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Theorem 7.17. Let \( R \) be a ring and let \( a \in R \) . Then \( {aR} \mathrel{\text{:=}} \{ {ar} : r \in R\} \) is an ideal of \( R \) . | Proof. This is an easy calculation. For all \( {ar}, a{r}^{\prime } \in {aR} \) and \( {r}^{\prime \prime } \in R \), we have \( {ar} + a{r}^{\prime } = a\left( {r + {r}^{\prime }}\right) \in {aR} \) and \( \left( {ar}\right) {r}^{\prime \prime } = a\left( {r{r}^{\prime \prime }}\right) \in {aR}. | Yes |
Theorem 7.18. If \( {I}_{1} \) and \( {I}_{2} \) are ideals of a ring \( R \), then so are \( {I}_{1} + {I}_{2} \) and \( {I}_{1} \cap {I}_{2} \) . | Proof. We already know that \( {I}_{1} + {I}_{2} \) and \( {I}_{1} \cap {I}_{2} \) are additive subgroups of \( R \), so it suffices to show that they are closed under multiplication by elements of \( R \) . The reader may easily verify that this is the case. | No |
Let \( n \) be a positive integer, and let \( x \) be any integer. Define \( I \mathrel{\text{:=}} \{ g \in \mathbb{Z}\left\lbrack X\right\rbrack : g\left( x\right) \equiv 0\left( {\;\operatorname{mod}\;n}\right) \} \). We claim that \( I \) is the ideal \( \left( {X - x, n}\right) \) of \( \mathbb{Z}\left\lbrack X\rig... | To see this, consider any fixed \( g \in \mathbb{Z}\left\lbrack X\right\rbrack \). Using Theorem 7.12, we have \( g = \left( {X - x}\right) q + g\left( x\right) \) for some \( q \in \mathbb{Z}\left\lbrack X\right\rbrack \). Using the division with remainder property for integers, we have \( g\left( x\right) = n{q}^{\pr... | Yes |
Theorem 7.19. Suppose \( I \) is an ideal of a ring \( R \) . For all \( a,{a}^{\prime }, b,{b}^{\prime } \in R \), if \( a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;I}\right) \) and \( b \equiv {b}^{\prime }\left( {\;\operatorname{mod}\;I}\right) \), then \( {ab} \equiv {a}^{\prime }{b}^{\prime }\left( {\;\ope... | Proof. If \( a = {a}^{\prime } + x \) for some \( x \in I \) and \( b = {b}^{\prime } + y \) for some \( y \in I \), then \( {ab} = {a}^{\prime }{b}^{\prime } + {a}^{\prime }y + {b}^{\prime }x + {xy} \) . Since \( I \) is closed under multiplication by elements of \( R \) , we see that \( {a}^{\prime }y,{b}^{\prime }x,... | Yes |
Let \( f \) be a polynomial over a ring \( R \) with \( \deg \left( f\right) = \ell \geq 0 \) and \( \operatorname{lc}\left( f\right) \in {R}^{ * } \), and consider the quotient ring \( E \mathrel{\text{:=}} R\left\lbrack X\right\rbrack /{fR}\left\lbrack X\right\rbrack \) . By the division with remainder property for p... | From this, it follows that every element of \( E \) can be written uniquely as \( {\left\lbrack h\right\rbrack }_{f} \), where \( h \in R\left\lbrack X\right\rbrack \) is a polynomial of degree less than \( \ell \) . Note that in this situation, we will generally prefer the more compact notation \( R\left\lbrack X\righ... | Yes |
Consider the polynomial \( f \mathrel{\text{:=}} {X}^{2} + X + 1 \in {\mathbb{Z}}_{2}\left\lbrack X\right\rbrack \) and the quotient ring \( E \mathrel{\text{:=}} {\mathbb{Z}}_{2}\left\lbrack X\right\rbrack /\left( f\right) \) . Let us name the elements of \( E \) as follows:\n\n\[ \n{00} \mathrel{\text{:=}} {\left\lbr... | This is the first example we have seen of a finite field whose cardinality is not prime. | No |
Suppose \( I \) is an ideal of a ring \( R \) . Analogous to Example 6.36, we may define the natural map from the ring \( R \) to the quotient ring \( R/I \) as follows:\n\n\[ \rho : \;R \rightarrow R/I \]\n\n\[ a \mapsto {\left\lbrack a\right\rbrack }_{I}. \] | Not only is this a surjective homomorphism of additive groups, with kernel \( I \), it is a ring homomorphism. Indeed, we have\n\n\[ \rho \left( {ab}\right) = {\left\lbrack ab\right\rbrack }_{I} = {\left\lbrack a\right\rbrack }_{I} \cdot {\left\lbrack b\right\rbrack }_{I} = \rho \left( a\right) \cdot \rho \left( b\righ... | Yes |
Let \( R \) be a subring of a ring \( E \), and fix \( \alpha \in E \). The polynomial evaluation map\n\n\[ \rho : \;R\left\lbrack X\right\rbrack \rightarrow E \]\n\n\[ g \mapsto g\left( \alpha \right) \]\n\nis a ring homomorphism. | The image of \( \rho \) consists of all polynomial expressions in \( \alpha \) with coefficients in \( R \), and is denoted \( R\left\lbrack \alpha \right\rbrack \). As the reader may verify, \( R\left\lbrack \alpha \right\rbrack \) is a subring of \( E \) containing \( \alpha \) and all of \( R \), and is the smallest... | No |
Let \( \rho : R \rightarrow {R}^{\prime } \) be a ring homomorphism. We can extend the domain of definition of \( \rho \) from \( R \) to \( R\left\lbrack X\right\rbrack \) by defining \( \rho \left( {\mathop{\sum }\limits_{i}{a}_{i}{X}^{i}}\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{i}\rho \left( {a}_{i}\right)... | To verify this, suppose \( g = \mathop{\sum }\limits_{i}{a}_{i}{X}^{i} \) and \( h = \mathop{\sum }\limits_{i}{b}_{i}{X}^{i} \) are polynomials in \( R\left\lbrack X\right\rbrack \) . Let \( s \mathrel{\text{:=}} g + h \in R\left\lbrack X\right\rbrack \) and \( p \mathrel{\text{:=}} {gh} \in R\left\lbrack X\right\rbrac... | Yes |
Consider the natural map that sends \( a \in \mathbb{Z} \) to \( \bar{a} \mathrel{\text{:=}} {\left\lbrack a\right\rbrack }_{n} \in {\mathbb{Z}}_{n} \) (see Example 7.43). As in the previous example, we may extend this to a ring homomorphism from \( \mathbb{Z}\left\lbrack X\right\rbrack \) to \( {\mathbb{Z}}_{n}\left\l... | Observe that if \( g = \mathop{\sum }\limits_{i}{a}_{i}{X}^{i} \), then \( \bar{g} = 0 \) if and only if \( n \mid {a}_{i} \) for each \( i \) ; therefore, the kernel is the ideal \( n\mathbb{Z}\left\lbrack X\right\rbrack \) of \( \mathbb{Z}\left\lbrack X\right\rbrack \) . | Yes |
Let \( R \) be a ring of prime characteristic \( p \) . For all \( a, b \in R \), we have (see Exercise 7.1)\n\n\[ \n{\left( a + b\right) }^{p} = \mathop{\sum }\limits_{{k = 0}}^{p}\left( \begin{array}{l} p \\ k \end{array}\right) {a}^{p - k}{b}^{k} \n\] | However, by Exercise 1.14, all of the binomial coefficients are multiples of \( p \) , except for \( k = 0 \) and \( k = p \), and hence in the ring \( R \), all of these terms vanish, leaving us with\n\n\[ \n{\left( a + b\right) }^{p} = {a}^{p} + {b}^{p}. \n\] | Yes |
In special situations, part (iii) of Definition 7.20 may be redundant. One such situation arises when \( \rho : R \rightarrow {R}^{\prime } \) is surjective. | In this case, we know that \( {1}_{{R}^{\prime }} = \rho \left( a\right) \) for some \( a \in R \), and by part (ii) of the definition, we have\n\n\[ \rho \left( {1}_{R}\right) = \rho \left( {1}_{R}\right) \cdot {1}_{{R}^{\prime }} = \rho \left( {1}_{R}\right) \rho \left( a\right) = \rho \left( {{1}_{R} \cdot a}\right)... | Yes |
Theorem 7.21. Let \( \rho : R \rightarrow {R}^{\prime } \) be a ring homomorphism.\n\n(i) If \( S \) is a subring of \( R \), then \( \rho \left( S\right) \) is a subring of \( {R}^{\prime } \) ; in particular (setting \( S \mathrel{\text{:=}} R),\operatorname{Im}\rho \) is a subring of \( {R}^{\prime } \) . | Proof. In each part, we already know that the relevant object is an additive subgroup, and so it suffices to show that the appropriate additional properties are satisfied.\n\n(i) For all \( a, b \in S \), we have \( {ab} \in S \), and hence \( \rho \left( S\right) \) contains \( \rho \left( {ab}\right) = \rho \left( a\... | Yes |
