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Theorem 9.14 (Lévy) The two processes \( {\left( {S}_{t},{S}_{t} - {B}_{t}\right) }_{t \geq 0} \) and \( {\left( {L}_{t}^{0}\left( B\right) ,\left| {B}_{t}\right| \right) }_{t \geq 0} \) have the same distribution. | Proof By Tanaka’s formula, for every \( t \geq 0 \) ,\n\n\[ \left| {B}_{t}\right| = - {\beta }_{t} + {L}_{t}^{0}\left( B\right) \]\n\n(9.18)\n\nwhere\n\n\[ {\beta }_{t} = - {\int }_{0}^{t}\operatorname{sgn}\left( {B}_{s}\right) \mathrm{d}{B}_{s} \]\n\nSince \( \langle \beta ,\beta {\rangle }_{t} = t \), Theorem 5.12 en... | Yes |
Proposition 9.15 We have a.s.\n\n\[ \n\\left\\{ {t \\geq 0 : {B}_{t} = 0}\\right\\} = \\left\\{ {{\\tau }_{s} : s \\geq 0}\\right\\} \\cup \\left\\{ {{\\tau }_{s - } : s \\in D}\\right\\} \n\]\n\nwhere \( D \) is the countable set of jump times of \( {\\left( {\\tau }_{s}\\right) }_{s \\geq 0} \) . | Proof We know from (9.17) that a.s.\n\n\[ \n\\operatorname{supp}\\left( {{\\mathrm{d}}_{t}{L}_{t}^{0}\\left( B\\right) }\\right) \\subset \\left\\{ {t \\geq 0 : {B}_{t} = 0}\\right\\} .\n\]\n\nIt follows that any time \( t \) of the form \( t = {\\tau }_{s} \) or \( t = {\\tau }_{s - } \) must belong to the zero set of... | Yes |
Theorem 1.1.1. Let \( \mu \) be a measure on \( \left( {\Omega ,\mathcal{F}}\right) \)\n\n(i) monotonicity. If \( A \subset B \) then \( \mu \left( A\right) \leq \mu \left( B\right) \).\n\n(ii) subadditivity. If \( A \subset { \cup }_{m = 1}^{\infty }{A}_{m} \) then \( \mu \left( A\right) \leq \mathop{\sum }\limits_{{m... | Proof. (i) Let \( B - A = B \cap {A}^{c} \) be the difference of the two sets. Using + to denote disjoint union, \( B = A + \left( {B - A}\right) \) so\n\n\[ \mu \left( B\right) = \mu \left( A\right) + \mu \left( {B - A}\right) \geq \mu \left( A\right) . \]\n\n(ii) Let \( {A}_{n}^{\prime } = {A}_{n} \cap A,{B}_{1} = {A... | Yes |
Theorem 1.1.2. Associated with each Stieltjes measure function \( F \) there is a unique measure \( \mu \) on \( \left( {\mathbf{R},\mathcal{R}}\right) \) with \( \mu (\left( {a, b\rbrack }\right) = F\left( b\right) - F\left( a\right) \) | The proof of Theorem 1.1.2 is a long and winding road, so we will content ourselves to describe the main ideas involved in this section and to hide the remaining details in the appendix in Section A.1. The choice of \ | No |
Lemma 1.1.3. If \( \mathcal{S} \) is a semialgebra then \( \overline{\mathcal{S}} = \{ \) finite disjoint unions of sets in \( \mathcal{S}\} \) is an algebra, called the algebra generated by \( \mathcal{S} \) . | Proof. Suppose \( A = { + }_{i}{S}_{i} \) and \( B = { + }_{j}{T}_{j} \), where + denotes disjoint union and we assume the index sets are finite. Then \( A \cap B = { + }_{i, j}{S}_{i} \cap {T}_{j} \in \overline{\mathcal{S}} \) . As for complements, if \( A = { + }_{i}{S}_{i} \) then \( {A}^{c} = { \cap }_{i}{S}_{i}^{c... | Yes |
Lemma 1.1.5. Suppose only that (i) holds.\n\n(a) If \( A,{B}_{i} \in \overline{\mathcal{S}} \) with \( A = { + }_{i = 1}^{n}{B}_{i} \) then \( \bar{\mu }\left( A\right) = \mathop{\sum }\limits_{i}\bar{\mu }\left( {B}_{i}\right) \) .\n\n(b) If \( A,{B}_{i} \in \overline{\mathcal{S}} \) with \( A \subset { \cup }_{i = 1}... | Proof. Observe that it follows from the definition that if \( A = { + }_{i}{B}_{i} \) is a finite disjoint union of sets in \( \overline{\mathcal{S}} \) and \( {B}_{i} = { + }_{j}{S}_{i, j} \), then\n\n\[ \bar{\mu }\left( A\right) = \mathop{\sum }\limits_{{i, j}}\mu \left( {S}_{i, j}\right) = \mathop{\sum }\limits_{i}\... | Yes |
Theorem 1.2.1. Any distribution function \( F \) has the following properties:\n\n(i) \( F \) is nondecreasing.\n\n(ii) \( \mathop{\lim }\limits_{{x \rightarrow \infty }}F\left( x\right) = 1,\mathop{\lim }\limits_{{x \rightarrow - \infty }}F\left( x\right) = 0 \) .\n\n(iii) \( F \) is right continuous, i.e. \( \mathop{... | Proof. To prove (i), note that if \( x \leq y \) then \( \{ X \leq x\} \subset \{ X \leq y\} \), and then use (i) in Theorem 1.1.1 to conclude that \( P\left( {X \leq x}\right) \leq P\left( {X \leq y}\right) \) .\n\nTo prove (ii), we observe that if \( x \uparrow \infty \), then \( \{ X \leq x\} \uparrow \Omega \), whi... | Yes |
