \mu \) . Let \( {Z}_{k} = {Y}_{k} - E{Y}_{k} \), so \( E{Z}_{k} = 0 \) . Now \( \operatorname{var}\l... | Yes |
Theorem 2.5.7. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. random variables with \( E{X}_{i} = 0 \) and \( E{X}_{i}^{2} = \) \( {\sigma }^{2} < \infty \) . Let \( {S}_{n} = {X}_{1} + \ldots + {X}_{n} \) . If \( \epsilon > 0 \) then\n\n\[ \n{S}_{n}/{n}^{1/2}{\left( \log n\right) }^{1/2 + \epsilon } \rightarrow 0\;\text{ ... | Proof. Let \( {a}_{n} = {n}^{1/2}{\left( \log n\right) }^{1/2 + \epsilon } \) for \( n \geq 2 \) and \( {a}_{1} > 0 \) .\n\n\[ \n\mathop{\sum }\limits_{{n = 1}}^{\infty }\operatorname{var}\left( {{X}_{n}/{a}_{n}}\right) = {\sigma }^{2}\left( {\frac{1}{{a}_{1}^{2}} + \mathop{\sum }\limits_{{n = 2}}^{\infty }\frac{1}{n{\... | Yes |
Theorem 2.5.9. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E\left| {X}_{1}\right| = \infty \) and let \( {S}_{n} = {X}_{1} + \) \( \cdots + {X}_{n} \) . Let \( {a}_{n} \) be a sequence of positive numbers with \( {a}_{n}/n \) increasing. Then \( \mathop{\limsup }\limits_{{n \rightarrow \infty }}\left| {S}_{n}\ri... | Proof. Since \( {a}_{n}/n \uparrow ,{a}_{kn} \geq k{a}_{n} \) for any integer \( k \) . Using this and \( {a}_{n} \uparrow \) ,\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }P\left( {\left| {X}_{1}\right| \geq k{a}_{n}}\right) \geq \mathop{\sum }\limits_{{n = 1}}^{\infty }P\left( {\left| {X}_{1}\right| \geq {a}_{kn}}\... | Yes |
Lemma 2.6.1. If \( {\gamma }_{m + n} \geq {\gamma }_{m} + {\gamma }_{n} \) then as \( n \rightarrow \infty ,{\gamma }_{n}/n \rightarrow \mathop{\sup }\limits_{m}{\gamma }_{m}/m \) . | Proof. Clearly, \( \lim \sup {\gamma }_{n}/n \leq \sup {\gamma }_{m}/m \) . To complete the proof, it suffices to prove that for any \( m \) liminf \( {\gamma }_{n}/n \geq {\gamma }_{m}/m \) . Writing \( n = {km} + \ell \) with \( 0 \leq \ell < m \) and making repeated use of the hypothesis gives \( {\gamma }_{n} \geq ... | Yes |
Lemma 2.6.2. If \( a > \mu \) and \( \theta > 0 \) is small then \( {a\theta } - \kappa \left( \theta \right) > 0 \) . | Proof. \( \kappa \left( 0\right) = \log \varphi \left( 0\right) = 0 \), so it suffices to show that (i) \( \kappa \) is continuous at 0,(ii) differentiable on \( \left( {0,{\theta }_{ + }}\right) \), and (iii) \( {\kappa }^{\prime }\left( \theta \right) \rightarrow \mu \) as \( \theta \rightarrow 0 \) . For then\n\n\[ ... | Yes |
\[ \int {e}^{\theta x}{\left( 2\pi \right) }^{-1/2}\exp \left( {-{x}^{2}/2}\right) {dx} = \exp \left( {{\theta }^{2}/2}\right) \int {\left( 2\pi \right) }^{-1/2}\exp \left( {-{\left( x - \theta \right) }^{2}/2}\right) {dx} \] | The integrand in the last integral is the density of a normal distribution with mean \( \theta \) and variance 1, so \( \varphi \left( \theta \right) = \exp \left( {{\theta }^{2}/2}\right) ,\theta \in \left( {-\infty ,\infty }\right) \) . In this case, \( {\varphi }^{\prime }\left( \theta \right) /\varphi \left( \theta... | Yes |
Example 2.6.2. Exponential distribution with parameter \( \lambda \) . If \( \theta < \lambda \) | \[ {\int }_{0}^{\infty }{e}^{\theta x}\lambda {e}^{-{\lambda x}}{dx} = \lambda /\left( {\lambda - \theta }\right) \] \( {\varphi }^{\prime }\left( \theta \right) \varphi \left( \theta \right) = 1/\left( {\lambda - \theta }\right) \) and \[ {F}_{\theta }\left( x\right) = \frac{\lambda }{\lambda - \theta }{\int }_{0}^{x}... | Yes |
Example 2.6.3. Coin flips. \( P\left( {{X}_{i} = 1}\right) = P\left( {{X}_{i} = - 1}\right) = 1/2 \) | \[ \varphi \left( \theta \right) = \left( {{e}^{\theta } + {e}^{-\theta }}\right) /2 \] \[ {\varphi }^{\prime }\left( \theta \right) /\varphi \left( \theta \right) = \left( {{e}^{\theta } - {e}^{-\theta }}\right) /\left( {{e}^{\theta } + {e}^{-\theta }}\right) \] \( {F}_{\theta }\left( {\{ x\} }\right) /F\left( {\{ x\}... | Yes |