Theorem 7.24. If \( \rho \) is a ring isomorphism of \( R \) with \( {R}^{\prime } \), then the inverse function \( {\rho }^{-1} \) is a ring isomorphism of \( {R}^{\prime } \) with \( R \) . | Proof. Exercise. | No |
Theorem 7.25. Let \( \rho : R \rightarrow {R}^{\prime } \) be a ring isomorphism.\n\n(i) For all \( a \in R, a \) is a zero divisor if and only if \( \rho \left( a\right) \) is a zero divisor.\n\n(ii) For all \( a \in R, a \) is a unit if and only if \( \rho \left( a\right) \) is a unit.\n\n(iii) The restriction of \( ... | ## Proof. Exercise. | No |
Theorem 7.26 (First isomorphism theorem). Let \( \rho : R \rightarrow {R}^{\prime } \) be a ring homomorphism with kernel \( K \) and image \( {S}^{\prime } \) . Then we have a ring isomorphism\n\n\[ R/K \cong {S}^{\prime }\text{.} \] | Specifically, the map\n\n\[ \bar{\rho } : \;R/K \rightarrow {R}^{\prime } \]\n\n\[ {\left\lbrack a\right\rbrack }_{K} \mapsto \rho \left( a\right) \]\n\nis an injective ring homomorphism whose image is \( {S}^{\prime } \) . | Yes |
Returning again to the Chinese remainder theorem and the discussion in Example 6.48, if \( {\left\{ {n}_{i}\right\} }_{i = 1}^{k} \) is a pairwise relatively prime family of positive integers, and \( n \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 1}}^{k}{n}_{i} \), then the map \[ \rho : \;\mathbb{Z} \rightarrow {\... | Applying Theorem 7.26, we get a ring isomorphism \[ \bar{\rho } : \;{\mathbb{Z}}_{n} \rightarrow {\mathbb{Z}}_{{n}_{1}} \times \cdots \times {\mathbb{Z}}_{{n}_{k}} \] \[ {\left\lbrack a\right\rbrack }_{n} \mapsto \left( {{\left\lbrack a\right\rbrack }_{{n}_{1}},\ldots ,{\left\lbrack a\right\rbrack }_{{n}_{k}}}\right) ,... | Yes |
For a ring \( R \), consider the map \( \rho : \mathbb{Z} \rightarrow R \) that sends \( m \in \mathbb{Z} \) to \( m \cdot {1}_{R} \) in \( R \). It is easily verified that \( \rho \) is a ring homomorphism. Since \( \operatorname{Ker}\rho \) is an ideal of \( \mathbb{Z} \), it is either \( \{ 0\} \) or of the form \( ... | In the first case, if \( \operatorname{Ker}\rho = \{ 0\} \), then \( \operatorname{Im}\rho \cong \mathbb{Z} \), and so the ring \( \mathbb{Z} \) is embedded in \( R \), and \( R \) has characteristic zero. In the second case, if \( \operatorname{Ker}\rho = n\mathbb{Z} \) for some \( n > 0 \), then by Theorem 7.26, \( \... | Yes |
We can generalize Example 7.44 by evaluating polynomials at several points. This is most fruitful when the underlying coefficient ring is a field, and the evaluation points belong to the same field. So let \( F \) be a field, and let \( {x}_{1},\ldots ,{x}_{k} \) be distinct elements of \( F \) . Define the map\n\n\[ \... | This is a ring homomorphism (as seen by applying Theorem 7.23 to the polynomial evaluation maps at the points \( \left. {{x}_{1},\ldots ,{x}_{k}}\right) \) . By Theorem 7.13, \( \operatorname{Ker}\rho = \left( f\right) \), where \( f \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 1}}^{k}\left( {X - {x}_{i}}\right) \)... | Yes |
As in Example 7.39, let \( f \) be a polynomial over a ring \( R \) with \( \deg \left( f\right) = \ell \) and \( \operatorname{lc}\left( f\right) \in {R}^{ * } \), but now assume that \( \ell > 0 \) . Consider the natural map \( \rho \) from \( R\left\lbrack X\right\rbrack \) to the quotient ring \( E \mathrel{\text{:... | \n\[
{\left\lbrack g\right\rbrack }_{f} = {\left\lbrack \mathop{\sum }\limits_{i}{a}_{i}{X}^{i}\right\rbrack }_{f} = \mathop{\sum }\limits_{i}{\left\lbrack {a}_{i}\right\rbrack }_{f}{\left( {\left\lbrack X\right\rbrack }_{f}\right) }^{i} = \mathop{\sum }\limits_{i}{a}_{i}{\xi }^{i} = g\left( \xi \right) ,
\]
where \( ... | Yes |
As a special case of Example 7.55, let \( f \mathrel{\text{:=}} {X}^{2} + 1 \in \mathbb{R}\left\lbrack X\right\rbrack \) , and consider the quotient ring \( \mathbb{R}\left\lbrack X\right\rbrack /\left( f\right) \) . If we set \( i \mathrel{\text{:=}} {\left\lbrack X\right\rbrack }_{f} \in \mathbb{R}\left\lbrack X\righ... | \[
\left( {a + {bi}}\right) + \left( {{a}^{\prime } + {b}^{\prime }i}\right) = \left( {a + {a}^{\prime }}\right) + \left( {b + {b}^{\prime }}\right) i
\]
and
\[
\left( {a + {bi}}\right) \cdot \left( {{a}^{\prime } + {b}^{\prime }i}\right) = \left( {a{a}^{\prime } - b{b}^{\prime }}\right) + \left( {a{b}^{\prime } + {a... | Yes |
Consider the polynomial evaluation map\n\n\\[ \rho : \\;\\mathbb{R}\\left\\lbrack X\\right\\rbrack \\rightarrow \\mathbb{C} = R\\left\\lbrack X\\right\\rbrack /\\left( {{X}^{2} + 1}\\right) \\]\n\n\\[ g \\mapsto g\\left( {-i}\\right) \\text{.} \\] | For every \\( g \\in \\mathbb{R}\\left\\lbrack X\\right\\rbrack \\), we may write \\( g = \\left( {{X}^{2} + 1}\\right) q + a + {bX} \\), where \\( q \\in \\mathbb{R}\\left\\lbrack X\\right\\rbrack \\) and \\( a, b \\in \\mathbb{R} \\) . Since \\( {\\left( -i\\right) }^{2} + 1 = {i}^{2} + 1 = 0 \\), we have\n\n\\[ g\\l... | Yes |
We defined the ring \( \mathbb{Z}\left\lbrack i\right\rbrack \) of Gaussian integers in Example 7.25 as a subring of \( \mathbb{C} \). Let us verify that the notation \( \mathbb{Z}\left\lbrack i\right\rbrack \) introduced in Example 7.25 is consistent with that introduced in Example 7.44. | Consider the polynomial evaluation map \( \rho : \mathbb{Z}\left\lbrack X\right\rbrack \rightarrow \mathbb{C} \) that sends \( g \in \mathbb{Z}\left\lbrack X\right\rbrack \) to \( g\left( i\right) \in \mathbb{C} \). For every \( g \in \mathbb{Z}\left\lbrack X\right\rbrack \), we may write \( g = \left( {{X}^{2} + 1}\ri... | Yes |
Let \( p \) be a prime, and consider the quotient ring \( E \mathrel{\text{:=}} {\mathbb{Z}}_{p}\left\lbrack X\right\rbrack /\left( f\right) \) , where \( f \mathrel{\text{:=}} {X}^{2} + 1 \) . If we set \( i \mathrel{\text{:=}} {\left\lbrack X\right\rbrack }_{f} \in E \), then \( E = {\mathbb{Z}}_{p}\left\lbrack i\rig... | If \( p = 2 \), then \( 0 = 1 + {i}^{2} = {\left( 1 + i\right) }^{2} \) (see Example 7.48), and so in this case, \( 1 + i \) is a zero divisor and \( E \) is not a field.\n\nNow suppose \( p \) is odd. There are two subcases to consider: \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) and \( p \equiv 3\left( {\... | Yes |
Example 7.60. Let \( p \) an odd prime, and let \( d \in {\mathbb{Z}}_{p}^{ * } \) . Let \( f \mathrel{\text{:=}} {X}^{2} - d \in {\mathbb{Z}}_{p}\left\lbrack X\right\rbrack \) , and consider the ring \( E \mathrel{\text{:=}} {\mathbb{Z}}_{p}\left\lbrack X\right\rbrack /\left( f\right) = {\mathbb{Z}}_{p}\left\lbrack \x... | Suppose that \( d \in {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \), so that \( d = {c}^{2} \) for some \( c \in {\mathbb{Z}}_{p}^{ * } \) . Then \( f = \left( {X - c}\right) \left( {X + c}\right) \) , and like in previous example, we have a ring isomorphism \( E \cong {\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p} \) (whic... | Yes |
Theorem 7.29. Let \( D \) be an integral domain and \( G \) a subgroup of \( {D}^{ * } \) of finite order. Then \( G \) is cyclic. | Proof. Suppose \( G \) is not cyclic. If \( m \) is the exponent of \( G \), then by Theorem 6.41, we know that \( m < \left| G\right| \) . Moreover, by definition, \( {a}^{m} = 1 \) for all \( a \in G \) ; that is, every element of \( G \) is a root of the polynomial \( {X}^{m} - 1 \in D\left\lbrack X\right\rbrack \) ... | Yes |
