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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 | \[ {K}_{1}\left( x\right) - {K}_{2}\left( x\right) = {\left( 2\pi \right) }^{-1}\int - {e}^{-{itx}}\frac{{\kappa }_{1}\left( t\right) - {\kappa }_{2}\left( t\right) }{it}{dt} \] Proof. Since the \( {\kappa }_{i} \) are integrable, the inversion formula, Theorem 3.3.4, implies that the density \( {k}_{i}\left( x\right) ... | 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 \ma... | 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\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 \(... | 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 wi... | 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} = \operatornam... | 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 \( \mu = E{X}_{i} \). | No |
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\[
{Z}_{n} = \left( {{X}_{1} + \cdots + {X}_{n} - {b}_{n}}\right) /{a}_{n}
\]\n\nand \( {S}_{n}^{j} = {X}_{\left( {j - 1}\right) n + 1} + \cdots + {X}_{jn} \) . A little arithmetic ... | Yes |
Example 3.7.5. 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 \( t \) . ... | \[ {X}_{t}\\overset{d}{ = }{t}^{3/2}{X}_{1} \] \n\n\( \\left( {3.7.13}\\right) \) \n\nIf we imagine thinning the Poisson process by rolling an \( n \) -sided die, then Exercise 3.6.12 implies \n\n\[ {X}_{t}\\overset{d}{ = }{X}_{t/n}^{1} + \\cdots + {X}_{t/n}^{n} \] \n\nwhere the random variables on the right-hand side ... | 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. (Let \( {X}_{n, j}{ =... | \[ E\exp \left( {itZ}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{e}^{-\lambda }\frac{{\lambda }^{n}}{n!}\varphi {\left( t\right) }^{n} = \exp \left( {-\lambda \left( {1 - \varphi \left( t\right) }\right) }\right) \] | Yes |
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. Comparing with (3.8.2) and recalling the proof of Theorem 3.7.2 suggests the following interpretation of \( \mu \) : If \( {\sigma }^{2} = 0 \) then \( Z \) can be built up by ma... | 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\[ \log \varphi \left( t\right) = \int \left( {{e}^{itx} - 1 - {itx}}\right) {x}^{-2}\nu \left( {dx}\right) \]\n\nHere the integrand is \( - {t}^{2}/2 \) at \( 0,\nu \) i... | To explain the formula, note that if \( {Z}_{\lambda } \) has a Poisson distribution with mean \( \lambda \n\n\[ E\exp \left( {{itx}\left( {{Z}_{\lambda } - \lambda }\right) }\right) = \exp \left( {\lambda \left( {{e}^{itx} - 1 - {itx}}\right) }\right) \]\n\nso the measure for \( Z = x\left( {{Z}_{\lambda } - \lambda }... | 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.\n\n(i) \( {S}_{n} = 0 \) for all \( n \) .\n\n(ii) \( {S}_{n} \\rightarrow \\infty \) .\n\n(iii) \( {S}_{n} \\rightarrow - \\infty \) .\n\n(iv) \( - \\infty = \\lim \\inf {S}_{n} < \\lim \\sup {S}... | 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 |
If we have any stopping time \( T \), we can define its iterates by \( {T}_{0} = 0 \) and\n\n\[ {T}_{n}\left( \omega \right) = {T}_{n - 1}\left( \omega \right) + T\left( {{\theta }^{{T}_{n - 1}}\omega }\right) \;\text{ for }n \geq 1 \]\n\nIf we assume \( P = \mu \times \mu \times \ldots \) then\n\n\[ P\left( {{T}_{n} <... | 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 |
Theorem 4.1.5. Wald’s equation. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E\left| {X}_{i}\right| < \infty \) . If \( N \) is a stopping time with \( {EN} < \infty \) then \( E{S}_{N} = E{X}_{1}{EN} \) . | Proof. First suppose the \( {X}_{i} \geq 0 \).\n\n\[ E{S}_{N} = \int {S}_{N}{dP} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\int {S}_{n}{1}_{\{ N = n\} }{dP} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\mathop{\sum }\limits_{{m = 1}}^{n}\int {X}_{m}{1}_{\{ N = n\} }{dP} \]\n\nSince the \( {X}_{i} \geq 0 \), we can inter... | Yes |
Simple random walk. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( P\left( {{X}_{i} = 1}\right) = \) \( 1/2 \) and \( P\left( {{X}_{i} = - 1}\right) = 1/2 \) . Let \( a < 0 < b \) be integers and let \( N = \inf \left\{ {n : {S}_{n} \notin }\right. \) \( \left( {a, b}\right) \} \) . To apply Theorem 4.1.5, we have ... | To do this, we observe that if \( x \in \left( {a, b}\right) \), then\n\n\[ P\left( {x + {S}_{b - a} \notin \left( {a, b}\right) }\right) \geq {2}^{-\left( {b - a}\right) } \]\n\nsince \( b - a \) steps of size +1 in a row will take us out of the interval. Iterating the last inequality, it follows that\n\n\[ P\left( {N... | Yes |
