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Theorem 2.5.6. The strong law of large numbers. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. random variables with \( E\left| {X}_{i}\right| < \infty \) . Let \( E{X}_{i} = \mu \) and \( {S}_{n} = {X}_{1} + \ldots + {X}_{n} \) . Then \( {S}_{n}/n \rightarrow \mu \) a.s. as \( n \rightarrow \infty \) . | Proof. Let \( {Y}_{k} = {X}_{k}{1}_{\left( \left| {X}_{k}\right| \leq k\right) } \) and \( {T}_{n} = {Y}_{1} + \cdots + {Y}_{n} \) . By (a) in the proof of Theorem 2.4.1 it suffices to show that \( {T}_{n}/n \rightarrow \mu \) . Let \( {Z}_{k} = {Y}_{k} - E{Y}_{k} \), so \( E{Z}_{k} = 0 \) . Now \( \operatorname{var}\l... | Yes |
Theorem 2.5.7. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. random variables with \( E{X}_{i} = 0 \) and \( E{X}_{i}^{2} = \) \( {\sigma }^{2} < \infty \) . Let \( {S}_{n} = {X}_{1} + \ldots + {X}_{n} \) . If \( \epsilon > 0 \) then\n\n\[ \n{S}_{n}/{n}^{1/2}{\left( \log n\right) }^{1/2 + \epsilon } \rightarrow 0\;\text{ ... | Proof. Let \( {a}_{n} = {n}^{1/2}{\left( \log n\right) }^{1/2 + \epsilon } \) for \( n \geq 2 \) and \( {a}_{1} > 0 \) .\n\n\[ \n\mathop{\sum }\limits_{{n = 1}}^{\infty }\operatorname{var}\left( {{X}_{n}/{a}_{n}}\right) = {\sigma }^{2}\left( {\frac{1}{{a}_{1}^{2}} + \mathop{\sum }\limits_{{n = 2}}^{\infty }\frac{1}{n{\... | Yes |
Theorem 2.5.9. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E\left| {X}_{1}\right| = \infty \) and let \( {S}_{n} = {X}_{1} + \) \( \cdots + {X}_{n} \) . Let \( {a}_{n} \) be a sequence of positive numbers with \( {a}_{n}/n \) increasing. Then \( \mathop{\limsup }\limits_{{n \rightarrow \infty }}\left| {S}_{n}\ri... | Proof. Since \( {a}_{n}/n \uparrow ,{a}_{kn} \geq k{a}_{n} \) for any integer \( k \) . Using this and \( {a}_{n} \uparrow \) ,\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }P\left( {\left| {X}_{1}\right| \geq k{a}_{n}}\right) \geq \mathop{\sum }\limits_{{n = 1}}^{\infty }P\left( {\left| {X}_{1}\right| \geq {a}_{kn}}\... | Yes |
Lemma 2.6.1. If \( {\gamma }_{m + n} \geq {\gamma }_{m} + {\gamma }_{n} \) then as \( n \rightarrow \infty ,{\gamma }_{n}/n \rightarrow \mathop{\sup }\limits_{m}{\gamma }_{m}/m \) . | Proof. Clearly, \( \lim \sup {\gamma }_{n}/n \leq \sup {\gamma }_{m}/m \) . To complete the proof, it suffices to prove that for any \( m \) liminf \( {\gamma }_{n}/n \geq {\gamma }_{m}/m \) . Writing \( n = {km} + \ell \) with \( 0 \leq \ell < m \) and making repeated use of the hypothesis gives \( {\gamma }_{n} \geq ... | Yes |
Lemma 2.6.2. If \( a > \mu \) and \( \theta > 0 \) is small then \( {a\theta } - \kappa \left( \theta \right) > 0 \) . | Proof. \( \kappa \left( 0\right) = \log \varphi \left( 0\right) = 0 \), so it suffices to show that (i) \( \kappa \) is continuous at 0,(ii) differentiable on \( \left( {0,{\theta }_{ + }}\right) \), and (iii) \( {\kappa }^{\prime }\left( \theta \right) \rightarrow \mu \) as \( \theta \rightarrow 0 \) . For then\n\n\[ ... | Yes |
\[ \int {e}^{\theta x}{\left( 2\pi \right) }^{-1/2}\exp \left( {-{x}^{2}/2}\right) {dx} = \exp \left( {{\theta }^{2}/2}\right) \int {\left( 2\pi \right) }^{-1/2}\exp \left( {-{\left( x - \theta \right) }^{2}/2}\right) {dx} \] | The integrand in the last integral is the density of a normal distribution with mean \( \theta \) and variance 1, so \( \varphi \left( \theta \right) = \exp \left( {{\theta }^{2}/2}\right) ,\theta \in \left( {-\infty ,\infty }\right) \) . In this case, \( {\varphi }^{\prime }\left( \theta \right) /\varphi \left( \theta... | Yes |
Example 2.6.2. Exponential distribution with parameter \( \lambda \) . If \( \theta < \lambda \)\n\n\[{\int }_{0}^{\infty }{e}^{\theta x}\lambda {e}^{-{\lambda x}}{dx} = \lambda /\left( {\lambda - \theta }\right)\] | \[{\varphi }^{\prime }\left( \theta \right) \varphi \left( \theta \right) = 1/\left( {\lambda - \theta }\right)\]\n\n\[{F}_{\theta }\left( x\right) = \frac{\lambda }{\lambda - \theta }{\int }_{0}^{x}{e}^{\theta y}\lambda {e}^{-{\lambda y}}{dy}\]\n\nis an exponential distribution with parameter \( \lambda - \theta \) an... | No |
Example 2.6.3. Coin flips. \( P\left( {{X}_{i} = 1}\right) = P\left( {{X}_{i} = - 1}\right) = 1/2 \) | \[ \varphi \left( \theta \right) = \left( {{e}^{\theta } + {e}^{-\theta }}\right) /2 \] \[ {\varphi }^{\prime }\left( \theta \right) /\varphi \left( \theta \right) = \left( {{e}^{\theta } - {e}^{-\theta }}\right) /\left( {{e}^{\theta } + {e}^{-\theta }}\right) \] \( {F}_{\theta }\left( {\{ x\} }\right) /F\left( {\{ x\}... | Yes |
Example 2.6.4. Perverted exponential. Let \( g\left( x\right) = C{x}^{-3}{e}^{-x} \) for \( x \geq 1, g\left( x\right) = 0 \) otherwise, and choose \( C \) so that \( g \) is a probability density. In this case, | \[ \varphi \left( \theta \right) = \int {e}^{\theta x}g\left( x\right) {dx} < \infty \] if and only if \( \theta \leq 1 \), and when \( \theta \leq 1 \), we have \[ \frac{{\varphi }^{\prime }\left( \theta \right) }{\varphi \left( \theta \right) } \leq \frac{{\varphi }^{\prime }\left( 1\right) }{\varphi \left( 1\right) ... | Yes |
Theorem 2.6.3. Suppose in addition to (H1) and (H2) that there is a \( {\theta }_{a} \in \left( {0,{\theta }_{ + }}\right) \) so that \( a = {\varphi }^{\prime }\left( {\theta }_{a}\right) /\varphi \left( {\theta }_{a}\right) \) . Then, as \( n \rightarrow \infty \) ,\n\n\[ \n{n}^{-1}\log P\left( {{S}_{n} \geq {na}}\ri... | Proof. The fact that the limsup of the left-hand side \( \leq \) the right-hand side follows from (2.6.2). To prove the other inequality, pick \( \lambda \in \left( {{\theta }_{a},{\theta }_{ + }}\right) \), let \( {X}_{1}^{\lambda },{X}_{2}^{\lambda },\ldots \) be i.i.d. with distribution \( {F}_{\lambda } \) and let ... | Yes |
Lemma 2.6.4. \( \frac{d{F}^{n}}{d{F}_{\lambda }^{n}} = {e}^{-{\lambda x}}\varphi {\left( \lambda \right) }^{n} \) . | Proof. We will prove this by induction. The result holds when \( n = 1 \) . For \( n > 1 \), we note that\n\n\[ \n{F}^{n} = {F}^{n - 1} * F\left( z\right) = {\int }_{-\infty }^{\infty }d{F}^{n - 1}\left( x\right) {\int }_{-\infty }^{z - x}{dF}\left( y\right) \n\]\n\n\[ \n= \int d{F}_{\lambda }^{n - 1}\left( x\right) \i... | Yes |
Lemma 3.1.1. If \( {c}_{j} \rightarrow 0,{a}_{j} \rightarrow \infty \) and \( {a}_{j}{c}_{j} \rightarrow \lambda \) then \( {\left( 1 + {c}_{j}\right) }^{{a}_{j}} \rightarrow {e}^{\lambda } \). | Proof. As \( x \rightarrow 0,\log \left( {1 + x}\right) /x \rightarrow 1 \), so \( {a}_{j}\log \left( {1 + {c}_{j}}\right) \rightarrow \lambda \) and the desired result follows. | No |
Theorem 3.1.3. The De Moivre-Laplace Theorem. If \( a < b \) then as \( m \rightarrow \infty \)\n\n\[ P\left( {a \leq {S}_{m}/\sqrt{m} \leq b}\right) \rightarrow {\int }_{a}^{b}{\left( 2\pi \right) }^{-1/2}{e}^{-{x}^{2}/2}{dx} \] | (To remove the restriction to even integers observe \( {S}_{{2n} + 1} = {S}_{2n} \pm 1 \) .) The last result is a special case of the central limit theorem given in Section 3.4, so further details are left to the reader. | No |