Theorem 1.2.2. If \( F \) satisfies (i),(ii), and (iii) in Theroem 1.2.1, then it is the distribution function of some random variable. | Proof. Let \( \Omega = \left( {0,1}\right) ,\mathcal{F} = \) the Borel sets, and \( P = \) Lebesgue measure. If \( \omega \in \left( {0,1}\right) \) , let\n\n\[ X\left( \omega \right) = \sup \{ y : F\left( y\right) < \omega \} \]\n\nOnce we show that\n\n\( \left( \star \right) \)\n\n\[ \{ \omega : X\left( \omega \right... | Yes |
Uniform distribution on \( \left( {0,1}\right) .f\left( x\right) = 1 \) for \( x \in \left( {0,1}\right) \) and 0 otherwise. Distribution function: | \[ F\left( x\right) = \left\{ \begin{array}{ll} 0 & x \leq 0 \\ x & 0 \leq x \leq 1 \\ 1 & x > 1 \end{array}\right. \] | Yes |
Example 1.2.2. Exponential distribution with rate \( \lambda .f\left( x\right) = \lambda {e}^{-{\lambda x}} \) for \( x \geq 0 \) and 0 otherwise. Distribution function: | \[ F\left( x\right) = \left\{ \begin{array}{ll} 0 & x \leq 0 \\ 1 - {e}^{-x} & x \geq 0 \end{array}\right. \] | Yes |
Theorem 1.2.3. For \( x > 0 \) ,\n\n\[ \n\left( {{x}^{-1} - {x}^{-3}}\right) \exp \left( {-{x}^{2}/2}\right) \leq {\int }_{x}^{\infty }\exp \left( {-{y}^{2}/2}\right) {dy} \leq {x}^{-1}\exp \left( {-{x}^{2}/2}\right) \n\] | Proof. Changing variables \( y = x + z \) and using \( \exp \left( {-{z}^{2}/2}\right) \leq 1 \) gives\n\n\[ \n{\int }_{x}^{\infty }\exp \left( {-{y}^{2}/2}\right) {dy} \leq \exp \left( {-{x}^{2}/2}\right) {\int }_{0}^{\infty }\exp \left( {-{xz}}\right) {dz} = {x}^{-1}\exp \left( {-{x}^{2}/2}\right) \n\]\n\nFor the oth... | Yes |
Uniform distribution on the Cantor set. The Cantor set \( C \) is defined by removing \( \left( {1/3,2/3}\right) \) from \( \left\lbrack {0,1}\right\rbrack \) and then removing the middle third of each interval that remains. We define an associated distribution function by setting \( F\left( x\right) = 0 \) for \( x \l... | There is no \( f \) for which (1.2.1) holds because such an \( f \) would be equal to 0 on a set of measure 1 . From the definition, it is immediate that the corresponding measure has \( \mu \left( {C}^{c}\right) = 0 \) . | Yes |
Theorem 1.3.1. If \( \{ \omega : X\left( \omega \right) \in A\} \in \mathcal{F} \) for all \( A \in \mathcal{A} \) and \( \mathcal{A} \) generates \( \mathcal{S} \) (i.e., \( \mathcal{S} \) is the smallest \( \sigma \) -field that contains \( \mathcal{A} \) ), then \( X \) is measurable. | Proof. Writing \( \{ X \in B\} \) as shorthand for \( \{ \omega : X\left( \omega \right) \in B\} \), we have\n\n\[ \left\{ {X \in { \cup }_{i}{B}_{i}}\right\} = { \cup }_{i}\left\{ {X \in {B}_{i}}\right\} \]\n\n\[ \left\{ {X \in {B}^{c}}\right\} = \{ X \in B{\} }^{c} \]\n\nSo the class of sets \( \mathcal{B} = \{ B : \... | Yes |
Theorem 1.3.2. If \( X : \left( {\Omega ,\mathcal{F}}\right) \rightarrow \left( {S,\mathcal{S}}\right) \) and \( f : \left( {S,\mathcal{S}}\right) \rightarrow \left( {T,\mathcal{T}}\right) \) are measurable maps, then \( f\left( X\right) \) is a measurable map from \( \left( {\Omega ,\mathcal{F}}\right) \) to \( \left(... | Proof. Let \( B \in \mathcal{T}.\;\{ \omega : f\left( {X\left( \omega \right) }\right) \in B\} = \left\{ {\omega : X\left( \omega \right) \in {f}^{-1}\left( B\right) }\right\} \in \mathcal{F} \), since by assumption \( {f}^{-1}\left( B\right) \in \mathcal{S} \) | Yes |
Theorem 1.3.3. If \( {X}_{1},\ldots {X}_{n} \) are random variables and \( f : \left( {{\mathbf{R}}^{n},{\mathcal{R}}^{n}}\right) \rightarrow \left( {\mathbf{R},\mathcal{R}}\right) \) is measurable, then \( f\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) is a random variable. | Proof. In view of Theorem 1.3.2, it suffices to show that \( \left( {{X}_{1},\ldots ,{X}_{n}}\right) \) is a random vector. To do this, we observe that if \( {A}_{1},\ldots ,{A}_{n} \) are Borel sets then\n\n\[ \left\{ {\left( {{X}_{1},\ldots ,{X}_{n}}\right) \in {A}_{1} \times \cdots \times {A}_{n}}\right\} = { \cap }... | Yes |