Example 2.6.4. Perverted exponential. Let \( g\left( x\right) = C{x}^{-3}{e}^{-x} \) for \( x \geq 1, g\left( x\right) = 0 \) otherwise, and choose \( C \) so that \( g \) is a probability density. In this case, | \[ \varphi \left( \theta \right) = \int {e}^{\theta x}g\left( x\right) {dx} < \infty \] if and only if \( \theta \leq 1 \), and when \( \theta \leq 1 \), we have \[ \frac{{\varphi }^{\prime }\left( \theta \right) }{\varphi \left( \theta \right) } \leq \frac{{\varphi }^{\prime }\left( 1\right) }{\varphi \left( 1\right) ... | Yes |
Theorem 2.6.3. Suppose in addition to (H1) and (H2) that there is a \( {\theta }_{a} \in \left( {0,{\theta }_{ + }}\right) \) so that \( a = {\varphi }^{\prime }\left( {\theta }_{a}\right) /\varphi \left( {\theta }_{a}\right) \) . Then, as \( n \rightarrow \infty \) ,\n\n\[ \n{n}^{-1}\log P\left( {{S}_{n} \geq {na}}\ri... | Proof. The fact that the limsup of the left-hand side \( \leq \) the right-hand side follows from (2.6.2). To prove the other inequality, pick \( \lambda \in \left( {{\theta }_{a},{\theta }_{ + }}\right) \), let \( {X}_{1}^{\lambda },{X}_{2}^{\lambda },\ldots \) be i.i.d. with distribution \( {F}_{\lambda } \) and let ... | Yes |
Lemma 2.6.4. \( \frac{d{F}^{n}}{d{F}_{\lambda }^{n}} = {e}^{-{\lambda x}}\varphi {\left( \lambda \right) }^{n} \) . | Proof. We will prove this by induction. The result holds when \( n = 1 \) . For \( n > 1 \), we note that\n\n\[ \n{F}^{n} = {F}^{n - 1} * F\left( z\right) = {\int }_{-\infty }^{\infty }d{F}^{n - 1}\left( x\right) {\int }_{-\infty }^{z - x}{dF}\left( y\right) \n\]\n\n\[ \n= \int d{F}_{\lambda }^{n - 1}\left( x\right) \i... | Yes |
Lemma 3.1.1. If \( {c}_{j} \rightarrow 0,{a}_{j} \rightarrow ∞ \) and \( {a}_{j}{c}_{j} \rightarrow \lambda \) then \( {\left( 1 + {c}_{j}\right) }^{{a}_{j}} \rightarrow {e}^{\lambda } \). | Proof. As \( x \rightarrow 0,\log \left( {1 + x}\right) /x \rightarrow 1 \), so \( {a}_{j}\log \left( {1 + {c}_{j}}\right) \rightarrow \lambda \) and the desired result follows. | No |
Theorem 3.1.3. The De Moivre-Laplace Theorem. If \( a < b \) then as \( m \rightarrow \infty \)\n\n\[ P\left( {a \leq {S}_{m}/\sqrt{m} \leq b}\right) \rightarrow {\int }_{a}^{b}{\left( 2\pi \right) }^{-1/2}{e}^{-{x}^{2}/2}{dx} \] | (To remove the restriction to even integers observe \( {S}_{{2n} + 1} = {S}_{2n} \pm 1 \) .) The last result is a special case of the central limit theorem given in Section 3.4, so further details are left to the reader. | No |
Example 3.2.1. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( P\left( {{X}_{i} = 1}\right) = P\left( {{X}_{i} = - 1}\right) = 1/2 \) and let \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) . Then Theorem 3.1.3 implies | \[ {F}_{n}\left( y\right) = P\left( {{S}_{n}/\sqrt{n} \leq y}\right) \rightarrow {\int }_{-\infty }^{y}{\left( 2\pi \right) }^{-1/2}{e}^{-{x}^{2}/2}{dx} \] | Yes |
Let \( X \) have distribution \( F \) . Then \( X + 1/n \) has distribution | \[ {F}_{n}\left( x\right) = P\left( {X + 1/n \leq x}\right) = F\left( {x - 1/n}\right) \] As \( n \rightarrow \infty ,{F}_{n}\left( x\right) \rightarrow F\left( {x - }\right) = \mathop{\lim }\limits_{{y \uparrow x}}F\left( y\right) \) so convergence only occurs at continuity points. | Yes |
Birthday problem. Let \( {X}_{1},{X}_{2},\ldots \) be independent and uniformly distributed on \( \{ 1,\ldots, N\} \), and let \( {T}_{N} = \min \left\{ {n : {X}_{n} = {X}_{m}}\right. \) for some \( \left. {m < n}\right\} \) . | \[ P\left( {{T}_{N} > n}\right) = \mathop{\prod }\limits_{{m = 2}}^{n}\left( {1 - \frac{m - 1}{N}}\right) \] | Yes |