Example 3.2.1. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( P\left( {{X}_{i} = 1}\right) = P\left( {{X}_{i} = - 1}\right) = 1/2 \) and let \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) . Then Theorem 3.1.3 implies | \[ {F}_{n}\left( y\right) = P\left( {{S}_{n}/\sqrt{n} \leq y}\right) \rightarrow {\int }_{-\infty }^{y}{\left( 2\pi \right) }^{-1/2}{e}^{-{x}^{2}/2}{dx} \] | Yes |
Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with distribution \( F \) . The Glivenko-Cantelli theorem (Theorem 2.4.7) implies that for almost every \( \omega \) , | \[ {F}_{n}\left( y\right) = {n}^{-1}\mathop{\sum }\limits_{{m = 1}}^{n}{1}_{\left( {X}_{m}\left( \omega \right) \leq y\right) } \rightarrow F\left( y\right) \text{ for all }y \] | Yes |
Example 3.2.3. Let \( X \) have distribution \( F \) . Then \( X + 1/n \) has distribution | \[ {F}_{n}\left( x\right) = P\left( {X + 1/n \leq x}\right) = F\left( {x - 1/n}\right) \] As \( n \rightarrow \infty ,{F}_{n}\left( x\right) \rightarrow F\left( {x - }\right) = \mathop{\lim }\limits_{{y \uparrow x}}F\left( y\right) \) so convergence only occurs at continuity points. | Yes |
Birthday problem. Let \( {X}_{1},{X}_{2},\ldots \) be independent and uniformly distributed on \( \{ 1,\ldots, N\} \), and let \( {T}_{N} = \min \left\{ {n : {X}_{n} = {X}_{m}}\right. \) for some \( \left. {m < n}\right\} \) . | \[ P\left( {{T}_{N} > n}\right) = \mathop{\prod }\limits_{{m = 2}}^{n}\left( {1 - \frac{m - 1}{N}}\right) \] When \( N = {365} \) this is the probability that two people in a group of size \( n \) do not have the same birthday (assuming all birthdays are equally likely). Using Exercise 3.1.1, it is easy to see that \[ ... | Yes |
Lemma 3.2.1. \( {V}_{n + 1} \) has density function\n\n\[ \n{f}_{{V}_{n + 1}}\left( x\right) = \left( {{2n} + 1}\right) \left( \begin{matrix} {2n} \\ n \end{matrix}\right) {x}^{n}{\left( 1 - x\right) }^{n} \n\] | Proof. There are \( {2n} + 1 \) ways to pick the observation that falls at \( x \), then we have to pick \( n \) indices for observations \( < x \), which can be done in \( \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \) ways. Once we have decided on the indices that will land \( < x \) and \( > x \), the probab... | Yes |
Theorem 3.2.2. If \( {F}_{n} \Rightarrow {F}_{\infty } \) then there are random variables \( {Y}_{n},1 \leq n \leq \infty \), with distribution \( {F}_{n} \) so that \( {Y}_{n} \rightarrow {Y}_{\infty } \) a.s. | Proof. Let \( \Omega = \left( {0,1}\right) ,\mathcal{F} = \) Borel sets, \( P = \) Lebesgue measure, and let \( {Y}_{n}\left( x\right) = \) \( \sup \left\{ {y : {F}_{n}\left( y\right) < x}\right\} \) . By Theorem 1.2.2, \( {Y}_{n} \) has distribution \( {F}_{n} \) . We will now show that \( {Y}_{n}\left( x\right) \righ... | Yes |
Theorem 3.2.3. \( {X}_{n} \Rightarrow {X}_{\infty } \) if and only if for every bounded continuous function \( g \) we have \( \operatorname{Eg}\left( {X}_{n}\right) \rightarrow \operatorname{Eg}\left( {X}_{\infty }\right) \) . | Proof. Let \( {Y}_{n} \) have the same distribution as \( {X}_{n} \) and converge a.s. Since \( g \) is continuous \( g\left( {Y}_{n}\right) \rightarrow g\left( {Y}_{\infty }\right) \) a.s. and the bounded convergence theorem implies\n\n\[ \n{Eg}\left( {X}_{n}\right) = {Eg}\left( {Y}_{n}\right) \rightarrow {Eg}\left( {... | Yes |
Theorem 3.2.4. Continuous mapping theorem. Let \( g \) be a measurable function and \( {D}_{g} = \{ x : g \) is discontinuous at \( x\} \) . If \( {X}_{n} \Rightarrow {X}_{\infty } \) and \( P\left( {{X}_{\infty } \in {D}_{g}}\right) = 0 \) then \( g\left( {X}_{n}\right) \Rightarrow g\left( X\right) \) . If in addition... | Proof. Let \( {Y}_{n}{ = }_{d}{X}_{n} \) with \( {Y}_{n} \rightarrow {Y}_{\infty } \) a.s. If \( f \) is continuous then \( {D}_{f \circ g} \subset {D}_{g} \) so \( P\left( {{Y}_{\infty } \in {D}_{f \circ g}}\right) = 0 \) and it follows that \( f\left( {g\left( {Y}_{n}\right) }\right) \rightarrow f\left( {g\left( {Y}_... | Yes |
Theorem 3.2.5. The following statements are equivalent: (i) \( {X}_{n} \Rightarrow {X}_{\infty } \)\n\n(ii) For all open sets \( G,\lim \mathop{\inf }\limits_{{n \rightarrow \infty }}P\left( {{X}_{n} \in G}\right) \geq P\left( {{X}_{\infty } \in G}\right) \) .\n\n(iii) For all closed sets \( K,\lim \mathop{\sup }\limit... | Proof. We will prove four things and leave it to the reader to check that we have proved the result given above.\n\n(i) implies (ii): Let \( {Y}_{n} \) have the same distribution as \( {X}_{n} \) and \( {Y}_{n} \rightarrow {Y}_{\infty } \) a.s. Since \( G \) is open\n\n\[ \mathop{\liminf }\limits_{{n \rightarrow \infty... | No |
Theorem 3.2.6. Helly’s selection theorem. For every sequence \( {F}_{n} \) of distribution functions, there is a subsequence \( {F}_{n\left( k\right) } \) and a right continuous nondecreasing function \( F \) so that \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{F}_{n\left( k\right) }\left( y\right) = F\left( y\ri... | Proof. The first step is a diagonal argument. Let \( {q}_{1},{q}_{2},\ldots \) be an enumeration of the rationals. Since for each \( k,{F}_{m}\left( {q}_{k}\right) \in \left\lbrack {0,1}\right\rbrack \) for all \( m \), there is a sequence \( {m}_{k}\left( i\right) \rightarrow \infty \) that is a subsequence of \( {m}_... | Yes |
Every subsequential limit is the distribution function of a probability measure if and only if the sequence \( {F}_{n} \) is \( \mathbf{{tight}} \), i.e., for all \( \epsilon > 0 \) there is an \( {M}_{\epsilon } \) so that \[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}1 - {F}_{n}\left( {M}_{\epsilon }\right) + ... | Proof. Suppose the sequence is tight and \( {F}_{n\left( k\right) }{ \Rightarrow }_{v}F \) . Let \( r < - {M}_{\epsilon } \) and \( s > {M}_{\epsilon } \) be continuity points of \( F \) . Since \( {F}_{n}\left( r\right) \rightarrow F\left( r\right) \) and \( {F}_{n}\left( s\right) \rightarrow F\left( s\right) \), we h... | Yes |
Theorem 3.2.8. If there is a \( \varphi \geq 0 \) so that \( \varphi \left( x\right) \rightarrow \infty \) as \( \left| x\right| \rightarrow \infty \) and\n\n\[ C = \mathop{\sup }\limits_{n}\int \varphi \left( x\right) d{F}_{n}\left( x\right) < \infty \]\n\nthen \( {F}_{n} \) is tight. | Proof. \( 1 - {F}_{n}\left( M\right) + {F}_{n}\left( {-M}\right) \leq C/\mathop{\inf }\limits_{{\left| x\right| \geq M}}\varphi \left( x\right) \) | Yes |