Lemma 7.30. Let \( p \) be a prime. For every positive integer \( e \), if \( a \equiv b\\left( {\\;\\operatorname{mod\\;}{p}^{e}}\\right) \), then \( {a}^{p} \equiv {b}^{p}\\left( {\\;\\operatorname{mod\\;}{p}^{e + 1}}\\right) \). | Proof. Suppose \( a \equiv b\\left( {\\;\\operatorname{mod\\;}{p}^{e}}\\right) \), so that \( a = b + c{p}^{e} \) for some \( c \\in \\mathbb{Z} \). Then \( {a}^{p} = {b}^{p} + p{b}^{p - 1}c{p}^{e} + d{p}^{2e} \) for some \( d \\in \\mathbb{Z} \), and it follows that \( {a}^{p} \equiv {b}^{p}\\left( {\\;\\operatorname{... | Yes |
Lemma 7.31. Let \( p \) be a prime, and let \( e \) be a positive integer such that \( {p}^{e} > 2 \) . If \( a \equiv 1 + {p}^{e}\left( {\;\operatorname{mod}\;{p}^{e + 1}}\right) \), then \( {a}^{p} \equiv 1 + {p}^{e + 1}\left( {\;\operatorname{mod}\;{p}^{e + 2}}\right) \) . | Proof. Suppose \( a \equiv 1 + {p}^{e}\left( {\;\operatorname{mod}\;{p}^{e + 1}}\right) \) . By Lemma 7.30, \( {a}^{p} \equiv {\left( 1 + {p}^{e}\right) }^{p}\left( {\;\operatorname{mod}\;{p}^{e + 2}}\right) \) . Expanding \( {\left( 1 + {p}^{e}\right) }^{p} \), we have\n\n\[ \n{\left( 1 + {p}^{e}\right) }^{p} = 1 + p ... | Yes |
Example 7.61. Let \( p \) be an odd prime, and let \( d \) be a positive integer dividing \( p - 1 \) . Since \( {\mathbb{Z}}_{p}^{ * } \) is a cyclic group of order \( p - 1 \), Theorem 6.32, implies that \( {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{d} \) is the unique subgroup of \( {\mathbb{Z}}_{p}^{ * } \) of order ... | \[ \alpha = {\beta }^{d}\text{ for some }\beta \in {\mathbb{Z}}_{p}^{ * } \Leftrightarrow {\alpha }^{\left( {p - 1}\right) /d} = 1. \] | Yes |
De Morgan’s law says that for all events \( \mathcal{A} \) and \( \mathcal{B} \) , | \[\overline{\mathcal{A} \cup \mathcal{B}} = \overline{\mathcal{A}} \cap \overline{\mathcal{B}}\text{ and }\overline{\mathcal{A} \cap \mathcal{B}} = \overline{\mathcal{A}} \cup \overline{\mathcal{B}}.\] | Yes |
Alice rolls two dice, and asks Bob to guess a value that appears on either of the two dice (without looking). Let us model this situation by considering the uniform distribution on \( \Omega \mathrel{\text{:=}} \{ 1,\ldots ,6\} \times \{ 1,\ldots ,6\} \), where for each pair \( \left( {s, t}\right) \in \Omega, s \) rep... | \n\[ \mathsf{P}\left\lbrack {\mathcal{C}}_{k}\right\rbrack = \mathsf{P}\left\lbrack {{\mathcal{A}}_{k} \cup {\mathcal{B}}_{k}}\right\rbrack = \mathsf{P}\left\lbrack {\mathcal{A}}_{k}\right\rbrack + \mathsf{P}\left\lbrack {\mathcal{B}}_{k}\right\rbrack - \mathsf{P}\left\lbrack {{\mathcal{A}}_{k} \cap {\mathcal{B}}_{k}}\... | Yes |
Suppose we have a coin that comes up heads with some probability \( p \), and tails with probability \( q \mathrel{\text{:=}} 1 - p \). We toss the coin \( n \) times, and record the outcomes. We can model this as the product distribution \( \mathrm{P} = {\mathrm{P}}_{1}^{n} \), where \( {\mathrm{P}}_{1} \) is the dist... | For each \( k = 0,\ldots, n \), let \( {\mathcal{A}}_{k} \) be the event that our coin comes up heads exactly \( k \) times. As a set, \( {\mathcal{A}}_{k} \) consists of all those tuples in the sample space with exactly \( k \) heads, and so\n\n\[ \left| {\mathcal{A}}_{k}\right| = \left( \begin{array}{l} n \\ k \end{a... | Yes |
Consider again Example 8.4, where \( \mathcal{A} \) is the event that the value on the die is odd, and \( \mathcal{B} \) is the event that the value of the die exceeds 2 . Then as we calculated, \( \mathrm{P}\left\lbrack \mathcal{A}\right\rbrack = 1/2,\mathrm{P}\left\lbrack \mathcal{B}\right\rbrack = 2/3 \), and \( \ma... | Indeed, \( \mathrm{P}\left\lbrack {\mathcal{A} \mid \mathcal{B}}\right\rbrack = \) \( \left( {1/3}\right) /\left( {2/3}\right) = 1/2 = \mathrm{P}\left\lbrack \mathcal{A}\right\rbrack \) ; intuitively, given the partial knowledge that the value on the die exceeds 2, we know it is equally likely to be either 3, 4, 5, or ... | Yes |
For example, suppose Alice tells Bob the sum is 4. Then what is Bob's best strategy in this case? Let \( {D}_{\ell } \) be the event that the sum is \( \ell \), for \( \ell = 2,\ldots ,{12} \), and consider the conditional distribution given \( {\mathcal{D}}_{4} \) . This conditional distribution is essentially the uni... | The numbers 1 and 3 both appear in two pairs, while the number 2 appears in just one pair. Therefore, \[ \mathrm{P}\left\lbrack {{\mathcal{C}}_{1} \mid {\mathcal{D}}_{4}}\right\rbrack = \mathrm{P}\left\lbrack {{\mathcal{C}}_{3} \mid {\mathcal{D}}_{4}}\right\rbrack = 2/3 \] while \[ \mathrm{P}\left\lbrack {{\mathcal{C}}... | Yes |
Let us continue with Example 8.11, and compute Bob's overall probability of winning, assuming he follows an optimal strategy. If the sum is 2 or 12, clearly there is only one sensible choice for Bob to make, and it will certainly be correct. If the sum is any other number \( \ell \), and there are \( {N}_{\ell } \) pai... | \[ \mathrm{P}\left\lbrack \mathcal{C}\right\rbrack = \mathop{\sum }\limits_{{\ell = 2}}^{{12}}\mathrm{P}\left\lbrack {\mathcal{C} \mid {\mathcal{D}}_{\ell }}\right\rbrack \mathrm{P}\left\lbrack {\mathcal{D}}_{\ell }\right\rbrack \] Moreover, \[ \mathrm{P}\left\lbrack {\mathcal{C} \mid {\mathcal{D}}_{2}}\right\rbrack \m... | Yes |
Suppose that the rate of incidence of disease \( X \) in the overall population is \( 1\% \) . Also suppose that there is a test for disease \( X \) ; however, the test is not perfect: it has a 5% false positive rate (i.e., 5% of healthy patients test positive for the disease), and a \( 2\% \) false negative rate (i.e.... | Let \( \mathcal{A} \) be the event that the test is positive and let \( \mathcal{B} \) be the event that the patient has disease \( X \) . The relevant quantity that we need to estimate is \( \mathrm{P}\left\lbrack {\mathcal{B} \mid \mathcal{A}}\right\rbrack \) ; that is, the probability that the patient has disease \(... | Yes |
In this game, a contestant chooses one of three doors. Behind two doors is a 'zonk,' and behind one of the doors is a 'grand prize.' After the contestant chooses a door, the host of the show, Monty Hall, always reveals a zonk behind one of the two doors not chosen by the contestant. The contestant is then given a choic... | Let us evaluate the two strategies. If the contestant always stays with his initial selection, then it is clear that his probability of success is exactly \( 1/3 \) .\n\nNow consider the strategy of always switching. Let \( \mathcal{B} \) be the event that the contestant’s initial choice was correct, and let \( \mathca... | Yes |
Suppose we toss a fair coin three times, which we formally model using the uniform distribution on the set of all 8 possible outcomes of the three coin tosses: (heads, heads, heads), (heads, heads, tails), etc., as in Example 8.8. For \( i = 1,2,3 \), let \( {\mathcal{A}}_{i} \) be the event that the \( i \) th toss co... | Now let \( {\mathcal{B}}_{12} \) be the event that the first and second tosses agree (i.e., both heads or both tails), let \( {\mathcal{B}}_{13} \) be the event that the first and third tosses agree, and let \( {\mathcal{B}}_{23} \) be the event that the second and third tosses agree. Then the family of events \( {\mat... | Yes |