Lemma 3.2.1. \( {V}_{n + 1} \) has density function\n\n\[ \n{f}_{{V}_{n + 1}}\left( x\right) = \left( {{2n} + 1}\right) \left( \begin{matrix} {2n} \\ n \end{matrix}\right) {x}^{n}{\left( 1 - x\right) }^{n} \n\] | Proof. There are \( {2n} + 1 \) ways to pick the observation that falls at \( x \), then we have to pick \( n \) indices for observations \( < x \), which can be done in \( \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \) ways. Once we have decided on the indices that will land \( < x \) and \( > x \), the probab... | Yes |
Theorem 3.2.2. If \( {F}_{n} \Rightarrow {F}_{\infty } \) then there are random variables \( {Y}_{n},1 \leq n \leq \infty \), with distribution \( {F}_{n} \) so that \( {Y}_{n} \rightarrow {Y}_{\infty } \) a.s. | Proof. Let \( \Omega = \left( {0,1}\right) ,\mathcal{F} = \) Borel sets, \( P = \) Lebesgue measure, and let \( {Y}_{n}\left( x\right) = \) \( \sup \left\{ {y : {F}_{n}\left( y\right) < x}\right\} \) . By Theorem 1.2.2, \( {Y}_{n} \) has distribution \( {F}_{n} \) . We will now show that \( {Y}_{n}\left( x\right) \righ... | Yes |
Theorem 3.2.3. \( {X}_{n} \Rightarrow {X}_{\infty } \) if and only if for every bounded continuous function \( g \) we have \( \operatorname{Eg}\left( {X}_{n}\right) \rightarrow \operatorname{Eg}\left( {X}_{\infty }\right) \) . | Proof. Let \( {Y}_{n} \) have the same distribution as \( {X}_{n} \) and converge a.s. Since \( g \) is continuous \( g\left( {Y}_{n}\right) \rightarrow g\left( {Y}_{\infty }\right) \) a.s. and the bounded convergence theorem implies\n\n\[ \n{Eg}\left( {X}_{n}\right) = {Eg}\left( {Y}_{n}\right) \rightarrow {Eg}\left( {... | Yes |
Theorem 3.2.4. Continuous mapping theorem. Let \( g \) be a measurable function and \( {D}_{g} = \{ x : g \) is discontinuous at \( x\} \) . If \( {X}_{n} \Rightarrow {X}_{\infty } \) and \( P\left( {{X}_{\infty } \in {D}_{g}}\right) = 0 \) then \( g\left( {X}_{n}\right) \Rightarrow g\left( X\right) \) . If in addition... | Proof. Let \( {Y}_{n}{ = }_{d}{X}_{n} \) with \( {Y}_{n} \rightarrow {Y}_{\infty } \) a.s. If \( f \) is continuous then \( {D}_{f \circ g} \subset {D}_{g} \) so \( P\left( {{Y}_{\infty } \in {D}_{f \circ g}}\right) = 0 \) and it follows that \( f\left( {g\left( {Y}_{n}\right) }\right) \rightarrow f\left( {g\left( {Y}_... | Yes |
Theorem 3.2.5. The following statements are equivalent: (i) \( {X}_{n} \Rightarrow {X}_{\infty } \)\n\n(ii) For all open sets \( G,\lim \mathop{\inf }\limits_{{n \rightarrow \infty }}P\left( {{X}_{n} \in G}\right) \geq P\left( {{X}_{\infty } \in G}\right) \) .\n\n(iii) For all closed sets \( K,\lim \mathop{\sup }\limit... | Proof. We will prove four things and leave it to the reader to check that we have proved the result given above.\n\n(i) implies (ii): Let \( {Y}_{n} \) have the same distribution as \( {X}_{n} \) and \( {Y}_{n} \rightarrow {Y}_{\infty } \) a.s. Since \( G \) is open\n\n\[ \mathop{\liminf }\limits_{{n \rightarrow \infty... | No |
Theorem 3.2.6. Helly’s selection theorem. For every sequence \( {F}_{n} \) of distribution functions, there is a subsequence \( {F}_{n\left( k\right) } \) and a right continuous nondecreasing function \( F \) so that \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{F}_{n\left( k\right) }\left( y\right) = F\left( y\ri... | Proof. The first step is a diagonal argument. Let \( {q}_{1},{q}_{2},\ldots \) be an enumeration of the rationals. Since for each \( k,{F}_{m}\left( {q}_{k}\right) \in \left\lbrack {0,1}\right\rbrack \) for all \( m \), there is a sequence \( {m}_{k}\left( i\right) \rightarrow \infty \) that is a subsequence of \( {m}_... | Yes |