Theorem 1.3.4. If \( {X}_{1},\ldots ,{X}_{n} \) are random variables then \( {X}_{1} + \ldots + {X}_{n} \) is a random variable. | Proof. In view of Theorem 1.3.3 it suffices to show that \( f\left( {{x}_{1},\ldots ,{x}_{n}}\right) = {x}_{1} + \ldots + {x}_{n} \) is measurable. To do this, we use Example 1.3.1 and note that \( \left\{ {x : {x}_{1} + \ldots + {x}_{n} < a}\right\} \) is an open set and hence is in \( {\mathcal{R}}^{n} \) . | Yes |
Theorem 1.3.5. If \( {X}_{1},{X}_{2},\ldots \) are random variables then so are\n\n\[ \mathop{\inf }\limits_{n}{X}_{n}\;\mathop{\sup }\limits_{n}{X}_{n}\;\mathop{\limsup }\limits_{n}{X}_{n}\;\mathop{\liminf }\limits_{n}{X}_{n} \] | Proof. Since the infimum of a sequence is \( < a \) if and only if some term is \( < a \) (if all terms are \( \geq a \) then the infimum is), we have\n\n\[ \left\{ {\mathop{\inf }\limits_{n}{X}_{n} < a}\right\} = { \cup }_{n}\left\{ {{X}_{n} < a}\right\} \in \mathcal{F} \]\n\nA similar argument shows \( \left\{ {\math... | Yes |
Lemma 1.4.1. Let \( \varphi \) and \( \psi \) be simple functions.\n\n(i) If \( \varphi \geq 0 \) a.e. then \( \int {\varphi d\mu } \geq 0 \) .\n\n(ii) For any \( a \in \mathbf{R},\int {a\varphi d\mu } = a\int {\varphi d\mu } \) .\n\n(iii) \( \int \varphi + {\psi d\mu } = \int {\varphi d\mu } + \int {\psi d\mu } \) . | Proof. (i) and (ii) are immediate consequences of the definition. To prove (iii), suppose\n\n\[ \varphi = \mathop{\sum }\limits_{{i = 1}}^{m}{a}_{i}{1}_{{A}_{i}}\;\text{ and }\;\psi = \mathop{\sum }\limits_{{j = 1}}^{n}{b}_{j}{1}_{{B}_{j}} \]\n\nTo make the supports of the two functions the same, we let \( {A}_{0} = { ... | Yes |
Lemma 1.4.2. If (i) and (iii) hold then we have:\n\n(iv) If \( \varphi \leq \psi \) a.e. then \( \int {\varphi d\mu } \leq \int {\psi d\mu } \).\n\n(v) If \( \varphi = \psi \) a.e. then \( \int {\varphi d\mu } = \int {\psi d\mu } \). | Proof. By (iii), \( \int {\psi d\mu } = \int {\varphi d\mu } + \int \left( {\psi - \varphi }\right) {d\mu } \) and the second integral is \( \geq 0 \) by (i), so (iv) holds. \( \varphi = \psi \) a.e. implies \( \varphi \leq \psi \) a.e. and \( \psi \leq \varphi \) a.e. so (v) follows from two applications of (iv). To p... | Yes |
Lemma 1.4.3. Let \( E \) be a set with \( \mu \left( E\right) < \infty \) . If \( f \) and \( g \) are bounded functions that vanish on \( {E}^{c} \) then:\n\n(i) If \( f \geq 0 \) a.e. then \( \int {fd\mu } \geq 0 \) .\n\n(ii) For any \( a \in \mathbf{R},\int {afd\mu } = a\int {fd\mu } \) .\n\n(iii) \( \int f + {gd\mu... | Proof. Since we can take \( \varphi \equiv 0 \) ,(i) is clear from the definition. To prove (ii), we observe that if \( a > 0 \), then \( {a\varphi } \leq {af} \) if and only if \( \varphi \leq f \), so\n\n\[ \n\int {afd\mu } = \mathop{\sup }\limits_{{\varphi \leq f}}\int {a\varphi d\mu } = \mathop{\sup }\limits_{{\var... | Yes |
Lemma 1.4.4. Let \( {E}_{n} \uparrow \Omega \) have \( \mu \left( {E}_{n}\right) < \infty \) and let \( a \land b = \min \left( {a, b}\right) \) . Then \[ {\int }_{{E}_{n}}f \land {nd\mu } \uparrow \int {fd\mu }\;\text{ as }n \uparrow \infty \] | Proof. It is clear that from (iv) in Lemma 1.4.3 that the left-hand side increases as \( n \) does. Since \( h = \left( {f \land n}\right) {1}_{{E}_{n}} \) is a possibility in the sup, each term is smaller than the integral on the right. To prove that the limit is \( \int {fd\mu } \), observe that if \( 0 \leq h \leq f... | Yes |
Lemma 1.4.5. Suppose \( f, g \geq 0 \) .\n\n(i) \( \int {fd\mu } \geq 0 \)\n\n(ii) If \( a > 0 \) then \( \int {afd\mu } = a\int {fd\mu } \)\n\n(iii) \( \int f + {gd\mu } = \int {fd\mu } + \int {gd\mu } \)\n\n(iv) If \( 0 \leq g \leq f \) a.e. then \( \int {gd\mu } \leq \int {fd\mu } \)\n\n(v) If \( 0 \leq g = f \) a.e... | Proof. (i) is trivial from the definition. (ii) is clear, since when \( a > 0,{ah} \leq {af} \) if and only if \( h \leq f \) and we have \( \int {ahd\mu } = a\int {hdu} \) for \( h \) in the defining class. For (iii), we observe that if \( f \geq h \) and \( g \geq k \), then \( f + g \geq h + k \) so taking the sup o... | No |