Theorem 3.3.1. All characteristic functions have the following properties:\n\n(a) \( \varphi \left( 0\right) = 1 \) ,\n\n(b) \( \varphi \left( {-t}\right) = \overline{\varphi \left( t\right) } \) ,\n\n(c) \( \left| {\varphi \left( t\right) }\right| = \left| {E{e}^{itX}}\right| \leq E\left| {e}^{itX}\right| = 1 \)\n\n(d... | Proof. (a) is obvious. For (b) we note that\n\n\[ \varphi \left( {-t}\right) = E\left( {\cos \left( {-{tX}}\right) + i\sin \left( {-{tX}}\right) }\right) = E\left( {\cos \left( {tX}\right) - i\sin \left( {tX}\right) }\right) \]\n\n(c) follows from Exercise 1.6.2 since \( \varphi \left( {x, y}\right) = {\left( {x}^{2} +... | No |
Theorem 3.3.2. If \( {X}_{1} \) and \( {X}_{2} \) are independent and have ch.f.’s \( {\varphi }_{1} \) and \( {\varphi }_{2} \) then \( {X}_{1} + {X}_{2} \) has ch.f. \( {\varphi }_{1}\left( t\right) {\varphi }_{2}\left( t\right) \) . | Proof.\n\n\[ E{e}^{{it}\left( {{X}_{1} + {X}_{2}}\right) } = E\left( {{e}^{{it}{X}_{1}}{e}^{{it}{X}_{2}}}\right) = E{e}^{{it}{X}_{1}}E{e}^{{it}{X}_{2}} \]\n\nsince \( {e}^{{it}{X}_{1}} \) and \( {e}^{{it}{X}_{2}} \) are independent. | Yes |
If \( P\left( {X = 1}\right) = P\left( {X = - 1}\right) = 1/2 \) then | \[ E{e}^{itX} = \left( {{e}^{it} + {e}^{-{it}}}\right) /2 = \cos t \] | Yes |
Example 3.3.2. Poisson distribution. If \( P\left( {X = k}\right) = {e}^{-\lambda }{\lambda }^{k}/k! \) for \( k = 0,1,2,\ldots \) | then\n\[\nE{e}^{itX} = \mathop{\sum }\limits_{{k = 0}}^{\infty }{e}^{-\lambda }\frac{{\lambda }^{k}{e}^{itk}}{k!} = \exp \left( {\lambda \left( {{e}^{it} - 1}\right) }\right)\n\] | Yes |
Example 3.3.3. Normal distribution\n\n\[ \text{Density}\;{\left( 2\pi \right) }^{-1/2}\exp \left( {-{x}^{2}/2}\right) \]\n\n\[ \text{Ch.f.}\exp \left( {-{t}^{2}/2}\right) \]\n\nCombining this result with (e) of Theorem 3.3.1, we see that a normal distribution with mean \( \mu \) and variance \( {\sigma }^{2} \) has ch.... | Physics Proof\n\n\[ \int {e}^{itx}{\left( 2\pi \right) }^{-1/2}{e}^{-{x}^{2}/2}{dx} = {e}^{-{t}^{2}/2}\int {\left( 2\pi \right) }^{-1/2}{e}^{-{\left( x - it\right) }^{2}/2}{dx} \]\n\nThe integral is 1 since the integrand is the normal density with mean \( {it} \) and variance 1.\n\nMath Proof. Now that we have cheated ... | Yes |
Example 3.3.4. Uniform distribution on \( \\left( {a, b}\\right) \) | Proof. Once you recall that \( {\\int }_{a}^{b}{e}^{\\lambda x}{dx} = \\left( {{e}^{\\lambda b} - {e}^{\\lambda a}}\\right) /\\lambda \) holds for complex \( \\lambda \), this is immediate. | No |
Example 3.3.5. Triangular distribution\n\n\\[ \n\\text{Density}\\;1 - \\left| x\\right| \\;x \\in \\left( {-1,1}\\right) \n\\]\n\n\\[ \n\\text{Ch.f.}\\;2\\left( {1 - \\cos t}\\right) /{t}^{2} \n\\] | Proof. To see this, notice that if \\( X \\) and \\( Y \\) are independent and uniform on \\( \\left( {-1/2,1/2}\\right) \\) then \\( X + Y \\) has a triangular distribution. Using Example 3.3.4 now and Theorem 3.3.2 it follows that the desired ch.f. is\n\n\\[ \n{\\left\\{ \\left( {e}^{{it}/2} - {e}^{-{it}/2}\\right) /... | Yes |
Example 3.3.6. Exponential distribution\n\n\[ \n\\begin{array}{ll} \\text{ Density } & {e}^{-x}\;x \\in \\left( {0,\\infty }\\right) \\\\ \\text{ Ch.f. } & 1/\\left( {1 - {it}}\\right) \\end{array} \n\] | Proof. Integrating gives\n\n\[ \n{\\int }_{0}^{\\infty }{e}^{itx}{e}^{-x}{dx} = {\\left. \\frac{{e}^{\\left( {{it} - 1}\\right) x}}{{it} - 1}\\right| }_{0}^{\\infty } = \\frac{1}{1 - {it}} \n\]\n\nsince \( \\exp \\left( {\\left( {{it} - 1}\\right) x}\\right) \\rightarrow 0 \) as \( x \\rightarrow \\infty \) . | Yes |
Example 3.3.7. Bilateral exponential\n\n\\[ \n\\begin{array}{ll} \\text{ Density } & \\frac{1}{2}{e}^{-\\left| x\\right| } \\\\ \\text{ Ch.f. } & 1/\\left( {1 + {t}^{2}}\\right) \\end{array}x \\in \\left( {-\\infty ,\\infty }\\right) \n\\] | Proof This follows from Lemma 3.3.3 with \\( {F}_{1} \\) the distribution of an exponential random variable \\( X,{F}_{2} \\) the distribution of \\( - X \\), and \\( {\\lambda }_{1} = {\\lambda }_{2} = 1/2 \\) then using (b) of Theorem 3.3.1 we see the desired ch.f. is\n\n\\[ \n\\frac{1}{2\\left( {1 - {it}}\\right) } ... | Yes |
Theorem 3.3.4. The inversion formula. Let \( \varphi \left( t\right) = \int {e}^{itx}\mu \left( {dx}\right) \) where \( \mu \) is a probability measure. If \( a < b \) then\n\n\[ \mathop{\lim }\limits_{{T \rightarrow \infty }}{\left( 2\pi \right) }^{-1}{\int }_{-T}^{T}\frac{{e}^{-{ita}} - {e}^{-{itb}}}{it}\varphi \left... | Proof. Let\n\n\[ {I}_{T} = {\int }_{-T}^{T}\frac{{e}^{-{ita}} - {e}^{-{itb}}}{it}\varphi \left( t\right) {dt} = {\int }_{-T}^{T}\int \frac{{e}^{-{ita}} - {e}^{-{itb}}}{it}{e}^{itx}\mu \left( {dx}\right) {dt} \]\n\nThe integrand may look bad near \( t = 0 \) but if we observe that\n\n\[ \frac{{e}^{-{ita}} - {e}^{-{itb}}... | Yes |
Theorem 3.3.5. If \( \int \left| {\varphi \left( t\right) }\right| {dt} < \infty \) then \( \mu \) has bounded continuous density | \[ f\left( y\right) = \frac{1}{2\pi }\int {e}^{-{ity}}\varphi \left( t\right) {dt} \] Proof. As we observed in the proof of Theorem 3.3.4 \[ \left| \frac{{e}^{-{ita}} - {e}^{-{itb}}}{it}\right| = \left| {{\int }_{a}^{b}{e}^{-{ity}}{dy}}\right| \leq \left| {b - a}\right| \] so the integral in Theorem 3.3.4 converges abs... | Yes |
Example 3.3.8. Polya's distribution\n\n\\[ \n\\begin{array}{ll} \\text{ Density } & \\left( {1 - \\cos x}\\right) /\\pi {x}^{2} \\\\ \\text{ Ch.f. } & {\\left( 1 - \\left| t\\right| \\right) }^{ + } \\end{array} \n\\] | Proof. Theorem 3.3.5 implies\n\n\\[ \n\\frac{1}{2\\pi }\\int \\frac{2\\left( {1 - \\cos s}\\right) }{{s}^{2}}{e}^{-{isy}}{ds} = {\\left( 1 - \\left| y\\right| \\right) }^{ + } \n\\]\n\nNow let \\( s = x, y = - t \\) . | Yes |
Example 3.3.9. The Cauchy distribution\n\n\[ \n\\begin{array}{ll} \\text{ Density } & 1/\\pi \\left( {1 + {x}^{2}}\\right) \\\\ \\text{ Ch.f. } & \\exp \\left( {-\\left| t\\right| }\\right) \\end{array} \n\] | Proof. Theorem 3.3.5 implies\n\n\[ \n\\frac{1}{2\\pi }\\int \\frac{1}{1 + {s}^{2}}{e}^{-{isy}}{ds} = \\frac{1}{2}{e}^{-\\left| y\\right| } \n\]\n\nNow let \( s = x, y = - t \) and multiply each side by 2 . | Yes |
Theorem 3.3.6. Continuity theorem. Let \( {\mu }_{n},1 \leq n \leq \infty \) be probability measures with ch.f. \( {\varphi }_{n} \) . (i) If \( {\mu }_{n} \Rightarrow {\mu }_{\infty } \) then \( {\varphi }_{n}\left( t\right) \rightarrow {\varphi }_{\infty }\left( t\right) \) for all \( t \) . (ii) If \( {\varphi }_{n}... | Proof. (i) is easy. \( {e}^{itx} \) is bounded and continuous so if \( {\mu }_{n} \Rightarrow {\mu }_{\infty } \) then Theorem 3.2.3 implies \( {\varphi }_{n}\left( t\right) \rightarrow {\varphi }_{\infty }\left( t\right) \) . To prove (ii), our first goal is to prove tightness. We begin with some calculations that may... | Yes |