Theorem 8.2. If \( \mathcal{A} \) and \( \mathcal{B} \) are independent events, then so are \( \mathcal{A} \) and \( \overline{\mathcal{B}} \) . | Proof. We have\n\n\[ \mathsf{P}\left\lbrack \mathcal{A}\right\rbrack = \mathsf{P}\left\lbrack {\mathcal{A} \cap \mathcal{B}}\right\rbrack + \mathsf{P}\left\lbrack {\mathcal{A} \cap \overline{\mathcal{B}}}\right\rbrack \;\text{(by total probability (8.9))} \]\n\n\[ = \mathsf{P}\left\lbrack \mathcal{A}\right\rbrack \math... | Yes |
Theorem 8.3. Let \( {\left\{ {\mathcal{A}}_{i}\right\} }_{i \in I} \) be a finite, \( k \) -wise independent family of events. Let \( J \) be a subset of \( I \), and for each \( i \in I \), define \( {\mathcal{A}}_{i}^{\prime } \mathrel{\text{:=}} {\mathcal{A}}_{i} \) if \( i \in J \), and \( {\mathcal{A}}_{i}^{\prime... | Proof. It suffices to prove the theorem for the case where \( J = I \smallsetminus \{ d\} \), for an arbitrary \( d \in I \) : this allows us to complement any single member of the family that we wish, without affecting independence; by repeating the procedure, we can complement any number of them.\n\nTo this end, it w... | Yes |
If \( \mathcal{A} \) is an event, we may define a random variable \( X \) as follows: \( X \mathrel{\text{:=}} 1 \) if the event \( \mathcal{A} \) occurs, and \( X \mathrel{\text{:=}} 0 \) otherwise. The variable \( X \) is called the indicator variable for \( \mathcal{A} \). Formally, \( X \) is the function that maps... | It is not hard to see that \( 1 - X \) is the indicator variable for \( \overline{\mathcal{A}} \). Now suppose \( \mathcal{B} \) is another event, with indicator variable \( Y \). Then it is also not hard to see that \( {XY} \) is the indicator variable for \( \mathcal{A} \cap \mathcal{B} \), and that \( X + Y - {XY} \... | Yes |
Consider again Example 8.8, where we have a coin that comes up heads with probability \( p \), and tails with probability \( q \mathrel{\text{:=}} 1 - p \), and we toss it \( n \) times. For each \( i = 1,\ldots, n \), let \( {\mathcal{A}}_{i} \) be the event that the \( i \) th toss comes up heads, and let \( {X}_{i} ... | By the calculations made in Example 8.8, for each \( k = 0,\ldots, n \), we have\n\n\[ \mathrm{P}\left\lbrack {X = k}\right\rbrack = \left( \begin{array}{l} n \\ k \end{array}\right) {p}^{k}{q}^{n - k} \] | Yes |
Theorem 8.4. Suppose \( f : S \rightarrow T \) is a surjective, regular function, and that \( X \) is a random variable that is uniformly distributed over \( S \) . Then \( f\left( X\right) \) is uniformly distributed over \( T \) . | Proof. The assumption that \( f \) is surjective and regular implies that for every \( t \in T \) , the set \( {S}_{t} \mathrel{\text{:=}} {f}^{-1}\left( {\{ t\} }\right) \) has size \( \left| S\right| /\left| T\right| \) . So, for each \( t \in T \), working directly from the definitions, we have\n\n\[ \mathrm{P}\left... | Yes |
Theorem 8.5. Suppose that \( \rho : G \rightarrow {G}^{\prime } \) is a surjective homomorphism of finite abelian groups \( G \) and \( {G}^{\prime } \), and that \( X \) is a random variable that is uniformly distributed over \( G \) . Then \( \rho \left( X\right) \) is uniformly distributed over \( {G}^{\prime } \) . | Proof. It suffices to show that \( \rho \) is regular. Recall that the kernel \( K \) of \( \rho \) is a subgroup of \( G \), and that for every \( {g}^{\prime } \in {G}^{\prime } \), the set \( {\rho }^{-1}\left( \left\{ {g}^{\prime }\right\} \right) \) is a coset of \( K \) (see Theorem 6.19); moreover, every coset o... | No |
We claim that \( {Z}^{\prime } \) is uniformly distributed over \( {\mathbb{Z}}_{6} \). | This follows immediately from the fact that the map that sends \( \left( {a, b}\right) \in {\mathbb{Z}}_{6} \times {\mathbb{Z}}_{6} \) to \( a + b \in {\mathbb{Z}}_{6} \) is a surjective group homomorphism (see Example 6.45). | Yes |
Let us continue with Examples 8.16 and 8.19. The random variables \( X \) and \( Y \) are independent: each is uniformly distributed over \( \{ 1,\ldots ,6\} \), and \( \left( {X, Y}\right) \) is uniformly distributed over \( \{ 1,\ldots ,6\} \times \{ 1,\ldots ,6\} \) . Let us calculate the conditional distribution of... | We have \( \mathrm{P}\left\lbrack {X = s \mid Z = 4}\right\rbrack = 1/3 \) for \( s = 1,2,3 \), and \( \mathrm{P}\left\lbrack {X = s \mid Z = 4}\right\rbrack = 0 \) for \( s = 4,5,6 \) . Thus, the conditional distribution of \( X \) given \( Z = 4 \) is essentially the uniform distribution on \( \{ 1,2,3\} \) . | Yes |
Returning again to Examples 8.16, 8.19, and 8.20, we see that the family of random variables \( {X}^{\prime },{Y}^{\prime },{Z}^{\prime } \) is pairwise independent, but not mutually independent; for example, | \[ \mathrm{P}\left\lbrack {\left( {{X}^{\prime } = {\left\lbrack 0\right\rbrack }_{6}}\right) \cap \left( {{Y}^{\prime } = {\left\lbrack 0\right\rbrack }_{6}}\right) \cap \left( {{Z}^{\prime } = {\left\lbrack 0\right\rbrack }_{6}}\right) }\right\rbrack = 1/{6}^{2}, \] but \[ \mathrm{P}\left\lbrack {{X}^{\prime } = {\le... | Yes |
Theorem 8.6. Let \( X \) be a random variable with image \( S \), and \( Y \) be a random variable with image \( T \). Further, suppose that \( f : S \rightarrow \left\lbrack {0,1}\right\rbrack \) and \( g : T \rightarrow \left\lbrack {0,1}\right\rbrack \) are functions such that\n\n\[ \mathop{\sum }\limits_{{s \in S}}... | Proof. Since \( \{ Y = t{\} }_{t \in T} \) is a partition of the sample space, making use of total probability (8.9), along with (8.15) and (8.14), we see that for all \( s \in S \), we have\n\n\[ \mathrm{P}\left\lbrack {X = s}\right\rbrack = \mathop{\sum }\limits_{{t \in T}}\mathrm{P}\left\lbrack {\left( {X = s}\right... | Yes |
Theorem 8.7. Let \( {\left\{ {X}_{i}\right\} }_{i \in I} \) be a finite family of random variables, where each \( {X}_{i} \) has image \( {S}_{i} \) . Also, let \( {\left\{ {f}_{i}\right\} }_{i \in I} \) be a family of functions, where for each \( i \in I \) , \( {f}_{i} : {S}_{i} \rightarrow \left\lbrack {0,1}\right\r... | Proof. To prove the theorem, it suffices to prove the following statement: for every subset \( J \) of \( I \), and every assignment \( {\left\{ {s}_{j}\right\} }_{j \in J} \) to \( {\left\{ {X}_{j}\right\} }_{j \in J} \), we have\n\n\[ \mathrm{P}\left\lbrack {\mathop{\bigcap }\limits_{{j \in J}}\left( {{X}_{j} = {s}_{... | Yes |
Theorem 8.12. Suppose \( {\left\{ {X}_{i}\right\} }_{i = 1}^{n} \) is a mutually independent family of random variables. Further, suppose that for \( i = 1,\ldots, n,{Y}_{i} \mathrel{\text{:=}} {g}_{i}\left( {X}_{i}\right) \) for some function \( {g}_{i} \) . Then \( {\left\{ {Y}_{i}\right\} }_{i = 1}^{n} \) is mutuall... | Proof. It suffices to prove the theorem for \( n = 2 \) . The general case follows easily by induction, using Theorem 8.9. For \( i = 1,2 \), let \( {t}_{i} \) be any value in the image of \( {Y}_{i} \), and let \( {S}_{i}^{\prime } \mathrel{\text{:=}} {g}_{i}^{-1}\left( \left\{ {t}_{i}\right\} \right) \) . We have\n\n... | Yes |