Lemma 1.4.6. If \( f = {f}_{1} - {f}_{2} \) where \( {f}_{1},{f}_{2} \geq 0 \) and \( \int {f}_{i}{d\mu } < \infty \) then\n\n\[ \int {fd\mu } = \int {f}_{1}{d\mu } - \int {f}_{2}{d\mu } \] | Proof. \( {f}_{1} + {f}^{ - } = {f}_{2} + {f}^{ + } \) and all four functions are \( \geq 0 \), so by (iii) of Lemma 1.4.5,\n\n\[ \int {f}_{1}{d\mu } + \int {f}^{ - }{d\mu } = \int {f}_{1} + {f}^{ - }{d\mu } = \int {f}_{2} + {f}^{ + }{d\mu } = \int {f}_{2}{d\mu } + \int {f}^{ + }{d\mu } \]\n\nRearranging gives the desi... | Yes |
Theorem 1.4.7. Suppose \( f \) and \( g \) are integrable.\n\n(i) If \( f \geq 0 \) a.e. then \( \int {fd\mu } \geq 0 \) . | Proof. (i) is trivial. | No |
Theorem 1.5.1. Jensen’s inequality. Suppose \( \varphi \) is convex, that is,\n\n\[ \n{\lambda \varphi }\left( x\right) + \left( {1 - \lambda }\right) \varphi \left( y\right) \geq \varphi \left( {{\lambda x} + \left( {1 - \lambda }\right) y}\right) \n\]\n\nfor all \( \lambda \in \left( {0,1}\right) \) and \( x, y \in \... | Proof. Let \( c = \int {fd\mu } \) and let \( \ell \left( x\right) = {ax} + b \) be a linear function that has \( \ell \left( c\right) = \varphi \left( c\right) \) and \( \varphi \left( x\right) \geq \ell \left( x\right) \) . To see that such a function exists, recall that convexity implies\n\n\[ \n\mathop{\lim }\limit... | Yes |
Theorem 1.5.2. Hölder’s inequality. If \( p, q \in \left( {1,\infty }\right) \) with \( 1/p + 1/q = 1 \) then\n\n\[ \int \left| {fg}\right| {d\mu } \leq \parallel f{\parallel }_{p}\parallel g{\parallel }_{q} \] | Proof. If \( \parallel f{\parallel }_{p} \) or \( \parallel g{\parallel }_{q} = 0 \) then \( \left| {fg}\right| = 0 \) a.e., so it suffices to prove the result when \( \parallel f{\parallel }_{p} \) and \( \parallel g{\parallel }_{q} > 0 \) or by dividing both sides by \( \parallel f{\parallel }_{p}\parallel g{\paralle... | Yes |
Theorem 1.5.3. Bounded convergence theorem. Let \( E \) be a set with \( \mu \left( E\right) < \infty \) . Suppose \( {f}_{n} \) vanishes on \( {E}^{c},\left| {{f}_{n}\left( x\right) }\right| \leq M \), and \( {f}_{n} \rightarrow f \) in measure. Then | \[ \int {fd\mu } = \mathop{\lim }\limits_{{n \rightarrow \infty }}\int {f}_{n}{d\mu } \] | Yes |
Consider the real line \( \mathbf{R} \) equipped with the Borel sets \( \mathcal{R} \) and Lebesgue measure \( \lambda \) . The functions \( {f}_{n}\left( x\right) = 1/n \) on \( \left\lbrack {0, n}\right\rbrack \) and 0 otherwise on show that the conclusion of Theorem 1.5.3 does not hold when \( \mu \left( E\right) = ... | Proof. Let \( \epsilon > 0,{G}_{n} = \left\{ {x : \left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| < \epsilon }\right\} \) and \( {B}_{n} = E - {G}_{n} \) . Using (iii) and (vi) from Theorem 1.4.7,\n\n\[ \left| {\int {fd\mu }-\int {f}_{n}{d\mu }}\right| = \left| {\int \left( {f - {f}_{n}}\right) {d\mu }}\righ... | No |
Example 1.5.1 shows that we may have strict inequality in Theorem 1.5.4. The functions \( {f}_{n}\left( x\right) = n{1}_{(0,1/n\rbrack }\left( x\right) \) on \( \left( {0,1}\right) \) equipped with the Borel sets and Lebesgue measure show that this can happen on a space of finite measure. | Proof. Let \( {g}_{n}\left( x\right) = \mathop{\inf }\limits_{{m \geq n}}{f}_{m}\left( x\right) .{f}_{n}\left( x\right) \geq {g}_{n}\left( x\right) \) and as \( n \uparrow \infty \) ,\n\n\[ \n{g}_{n}\left( x\right) \uparrow g\left( x\right) = \mathop{\liminf }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) \n\]... | Yes |
Theorem 1.5.5. Monotone convergence theorem. If \( {f}_{n} \geq 0 \) and \( {f}_{n} \uparrow f \) then\n\n\[ \int {f}_{n}{d\mu } \uparrow \int {fd\mu } \] | Proof. Fatou’s lemma, Theorem 1.5.4, implies liminf \( \int {f}_{n}{d\mu } \geq \int {fd\mu } \) . On the other hand, \( {f}_{n} \leq f \) implies \( \lim \sup \int {f}_{n}{d\mu } \leq \int {fd\mu } \). | Yes |