Lemma 3.3.7.\n\[ \left| {{e}^{ix} - \mathop{\sum }\limits_{{m = 0}}^{n}\frac{{\left( ix\right) }^{m}}{m!}}\right| \leq \min \left( {\frac{{\left| x\right| }^{n + 1}}{\left( {n + 1}\right) !},\frac{2{\left| x\right| }^{n}}{n!}}\right) \] | Proof. Integrating by parts gives\n\n\[ {\int }_{0}^{x}{\left( x - s\right) }^{n}{e}^{is}{ds} = \frac{{x}^{n + 1}}{n + 1} + \frac{i}{n + 1}{\int }_{0}^{x}{\left( x - s\right) }^{n + 1}{e}^{is}{ds} \]\n\nWhen \( n = 0 \), this says\n\n\[ {\int }_{0}^{x}{e}^{is}{ds} = x + i{\int }_{0}^{x}\left( {x - s}\right) {e}^{is}{ds... | No |
Theorem 3.3.8. If \( E{\left| X\right| }^{2} < \infty \) then\n\n\[ \varphi \left( t\right) = 1 + {itEX} - {t}^{2}E\left( {X}^{2}\right) /2 + o\left( {t}^{2}\right) \] | Proof. The error term is \( \leq {t}^{2}E\left( {\left| t\right| \cdot {\left| X\right| }^{3} \land 2{\left| X\right| }^{2}}\right) \) . The variable in parentheses is smaller than \( 2{\left| X\right| }^{2} \) and converges to 0 as \( t \rightarrow 0 \), so the desired conclusion follows from the dominated convergence... | Yes |
Theorem 3.3.9. If \( \lim \mathop{\sup }\limits_{{h \downarrow 0}}\{ \varphi \left( h\right) - {2\varphi }\left( 0\right) + \varphi \left( {-h}\right) \} /{h}^{2} > - \infty \), then \( E{\left| X\right| }^{2} < \infty \) . | Proof. \( \left( {{e}^{ihx} - 2 + {e}^{-{ihx}}}\right) /{h}^{2} = - 2\left( {1 - \cos {hx}}\right) /{h}^{2} \leq 0 \) and \( 2\left( {1 - \cos {hx}}\right) /{h}^{2} \rightarrow {x}^{2} \) as \( h \rightarrow 0 \) so Fatou’s lemma and Fubini’s theorem imply\n\n\[ \n\int {x}^{2}{dF}\left( x\right) \leq 2\mathop{\liminf }... | Yes |
Theorem 3.3.10. Polya’s criterion. Let \( \varphi \left( t\right) \) be real nonnegative and have \( \varphi \left( 0\right) = \) \( 1,\varphi \left( t\right) = \varphi \left( {-t}\right) \), and \( \varphi \) is decreasing and convex on \( \left( {0,\infty }\right) \) with\n\n\[ \mathop{\lim }\limits_{{t \downarrow 0}... | Proof. Let \( {\varphi }^{\prime } \) be the right derivative of \( \phi \), i.e.,\n\n\[ {\varphi }^{\prime }\left( t\right) = \mathop{\lim }\limits_{{h \downarrow 0}}\frac{\varphi \left( {t + h}\right) - \varphi \left( t\right) }{h} \]\n\nSince \( \varphi \) is convex this exists and is right continuous and increasing... | Yes |
Example 3.3.10. \( \exp \left( {-{\left| t\right| }^{\alpha }}\right) \) is a characteristic function for \( 0 < \alpha < 2 \) . | Proof. A little calculus shows that for any \( \beta \) and \( \left| x\right| < 1 \)\n\n\[{\left( 1 - x\right) }^{\beta } = \mathop{\sum }\limits_{{n = 0}}^{\infty }\left( \begin{array}{l} \beta \\ n \end{array}\right) {\left( -x\right) }^{n}\]\n\nwhere\n\n\[ \left( \begin{array}{l} \beta \\ n \end{array}\right) = \fr... | Yes |
Example 3.3.11. For some purposes, it is nice to have an explicit example of two ch.f.’s that agree on \( \left\lbrack {-1,1}\right\rbrack \) . From Example 3.3.8, we know that \( {\left( 1 - \left| t\right| \right) }^{ + } \) is the ch.f. of the density \( \left( {1 - \cos x}\right) /\pi {x}^{2} \) . Define \( \psi \l... | The Fourier series for \( \psi \) is\n\n\[ \psi \left( u\right) = \frac{1}{2} + \mathop{\sum }\limits_{{n = - \infty }}^{\infty }\frac{2}{{\pi }^{2}{\left( 2n - 1\right) }^{2}}\exp \left( {i\left( {{2n} - 1}\right) {\pi u}}\right) \]\n\nThe right-hand side is the ch.f. of a discrete distribution with\n\n\[ P\left( {X =... | Yes |
Theorem 3.3.11. If \( \mathop{\limsup }\limits_{{k \rightarrow \infty }}{\mu }_{2k}^{1/{2k}}/{2k} = r < \infty \) then there is at most one d.f. \( F \) with \( {\mu }_{k} = \int {x}^{k}{dF}\left( x\right) \) for all positive integers \( k \) . | Proof. Let \( F \) be any d.f. with the moments \( {\mu }_{k} \) and let \( {\nu }_{k} = \int {\left| x\right| }^{k}{dF}\left( x\right) \) . The Cauchy-Schwarz inequality implies \( {\nu }_{{2k} + 1}^{2} \leq {\mu }_{2k}{\mu }_{{2k} + 2} \) so\n\n\[ \mathop{\limsup }\limits_{{k \rightarrow \infty }}\left( {\nu }_{k}^{1... | Yes |
Theorem 3.4.1. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E{X}_{i} = \mu \) , \( \operatorname{var}\left( {X}_{i}\right) = {\sigma }^{2} \in \left( {0,\infty }\right) \) . If \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) then\n\n\[\n\left( {{S}_{n} - {n\mu }}\right) /\sigma {n}^{1/2} \Rightarrow \chi\n\]\n\nwhere ... | Proof By considering \( {X}_{i}^{\prime } = {X}_{i} - \mu \), it suffices to prove the result when \( \mu = 0 \) . From\n\nTheorem 3.3.8\n\[\n\varph | No |
Theorem 3.3.8\n\[ \varphi \left( t\right) = E\exp \left( {{it}{X}_{1}}\right) = 1 - \frac{{\sigma }^{2}{t}^{2}}{2} + o\left( {t}^{2}\right) \] | so\n\[ E\exp \left( {{it}{S}_{n}/\sigma {n}^{1/2}}\right) = {\left( 1 - \frac{{t}^{2}}{2n} + o\left( {n}^{-1}\right) \right) }^{n} \]\n\nFrom Lemma 3.1.1 it should be clear that the last quantity \( \rightarrow \exp \left( {-{t}^{2}/2}\right) \) as \( n \rightarrow \infty \) , which with Theorem 3.3.6 completes the pro... | No |
Theorem 3.4.2. If \( {c}_{n} \rightarrow c \in \mathbf{C} \) then \( {\left( 1 + {c}_{n}/n\right) }^{n} \rightarrow {e}^{c} \). | Proof. The proof is based on two simple facts: | No |
Lemma 3.4.3. Let \( {z}_{1},\ldots ,{z}_{n} \) and \( {w}_{1},\ldots ,{w}_{n} \) be complex numbers of modulus \( \leq \theta \) .\n\nThen\n\[ \left| {\mathop{\prod }\limits_{{m = 1}}^{n}{z}_{m} - \mathop{\prod }\limits_{{m = 1}}^{n}{w}_{m}}\right| \leq {\theta }^{n - 1}\mathop{\sum }\limits_{{m = 1}}^{n}\left| {{z}_{m... | Proof. The result is true for \( n = 1 \) . To prove it for \( n > 1 \) observe that\n\n\[ \left| {\mathop{\prod }\limits_{{m = 1}}^{n}{z}_{m} - \mathop{\prod }\limits_{{m = 1}}^{n}{w}_{m}}\right| \leq \left| {{z}_{1}\mathop{\prod }\limits_{{m = 2}}^{n}{z}_{m} - {z}_{1}\mathop{\prod }\limits_{{m = 2}}^{n}{w}_{m}}\right... | Yes |
Lemma 3.4.4. If \( b \) is a complex number with \( \left| b\right| \leq 1 \) then \( \left| {{e}^{b} - \left( {1 + b}\right) }\right| \leq {\left| b\right| }^{2} \) . | Proof. \( {e}^{b} - \left( {1 + b}\right) = {b}^{2}/2! + {b}^{3}/3! + {b}^{4}/4! + \ldots \) so if \( \left| b\right| \leq 1 \) then\n\n\[ \left| {{e}^{b} - \left( {1 + b}\right) }\right| \leq \frac{{\left| b\right| }^{2}}{2}\left( {1 + 1/2 + 1/{2}^{2} + \ldots }\right) = {\left| b\right| }^{2} \]\n | Yes |
A roulette wheel has slots numbered 1-36 (18 red and 18 black) and two slots numbered 0 and 00 that are painted green. Players can bet \$1 that the ball will land in a red (or black) slot and win \$1 if it does. If we let \( {X}_{i} \) be the winnings on the \( i \) th play then \( {X}_{1},{X}_{2},\ldots \) are i.i.d. ... | \[ E{X}_{i} = - 1/{19}\text{ and }\operatorname{var}\left( X\right) = E{X}^{2} - {\left( EX\right) }^{2} = 1 - {\left( 1/{19}\right) }^{2} = {0.9972} \] We are interested in \[ P\left( {{S}_{n} \geq 0}\right) = P\left( {\frac{{S}_{n} - {n\mu }}{\sigma \sqrt{n}} \geq \frac{-{n\mu }}{\sigma \sqrt{n}}}\right) \] Taking \(... | Yes |
Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( P\left( {{X}_{i} = 0}\right) = P\left( {{X}_{i} = 1}\right) = \) \( 1/2 \) . If \( {X}_{i} = 1 \) indicates that a heads occured on the \( i \) th toss then \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) is the total number of heads at time \( n \) . | \[ E{X}_{i} = 1/2\;\text{ and }\;\operatorname{var}\left( X\right) = E{X}^{2} - {\left( EX\right) }^{2} = 1/2 - 1/4 = 1/4 \] So the central limit theorem tells us \( \left( {{S}_{n} - n/2}\right) /\sqrt{n/4} \Rightarrow \chi \) . Our table of the normal distribution tells us that \[ P\left( {\chi > 2}\right) = 1 - {0.9... | Yes |
To estimate \( P\left( {{S}_{16} = 8}\right) \) using the central limit theorem, we regard 8 as the interval \( \left\lbrack {{7.5},{8.5}}\right\rbrack \) . Since \( \mu = 1/2 \), and \( \sigma \sqrt{n} = 2 \) for \( n = {16} \) | \[ P\left( {\left| {{S}_{16} - 8}\right| \leq {0.5}}\right) = P\left( {\frac{\left| {S}_{n} - n\mu \right| }{\sigma \sqrt{n}} \leq {0.25}}\right) \] \[ \approx P\left( {\left| \chi \right| \leq {0.25}}\right) = 2\left( {{0.5987} - {0.5}}\right) = {0.1974} \] Even though \( n \) is small, this agrees well with the exact... | Yes |
Let \( {Z}_{\lambda } \) have a Poisson distribution with mean \( \lambda \) . If \( {X}_{1},{X}_{2},\ldots \) are independent and have Poisson distributions with mean 1, then \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) has a Poisson distribution with mean \( n \) . Since \( \operatorname{var}\left( {X}_{i}\right) = 1 \... | \[ \left( {{S}_{n} - n}\right) /{n}^{1/2} \Rightarrow \chi \;\text{ as }n \rightarrow \infty \] | No |
Pairwise independence is good enough for the strong law of large numbers (see Theorem 2.4.1). It is not good enough for the central limit theorem. Let \( {\xi }_{1},{\xi }_{2},\ldots \) be i.i.d. with \( P\left( {{\xi }_{i} = 1}\right) = P\left( {{\xi }_{i} = - 1}\right) = 1/2 \) . We will arrange things so that for \(... | \[ {S}_{{2}^{n}} = {\xi }_{1}\left( {1 + {\xi }_{2}}\right) \cdots \left( {1 + {\xi }_{n + 1}}\right) = \left\{ \begin{array}{ll} \pm {2}^{n} & \text{ with prob }{2}^{-n - 1} \\ 0 & \text{ with prob }1 - {2}^{-n} \end{array}\right. \] To do this we let \( {X}_{1} = {\xi }_{1},{X}_{2} = {\xi }_{1}{\xi }_{2} \), and for ... | Yes |
Theorem 3.4.5. The Lindeberg-Feller theorem. For each \( n \), let \( {X}_{n, m},1 \leq m \leq \) \( n \), be independent random variables with \( E{X}_{n, m} = 0 \) . Suppose\n\n(i) \( \mathop{\sum }\limits_{{m = 1}}^{n}E{X}_{n, m}^{2} \rightarrow {\sigma }^{2} > 0 \)\n\n(ii) For all \( \epsilon > 0,\mathop{\lim }\lim... | Proof. Let \( {\varphi }_{n, m}\left( t\right) = E\exp \left( {{it}{X}_{n, m}}\right) ,{\sigma }_{n, m}^{2} = E{X}_{n, m}^{2} \) . By Theorem 3.3.6, it suffices to show that\n\n\[ \mathop{\prod }\limits_{{m = 1}}^{n}{\varphi }_{n, m}\left( t\right) \rightarrow \exp \left( {-{t}^{2}{\sigma }^{2}/2}\right) \]\n\nLet \( {... | Yes |
Example 3.4.6. Cycles in a random permutation and record values. Continuing the analysis of Examples 2.2.4 and 2.3.2, let \( {Y}_{1},{Y}_{2},\ldots \) be independent with \( P\left( {{Y}_{m} = 1}\right) = 1/m \), and \( P\left( {{Y}_{m} = 0}\right) = 1 - 1/m.E{Y}_{m} = 1/m \) and \( \operatorname{var}\left( {Y}_{m}\rig... | \[ {X}_{n, m} = \left( {{Y}_{m} - 1/m}\right) /{\left( \log n\right) }^{1/2} \] \( E{X}_{n, m} = 0,\mathop{\sum }\limits_{{m = 1}}^{n}E{X}_{n, m}^{2} \rightarrow 1 \), and for any \( \epsilon > 0 \) \[ \mathop{\sum }\limits_{{m = 1}}^{n}E\left( {{\left| {X}_{n, m}\right| }^{2};\left| {X}_{n, m}\right| > \epsilon }\righ... | Yes |
Example 3.4.7. The converse of the three series theorem. Recall the set up of Theorem 2.5.4. Let \( {X}_{1},{X}_{2},\ldots \) be independent, let \( A > 0 \), and let \( {Y}_{m} = {X}_{m}{1}_{\left( \left| {X}_{m}\right| \leq A\right) } \) . In order that \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{X}_{n} \) converges... | Proof. The necessity of the first condition is clear. For if that sum is infinite, \( P\left( {\left| {X}_{n}\right| > }\right. \) \( A \) i.o. \( ) > 0 \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sum }\limits_{{m = 1}}^{n}{X}_{m} \) cannot exist. Suppose next that the sum in (i) is finite but the... | Yes |
Infinite variance. Suppose \( {X}_{1},{X}_{2},\ldots \) are i.i.d. and have \( P\left( {{X}_{1} > }\right. \) \( x) = P\left( {{X}_{1} < - x}\right) \) and \( P\left( {\left| {X}_{1}\right| > x}\right) = {x}^{-2} \) for \( x \geq 1 \) . | \[ E{\left| {X}_{1}\right| }^{2} = {\int }_{0}^{\infty }{2xP}\left( {\left| {X}_{1}\right| > x}\right) {dx} = \infty \] | Yes |
Theorem 3.4.6. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. and \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \) . In order that there exist constants \( {a}_{n} \) and \( {b}_{n} > 0 \) so that \( \left( {{S}_{n} - {a}_{n}}\right) /{b}_{n} \Rightarrow \chi \), it is necessary and sufficient that | \[ {y}^{2}P\left( {\left| {X}_{1}\right| > y}\right) /E\left( {{\left| {X}_{1}\right| }^{2};\left| {X}_{1}\right| \leq y}\right) \rightarrow 0. \] A proof can be found in Gnedenko and Kolmogorov (1954), a reference that contains the last word on many results about sums of independent random variables. | Yes |
Lemma 3.4.7. \( {h}_{n}\left( \epsilon \right) \rightarrow 0 \) for each fixed \( \epsilon > 0 \) so we can pick \( {\epsilon }_{n} \rightarrow 0 \) so that \( {h}_{n}\left( {\epsilon }_{n}\right) \rightarrow 0 \) | Proof. Let \( {N}_{m} \) be chosen so that \( {h}_{n}\left( {1/m}\right) \leq 1/m \) for \( n \geq {N}_{m} \) and \( m \rightarrow {N}_{m} \) is increasing. Let \( {\epsilon }_{n} = 1/m \) for \( {N}_{m} \leq n < {N}_{m + 1} \), and \( = 1 \) for \( n < {N}_{1} \) . When \( {N}_{m} \leq n < {N}_{m + 1},{\epsilon }_{n} ... | Yes |
Theorem 3.4.8. Erdös-Kac central limit theorem. As \( n \rightarrow \infty \)\n\n\[ {P}_{n}\left( {m \leq n : g\left( m\right) - \log \log n \leq x{\left( \log \log n\right) }^{1/2}}\right) \rightarrow P\left( {\chi \leq x}\right) \] | Proof. We begin by showing that we can ignore the primes \ | No |
Theorem 3.4.9. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E{X}_{i} = 0, E{X}_{i}^{2} = {\sigma }^{2} \), and \( E{\left| {X}_{i}\right| }^{3} = \) \( \rho < \infty \) . If \( {F}_{n}\left( x\right) \) is the distribution of \( \left( {{X}_{1} + \cdots + {X}_{n}}\right) /\sigma \sqrt{n} \) and \( \mathcal{N}\lef... | Proof. Since neither side of the inequality is affected by scaling, we can suppose without loss of generality that \( {\sigma }^{2} = 1 \) . The first phase of the argument is to derive an inequality, Lemma 3.4.11, that relates the difference between the two distributions to the distance between their ch.f.'s. Polya's ... | No |