Theorem 8.13. Suppose that \( G \) is a finite abelian group, and that \( W \) is a random variable uniformly distributed over \( G \) . Let \( Z \) be another random variable, taking values in some finite set \( U \), and suppose that \( W \) and \( Z \) are independent. Let \( \sigma : U \rightarrow G \) be some func... | Proof. Consider any fixed values \( t \in G \) and \( u \in U \) . Evidently, the events \( \left( {Y = t}\right) \cap \left( {Z = u}\right) \) and \( \left( {W = t - \sigma \left( u\right) }\right) \cap \left( {Z = u}\right) \) are the same, and therefore, because \( W \) and \( Z \) are independent, we have\n\n\[ \ma... | Yes |
Let \( k \) be a positive integer. This example shows how we can take a mutually independent family of \( k \) random variables, and, from it, construct a much larger, \( k \) -wise independent family of random variables. | Let \( p \) be a prime, with \( p \geq k \) . Let \( {\left\{ {H}_{i}\right\} }_{i = 0}^{k - 1} \) be a mutually independent family of random variables, each of which is uniformly distributed over \( {\mathbb{Z}}_{p} \) . Let us set \( H \mathrel{\text{:=}} \left( {{H}_{0},\ldots ,{H}_{k - 1}}\right) \), which, by assu... | Yes |
Consider again the secret sharing scenario of Example 8.26. Suppose at the critical moment, one of the officers is missing in action. The military planners would perhaps like a more flexible secret sharing scheme; for example, perhaps shares of the launch code should be distributed to three officers, in such a way that... | First, we show how any coalition of \( k + 1 \) officers can reconstruct the launch code from their collection of shares, say, \( {Y}_{{s}_{1}},\ldots ,{Y}_{{s}_{k + 1}} \) . This is easily done by means of the Lagrange interpolation formula (again, Theorem 7.15). Indeed, we only need to recover the high-order coeffici... | Yes |
Theorem 8.14 (Linearity of expectation). If \( X \) and \( Y \) are real-valued random variables, and a is a real number, then\n\n\[ \mathrm{E}\left\lbrack {X + Y}\right\rbrack = \mathrm{E}\left\lbrack X\right\rbrack + \mathrm{E}\left\lbrack Y\right\rbrack \text{ and }\mathrm{E}\left\lbrack {aX}\right\rbrack = a\mathrm... | Proof. It is easiest to prove this using the defining equation (8.18) for expectation. For \( \omega \in \Omega \), the value of the random variable \( X + Y \) at \( \omega \) is by definition \( X\left( \omega \right) + Y\left( \omega \right) \), and so we have\n\n\[ \mathrm{E}\left\lbrack {X + Y}\right\rbrack = \mat... | Yes |
Theorem 8.15. If \( X \) and \( Y \) are independent, real-valued random variables, then \( \mathrm{E}\left\lbrack {XY}\right\rbrack = \mathrm{E}\left\lbrack X\right\rbrack \mathrm{E}\left\lbrack Y\right\rbrack \) | Proof. It is easiest to prove this using (8.20), with the function \( f\left( {s, t}\right) \mathrel{\text{:=}} {st} \) applied to the random variable \( \left( {X, Y}\right) \) . We have\n\n\[ \mathrm{E}\left\lbrack {XY}\right\rbrack = \mathop{\sum }\limits_{{s, t}}{st}\mathrm{P}\left\lbrack {\left( {X = s}\right) \ca... | Yes |
Theorem 8.16. Let \( X \) be a \( 0/1 \) -valued random variable. Then \( \mathrm{E}\left\lbrack X\right\rbrack = \mathrm{P}\left\lbrack {X = 1}\right\rbrack \) . | Proof. \( \mathrm{E}\left\lbrack X\right\rbrack = 0 \cdot \mathrm{P}\left\lbrack {X = 0}\right\rbrack + 1 \cdot \mathrm{P}\left\lbrack {X = 1}\right\rbrack = \mathrm{P}\left\lbrack {X = 1}\right\rbrack \) . | Yes |
Theorem 8.17. If \( X \) is a random variable that takes only non-negative integer values, then\n\n\[ \mathrm{E}\left\lbrack X\right\rbrack = \mathop{\sum }\limits_{{i \geq 1}}\mathrm{P}\left\lbrack {X \geq i}\right\rbrack \]\n\nNote that since \( X \) has a finite image, the sum appearing above is finite. | Proof. Suppose that the image of \( X \) is contained in \( \{ 0,\ldots, n\} \), and for \( i = 1,\ldots, n \) , let \( {X}_{i} \) be the indicator variable for the event \( X \geq i \) . Then \( X = {X}_{1} + \cdots + {X}_{n} \), and by linearity of expectation and Theorem 8.16, we have\n\n\[ \mathrm{E}\left\lbrack X\... | Yes |
Theorem 8.18. Let \( X \) be a real-valued random variable, with \( \mu \mathrel{\text{:=}} \mathrm{E}\left\lbrack X\right\rbrack \), and let \( a \) and \( b \) be real numbers. Then we have\n\n(i) \( \operatorname{Var}\left\lbrack X\right\rbrack = \mathrm{E}\left\lbrack {X}^{2}\right\rbrack - {\mu }^{2} \) | Proof. For part (i), observe that\n\n\[ \operatorname{Var}\left\lbrack X\right\rbrack = \mathrm{E}\left\lbrack {\left( X - \mu \right) }^{2}\right\rbrack = \mathrm{E}\left\lbrack {{X}^{2} - {2\mu X} + {\mu }^{2}}\right\rbrack \]\n\n\[ = \mathsf{E}\left\lbrack {X}^{2}\right\rbrack - {2\mu }\;\mathsf{E}\left\lbrack X\rig... | Yes |
Theorem 8.20. If \( {\left\{ {X}_{i}\right\} }_{i \in I} \) is a finite, pairwise independent family of real-valued random variables, then\n\n\[ \operatorname{Var}\left\lbrack {\mathop{\sum }\limits_{{i \in I}}{X}_{i}}\right\rbrack = \mathop{\sum }\limits_{{i \in I}}\operatorname{Var}\left\lbrack {X}_{i}\right\rbrack \... | Proof. We have\n\n\[ \operatorname{Var}\left\lbrack {\mathop{\sum }\limits_{{i \in I}}{X}_{i}}\right\rbrack = \mathrm{E}\left\lbrack {\left( \mathop{\sum }\limits_{{i \in I}}{X}_{i}\right) }^{2}\right\rbrack - {\left( \mathrm{E}\left\lbrack \mathop{\sum }\limits_{{i \in I}}{X}_{i}\right\rbrack \right) }^{2} \]\n\n\[ = ... | Yes |
Theorem 8.21. Let \( X \) be a \( 0/1 \) -valued random variable, with \( p \mathrel{\text{:=}} \mathrm{P}\left\lbrack {X = 1}\right\rbrack \) and \( q \mathrel{\text{:=}} \mathrm{P}\left\lbrack {X = 0}\right\rbrack = 1 - p \) . Then \( \operatorname{Var}\left\lbrack X\right\rbrack = {pq} \) . | Proof. We have \( \mathrm{E}\left\lbrack X\right\rbrack = p \) and \( \mathrm{E}\left\lbrack {X}^{2}\right\rbrack = \mathrm{P}\left\lbrack {{X}^{2} = 1}\right\rbrack = \mathrm{P}\left\lbrack {X = 1}\right\rbrack = p \) . Therefore,\n\n\[ \operatorname{Var}\left\lbrack X\right\rbrack = \mathrm{E}\left\lbrack {X}^{2}\rig... | Yes |
Let \( X \) be uniformly distributed over \( \{ 1,\ldots, m\} \) . Let us compute \( \mathrm{E}\left\lbrack X\right\rbrack \) and \( \operatorname{Var}\left\lbrack X\right\rbrack \) . | We have\n\n\[ \mathrm{E}\left\lbrack X\right\rbrack = \mathop{\sum }\limits_{{s = 1}}^{m}s \cdot \frac{1}{m} = \frac{m\left( {m + 1}\right) }{2} \cdot \frac{1}{m} = \frac{m + 1}{2}. \]\n\nWe also have\n\n\[ \mathrm{E}\left\lbrack {X}^{2}\right\rbrack = \mathop{\sum }\limits_{{s = 1}}^{m}{s}^{2} \cdot \frac{1}{m} = \fra... | Yes |
Let \( X \) denote the value of a roll of a die. Let \( \mathcal{A} \) be the event that \( X \) is even. Then the conditional distribution of \( X \) given \( \mathcal{A} \) is essentially the uniform distribution on \( \{ 2,4,6\} \), and hence | \[ \mathrm{E}\left\lbrack {X \mid \mathcal{A}}\right\rbrack = \frac{2 + 4 + 6}{3} = 4. \] Similarly, the conditional distribution of \( X \) given \( \overline{\mathcal{A}} \) is essentially the uniform distribution on \( \{ 1,3,5\} \), and so \[ \mathrm{E}\left\lbrack {X \mid \overline{\mathcal{A}}}\right\rbrack = \fr... | Yes |