Theorem 1.5.6. Dominated convergence theorem. If \( {f}_{n} \rightarrow f \) a.e., \( \left| {f}_{n}\right| \leq g \) for all \( n \), and \( g \) is integrable, then \( \int {f}_{n}{d\mu } \rightarrow \int {fd\mu } \) . | Proof. \( {f}_{n} + g \geq 0 \) so Fatou’s lemma implies\n\n\[ \mathop{\liminf }\limits_{{n \rightarrow \infty }}\int {f}_{n} + {gd\mu } \geq \int f + {gd\mu } \]\n\nSubtracting \( \int {gd\mu } \) from both sides gives\n\n\[ \mathop{\liminf }\limits_{{n \rightarrow \infty }}\int {f}_{n}{d\mu } \geq \int {fd\mu } \]\n\... | Yes |
Theorem 1.6.2. Jensen’s inequality. Suppose \( \varphi \) is convex, that is,\n\n\[ \n{\lambda \varphi }\left( x\right) + \left( {1 - \lambda }\right) \varphi \left( y\right) \geq \varphi \left( {{\lambda x} + \left( {1 - \lambda }\right) y}\right) \n\]\n\nfor all \( \lambda \in \left( {0,1}\right) \) and \( x, y \in \... | To recall the direction in which the inequality goes note that if \( P\left( {X = x}\right) = \lambda \) and \( P\left( {X = y}\right) = 1 - \lambda \) then\n\n\[ \n{E\varphi }\left( X\right) = {\lambda \varphi }\left( x\right) + \left( {1 - \lambda }\right) \varphi \left( y\right) \geq \varphi \left( {{\lambda x} + \l... | Yes |
Theorem 1.6.4. Chebyshev’s inequality. Suppose \( \varphi : \mathbf{R} \rightarrow \mathbf{R} \) has \( \varphi \geq 0 \), let \( A \in \mathcal{R} \) and let \( {i}_{A} = \inf \{ \varphi \left( y\right) : y \in A\} \) . \[ {i}_{A}P\left( {X \in A}\right) \leq E\left( {\varphi \left( X\right) ;X \in A}\right) \leq {E\v... | Proof. The definition of \( {i}_{A} \) and the fact that \( \varphi \geq 0 \) imply that \[ {i}_{A}{1}_{\left( X \in A\right) } \leq \varphi \left( X\right) {1}_{\left( X \in A\right) } \leq \varphi \left( X\right) \] So taking expected values and using part (c) of Theorem 1.6.1 gives the desired result. | Yes |
Theorem 1.6.8. Suppose \( {X}_{n} \rightarrow X \) a.s. Let \( g, h \) be continuous functions with\n\n(i) \( g \geq 0 \) and \( g\left( x\right) \rightarrow \infty \) as \( \left| x\right| \rightarrow \infty \) ,\n\n(ii) \( \left| {h\left( x\right) }\right| /g\left( x\right) \rightarrow 0 \) as \( \left| x\right| \rig... | Proof. By subtracting a constant from \( h \), we can suppose without loss of generality that \( h\left( 0\right) = 0 \) . Pick \( M \) large so that \( P\left( {\left| X\right| = M}\right) = 0 \) and \( g\left( x\right) > 0 \) when \( \left| x\right| \geq M \) . Given a random variable \( Y \), let \( \bar{Y} = Y{1}_{... | Yes |
Theorem 1.6.9. Change of variables formula. Let \( X \) be a random element of \( \left( {S,\mathcal{S}}\right) \) with distribution \( \mu \), i.e., \( \mu \left( A\right) = P\left( {X \in A}\right) \) . If \( f \) is a measurable function from \( \left( {S,\mathcal{S}}\right) \) to \( \left( {\mathbf{R},\mathcal{R}}\... | Proof. We will prove this result by verifying it in four increasingly more general special cases that parallel the way that the integral was defined in Section 1.4. The reader should note the method employed, since it will be used several times below.\n\nCASE 1: INDICATOR FUNCTIONS. If \( B \in \mathcal{S} \) and \( f ... | Yes |
If \( X \) has an exponential distribution with rate 1 then | \[ E{X}^{k} = {\int }_{0}^{\infty }{x}^{k}{e}^{-x}{dx} = k! \] So the mean of \( X \) is 1 and variance is \( E{X}^{2} - {\left( EX\right) }^{2} = 2 - {1}^{2} = 1 \) . If we let \( Y = X/\lambda \) , then by Exercise 1.2.5, \( Y \) has density \( \lambda {e}^{-{\lambda y}} \) for \( y \geq 0 \), the exponential density... | No |
If \( X \) has a standard normal distribution, | \[ {EX} = \int x{\left( 2\pi \right) }^{-1/2}\exp \left( {-{x}^{2}/2}\right) {dx} = 0\;\text{ (by symmetry) } \] \[ \operatorname{var}\left( X\right) = E{X}^{2} = \int {x}^{2}{\left( 2\pi \right) }^{-1/2}\exp \left( {-{x}^{2}/2}\right) {dx} = 1 \] | Yes |
We say that \( X \) has a Bernoulli distribution with parameter \( p \) if \( P\left( {X = 1}\right) = p \) and \( P\left( {X = 0}\right) = 1 - p \). | Clearly, \[ {EX} = p \cdot 1 + \left( {1 - p}\right) \cdot 0 = p \] Since \( {X}^{2} = X \), we have \( E{X}^{2} = {EX} = p \) and \[ \operatorname{var}\left( X\right) = E{X}^{2} - {\left( EX\right) }^{2} = p - {p}^{2} = p\left( {1 - p}\right) \] | Yes |