Every subsequential limit is the distribution function of a probability measure if and only if the sequence \( {F}_{n} \) is \( \mathbf{{tight}} \), i.e., for all \( \epsilon > 0 \) there is an \( {M}_{\epsilon } \) so that \[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}1 - {F}_{n}\left( {M}_{\epsilon }\right) + ... | Proof. Suppose the sequence is tight and \( {F}_{n\left( k\right) }{ \Rightarrow }_{v}F \) . Let \( r < - {M}_{\epsilon } \) and \( s > {M}_{\epsilon } \) be continuity points of \( F \) . Since \( {F}_{n}\left( r\right) \rightarrow F\left( r\right) \) and \( {F}_{n}\left( s\right) \rightarrow F\left( s\right) \), we h... | Yes |
Theorem 3.2.8. If there is a \( \varphi \geq 0 \) so that \( \varphi \left( x\right) \rightarrow \infty \) as \( \left| x\right| \rightarrow \infty \) and\n\n\[ C = \mathop{\sup }\limits_{n}\int \varphi \left( x\right) d{F}_{n}\left( x\right) < \infty \]\n\nthen \( {F}_{n} \) is tight. | Proof. \( 1 - {F}_{n}\left( M\right) + {F}_{n}\left( {-M}\right) \leq C/\mathop{\inf }\limits_{{\left| x\right| \geq M}}\varphi \left( x\right) \). | Yes |
Theorem 3.3.1. All characteristic functions have the following properties:\n\n(a) \( \varphi \left( 0\right) = 1 \) ,\n\n(b) \( \varphi \left( {-t}\right) = \overline{\varphi \left( t\right) } \) ,\n\n(c) \( \left| {\varphi \left( t\right) }\right| = \left| {E{e}^{itX}}\right| \leq E\left| {e}^{itX}\right| = 1 \)\n\n(d... | Proof. (a) is obvious. For (b) we note that\n\n\[ \varphi \left( {-t}\right) = E\left( {\cos \left( {-{tX}}\right) + i\sin \left( {-{tX}}\right) }\right) = E\left( {\cos \left( {tX}\right) - i\sin \left( {tX}\right) }\right) \]\n\n(c) follows from Exercise 1.6.2 since \( \varphi \left( {x, y}\right) = {\left( {x}^{2} +... | No |
Theorem 3.3.2. If \( {X}_{1} \) and \( {X}_{2} \) are independent and have ch.f.’s \( {\varphi }_{1} \) and \( {\varphi }_{2} \) then \( {X}_{1} + {X}_{2} \) has ch.f. \( {\varphi }_{1}\left( t\right) {\varphi }_{2}\left( t\right) \) . | Proof.\n\n\[ E{e}^{{it}\left( {{X}_{1} + {X}_{2}}\right) } = E\left( {{e}^{{it}{X}_{1}}{e}^{{it}{X}_{2}}}\right) = E{e}^{{it}{X}_{1}}E{e}^{{it}{X}_{2}} \]\n\nsince \( {e}^{{it}{X}_{1}} \) and \( {e}^{{it}{X}_{2}} \) are independent. | Yes |
If \( P\left( {X = 1}\right) = P\left( {X = - 1}\right) = 1/2 \) then | \[ E{e}^{itX} = \left( {{e}^{it} + {e}^{-{it}}}\right) /2 = \cos t \] | Yes |
Example 3.3.2. Poisson distribution. If \( P\left( {X = k}\right) = {e}^{-\lambda }{\lambda }^{k}/k \) for \( k = 0,1,2,\ldots \) | then \[ E{e}^{itX} = \mathop{\sum }\limits_{{k = 0}}^{\infty }{e}^{-\lambda }\frac{{\lambda }^{k}{e}^{itk}}{k!} = \exp \left( {\lambda \left( {{e}^{it} - 1}\right) }\right) \] | Yes |
Example 3.3.3. Normal distribution\n\n\[ \text{Density}\;{\left( 2\pi \right) }^{-1/2}\exp \left( {-{x}^{2}/2}\right) \]\n\n\[ \text{Ch.f.}\exp \left( {-{t}^{2}/2}\right) \]\n\nCombining this result with (e) of Theorem 3.3.1, we see that a normal distribution with mean \( \mu \) and variance \( {\sigma }^{2} \) has ch.... | Physics Proof\n\n\[ \int {e}^{itx}{\left( 2\pi \right) }^{-1/2}{e}^{-{x}^{2}/2}{dx} = {e}^{-{t}^{2}/2}\int {\left( 2\pi \right) }^{-1/2}{e}^{-{\left( x - it\right) }^{2}/2}{dx} \]\n\nThe integral is 1 since the integrand is the normal density with mean \( {it} \) and variance 1.\n\nMath Proof. Now that we have cheated ... | Yes |