Lemma 3.4.10. Let \( F \) and \( G \) be distribution functions with \( {G}^{\prime }\left( x\right) \leq \lambda < \infty \) . Let \( \Delta \left( x\right) = F\left( x\right) - G\left( x\right) ,\eta = \sup \left| {\Delta \left( x\right) }\right| ,{\Delta }_{L} = \Delta * {H}_{L} \), and \( {\eta }_{L} = \sup \left| ... | Proof. \( \Delta \) goes to 0 at \( \pm \infty, G \) is continuous, and \( F \) is a d.f., so there is an \( {x}_{0} \) with \( \Delta \left( {x}_{0}\right) = \eta \) or \( \Delta \left( {{x}_{0} - }\right) = - \eta \) . By looking at the d.f.’s of (-1) times the r.v.’s in the second case, we can suppose without loss o... | Yes |
Lemma 3.4.11. Let \( {K}_{1} \) and \( {K}_{2} \) be d.f. with mean 0 whose ch.f. \( {\kappa }_{i} \) are integrable | Proof. Since the \( {\kappa }_{i} \) are integrable, the inversion formula, Theorem 3.3.4, implies that the density \( {k}_{i}\left( x\right) \) has\n\n\[ {k}_{i}\left( y\right) = {\left( 2\pi \right) }^{-1}\int {e}^{-{ity}}{\kappa }_{i}\left( t\right) {dt} \]\n\nSubtracting the last expression with \( i = 2 \) from th... | No |
Theorem 3.5.1. Let \( \varphi \left( t\right) = E{e}^{itX} \) . There are only three possibilities.\n\n(i) \( \left| {\varphi \left( t\right) }\right| < 1 \) for all \( t \neq 0 \) .\n\n(ii) There is a \( \lambda > 0 \) so that \( \left| {\varphi \left( \lambda \right) }\right| = 1 \) and \( \left| {\varphi \left( t\ri... | Proof. We begin with (ii). It suffices to show that \( \left| {\varphi \left( t\right) }\right| = 1 \) if and only if \( P(X \in \) \( b + \left( {{2\pi }/t}\right) \mathbf{Z}) = 1 \) for some \( b \) . First, if \( P\left( {X \in b + \left( {{2\pi }/t}\right) \mathbf{Z}}\right) = 1 \) then\n\n\[ \varphi \left( t\right... | Yes |
Theorem 3.5.2. Under the hypotheses above, as \( n \rightarrow \infty \)\n\n\[ \mathop{\sup }\limits_{{x \in {\mathcal{L}}_{n}}}\left| {\frac{{n}^{1/2}}{h}{p}_{n}\left( x\right) - n\left( x\right) }\right| \rightarrow 0 \] | Proof. Let \( Y \) be a random variable with \( P\left( {Y \in a + \theta \mathbf{Z}}\right) = 1 \) and \( \psi \left( t\right) = E\exp \left( {itY}\right) \) . It follows from part (iii) of Exercise 3.3.2 that\n\n\[ P\left( {Y = x}\right) = \frac{1}{{2\pi }/\theta }{\int }_{-\pi /\theta }^{\pi /\theta }{e}^{-{itx}}\ps... | Yes |
Theorem 3.6.1. For each \( n \) let \( {X}_{n, m},1 \leq m \leq n \) be independent random variables with \( P\left( {{X}_{n, m} = 1}\right) = {p}_{n, m}, P\left( {{X}_{n, m} = 0}\right) = 1 - {p}_{n, m} \) . Suppose\n\n(i) \( \mathop{\sum }\limits_{{m = 1}}^{n}{p}_{n, m} \rightarrow \lambda \in \left( {0,\infty }\righ... | First proof. Let \( {\varphi }_{n, m}\left( t\right) = E\left( {\exp \left( {{it}{X}_{n, m}}\right) }\right) = \left( {1 - {p}_{n, m}}\right) + {p}_{n, m}{e}^{it} \) and let \( {S}_{n} = \) \( {X}_{n,1} + \cdots + {X}_{n, n} \) . Then\n\n\[ E\exp \left( {{it}{S}_{n}}\right) = \mathop{\prod }\limits_{{m = 1}}^{n}\left( ... | Yes |
In a calculus class with 400 students, the number of students who have their birthday on the day of the final exam has approximately a Poisson distribution with mean \( {400}/{365} = {1.096} \) . This means that the probability no one was born on that date is about \( {e}^{-{1.096}} = {0.334} \) . | Similar reasoning shows that the number of babies born on a given day or the number of people who arrive at a bank between 1:15 and 1:30 should have a Poisson distribution. | No |
Lemma 3.6.2. If \( {\mu }_{1} \times {\mu }_{2} \) denotes the product measure on \( \mathbf{Z} \times \mathbf{Z} \) that has \( \left( {{\mu }_{1} \times }\right. \) \( \left. {\mu }_{2}\right) \left( {x, y}\right) = {\mu }_{1}\left( x\right) {\mu }_{2}\left( y\right) \) then\n\n\[ \begin{Vmatrix}{{\mu }_{1} \times {\... | Proof. \( 2\begin{Vmatrix}{{\mu }_{1} \times {\mu }_{2} - {\nu }_{1} \times {\nu }_{2}}\end{Vmatrix} = \mathop{\sum }\limits_{{x, y}}\left| {{\mu }_{1}\left( x\right) {\mu }_{2}\left( y\right) - {\nu }_{1}\left( x\right) {\nu }_{2}\left( y\right) }\right| \)\n\n\[ \leq \mathop{\sum }\limits_{{x, y}}\left| {{\mu }_{1}\l... | Yes |
Lemma 3.6.3. If \( {\mu }_{1} * {\mu }_{2} \) denotes the convolution of \( {\mu }_{1} \) and \( {\mu }_{2} \), that is,\n\n\[ \n{\mu }_{1} * {\mu }_{2}\left( x\right) = \mathop{\sum }\limits_{y}{\mu }_{1}\left( {x - y}\right) {\mu }_{2}\left( y\right) \n\]\n\nthen \( \begin{Vmatrix}{{\mu }_{1} * {\mu }_{2} - {\nu }_{1... | Proof. \( 2\begin{Vmatrix}{{\mu }_{1} * {\mu }_{2} - {\nu }_{1} * {\nu }_{2}}\end{Vmatrix} = \mathop{\sum }\limits_{x}\left| {\mathop{\sum }\limits_{y}{\mu }_{1}\left( {x - y}\right) {\mu }_{2}\left( y\right) - \mathop{\sum }\limits_{y}{\nu }_{1}\left( {x - y}\right) {\nu }_{2}\left( y\right) }\right| \)\n\n\[ \n\leq \... | Yes |
Lemma 3.6.4. Let \( \mu \) be the measure with \( \mu \left( 1\right) = p \) and \( \mu \left( 0\right) = 1 - p \) . Let \( \nu \) be a Poisson distribution with mean \( p \) . Then \( \parallel \mu - \nu \parallel \leq {p}^{2} \) . | Proof. \( 2\parallel \mu - \nu \parallel = \left| {\mu \left( 0\right) - \nu \left( 0\right) }\right| + \left| {\mu \left( 1\right) - \nu \left( 1\right) }\right| + \mathop{\sum }\limits_{{n \geq 2}}\nu \left( n\right) \n\n\[ \n= \left| {1 - p - {e}^{-p}}\right| + \left| {p - p{e}^{-p}}\right| + 1 - {e}^{-p}\left( {1 +... | Yes |
Let \( \pi \) be a random permutation of \( \{ 1,2,\ldots, n\} \), let \( {X}_{n, m} = 1 \) if \( m \) is a fixed point ( 0 otherwise), and let \( {S}_{n} = {X}_{n,1} + \cdots + {X}_{n, n} \) be the number of fixed points. We want to compute \( P\left( {{S}_{n} = 0}\right) \). | Let \( {A}_{n, m} = \left\{ {{X}_{n, m} = 1}\right\} \). The inclusion-exclusion formula implies\n\n\[ P\left( {{ \cup }_{m = 1}^{n}{A}_{m}}\right) = \mathop{\sum }\limits_{m}P\left( {A}_{m}\right) - \mathop{\sum }\limits_{{\ell < m}}P\left( {{A}_{\ell } \cap {A}_{m}}\right) + \mathop{\sum }\limits_{{k < \ell < m}}P\le... | Yes |
Coupon collector’s problem. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. uniform on \( \{ 1,2,\ldots, n\} \) and \( {T}_{n} = \inf \left\{ {m : \left\{ {{X}_{1},\ldots {X}_{m}}\right\} = \{ 1,2,\ldots, n\} }\right\} \) . Since \( {T}_{n} \leq m \) if and only if \( m \) balls fill up all \( n \) boxes, it follows from Th... | Proof. If \( r = n\log n + {nx} \) then \( n{e}^{-r/n} \rightarrow {e}^{-x} \). | No |
Theorem 3.6.6. Let \( {X}_{n, m},1 \leq m \leq n \) be independent nonnegative integer valued random variables with \( P\left( {{X}_{n, m} = 1}\right) = {p}_{n, m}, P\left( {{X}_{n, m} \geq 2}\right) = {\epsilon }_{n, m} \) . (i) \( \mathop{\sum }\limits_{{m = 1}}^{n}{p}_{n, m} \rightarrow \lambda \in \left( {0,\infty ... | Proof. Let \( {X}_{n, m}^{\prime } = 1 \) if \( {X}_{n, m} = 1 \), and 0 otherwise. Let \( {S}_{n}^{\prime } = {X}_{n,1}^{\prime } + \cdots + {X}_{n, n}^{\prime } \) . (i)-(ii) and Theorem 3.6.1 imply \( {S}_{n}^{\prime } \Rightarrow Z \) ,(iii) tells us \( P\left( {{S}_{n} \neq {S}_{n}^{\prime }}\right) \rightarrow 0 ... | No |