We say that \( X \) has a Poisson distribution with parameter \( \lambda \) if\n\n\[ P\left( {X = k}\right) = {e}^{-\lambda }{\lambda }^{k}/k!\text{ for }k = 0,1,2,\ldots \] | To evaluate the moments of the Poisson random variable, we use a little inspiration to observe that for \( k \geq 1 \)\n\n\[ E\left( {X\left( {X - 1}\right) \cdots \left( {X - k + 1}\right) }\right) = \mathop{\sum }\limits_{{j = k}}^{\infty }j\left( {j - 1}\right) \cdots \left( {j - k + 1}\right) {e}^{-\lambda }\frac{{... | Yes |
Theorem 1.7.1. There is a unique measure \( \mu \) on \( \mathcal{F} \) with\n\n\[ \mu \left( {A \times B}\right) = {\mu }_{1}\left( A\right) {\mu }_{2}\left( B\right) \] | Proof. By Theorem 1.1.4 it is enough to show that if \( A \times B = { + }_{i}\left( {{A}_{i} \times {B}_{i}}\right) \) is a finite or countable disjoint union then\n\n\[ \mu \left( {A \times B}\right) = \mathop{\sum }\limits_{i}\mu \left( {{A}_{i} \times {B}_{i}}\right) \]\n\nFor each \( x \in A \), let \( I\left( x\r... | No |
Theorem 1.7.2. Fubini’s theorem. If \( f \geq 0 \) or \( \int \left| f\right| {d\mu } < \infty \) then\n\n\[{\int }_{X}{\int }_{Y}f\left( {x, y}\right) {\mu }_{2}\left( {dy}\right) {\mu }_{1}\left( {dx}\right) = {\int }_{X \times Y}{fd\mu } = {\int }_{Y}{\int }_{X}f\left( {x, y}\right) {\mu }_{1}\left( {dx}\right) {\mu... | Proof. We will prove only the first equality, since the second follows by symmetry. Two technical things that need to be proved before we can assert that the first integral makes sense are:\n\nWhen \( x \) is fixed, \( y \rightarrow f\left( {x, y}\right) \) is \( \mathcal{B} \) measurable.\n\n\( x \rightarrow {\int }_{... | Yes |
Lemma 1.7.3. If \( E \in \mathcal{F} \) then \( {E}_{x} \in \mathcal{B} \). | Proof. \( {\left( {E}^{c}\right) }_{x} = {\left( {E}_{x}\right) }^{c} \) and \( {\left( { \cup }_{i}{E}_{i}\right) }_{x} = { \cup }_{i}{\left( {E}_{i}\right) }_{x} \), so if \( \mathcal{E} \) is the collection of sets \( E \) for which \( {E}_{x} \in \mathcal{B} \), then \( \mathcal{E} \) is a \( \sigma \) -algebra. Si... | Yes |
Lemma 1.7.4. If \( E \in \mathcal{F} \) then \( g\left( x\right) \equiv {\mu }_{2}\left( {E}_{x}\right) \) is \( \mathcal{A} \) measurable and \[ {\int }_{X}{gd}{\mu }_{1} = \mu \left( E\right) \] | Proof. If conclusions hold for \( {E}_{n} \) and \( {E}_{n} \uparrow E \), then Theorem 1.3.5 and the monotone convergence theorem imply that they hold for \( E \) . Since \( {\mu }_{1} \) and \( {\mu }_{2} \) are \( \sigma \) -finite, it is enough then to prove the result for \( E \subset F \times G \) with \( {\mu }_... | Yes |
Let \( X = Y = \{ 1,2,\ldots \} \) with \( \mathcal{A} = \mathcal{B} = \) all subsets and \( {\mu }_{1} = {\mu }_{2} = \) counting measure. For \( m \geq 1 \), let \( f\left( {m, m}\right) = 1 \) and \( f\left( {m + 1, m}\right) = - 1 \), and let \( f\left( {m, n}\right) = 0 \) otherwise. We claim that\n\n\[ \mathop{\s... | In words, if we sum the columns first, the first one gives us a 1 and the others 0 , while if we sum the rows each one gives us a 0 . | Yes |
Let \( X = \left( {0,1}\right), Y = \left( {1,\infty }\right) \), both equipped with the Borel sets and Lebesgue measure. Let \( f\left( {x, y}\right) = {e}^{-{xy}} - 2{e}^{-{2xy}} \). | \[ {\int }_{0}^{1}{\int }_{1}^{\infty }f\left( {x, y}\right) {dydx} = {\int }_{0}^{1}{x}^{-1}\left( {{e}^{-x} - {e}^{-{2x}}}\right) {dx} > 0 \] \[ {\int }_{1}^{\infty }{\int }_{0}^{1}f\left( {x, y}\right) {dxdy} = {\int }_{1}^{\infty }{y}^{-1}\left( {{e}^{-{2y}} - {e}^{-y}}\right) {dy} < 0 \] | Yes |
Let \( X = \left( {0,1}\right) \) with \( \mathcal{A} = \) the Borel sets and \( {\mu }_{1} = \) Lebesgue measure. Let \( Y = \left( {0,1}\right) \) with \( \mathcal{B} = \) all subsets and \( {\mu }_{2} = \) counting measure. Let \( f\left( {x, y}\right) = 1 \) if \( x = y \) and 0 otherwise | \[ {\int }_{Y}f\left( {x, y}\right) {\mu }_{2}\left( {dy}\right) = 1\;\text{ for all }x\text{ so }\;{\int }_{X}{\int }_{Y}f\left( {x, y}\right) {\mu }_{2}\left( {dy}\right) {\mu }_{1}\left( {dx}\right) = 1 \] \[ {\int }_{X}f\left( {x, y}\right) {\mu }_{1}\left( {dx}\right) = 0\;\text{ for all }y\text{ so }\;{\int }_{Y}... | Yes |