Example 3.3.4. Uniform distribution on \( \\left( {a, b}\\right) \)\n\n\[ \n\\begin{array}{ll} \\text{ Density } & 1/\\left( {b - a}\\right) \\;x \\in \\left( {a, b}\\right) \\\\ \\text{ Ch.f. } & \\left( {{e}^{itb} - {e}^{ita}}\\right) /{it}\\left( {b - a}\\right) \\end{array} \n\]\n\nIn the special case \( a = - c, b... | Proof. Once you recall that \( {\\int }_{a}^{b}{e}^{\\lambda x}{dx} = \\left( {{e}^{\\lambda b} - {e}^{\\lambda a}}\\right) /\\lambda \) holds for complex \( \\lambda \), this is immediate. | Yes |
Example 3.3.5. Triangular distribution\n\n\\[ \n\\text{Density}\\;1 - \\left| x\\right| \\;x \\in \\left( {-1,1}\\right) \n\\]\n\n\\[ \n\\text{Ch.f.}\\;2\\left( {1 - \\cos t}\\right) /{t}^{2} \n\\] | Proof. To see this, notice that if \\( X \\) and \\( Y \\) are independent and uniform on \\( \\left( {-1/2,1/2}\\right) \\) then \\( X + Y \\) has a triangular distribution. Using Example 3.3.4 now and Theorem 3.3.2 it follows that the desired ch.f. is\n\n\\[ \n{\\left\\{ \\left( {e}^{{it}/2} - {e}^{-{it}/2}\\right) /... | Yes |
Example 3.3.6. Exponential distribution\n\n\\[ \n\\begin{array}{ll} \\text{ Density } & {e}^{-x}\\;x \\in \\left( {0,\\infty }\\right) \\\\ \\text{ Ch.f. } & 1/\\left( {1 - {it}}\\right) \\end{array} \n\\] | Proof. Integrating gives\n\n\\[ \n{\\int }_{0}^{\infty }{e}^{itx}{e}^{-x}{dx} = {\\left. \\frac{{e}^{\\left( {{it} - 1}\\right) x}}{{it} - 1}\\right| }_{0}^{\infty } = \\frac{1}{1 - {it}} \n\\]\n\nsince \\( \\exp \\left( {\\left( {{it} - 1}\\right) x}\\right) \\rightarrow 0 \\) as \\( x \\rightarrow \\infty \\) . | Yes |
Example 3.3.7. Bilateral exponential\n\n\\[ \n\\begin{array}{ll} \\text{ Density } & \\frac{1}{2}{e}^{-\\left| x\\right| } \\\\ \\text{ Ch.f. } & 1/\\left( {1 + {t}^{2}}\\right) \\end{array}x \\in \\left( {-\\infty ,\\infty }\\right) \n\\] | Proof This follows from Lemma 3.3.3 with \\( {F}_{1} \\) the distribution of an exponential random variable \\( X,{F}_{2} \\) the distribution of \\( - X \\), and \\( {\\lambda }_{1} = {\\lambda }_{2} = 1/2 \\) then using (b) of Theorem 3.3.1 we see the desired ch.f. is\n\n\\[ \n\\frac{1}{2\\left( {1 - {it}}\\right) } ... | Yes |
Theorem 3.3.4. The inversion formula. Let \( \varphi \left( t\right) = \int {e}^{itx}\mu \left( {dx}\right) \) where \( \mu \) is a probability measure. If \( a < b \) then\n\n\[ \mathop{\lim }\limits_{{T \rightarrow \infty }}{\left( 2\pi \right) }^{-1}{\int }_{-T}^{T}\frac{{e}^{-{ita}} - {e}^{-{itb}}}{it}\varphi \left... | Proof. Let\n\n\[ {I}_{T} = {\int }_{-T}^{T}\frac{{e}^{-{ita}} - {e}^{-{itb}}}{it}\varphi \left( t\right) {dt} = {\int }_{-T}^{T}\int \frac{{e}^{-{ita}} - {e}^{-{itb}}}{it}{e}^{itx}\mu \left( {dx}\right) {dt} \]\n\nThe integrand may look bad near \( t = 0 \) but if we observe that\n\n\[ \frac{{e}^{-{ita}} - {e}^{-{itb}}... | Yes |
Theorem 3.3.5. If \( \int \left| {\varphi \left( t\right) }\right| {dt} < \infty \) then \( \mu \) has bounded continuous density | \[ f\left( y\right) = \frac{1}{2\pi }\int {e}^{-{ity}}\varphi \left( t\right) {dt} \] Proof. As we observed in the proof of Theorem 3.3.4 \[ \left| \frac{{e}^{-{ita}} - {e}^{-{itb}}}{it}\right| = \left| {{\int }_{a}^{b}{e}^{-{ity}}{dy}}\right| \leq \left| {b - a}\right| \] so the integral in Theorem 3.3.4 converges abs... | Yes |
Example 3.3.8. Polya's distribution\n\n\\[ \n\\begin{array}{ll} \\text{ Density } & \\left( {1 - \\cos x}\\right) /\\pi {x}^{2} \\\\\n\\text{ Ch.f. } & {\\left( 1 - \\left| t\\right| \\right) }^{ + } \\end{array} \n\\] | Proof. Theorem 3.3.5 implies\n\n\\[ \n\\frac{1}{2\\pi }\\int \\frac{2\\left( {1 - \\cos s}\\right) }{{s}^{2}}{e}^{-{isy}}{ds} = {\\left( 1 - \\left| y\\right| \\right) }^{ + } \n\\]\n\nNow let \\( s = x, y = - t \\) . | Yes |