Let \( X \) be a random variable with a binomial distribution, as in Example 8.18, that counts the number of successes among \( n \) Bernoulli trials, each of which succeeds with probability \( p \). Let us compute \( \mathrm{E}\left\lbrack X\right\rbrack \) and \( \operatorname{Var}\left\lbrack X\right\rbrack \). | We can write \( X \) as the sum of indicator variables, \( X = \mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} \), where \( {X}_{i} \) is the indicator variable for the event that the \( i \) th trial succeeds; each \( {X}_{i} \) takes the value 1 with probability \( p \) and 0 with probability \( q \mathrel{\text{:=}} 1 - ... | Yes |
Example 8.32. Our proof of Theorem 8.1 could be elegantly recast in terms of indicator variables. For \( \mathcal{B} \subseteq \Omega \), let \( {X}_{\mathcal{B}} \) be the indicator variable for \( \mathcal{B} \), so that \( {X}_{\mathcal{B}}\left( \omega \right) = {\delta }_{\omega }\left\lbrack \mathcal{B}\right\rbr... | and by Theorem 8.16 and linearity of expectation, we have\n\n\[ \n\mathrm{P}\left\lbrack \mathcal{A}\right\rbrack = \mathrm{E}\left\lbrack {X}_{\mathcal{A}}\right\rbrack = \mathop{\sum }\limits_{{\varnothing \varsubsetneq J \subseteq I}}{\left( -1\right) }^{\left| J\right| - 1}\mathrm{E}\left\lbrack {X}_{{\mathcal{A}}_... | Yes |
Theorem 8.22 (Markov’s inequality). Let \( X \) be a random variable that takes only non-negative real values. Then for every \( \alpha > 0 \), we have\n\n\[ \mathrm{P}\left\lbrack {X \geq \alpha }\right\rbrack \leq \mathrm{E}\left\lbrack X\right\rbrack /\alpha . \] | Proof. We have\n\n\[ \mathrm{E}\left\lbrack X\right\rbrack = \mathop{\sum }\limits_{s}s\mathrm{P}\left\lbrack {X = s}\right\rbrack = \mathop{\sum }\limits_{{s < \alpha }}s\mathrm{P}\left\lbrack {X = s}\right\rbrack + \mathop{\sum }\limits_{{s \geq \alpha }}s\mathrm{P}\left\lbrack {X = s}\right\rbrack ,\]\n\nwhere the s... | Yes |
Theorem 8.23 (Chebyshev’s inequality). Let \( X \) be a real-valued random variable, with \( \mu \mathrel{\text{:=}} \mathrm{E}\left\lbrack X\right\rbrack \) and \( \nu \mathrel{\text{:=}} \operatorname{Var}\left\lbrack X\right\rbrack \) . Then for every \( \alpha > 0 \), we have\n\n\[ \mathrm{P}\left\lbrack {\left| {X... | Proof. Let \( Y \mathrel{\text{:=}} {\left( X - \mu \right) }^{2} \) . Then \( Y \) is always non-negative, and \( \mathrm{E}\left\lbrack Y\right\rbrack = v \) . Applying Markov’s inequality to \( Y \), we have\n\n\[ \mathrm{P}\left\lbrack {\left| {X - \mu }\right| \geq \alpha }\right\rbrack = \mathrm{P}\left\lbrack {Y... | Yes |
Theorem 8.20 (along with part (ii) of Theorem 8.18) that \( \operatorname{Var}\left\lbrack \bar{X}\right\rbrack = v/n \) . Applying Chebyshev’s inequality, for every \( \varepsilon > 0 \), we have | \[ \mathrm{P}\left\lbrack {\left| {\bar{X} - \mu }\right| \geq \varepsilon }\right\rbrack \leq \frac{v}{n{\varepsilon }^{2}} \] | Yes |
Suppose we toss a fair coin 10,000 times. The expected number of heads is 5,000. What is an upper bound on the probability \( \alpha \) that we get 6,000 or more heads? | Using Markov’s inequality, we get \( \alpha \leq 5/6 \). Using Chebyshev’s inequality, and in particular, the inequality (8.27), we get\n\n\[ \alpha \leq \frac{1/4}{{10}^{4}{10}^{-2}} = \frac{1}{400} \]\n\nFinally, using the Chernoff bound, we obtain\n\n\[ \alpha \leq {e}^{-{10}^{4}{10}^{-2}/2\left( {0.5}\right) } = {e... | Yes |
Theorem 8.25. Suppose \( {\left\{ {X}_{i}\right\} }_{i \in I} \) is pairwise independent. Then for all \( i, j \in I \) with \( i \neq j \), we have \( \mathrm{P}\left\lbrack {{X}_{i} = {X}_{j}}\right\rbrack = 1/m \) . | Proof. The event \( {X}_{i} = {X}_{j} \) occurs if and only if \( {X}_{i} = s \) and \( {X}_{j} = s \) for some \( s \in S \) . Therefore,\n\n\[ \mathrm{P}\left\lbrack {{X}_{i} = {X}_{j}}\right\rbrack = \mathop{\sum }\limits_{{s \in S}}\mathrm{P}\left\lbrack {\left( {{X}_{i} = s}\right) \cap \left( {{X}_{j} = s}\right)... | Yes |
Theorem 8.26. Suppose \( {\left\{ {X}_{i}\right\} }_{i \in I} \) is pairwise independent. Then\n\n\[ \mathrm{P}\left\lbrack \mathcal{C}\right\rbrack \leq \frac{n\left( {n - 1}\right) }{2m} \] | Proof. Let \( {I}^{\left( 2\right) } \mathrel{\text{:=}} \{ J \subseteq I : \left| J\right| = 2\} \) . Then using Boole’s inequality (8.6) and Theorem 8.25, we have\n\n\[ \mathrm{P}\left\lbrack C\right\rbrack \leq \mathop{\sum }\limits_{{\{ i, j\} \in {I}^{\left( 2\right) }}}\mathrm{P}\left\lbrack {{X}_{i} = {X}_{j}}\r... | Yes |
Theorem 8.27. Suppose \( {\left\{ {X}_{i}\right\} }_{i \in I} \) is pairwise independent. Then\n\n\[ \mathrm{E}\left\lbrack M\right\rbrack \leq \sqrt{{n}^{2}/m + n} \] | Proof. To prove this, we use the fact that \( \mathrm{E}{\left\lbrack M\right\rbrack }^{2} \leq \mathrm{E}\left\lbrack {M}^{2}\right\rbrack \) (see Theorem 8.19), and that \( {M}^{2} \leq Z \mathrel{\text{:=}} \mathop{\sum }\limits_{{s \in S}}{N}_{s}^{2} \) . It will therefore suffice to show that\n\n\[ \mathrm{E}\left... | Yes |
Theorem 8.28. Suppose \( {\left\{ {X}_{i}\right\} }_{i \in I} \) is mutually independent. Then\n\n\[ \mathrm{P}\left\lbrack \mathcal{C}\right\rbrack \geq 1 - {e}^{-n\left( {n - 1}\right) /{2m}}. \] | Proof. Let \( \alpha \mathrel{\text{:=}} \mathrm{P}\left\lbrack \bar{C}\right\rbrack \) . We want to show \( \alpha \leq {e}^{-n\left( {n - 1}\right) /{2m}} \) . We may assume that \( I = \{ 1,\ldots, n\} \) (the labels make no difference) and that \( n \leq m \) (otherwise, \( \alpha = 0 \) ). Under the hypothesis of ... | Yes |
Suppose Alice wants to send a message to Bob in such a way that Bob can be reasonably sure that the message he receives really came from Alice, and was not modified in transit by some malicious adversary. We present a solution to this problem here that works assuming that Alice and Bob share a randomly generated secret... | Suppose that \( {\left\{ {\Phi }_{r}\right\} }_{r \in R} \) is a pairwise independent family of hash functions from \( S \) to \( T \) . We model the shared random key as a random variable \( H \), uniformly distributed over \( R \) . We also model Alice’s message as a random variable \( X \), taking values in the set ... | No |
Example 8.36. By setting \( k \mathrel{\text{:=}} 2 \) in Example 8.27, for each prime \( p \), we immediately get a pairwise independent family of hash functions \( {\left\{ {\Phi }_{r}\right\} }_{r \in R} \) from \( {\mathbb{Z}}_{p} \) to \( {\mathbb{Z}}_{p} \) , where \( R = {\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p} ... | \[ {\Phi }_{r} : \;{\mathbb{Z}}_{p} \rightarrow {\mathbb{Z}}_{p} \] \[ s \mapsto {r}_{0} + {r}_{1}s. \] | Yes |
Let \( p \) be a prime, and let \( \ell \) be a positive integer. Let \( S \mathrel{\text{:=}} {\mathbb{Z}}_{p}^{\times \ell } \) and \( R \mathrel{\text{:=}} {\mathbb{Z}}_{p}^{\times \left( {\ell + 1}\right) } \) . For each \( r = \left( {{r}_{0},{r}_{1},\ldots ,{r}_{\ell }}\right) \in R \), we define the hash functio... | To this end, let \( H \) be a random variable uniformly distributed over \( R \) . We want to show that for each \( s,{s}^{\prime } \in S \) with \( s \neq {s}^{\prime } \), the random variable \( \left( {{\Phi }_{H}\left( s\right) ,{\Phi }_{H}\left( {s}^{\prime }\right) }\right) \) is uniformly distributed over \( {\m... | Yes |