Example 1.7.4. By the axiom of choice and the continuum hypothesis one can define an order relation \( { < }^{\prime } \) on \( \left( {0,1}\right) \) so that \( \left\{ {x : x{ < }^{\prime }y}\right\} \) is countable for each \( y \) . Let \( X = Y = \left( {0,1}\right) \), let \( \mathcal{A} = \mathcal{B} = \) the Bo... | \[ {\int }_{X}f\left( {x, y}\right) {\mu }_{1}\left( {dx}\right) = 0\;\text{ for all }y \] \[ {\int }_{Y}f\left( {x, y}\right) {\mu }_{2}\left( {dy}\right) = 1\;\text{ for all }x \] | Yes |
Lemma 2.1.1. Without loss of generality we can suppose each \( {\mathcal{A}}_{i} \) contains \( \Omega \) . In this case the condition is equivalent to\n\n\[ P\left( {{ \cap }_{i = 1}^{n}{A}_{i}}\right) = \mathop{\prod }\limits_{{i = 1}}^{n}P\left( {A}_{i}\right) \;\text{ whenever }{A}_{i} \in {\mathcal{A}}_{i} \]\n\ns... | Proof. If \( {\mathcal{A}}_{1},{\mathcal{A}}_{2},\ldots ,{\mathcal{A}}_{n} \) are independent and \( {\overline{\mathcal{A}}}_{i} = {\mathcal{A}}_{i} \cup \{ \Omega \} \) then \( {\overline{\mathcal{A}}}_{1},{\overline{\mathcal{A}}}_{2},\ldots ,{\overline{\mathcal{A}}}_{n} \) are independent, since if \( {A}_{i} \in {\... | No |
Theorem 2.1.3. Suppose \( {\mathcal{A}}_{1},{\mathcal{A}}_{2},\ldots ,{\mathcal{A}}_{n} \) are independent and each \( {\mathcal{A}}_{i} \) is a \( \pi \) -system. Then \( \sigma \left( {\mathcal{A}}_{1}\right) ,\sigma \left( {\mathcal{A}}_{2}\right) ,\ldots ,\sigma \left( {\mathcal{A}}_{n}\right) \) are independent. | Proof. Let \( {A}_{2},\ldots ,{A}_{n} \) be sets with \( {A}_{i} \in {\mathcal{A}}_{i} \), let \( F = {A}_{2} \cap \cdots \cap {A}_{n} \) and let \( \mathcal{L} = \) \( \{ A : P\left( {A \cap F}\right) = P\left( A\right) P\left( F\right) \} \) . Since \( P\left( {\Omega \cap F}\right) = P\left( \Omega \right) P\left( F... | Yes |
In order for \( {X}_{1},\ldots ,{X}_{n} \) to be independent, it is sufficient that for all \( {x}_{1},\ldots ,{x}_{n} \in ( - \infty ,\infty \rbrack \)\n\n\[ P\left( {{X}_{1} \leq {x}_{1},\ldots ,{X}_{n} \leq {x}_{n}}\right) = \mathop{\prod }\limits_{{i = 1}}^{n}P\left( {{X}_{i} \leq {x}_{i}}\right) \] | Proof. Let \( {\mathcal{A}}_{i} = \) the sets of the form \( \left\{ {{X}_{i} \leq {x}_{i}}\right\} \) . Since\n\n\[ \left\{ {{X}_{i} \leq x}\right\} \cap \left\{ {{X}_{i} \leq y}\right\} = \left\{ {{X}_{i} \leq x \land y}\right\} \]\n\nwhere \( {\left( x \land y\right) }_{i} = {x}_{i} \land {y}_{i} = \min \left\{ {{x}... | No |
Theorem 2.1.5. Suppose \( {\mathcal{F}}_{i, j},1 \leq i \leq n,1 \leq j \leq m\left( i\right) \) are independent and let \( {\mathcal{G}}_{i} = \sigma \left( {{ \cup }_{j}{\mathcal{F}}_{i, j}}\right) \) . Then \( {\mathcal{G}}_{1},\ldots ,{\mathcal{G}}_{n} \) are independent. | Proof. Let \( {\mathcal{A}}_{i} \) be the collection of sets of the form \( { \cap }_{j}{A}_{i, j} \) where \( {A}_{i, j} \in {\mathcal{F}}_{i, j}.{\mathcal{A}}_{i} \) is a \( \pi \) -system that contains \( \Omega \) and contains \( { \cup }_{j}{\mathcal{F}}_{i, j} \) so Theorem 2.1.3 implies \( \sigma \left( {\mathca... | Yes |
Theorem 2.1.6. If for \( 1 \leq i \leq n,1 \leq j \leq m\left( i\right) ,{X}_{i, j} \) are independent and \( {f}_{i} \) : \( {\mathbf{R}}^{m\left( i\right) } \rightarrow \mathbf{R} \) are measurable then \( {f}_{i}\left( {{X}_{i,1},\ldots ,{X}_{i, m\left( i\right) }}\right) \) are independent. | Proof. Let \( {\mathcal{F}}_{i, j} = \sigma \left( {X}_{i, j}\right) \) and \( {\mathcal{G}}_{i} = \sigma \left( {{ \cup }_{j}{\mathcal{F}}_{i, j}}\right) \) . Since \( {f}_{i}\left( {{X}_{i,1},\ldots ,{X}_{i, m\left( i\right) }}\right) \in {\mathcal{G}}_{i} \), the desired result follows from Theorem 2.1.5 and Exercis... | No |