Example 3.3.9. The Cauchy distribution\n\n\[ \n\begin{array}{ll} \text{ Density } & 1/\pi \left( {1 + {x}^{2}}\right) \\ \text{ Ch.f. } & \exp \left( {-\left| t\right| }\right) \end{array} \n\] | Proof. Theorem 3.3.5 implies\n\n\[ \n\frac{1}{2\pi }\int \frac{1}{1 + {s}^{2}}{e}^{-{isy}}{ds} = \frac{1}{2}{e}^{-\left| y\right| } \n\]\n\nNow let \( s = x, y = - t \) and multiply each side by 2 . | No |
Theorem 3.3.6. Continuity theorem. Let \( {\mu }_{n},1 \leq n \leq \infty \) be probability measures with ch.f. \( {\varphi }_{n} \) . (i) If \( {\mu }_{n} \Rightarrow {\mu }_{\infty } \) then \( {\varphi }_{n}\left( t\right) \rightarrow {\varphi }_{\infty }\left( t\right) \) for all \( t \) . (ii) If \( {\varphi }_{n}... | Proof. (i) is easy. \( {e}^{itx} \) is bounded and continuous so if \( {\mu }_{n} \Rightarrow {\mu }_{\infty } \) then Theorem 3.2.3 implies \( {\varphi }_{n}\left( t\right) \rightarrow {\varphi }_{\infty }\left( t\right) \) . To prove (ii), our first goal is to prove tightness. We begin with some calculations that may... | Yes |
Lemma 3.3.7.\n\\[ \n\\left| {{e}^{ix} - \\mathop{\\sum }\\limits_{{m = 0}}^{n}\\frac{{\\left( ix\\right) }^{m}}{m!}}\\right| \\leq \\min \\left( {\\frac{{\\left| x\\right| }^{n + 1}}{\\left( {n + 1}\\right) !},\\frac{2{\\left| x\\right| }^{n}}{n!}}\\right) \n\\] | Proof. Integrating by parts gives\n\n\\[ \n{\\int }_{0}^{x}{\\left( x - s\\right) }^{n}{e}^{is}{ds} = \\frac{{x}^{n + 1}}{n + 1} + \\frac{i}{n + 1}{\\int }_{0}^{x}{\\left( x - s\\right) }^{n + 1}{e}^{is}{ds} \n\\]\n\nWhen \\( n = 0 \\), this says\n\n\\[ \n{\\int }_{0}^{x}{e}^{is}{ds} = x + i{\\int }_{0}^{x}\\left( {x -... | No |
Theorem 3.3.8. If \( E{\left| X\right| }^{2} < \infty \) then\n\n\[ \varphi \left( t\right) = 1 + {itEX} - {t}^{2}E\left( {X}^{2}\right) /2 + o\left( {t}^{2}\right) \] | Proof. The error term is \( \leq {t}^{2}E\left( {\left| t\right| \cdot {\left| X\right| }^{3} \land 2{\left| X\right| }^{2}}\right) \) . The variable in parentheses is smaller than \( 2{\left| X\right| }^{2} \) and converges to 0 as \( t \rightarrow 0 \), so the desired conclusion follows from the dominated convergence... | Yes |
Theorem 3.3.9. If \( \lim \mathop{\sup }\limits_{{h \downarrow 0}}\{ \varphi \left( h\right) - {2\varphi }\left( 0\right) + \varphi \left( {-h}\right) \} /{h}^{2} > - \infty \), then \( E{\left| X\right| }^{2} < \infty \). | Proof. \( \left( {{e}^{ihx} - 2 + {e}^{-{ihx}}}\right) /{h}^{2} = - 2\left( {1 - \cos {hx}}\right) /{h}^{2} \leq 0 \) and \( 2\left( {1 - \cos {hx}}\right) /{h}^{2} \rightarrow {x}^{2} \) as \( h \rightarrow 0 \) so Fatou’s lemma and Fubini’s theorem imply\n\n\[ \n\int {x}^{2}{dF}\left( x\right) \leq 2\mathop{\liminf }... | Yes |
Theorem 3.3.10. Polya’s criterion. Let \( \varphi \left( t\right) \) be real nonnegative and have \( \varphi \left( 0\right) = \) \( 1,\varphi \left( t\right) = \varphi \left( {-t}\right) \), and \( \varphi \) is decreasing and convex on \( \left( {0,\infty }\right) \) with\n\n\[ \mathop{\lim }\limits_{{t \downarrow 0}... | Proof. Let \( {\varphi }^{\prime } \) be the right derivative of \( \phi \), i.e.,\n\n\[ {\varphi }^{\prime }\left( t\right) = \mathop{\lim }\limits_{{h \downarrow 0}}\frac{\varphi \left( {t + h}\right) - \varphi \left( t\right) }{h} \]\n\nSince \( \varphi \) is convex this exists and is right continuous and increasing... | Yes |
Example 3.3.10. \( \exp \left( {-{\left| t\right| }^{\alpha }}\right) \) is a characteristic function for \( 0 < \alpha < 2 \) . | Proof. A little calculus shows that for any \( \beta \) and \( \left| x\right| < 1 \)\n\n\[{\left( 1 - x\right) }^{\beta } = \mathop{\sum }\limits_{{n = 0}}^{\infty }\left( \begin{array}{l} \beta \\ n \end{array}\right) {\left( -x\right) }^{n}\]\n\nwhere\n\n\[ \left( \begin{array}{l} \beta \\ n \end{array}\right) = \fr... | Yes |