Our goal is to show that \( {\left\{ {\Phi }_{r}\right\} }_{r \in R} \) is a universal family of hash functions from \( {\mathbb{Z}}_{p} \) to \( {\mathbb{Z}}_{m} \) . So let \( s,{s}^{\prime } \in {\mathbb{Z}}_{p} \) with \( s \neq {s}^{\prime } \), let \( {H}_{0} \) and \( {H}_{1} \) be independent random variables, ... | Let us define random variables \( Y \mathrel{\text{:=}} {H}_{0} + {H}_{1}s \) and \( {Y}^{\prime } \mathrel{\text{:=}} {H}_{0} + {H}_{1}{s}^{\prime } \) . Also, let \( \widehat{s} \mathrel{\text{:=}} {s}^{\prime } - s \neq 0 \) . Then we have\n\n\[ \mathrm{P}\left\lbrack \mathcal{C}\right\rbrack = \mathrm{P}\left\lbrac... | Yes |
Our goal is to show that \( {\left\{ {\Phi }_{r}\right\} }_{r \in R} \) is a universal family of hash functions from \( S \) to \( {\mathbb{Z}}_{p} \) . So let \( s,{s}^{\prime } \in S \) with \( s \neq {s}^{\prime } \), and let \( H \) be a random variable that is uniformly distributed over \( R \) . We want to show t... | Let \( s = \left( {{s}_{0},{s}_{1},\ldots ,{s}_{\ell }}\right) \) and \( {s}^{\prime } = \left( {{s}_{0}^{\prime },{s}_{1}^{\prime },\ldots ,{s}_{\ell }^{\prime }}\right) \), and set \( {\widehat{s}}_{i} \mathrel{\text{:=}} {s}_{i}^{\prime } - {s}_{i} \) for \( i = 0,1,\ldots ,\ell \) . Let us define the function\n\n\[... | Yes |
Theorem 8.30. For random variables \( X, Y, Z \), we have\n\n(i) \( 0 \leq \Delta \left\lbrack {X;Y}\right\rbrack \leq 1 \) ,\n\n(ii) \( \Delta \left\lbrack {X;X}\right\rbrack = 0 \) ,\n\n(iii) \( \Delta \left\lbrack {X;Y}\right\rbrack = \Delta \left\lbrack {Y;X}\right\rbrack \), and\n\n(iv) \( \Delta \left\lbrack {X;Z... | Proof. Exercise. | No |
Suppose \( X \) has the uniform distribution on \( \{ 1,\ldots, m\} \), and \( Y \) has the uniform distribution on \( \{ 1,\ldots, m - \delta \} \), where \( \delta \in \{ 0,\ldots, m - 1\} \). Let us compute \( \Delta \left\lbrack {X;Y}\right\rbrack \). | The statistical distance between \( X \) and \( Y \) is just \( 1/2 \) times the area of regions \( A \) and \( C \) in the diagram. Moreover, because probability distributions sum to 1, we must have\n\n\[ \text{area of}B + \text{area of}A = 1 = \text{area of}B + \text{area of}C\text{,}\]\n\nand hence, the areas of reg... | Yes |
Theorem 8.31. Let \( X \) and \( Y \) be random variables taking values in a set \( S \). For every \( {S}^{\prime } \subseteq S \), we have\n\n\[ \Delta \left\lbrack {X;Y}\right\rbrack \geq \left| {\mathrm{P}\left\lbrack {X \in {S}^{\prime }}\right\rbrack - \mathrm{P}\left\lbrack {Y \in {S}^{\prime }}\right\rbrack }\r... | Proof. Suppose we split the set \( S \) into two disjoint subsets: the set \( {S}_{0} \) consisting of those \( s \in S \) such that \( \mathrm{P}\left\lbrack {X = s}\right\rbrack < \mathrm{P}\left\lbrack {Y = s}\right\rbrack \), and the set \( {S}_{1} \) consisting of those \( s \in S \) such that \( \mathrm{P}\left\l... | Yes |
Theorem 8.32. If \( S \) and \( T \) are finite sets, \( X \) and \( Y \) are random variables taking values in \( S \), and \( f : S \rightarrow T \) is a function, then \( \Delta \left\lbrack {f\left( X\right) ;f\left( Y\right) }\right\rbrack \leq \Delta \left\lbrack {X;Y}\right\rbrack \) . | Proof. We have\n\n\[ \Delta \left\lbrack {f\left( X\right) ;f\left( Y\right) }\right\rbrack = \left| {\mathrm{P}\left\lbrack {f\left( X\right) \in {T}^{\prime }}\right\rbrack - \mathrm{P}\left\lbrack {f\left( Y\right) \in {T}^{\prime }}\right\rbrack }\right| \text{ for some }{T}^{\prime } \subseteq T \]\n\n\[ \text{(by... | Yes |
Let \( X \) be uniformly distributed over the set \( \{ 0,\ldots, m - 1\} \), and let \( Y \) be uniformly distributed over the set \( \{ 0,\ldots, n - 1\} \), for \( n \geq m \). Let \( f\left( t\right) \mathrel{\text{:=}} t{\;\operatorname{mod}\;m} \). We want to compute an upper bound on the statistical distance bet... | We can do this as follows. Let \( n = {qm} - r \), where \( 0 \leq r < m \), so that \( q = \lceil n/m\rceil \). Also, let \( Z \) be uniformly distributed over \( \{ 0,\ldots ,{qm} - 1\} \). Then \( f\left( Z\right) \) is uniformly distributed over \( \{ 0,\ldots, m - 1\} \), since every element of \( \{ 0,\ldots, m -... | Yes |
Theorem 8.33. Suppose \( X, Y \), and \( Z \) are random variables, where \( X \) and \( Z \) are independent, and \( Y \) and \( Z \) are independent. Then \( \Delta \left\lbrack {X, Z;Y, Z}\right\rbrack = \Delta \left\lbrack {X, Y}\right\rbrack \) . | Proof. Suppose \( X \) and \( Y \) take values in a finite set \( S \), and \( Z \) takes values in a finite set \( T \). From the definition of statistical distance,\n\n\[ \n{2\Delta }\left\lbrack {X, Z;Y, Z}\right\rbrack = \mathop{\sum }\limits_{{s, t}}\left| {\mathrm{P}\left\lbrack {\left( {X = s}\right) \cap \left(... | Yes |
Theorem 8.34. Let \( {X}_{1},\ldots ,{X}_{n},{Y}_{1},\ldots ,{Y}_{n} \) be random variables, where \( {\left\{ {X}_{i}\right\} }_{i = 1}^{n} \) is mutually independent, and \( {\left\{ {Y}_{i}\right\} }_{i = 1}^{n} \) is mutually independent. Then we have\n\n\[ \Delta \left\lbrack {{X}_{1},\ldots ,{X}_{n};{Y}_{1},\ldot... | Proof. Since \( \Delta \left\lbrack {{X}_{1},\ldots ,{X}_{n};{Y}_{1},\ldots ,{Y}_{n}}\right\rbrack \) depends only on the individual distributions of the random variables \( \left( {{X}_{1},\ldots ,{X}_{n}}\right) \) and \( \left( {{Y}_{1},\ldots ,{Y}_{n}}\right) \), without loss of generality, we may assume that \( \l... | Yes |
Theorem 8.35. Suppose \( X \) is a random variable that takes values in a finite set \( S \) of size \( m \) . If \( X \) has collision probability \( \beta \), guessing probability \( \gamma \), and distance \( \delta \) from uniform on \( S \), then:\n\n(i) \( \beta \geq 1/m \) ;\n\n(ii) \( {\gamma }^{2} \leq \beta \... | Proof. Part (i) is immediate from Exercise 8.37. The other inequalities are left as easy exercises. | No |
Theorem 8.36. Suppose \( X \) is a random variable that takes values in a finite set \( S \) of size \( m \) . If \( X \) has collision probability \( \beta \), and distance \( \delta \) from uniform on \( S \), then \( \delta \leq \frac{1}{2}\sqrt{{m\beta } - 1}. \) | Proof. We may assume that \( \delta > 0 \), since otherwise the theorem is already true, simply from the fact that \( \beta \geq 1/m \) . For \( s \in S \), let \( {p}_{s} \mathrel{\text{:=}} \mathrm{P}\left\lbrack {X = s}\right\rbrack \) . We have \( \delta = \frac{1}{2}\mathop{\sum }\limits_{s}\left| {{p}_{s} - 1/m}\... | No |
Theorem 8.37 (Leftover hash lemma). Let \( {\left\{ {\Phi }_{r}\right\} }_{r \in R} \) be a \( \left( {1 + \alpha }\right) /m \) -almost universal family of hash functions from \( S \) to \( T \), where \( m \mathrel{\text{:=}} \left| T\right| \) . Let \( H \) and \( X \) be independent random variables, where \( H \) ... | Proof. Let \( {\beta }^{\prime } \) be the collision probability of \( \left( {H,{\Phi }_{H}\left( X\right) }\right) \) . Our goal is to bound \( {\beta }^{\prime } \) from above, and then apply Theorem 8.36 to the random variable \( \left( {H,{\Phi }_{H}\left( X\right) }\right) \) . To this end, let \( \ell \mathrel{\... | Yes |