Theorem 3.6.7. If (i)-(iv) hold then \( N\left( {0, t}\right) \) has a Poisson distribution with mean \( {\lambda t} \) . | Proof. Let \( {X}_{n, m} = N\left( {\left( {m - 1}\right) t/n,{mt}/n}\right) \) for \( 1 \leq m \leq n \) and apply Theorem 3.6.6. | No |
A Poisson process on a measure space \( \left( {S,\mathcal{S},\mu }\right) \) is a random map \( m : \mathcal{S} \rightarrow \{ 0,1,\ldots \} \) that for each \( \omega \) is a measure on \( \mathcal{S} \) and has the following property: if \( {A}_{1},\ldots ,{A}_{n} \) are disjoint sets with \( \mu \left( {A}_{i}\righ... | Exercise 3.6.12 implies that if \( \mu \left( S\right) < \infty \) we can construct \( m \) by the following recipe: let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. elements of \( S \) with distribution \( \nu \left( \cdot \right) = \mu \left( \cdot \right) /\mu \left( S\right) \), let \( N \) be an independent Poisson rand... | No |
Lemma 3.7.1. If \( {h}_{n}\left( \epsilon \right) \rightarrow g\left( \epsilon \right) \) for each \( \epsilon > 0 \) and \( g\left( \epsilon \right) \rightarrow g\left( 0\right) \) as \( \epsilon \rightarrow 0 \) then we can pick \( {\epsilon }_{n} \rightarrow 0 \) so that \( {h}_{n}\left( {\epsilon }_{n}\right) \righ... | Proof. Let \( {N}_{m} \) be chosen so that \( \left| {{h}_{n}\left( {1/m}\right) - g\left( {1/m}\right) }\right| \leq 1/m \) for \( n \geq {N}_{m} \) and \( m \rightarrow {N}_{m} \) is increasing. Let \( {\epsilon }_{n} = 1/m \) for \( {N}_{m} \leq n < {N}_{m + 1} \) and \( = 1 \) for \( n < {N}_{1} \) . When \( {N}_{m... | Yes |
Theorem 3.7.2. Suppose \( {X}_{1},{X}_{2},\ldots \) are i.i.d. with a distribution that satisfies\n\n(i) \( \mathop{\lim }\limits_{{x \rightarrow \infty }}P\left( {{X}_{1} > x}\right) /P\left( {\left| {X}_{1}\right| > x}\right) = \theta \in \left\lbrack {0,1}\right\rbrack \)\n\n(ii) \( P\left( {\left| {X}_{1}\right| > ... | Proof. It is not hard to see that (ii) implies\n\n\[ \n{nP}\left( {\left| {X}_{1}\right| > {a}_{n}}\right) \rightarrow 1 \]\n\n(3.7.6)\n\nTo prove this, note that \( {nP}\left( {\left| {X}_{1}\right| > {a}_{n}}\right) \leq 1 \) and let \( \epsilon > 0 \) . Taking \( x = {a}_{n}/\left( {1 + \epsilon }\right) \) and \( t... | Yes |
Lemma 3.7.3. For any \( \delta > 0 \) there is \( C \) so that for all \( t \geq {t}_{0} \) and \( y \leq 1 \)\n\n\[ P\left( {\left| {X}_{1}\right| > {yt}}\right) /P\left( {\left| {X}_{1}\right| > t}\right) \leq C{y}^{-\alpha - \delta } \] | Proof. (ii) implies that as \( t \rightarrow \infty \)\n\n\[ P\left( {\left| {X}_{1}\right| > t/2}\right) /P\left( {\left| {X}_{1}\right| > t}\right) \rightarrow {2}^{\alpha } \]\n\nso for \( t \geq {t}_{0} \) we have\n\n\[ P\left( {\left| {X}_{1}\right| > t/2}\right) /P\left( {\left| {X}_{1}\right| > t}\right) \leq {2... | Yes |
Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with a density that is symmetric about 0, and continuous and positive at 0 . We claim that\n\n\[ \frac{1}{n}\left( {\frac{1}{{X}_{1}} + \cdots + \frac{1}{{X}_{n}}}\right) \Rightarrow \text{a Cauchy distribution}\left( {\alpha = 1,\kappa = 0}\right) \] | To verify this, note that\n\n\[ P\left( {1/{X}_{i} > x}\right) = P\left( {0 < {X}_{i} < {x}^{-1}}\right) = {\int }_{0}^{{x}^{-1}}f\left( y\right) {dy} \sim f\left( 0\right) /x \] \nas \( x \rightarrow \infty \) . A similar calculation shows \( P\left( {1/{X}_{i} < - x}\right) \sim f\left( 0\right) /x \) so in (i) in Th... | Yes |
Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( P\left( {{X}_{i} = 1}\right) = P\left( {{X}_{i} = - 1}\right) = 1/2 \), let \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \), and let \( \tau = \inf \left\{ {n \geq 1 : {S}_{n} = 1}\right\} \). Let \( {\tau }_{1},{\tau }_{2},\ldots \) be independent with the same distributio... | To prove the claim, note that in (i) in Theorem 3.7.2 holds with \( \theta = 1 \) and (ii) holds with \( \alpha = 1/2 \). The scaling constant \( {a}_{n} \sim C{n}^{2} \). Since \( \alpha < 1 \), Exercise 3.7.2 implies the centering constant is unnecessary. | No |
Assume \( n \) objects \( {X}_{n,1},\ldots ,{X}_{n, n} \) are placed independently and at random in \( \left\lbrack {-n, n}\right\rbrack \) . Let\n\n\[ \n{F}_{n} = \mathop{\sum }\limits_{{m = 1}}^{n}\operatorname{sgn}\left( {X}_{n, m}\right) /{\left| {X}_{n, m}\right| }^{p}\n\]\n\nbe the net force exerted on 0 . We wil... | To do this, it is convenient to let \( {X}_{n, m} = n{Y}_{m} \) where the \( {Y}_{i} \) are i.i.d. on \( \left\lbrack {-1,1}\right\rbrack \) . Then\n\n\[ \n{F}_{n} = {n}^{-p}\mathop{\sum }\limits_{{m = 1}}^{n}\operatorname{sgn}\left( {Y}_{m}\right) /{\left| {Y}_{m}\right| }^{p}\n\]\n\nLetting \( {Z}_{m} = \operatorname... | Yes |
In the examples above, we have had \( {b}_{n} = 0 \) . To get a feel for the centering constants consider \( {X}_{1},{X}_{2},\ldots \) i.i.d. with\n\n\[ P\left( {{X}_{i} > x}\right) = \theta {x}^{-\alpha }\;P\left( {{X}_{i} < - x}\right) = \left( {1 - \theta }\right) {x}^{-\alpha } \]\n\nwhere \( 0 < \alpha < 2 \) . In... | When \( \alpha < 1 \) the centering is the same size as the scaling and can be ignored. When \( \alpha > 1,{b}_{n} \sim {n\mu } \) where \( \mu = E{X}_{i} \) . | Yes |
Theorem 3.7.4. \( Y \) is the limit of \( \left( {{X}_{1} + \cdots + {X}_{k} - {b}_{k}}\right) /{a}_{k} \) for some i.i.d. sequence \( {X}_{i} \) if and only if \( Y \) has a stable law. | Proof. If \( Y \) has a stable law we can take \( {X}_{1},{X}_{2},\ldots \) i.i.d. with distribution \( Y \) . To go the other way, let\n\n\[ \n{Z}_{n} = \left( {{X}_{1} + \cdots + {X}_{n} - {b}_{n}}\right) /{a}_{n} \]\n\nand \( {S}_{n}^{j} = {X}_{\left( {j - 1}\right) n + 1} + \cdots + {X}_{jn} \) . A little arithmeti... | Yes |
The Holtsmark distribution. \( \left( {\alpha = 3/2,\kappa = 0}\right) \) . Suppose stars are distributed in space according to a Poisson process with density \( t \) and their masses are i.i.d. Let \( {X}_{t} \) be the \( x \) -component of the gravitational force at 0 when the density is \( t \) . A change of density... | If we imagine thinning the Poisson process by rolling an \( n \) -sided die, then Exercise 3.6.12 implies\n\n\[ \n{X}_{t}\overset{d}{ = }{X}_{t/n}^{1} + \cdots + {X}_{t/n}^{n} \n\]\n\nwhere the random variables on the right-hand side are independent and have the same distribution as \( {X}_{t/n} \) . It follows from Th... | No |
Theorem 3.8.1. \( Z \) is a limit of sums of type \( \left( *\right) \) if and only if \( Z \) has an infinitely divisible distribution. | Proof. As remarked above, we only have to prove necessity. Write\n\n\[ \n{S}_{2n} = \left( {{X}_{{2n},1} + \cdots + {X}_{{2n}, n}}\right) + \left( {{X}_{{2n}, n + 1} + \cdots + {X}_{{2n},{2n}}}\right) \equiv {Y}_{n} + {Y}_{n}^{\prime } \n\]\n\nThe random variables \( {Y}_{n} \) and \( {Y}_{n}^{\prime } \) are independe... | Yes |