Example 3.3.11. For some purposes, it is nice to have an explicit example of two ch.f.’s that agree on \( \left\lbrack {-1,1}\right\rbrack \) . From Example 3.3.8, we know that \( {\left( 1 - \left| t\right| \right) }^{ + } \) is the ch.f. of the density \( \left( {1 - \cos x}\right) /\pi {x}^{2} \) . Define \( \psi \l... | The Fourier series for \( \psi \) is\n\n\[ \psi \left( u\right) = \frac{1}{2} + \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\frac{2}{{\pi }^{2}{\left( 2n - 1\right) }^{2}}\exp \left( {i\left( {{2n} - 1}\right) {\pi u}}\right) \]\n\nThe right-hand side is the ch.f. of a discrete distribution with\n\n\[ P\left( {X =... | Yes |
Theorem 3.3.11. If \( \mathop{\limsup }\limits_{{k \rightarrow \infty }}{\mu }_{2k}^{1/{2k}}/{2k} = r < \infty \) then there is at most one d.f. \( F \) with \( {\mu }_{k} = \int {x}^{k}{dF}\left( x\right) \) for all positive integers \( k \) . | Proof. Let \( F \) be any d.f. with the moments \( {\mu }_{k} \) and let \( {\nu }_{k} = \int {\left| x\right| }^{k}{dF}\left( x\right) \) . The Cauchy-Schwarz inequality implies \( {\nu }_{{2k} + 1}^{2} \leq {\mu }_{2k}{\mu }_{{2k} + 2} \) so\n\n\[ \mathop{\limsup }\limits_{{k \rightarrow \infty }}\left( {\nu }_{k}^{1... | Yes |
Theorem 3.4.1. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E{X}_{i} = \mu \) , \( \operatorname{var}\left( {X}_{i}\right) = {\sigma }^{2} \in \left( {0,\infty }\right) \) . If \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) then\n\n\[ \left( {{S}_{n} - {n\mu }}\right) /\sigma {n}^{1/2} \Rightarrow \chi \]\n\nwhere \(... | Proof By considering \( {X}_{i}^{\prime } = {X}_{i} - \mu \), it suffices to prove the result when \( \mu = 0 \) . From\n\nTheorem 3.3.8\n\[ \varph | No |
Theorem 3.3.8\n\[ \varphi \left( t\right) = E\exp \left( {{it}{X}_{1}}\right) = 1 - \frac{{\sigma }^{2}{t}^{2}}{2} + o\left( {t}^{2}\right) \] | so\n\[ E\exp \left( {{it}{S}_{n}/\sigma {n}^{1/2}}\right) = {\left( 1 - \frac{{t}^{2}}{2n} + o\left( {n}^{-1}\right) \right) }^{n} \]\n\nFrom Lemma 3.1.1 it should be clear that the last quantity \( \rightarrow \exp \left( {-{t}^{2}/2}\right) \) as \( n \rightarrow \infty \) , which with Theorem 3.3.6 completes the pro... | No |
Theorem 3.4.2. If \( {c}_{n} \rightarrow c \in \mathbf{C} \) then \( {\left( 1 + {c}_{n}/n\right) }^{n} \rightarrow {e}^{c} \) . | Proof. The proof is based on two simple facts: | No |
Lemma 3.4.3. Let \( {z}_{1},\ldots ,{z}_{n} \) and \( {w}_{1},\ldots ,{w}_{n} \) be complex numbers of modulus \( \leq \theta \) .\n\nThen\n\[ \left| {\mathop{\prod }\limits_{{m = 1}}^{n}{z}_{m} - \mathop{\prod }\limits_{{m = 1}}^{n}{w}_{m}}\right| \leq {\theta }^{n - 1}\mathop{\sum }\limits_{{m = 1}}^{n}\left| {{z}_{m... | Proof. The result is true for \( n = 1 \) . To prove it for \( n > 1 \) observe that\n\n\[ \left| {\mathop{\prod }\limits_{{m = 1}}^{n}{z}_{m} - \mathop{\prod }\limits_{{m = 1}}^{n}{w}_{m}}\right| \leq \left| {{z}_{1}\mathop{\prod }\limits_{{m = 2}}^{n}{z}_{m} - {z}_{1}\mathop{\prod }\limits_{{m = 2}}^{n}{w}_{m}}\right... | Yes |
Lemma 3.4.4. If \( b \) is a complex number with \( \left| b\right| \leq 1 \) then \( \left| {{e}^{b} - \left( {1 + b}\right) }\right| \leq {\left| b\right| }^{2} \) . | Proof. \( {e}^{b} - \left( {1 + b}\right) = {b}^{2}/2! + {b}^{3}/3! + {b}^{4}/4! + \ldots \) so if \( \left| b\right| \leq 1 \) then\n\n\[ \left| {{e}^{b} - \left( {1 + b}\right) }\right| \leq \frac{{\left| b\right| }^{2}}{2}\left( {1 + 1/2 + 1/{2}^{2} + \ldots }\right) = {\left| b\right| }^{2} \]\n | Yes |