Theorem 8.38. Let \( {\left\{ {\Phi }_{r}\right\} }_{r \in R} \) be a \( \left( {1 + \alpha }\right) /m \) -almost universal family of hash functions from \( S \) to \( T \), where \( m \mathrel{\text{:=}} \left| T\right| \) . Let \( H,{X}_{1},\ldots ,{X}_{n} \) be random variables, where \( H \) is uniformly distribut... | Proof. Let \( {Y}_{1},\ldots ,{Y}_{n} \) be random variables, each uniformly distributed over \( T \), and assume that \( H,{X}_{1},\ldots ,{X}_{n},{Y}_{1},\ldots ,{Y}_{n} \) form a mutually independent family of random variables. We shall make a hybrid argument (as in the proof of Theorem 8.34). Define random variable... | Yes |
Example 8.43. Suppose we toss a fair coin repeatedly until it comes up heads, and let \( k \) be the total number of tosses. We can model this experiment as a discrete probability distribution \( \mathrm{P} \), where the sample space consists of the set of all positive integers: for each positive integer \( k,\mathrm{P... | We can check that indeed \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{2}^{-k} = 1 \), as required. | Yes |
More generally, suppose we repeatedly execute a Bernoulli trial until it succeeds, where each execution succeeds with probability \( p > 0 \) independently of the previous trials, and let \( k \) be the total number of trials executed. Then we associate the probability \( \mathrm{P}\left( k\right) \mathrel{\text{:=}} {... | One can easily check that these probabilities sum to 1 . Such a distribution is called a geometric distribution. | No |
The series \( \mathop{\sum }\limits_{{k = 1}}^{\infty }1/{k}^{3} \) converges to some positive number \( c \) . Therefore, we can define a probability distribution on the set of positive integers, where we associate with each \( k \geq 1 \) the probability \( 1/c{k}^{3} \) . | As in the finite case, an event is an arbitrary subset \( \mathcal{A} \) of \( \Omega \) . The probability \( \mathrm{P}\left\lbrack \mathcal{A}\right\rbrack \) of \( \mathcal{A} \) is defined as the sum of the probabilities associated with the elements of \( \mathcal{A} \) . This sum is treated as an infinite series w... | No |
Consider the geometric distribution discussed in Example 8.44, where \( p \) is the success probability of each Bernoulli trial, and \( q \mathrel{\text{:=}} 1 - p \). For a given integer \( i \geq 1 \), consider the event \( \mathcal{A} \) that the number of trials executed is at least \( i \). Formally, \( \mathcal{A... | \[ \mathrm{P}\left\lbrack \mathcal{A}\right\rbrack = \mathop{\sum }\limits_{{k \geq i}}\mathrm{P}\left( k\right) = \mathop{\sum }\limits_{{k \geq i}}{q}^{k - 1}p = {q}^{i - 1}p\mathop{\sum }\limits_{{k \geq 0}}{q}^{k} = {q}^{i - 1}p \cdot \frac{1}{1 - q} = {q}^{i - 1}. \] | Yes |
Theorem 8.39. Suppose \( \mathcal{A} \mathrel{\text{:=}} \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{\mathcal{A}}_{i} \), where \( {\left\{ {\mathcal{A}}_{i}\right\} }_{i = 1}^{\infty } \) is an infinite sequence of events. Then\n\n(i) \( \mathrm{P}\left\lbrack \mathcal{A}\right\rbrack \leq \mathop{\sum }\limits_{{i =... | Proof. As in the proof of Theorem 8.1, for \( \omega \in \Omega \) and \( \mathcal{B} \subseteq \Omega \), define \( {\delta }_{\omega }\left\lbrack \mathcal{B}\right\rbrack \mathrel{\text{:=}} 1 \) if \( \omega \in \mathcal{B} \), and \( {\delta }_{\omega }\left\lbrack \mathcal{B}\right\rbrack \mathrel{\text{:=}} 0 \)... | Yes |
Theorem 8.40. Let \( {\left\{ {X}_{i}\right\} }_{i = 1}^{\infty } \) be an infinite sequence of random variables. Suppose that for each \( i \geq 1,{X}_{i} \) takes non-negative values only, and has finite expectation. Also suppose that \( \mathop{\sum }\limits_{{i = 1}}^{\infty }{X}_{i}\left( \omega \right) \) converg... | Proof. This is a calculation just like the one made in the proof of Theorem 8.39, where, again, we use the fact that we may reverse the order of summation in an infinite double summation of non-negative terms:\n\n\[ \mathrm{E}\left\lbrack X\right\rbrack = \mathop{\sum }\limits_{{\omega \in \Omega }}\mathrm{P}\left( \om... | Yes |
Suppose \( X \) is a random variable with a geometric distribution, as in Example 8.44, with an associated success probability \( p \) and failure probability \( q \mathrel{\text{:=}} 1 - p \). As we saw in Example 8.46, for every integer \( i \geq 1 \), we have \( \mathrm{P}\left\lbrack {X \geq i}\right\rbrack = {q}^{... | \[ \mathrm{E}\left\lbrack X\right\rbrack = \mathop{\sum }\limits_{{i = 1}}^{\infty }\mathrm{P}\left\lbrack {X \geq i}\right\rbrack = \mathop{\sum }\limits_{{i = 1}}^{\infty }{q}^{i - 1} = \frac{1}{1 - q} = \frac{1}{p} \] | Yes |
Theorem 9.1. Let \( \Omega \) be the set of all exact execution paths for \( A \) on input \( x \) . Then \( \mathop{\sum }\limits_{{\omega \in \Omega }}{2}^{-\left| \omega \right| } \leq 1 \) | Proof. Let \( k \) be a non-negative integer. Let \( {\Omega }_{k} \subseteq \Omega \) be the set of all exact execution paths of length at most \( k \), and let \( {\alpha }_{k} \mathrel{\text{:=}} \mathop{\sum }\limits_{{\omega \in {\Omega }_{k}}}{2}^{-\left| \omega \right| } \) . We shall show below that\n\n\[{\alph... | Yes |
Suppose that on input \( x, A \) always halts within a finite number of steps, regardless of its random choices. More precisely, this means that there is a bound \( \ell \) (depending on \( A \) and \( x \) ), such that all execution paths of length \( \ell \) are complete. | In this case, we say that \( A \)’s running time on input \( x \) is strictly bounded by \( \ell \), and it is clear that \( A \) halts with probability 1 on input \( x \) . Moreover, one can much more simply model \( A \)’s computation on input \( x \) by working with the uniform distribution on execution paths of len... | Yes |
Suppose \( A \) and \( B \) are probabilistic algorithms that both halt with probability 1 on all inputs. Using \( A \) and \( B \) as subroutines, we can form their serial composition; that is, we can construct the algorithm \[ C\left( x\right) : \;\text{ output }B\left( {A\left( x\right) }\right) , \] which on input ... | For simplicity, we may assume that \( A \) places its output \( y \) in a location in memory where \( B \) expects to find its input, and that \( B \) places its output in a location in memory where \( C \) ’s output should go. With these assumptions, the program for \( C \) is obtained by simply concatenating the prog... | Yes |
Example 9.3. Suppose \( A, B \), and \( C \) are probabilistic algorithms that halt with probability 1 on all inputs, and that \( A \) always outputs either true or false. Then we can form the conditional construct \[ D\left( x\right) : \;\text{if}A\left( x\right) \text{then output}B\left( x\right) \text{else output}C\... | By a calculation similar to that in the previous example, it is easy to see that \( D \) halts with probability 1 on all inputs. | No |
Suppose \( A \) and \( B \) are probabilistic algorithms that halt with probability 1 on all inputs, and that \( A \) always outputs either true or false. We can form the iterative construct\n\n\[ C\left( x\right) : \;\text{ while }A\left( x\right) \text{ do }x \leftarrow B\left( x\right) \]\n\noutput \( x \) . | Algorithm \( C \) may or may not halt with probability 1 . To analyze \( C \), we define an infinite sequence of algorithms \( {\left\{ {C}_{n}\right\} }_{n = 0}^{\infty } \) ; namely, we define \( {C}_{0} \) as\n\n\[ {C}_{0}\left( x\right) : \text{ halt,}\]\n\nand for \( n > 0 \), we define \( {C}_{n} \) as\n\n\[ {C}_... | Yes |
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