Theorem 2.1.7. Suppose \( {X}_{1},\ldots ,{X}_{n} \) are independent random variables and \( {X}_{i} \) has distribution \( {\mu }_{i} \), then \( \left( {{X}_{1},\ldots ,{X}_{n}}\right) \) has distribution \( {\mu }_{1} \times \cdots \times {\mu }_{n} \) . | Proof. Using the definitions of (i) \( {A}_{1} \times \cdots \times {A}_{n} \) ,(ii) independence,(iii) \( {\mu }_{i} \), and (iv) \( {\mu }_{1} \times \cdots \times {\mu }_{n} \)\n\n\[ P\left( {\left( {{X}_{1},\ldots ,{X}_{n}}\right) \in {A}_{1} \times \cdots \times {A}_{n}}\right) = P\left( {{X}_{1} \in {A}_{1},\ldot... | Yes |
Theorem 2.1.8. Suppose \( X \) and \( Y \) are independent and have distributions \( \mu \) and \( \nu \) . If \( h : {\mathbf{R}}^{2} \rightarrow \mathbf{R} \) is a measurable function with \( h \geq 0 \) or \( E\left| {h\left( {X, Y}\right) }\right| < \infty \) then\n\n\[ \n{Eh}\left( {X, Y}\right) = \iint h\left( {x... | Proof. Using Theorem 1.6.9 and then Fubini's theorem (Theorem 1.7.2) we have\n\n\[ \n{Eh}\left( {X, Y}\right) = {\int }_{{\mathbf{R}}^{2}}{hd}\left( {\mu \times \nu }\right) = \iint h\left( {x, y}\right) \mu \left( {dx}\right) \nu \left( {dy}\right) \n\]\n\nTo prove the second result, we start with the result when \( f... | Yes |
Theorem 2.1.9. If \( {X}_{1},\ldots ,{X}_{n} \) are independent and have (a) \( {X}_{i} \geq 0 \) for all \( i \), or (b) \( E\left| {X}_{i}\right| < \infty \) for all \( i \) then \[ E\left( {\mathop{\prod }\limits_{{i = 1}}^{n}{X}_{i}}\right) = \mathop{\prod }\limits_{{i = 1}}^{n}E{X}_{i} \] i.e., the expectation on ... | Proof. \( X = {X}_{1} \) and \( Y = {X}_{2}\cdots {X}_{n} \) are independent by Theorem 2.1.6 so taking \( f\left( x\right) = \left| x\right| \) and \( g\left( y\right) = \left| y\right| \) we have \( E\left| {{X}_{1}\cdots {X}_{n}}\right| = E\left| {X}_{1}\right| E\left| {{X}_{2}\cdots {X}_{n}}\right| \), and it follo... | Yes |
It can happen that \( E\left( {XY}\right) = {EX} \cdot {EY} \) without the variables being independent. | Suppose the joint distribution of \( X \) and \( Y \) is given by the following table\n\n\[ \n\begin{matrix} & & & & Y & \\ & & & 1 & 0 & - 1 \\ & & 1 & 0 & a & 0 \\ X & 0 & b & c & b & \\ & - 1 & 0 & a & 0 & \end{matrix} \]\n\nwhere \( a, b > 0, c \geq 0 \), and \( {2a} + {2b} + c = 1 \) . Things are arranged so that ... | Yes |
Theorem 2.1.10. If \( X \) and \( Y \) are independent, \( F\left( x\right) = P\left( {X \leq x}\right) \), and \( G\left( y\right) = \) \( P\left( {Y \leq y}\right) \), then\n\n\[ P\left( {X + Y \leq z}\right) = \int F\left( {z - y}\right) {dG}\left( y}\right) \] | Proof. Let \( h\left( {x, y}\right) = {1}_{\left( x + y \leq z\right) } \) . Let \( \mu \) and \( \nu \) be the probability measures with distribution functions \( F \) and \( G \) . Since for fixed \( y \)\n\n\[ \int h\left( {x, y}\right) \mu \left( {dx}\right) = \int {1}_{( - \infty, z - y\rbrack }\left( x\right) \mu... | Yes |
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) = \int {\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... | Example 1.6.5 tells us that if \( X \) has a geometric distribution with parameter \( p \) then \( {EX} = 1/p \) and \( \operatorname{var}\left( X\right) \leq 1/{p}^{2} \) . Using the linearity of expected value, bounds on \( \mathop{\sum }\limits_{{m = 1}}^{n}1/m \) in (2.2.1), and Theorem 2.2.1 we see that\n\n\[ E{T}... | 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 |
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}_{i}\right) ... | 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\nLet \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) and let \( {\mu }_{n} = E\left( {{X}_{1}{1}_{\left( \left| {X}_{1}\right| \l... | 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\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}... | \[ \nP\left( {{\ell }_{n} \geq \left( {1 + \epsilon }\right) {\log }_{2}n}\right) \leq {n}^{-\left( {1 + \epsilon }\right) } \n\]\n\nfor any \( \epsilon > 0 \), so it follows from the Borel-Cantelli lemma that \( {\ell }_{n} \leq \left( {1 + \epsilon }\right) {\log }_{2}n \) for \( n \geq {N}_{\epsilon } \) . Since \( ... | 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 |
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 |
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