A roulette wheel has slots numbered 1-36 (18 red and 18 black) and two slots numbered 0 and 00 that are painted green. Players can bet \$1 that the ball will land in a red (or black) slot and win \$1 if it does. If we let \( {X}_{i} \) be the winnings on the \( i \) th play then \( {X}_{1},{X}_{2},\ldots \) are i.i.d. ... | \[ E{X}_{i} = - 1/{19}\text{ and }\operatorname{var}\left( X\right) = E{X}^{2} - {\left( EX\right) }^{2} = 1 - {\left( 1/{19}\right) }^{2} = {0.9972} \] We are interested in \[ P\left( {{S}_{n} \geq 0}\right) = P\left( {\frac{{S}_{n} - {n\mu }}{\sigma \sqrt{n}} \geq \frac{-{n\mu }}{\sigma \sqrt{n}}}\right) \] Taking \(... | Yes |
Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( P\left( {{X}_{i} = 0}\right) = P\left( {{X}_{i} = 1}\right) = 1/2 \). If \( {X}_{i} = 1 \) indicates that a heads occured on the \( i \) th toss then \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) is the total number of heads at time \( n \). | \[ E{X}_{i} = 1/2\;\text{ and }\;\operatorname{var}\left( X\right) = E{X}^{2} - {\left( EX\right) }^{2} = 1/2 - 1/4 = 1/4 \] So the central limit theorem tells us \( \left( {{S}_{n} - n/2}\right) /\sqrt{n/4} \Rightarrow \chi \). Our table of the normal distribution tells us that \[ P\left( {\chi > 2}\right) = 1 - {0.97... | Yes |
To estimate \( P\left( {{S}_{16} = 8}\right) \) using the central limit theorem, we regard 8 as the interval \( \left\lbrack {{7.5},{8.5}}\right\rbrack \) . Since \( \mu = 1/2 \), and \( \sigma \sqrt{n} = 2 \) for \( n = {16} \) | \[ P\left( {\left| {{S}_{16} - 8}\right| \leq {0.5}}\right) = P\left( {\frac{\left| {S}_{n} - n\mu \right| }{\sigma \sqrt{n}} \leq {0.25}}\right) \] \[ \approx P\left( {\left| \chi \right| \leq {0.25}}\right) = 2\left( {{0.5987} - {0.5}}\right) = {0.1974} \] Even though \( n \) is small, this agrees well with the exact... | Yes |
Let \( {Z}_{\lambda } \) have a Poisson distribution with mean \( \lambda \) . If \( {X}_{1},{X}_{2},\ldots \) are independent and have Poisson distributions with mean 1, then \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) has a Poisson distribution with mean \( n \) . Since \( \operatorname{var}\left( {X}_{i}\right) = 1 \... | \[ \left( {{S}_{n} - n}\right) /{n}^{1/2} \Rightarrow \chi \;\text{ as }n \rightarrow \infty \] | No |
Pairwise independence is good enough for the strong law of large numbers (see Theorem 2.4.1). It is not good enough for the central limit theorem. Let \( {\xi }_{1},{\xi }_{2},\ldots \) be i.i.d. with \( P\left( {{\xi }_{i} = 1}\right) = P\left( {{\xi }_{i} = - 1}\right) = 1/2 \) . We will arrange things so that for \(... | \[ {S}_{{2}^{n}} = {\xi }_{1}\left( {1 + {\xi }_{2}}\right) \cdots \left( {1 + {\xi }_{n + 1}}\right) = \left\{ \begin{array}{ll} \pm {2}^{n} & \text{ with prob }{2}^{-n - 1} \\ 0 & \text{ with prob }1 - {2}^{-n} \end{array}\right. \] To do this we let \( {X}_{1} = {\xi }_{1},{X}_{2} = {\xi }_{1}{\xi }_{2} \), and for ... | No |
Theorem 3.4.5. The Lindeberg-Feller theorem. For each \( n \), let \( {X}_{n, m},1 \leq m \leq \) \( n \), be independent random variables with \( E{X}_{n, m} = 0 \) . Suppose\n\n(i) \( \mathop{\sum }\limits_{{m = 1}}^{n}E{X}_{n, m}^{2} \rightarrow {\sigma }^{2} > 0 \)\n\n(ii) For all \( \epsilon > 0,\mathop{\lim }\lim... | Proof. Let \( {\varphi }_{n, m}\left( t\right) = E\exp \left( {{it}{X}_{n, m}}\right) ,{\sigma }_{n, m}^{2} = E{X}_{n, m}^{2} \) . By Theorem 3.3.6, it suffices to show that\n\n\[ \mathop{\prod }\limits_{{m = 1}}^{n}{\varphi }_{n, m}\left( t\right) \rightarrow \exp \left( {-{t}^{2}{\sigma }^{2}/2}\right) \]\n\nLet \( {... | Yes |
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