Example 3.8.4. Compound Poisson distribution. Let \( {\xi }_{1},{\xi }_{2},\ldots \) be i.i.d. and \( N\left( \lambda \right) \) be an independent Poisson r.v. with mean \( \lambda \) . Then \( Z = {\xi }_{1} + \cdots + {\xi }_{N\left( \lambda \right) } \) has an infinitely divisible distribution. | For developments below, we would like to observe that if \( \varphi \left( t\right) = E\exp \left( {{it}{\xi }_{i}}\right) \) then\n\n\[ E\exp \left( {itZ}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{e}^{-\lambda }\frac{{\lambda }^{n}}{n!}\varphi {\left( t\right) }^{n} = \exp \left( {-\lambda \left( {1 - \varphi... | No |
Theorem 3.8.2. Lévy-Khinchin Theorem. Z has an infinitely divisible distribution if and only if its characteristic function has\n\n\[ \log \varphi \left( t\right) = {ict} - \frac{{\sigma }^{2}{t}^{2}}{2} + \int \left( {{e}^{itx} - 1 - \frac{itx}{1 + {x}^{2}}}\right) \mu \left( {dx}\right) \]\n\nwhere \( \mu \) is a mea... | For a proof, see Breiman (1968), Section 9.5., or Feller II (1971), Section XVII.2. \( \mu \) is called the Lévy measure of the distribution. | No |
Theorem 3.8.3. Kolmogorov’s Theorem. Z has an infinitely divisible distribution with mean 0 and finite variance if and only if its ch.f. has\n\n\[ \log \varphi \left( t\right) = \int \left( {{e}^{itx} - 1 - {itx}}\right) {x}^{-2}\nu \left( {dx}\right) \] | Here the integrand is \( - {t}^{2}/2 \) at \( 0,\nu \) is called the canonical measure and \( \operatorname{var}\left( Z\right) = \) \( \nu \left( \mathbf{R}\right) \) .\n\nTo explain the formula, note that if \( {Z}_{\lambda } \) has a Poisson distribution with mean \( \lambda \n\n\[ E\exp \left( {{itx}\left( {{Z}_{\l... | No |
Theorem 3.9.1. The following statements are equivalent to \( {X}_{n} \Rightarrow {X}_{\infty } \) . | Proof. We will begin by showing that (i)-(vi) are equivalent.\n\n(i) implies (ii): Trivial.\n\n(ii) implies (iii): Let \( \rho \left( {x, K}\right) = \inf \{ \rho \left( {x, y}\right) : y \in K\} ,{\varphi }_{j}\left( r\right) = {\left( 1 - jr\right) }^{ + } \), and \( {f}_{j}\left( x\right) = \) \( {\varphi }_{j}\left... | Yes |
Theorem 3.9.2. If \( {\mu }_{n} \) is tight, then there is a weakly convergent subsequence. | Proof. Let \( {F}_{n} \) be the associated distribution functions, and let \( {q}_{1},{q}_{2},\ldots \) be an enumeration of \( {\mathbf{Q}}^{d} = \) the points in \( {\mathbf{R}}^{d} \) with rational coordinates. By a diagonal argument like the one in the proof of Theorem 3.2.6, we can pick a subsequence so that \( {F... | No |
Theorem 3.9.3. Inversion formula. If \( A = \left\lbrack {{a}_{1},{b}_{1}}\right\rbrack \times \ldots \times \left\lbrack {{a}_{d},{b}_{d}}\right\rbrack \) with \( \mu \left( {\partial A}\right) = 0 \) then\n\n\[ \mu \left( A\right) = \mathop{\lim }\limits_{{T \rightarrow \infty }}{\left( 2\pi \right) }^{-d}{\int }_{{\... | Proof. Fubini's theorem implies\n\n\[ {\int }_{{\left\lbrack -T, T\right\rbrack }^{d}}\int \mathop{\prod }\limits_{{j = 1}}^{d}{\psi }_{j}\left( {t}_{j}\right) \exp \left( {i{t}_{j}{x}_{j}}\right) \mu \left( {dx}\right) {dt} \]\n\n\[ = \int \mathop{\prod }\limits_{{j = 1}}^{d}{\int }_{-T}^{T}{\psi }_{j}\left( {t}_{j}\r... | Yes |
Theorem 3.9.4. Convergence theorem. Let \( {X}_{n},1 \leq n \leq \infty \) be random vectors with ch.f. \( {\varphi }_{n} \) . A necessary and sufficient condition for \( {X}_{n} \Rightarrow {X}_{\infty } \) is that \( {\varphi }_{n}\left( t\right) \rightarrow \) \( {\varphi }_{\infty }\left( t\right) \) . | Proof. \( \exp \left( {{it} \cdot x}\right) \) is bounded and continuous, so if \( {X}_{n} \Rightarrow {X}_{\infty } \) then \( {\varphi }_{n}\left( t\right) \rightarrow {\varphi }_{\infty }\left( t\right) \) . To prove the other direction it suffices, as in the proof of Theorem 3.3.6, to prove that the sequence is tig... | Yes |
Theorem 3.9.5. Cramér-Wold device. A sufficient condition for \( {X}_{n} \Rightarrow {X}_{\infty } \) is that \( \theta \cdot {X}_{n} \Rightarrow \theta \cdot {X}_{\infty } \) for all \( \theta \in {\mathbf{R}}^{d} \) . | Proof. The indicated condition implies \( E\exp \left( {{i\theta } \cdot {X}_{n}}\right) \rightarrow E\exp \left( {{i\theta } \cdot {X}_{\infty }}\right) \) for all \( \theta \in \) \( {\mathbf{R}}^{d} \) . | Yes |
Theorem 3.9.6. The central limit theorem in \( {\mathbf{R}}^{d} \) . Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. random vectors with \( E{X}_{n} = \mu \), and finite covariances\n\n\[ \n{\Gamma }_{ij} = E\left( {\left( {{X}_{n, i} - {\mu }_{i}}\right) \left( {{X}_{n, j} - {\mu }_{j}}\right) }\right) \n\]\n\nIf \( {S}_{n... | Proof. By considering \( {X}_{n}^{\prime } = {X}_{n} - \mu \), we can suppose without loss of generality that \( \mu = 0 \) . Let \( \theta \in {\mathbf{R}}^{d}.\theta \cdot {X}_{n} \) is a random variable with mean 0 and variance\n\n\[ \nE{\left( \mathop{\sum }\limits_{i}{\theta }_{i}{X}_{n, i}\right) }^{2} = \mathop{... | Yes |
Theorem 4.1.2. For a random walk on \( \\mathbf{R} \), there are only four possibilities, one of which has probability one. | Proof. Theorem 4.1.1 implies \( \\lim \\sup {S}_{n} \) is a constant \( c \\in \\left\\lbrack {-\\infty ,\\infty }\\right\\rbrack \) . Let \( {S}_{n}^{\\prime } = {S}_{n + 1} - \) \( {X}_{1} \) . Since \( {S}_{n}^{\\prime } \) has the same distribution as \( {S}_{n} \), it follows that \( c = c - {X}_{1} \) . If \( c \... | Yes |
Theorem 4.1.3. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d., \( {\mathcal{F}}_{n} = \sigma \left( {{X}_{1},\ldots ,{X}_{n}}\right) \) and \( N \) be a stopping time with \( P\left( {N < \infty }\right) > 0 \) . Conditional on \( \{ N < \infty \} ,\left\{ {{X}_{N + n}, n \geq 1}\right\} \) is independent of \( {\mathcal{F... | Proof. By Theorem A.1.5 it is enough to show that if \( A \in {\mathcal{F}}_{N} \) and \( {B}_{j} \in \mathcal{S} \) for \( 1 \leq j \leq k \) then\n\n\[ P\left( {A, N < \infty ,{X}_{N + j} \in {B}_{j},1 \leq j \leq k}\right) = P\left( {A\cap \{ N < \infty \} }\right) \mathop{\prod }\limits_{{j = 1}}^{k}\mu \left( {B}_... | Yes |
For a concrete example of the use of \( \theta \), suppose \( S = {\mathbf{R}}^{d} \) and let\n\n\[ \tau \left( \omega \right) = \inf \left\{ {n : {\omega }_{1} + \cdots + {\omega }_{n} = 0}\right\} \]\n\nwhere \( \inf \varnothing = \infty \), and we set \( \tau \left( \Delta \right) = \infty \) . If we let \( {\tau }_... | Proof. We will prove this by induction. The result is trivial when \( n = 1 \) . Suppose now that it is valid for \( n - 1 \) . Applying Theorem 4.1.3 to \( N = {T}_{n - 1} \), we see that conditional on \( {T}_{n - 1} < \infty, T\left( {\theta }^{{T}_{n - 1}}\right) < \infty \) has the same probability as \( T < \inft... | Yes |
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