Theorem 4.1.6. Wald’s second equation. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E{X}_{n} = 0 \) and \( E{X}_{n}^{2} = {\sigma }^{2} < \infty \) . If \( T \) is a stopping time with \( {ET} < \infty \) then \( E{S}_{T}^{2} = {\sigma }^{2}{ET} \) . | Proof. Using the definitions and then taking expected value\n\n\[ \n{S}_{T \land n}^{2} = {S}_{T \land \left( {n - 1}\right) }^{2} + \left( {2{X}_{n}{S}_{n - 1} + {X}_{n}^{2}}\right) {1}_{\left( T \geq n\right) } \n\]\n\n\[ \nE{S}_{T \land n}^{2} = E{S}_{T \land \left( {n - 1}\right) }^{2} + {\sigma }^{2}P\left( {T \ge... | Yes |
Example 4.1.6. Simple random walk, II. Continuing Example 4.1.5 we investigate \( N = \inf \left\{ {{S}_{n} \notin \left( {a, b}\right) }\right\} \) . We have shown that \( {EN} < \infty \) . | Since \( {\sigma }^{2} = 1 \) it follows from Theorem 4.1.6 and (4.1.2) that\n\n\[ \n{EN} = E{S}_{N}^{2} = {a}^{2}\frac{b}{b - a} + {b}^{2}\frac{-a}{b - a} = - {ab} \n\] \n\nIf \( b = L \) and \( a = - L,{EN} = {L}^{2} \). | Yes |
Theorem 4.1.7. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with \( E{X}_{n} = 0 \) and \( E{X}_{n}^{2} = 1 \), and let \( {T}_{c} = \inf \left\{ {n \geq 1 : \left| {S}_{n}\right| > c{n}^{1/2}}\right\} . \n\n\[ \nE{T}_{c}\;\left\{ \begin{array}{ll} < \infty & \text{ for }c < 1 \\ = \infty & \text{ for }c \geq 1 \end{arra... | Proof. One half of this is easy. If \( E{T}_{c} < \infty \) then the previous exercise implies \( E{T}_{c} = E\left( {S}_{{T}_{c}}^{2}\right) > {c}^{2}E{T}_{c} \) a contradiction if \( c \geq 1 \) . To prove the other direction, we let \( \tau = {T}_{c} \land n \) and observe \( {S}_{\tau - 1}^{2} \leq {c}^{2}\left( {\... | Yes |
Lemma 4.1.8. If \( T \) is a stopping time with \( {ET} = \infty \) then\n\n\[ E{X}_{T \land n}^{2}/E\left( {T \land n}\right) \rightarrow 0 \] | Proof. We begin by writing\n\n\[ E\left( {X}_{T \land n}^{2}\right) = E\left( {{X}_{T \land n}^{2};{X}_{T \land n}^{2} \leq \epsilon \left( {T \land n}\right) }\right) + \mathop{\sum }\limits_{{j = 1}}^{n}E\left( {{X}_{j}^{2};T \land n = j,{X}_{j}^{2} > {\epsilon j}}\right) \]\n\nThe first term is \( \leq {\epsilon E}\... | Yes |
Theorem 4.2.1. The set \( \mathcal{V} \) of recurrent values is either \( \varnothing \) or a closed subgroup of \( {\mathbf{R}}^{d} \) . In the second case, \( \mathcal{V} = \mathcal{U} \), the set of possible values. | Proof. Suppose \( \mathcal{V} \neq \varnothing \) . It is clear that \( {\mathcal{V}}^{c} \) is open, so \( \mathcal{V} \) is closed. To prove that \( \mathcal{V} \) is a group, we will first show that\n\n\( \left( *\right) \) if \( x \in \mathcal{U} \) and \( y \in \mathcal{V} \) then \( y - x \in \mathcal{V} \) .\n\n... | Yes |
Example 4.2.1. Simple random walk on \( {\mathbf{Z}}^{d} \) . | \[ P\left( {{X}_{i} = {e}_{j}}\right) = P\left( {{X}_{i} = - {e}_{j}}\right) = 1/{2d} \] for each of the \( d \) unit vectors \( {e}_{j} \). To analyze this case, we begin with a result that is valid for any random walk. Let \( {\tau }_{0} = 0 \) and \( {\tau }_{n} = \inf \left\{ {m > {\tau }_{n - 1} : {S}_{m} = 0}\rig... | Yes |
Theorem 4.2.2. For any random walk, the following are equivalent:\n\n(i) \( P\left( {{\tau }_{1} < \infty }\right) = 1 \) ,(ii) \( P\left( {{S}_{m} = 0\text{i.o.}}\right) = 1 \), and (iii) \( \mathop{\sum }\limits_{{m = 0}}^{\infty }P\left( {{S}_{m} = 0}\right) = \infty \) . | Proof. If \( P\left( {{\tau }_{1} < \infty }\right) = 1 \), then \( P\left( {{\tau }_{n} < \infty }\right) = 1 \) for all \( n \) and \( P\left( {{S}_{m} = 0\text{i.o.}}\right) = 1 \) . Let\n\n\[ V = \mathop{\sum }\limits_{{m = 0}}^{\infty }{1}_{\left( {S}_{m} = 0\right) } = \mathop{\sum }\limits_{{n = 0}}^{\infty }{1}... | Yes |
Theorem 4.2.3. Simple random walk is recurrent in \( d \leq 2 \) and transient in \( d \geq 3 \) . | Proof. Let \( {\rho }_{d}\left( m\right) = P\left( {{S}_{m} = 0}\right) .{\rho }_{d}\left( m\right) \) is 0 if \( m \) is odd. From Theorem 3.1.3, we get \( {\rho }_{1}\left( {2n}\right) \sim {\left( \pi n\right) }^{-1/2} \) as \( n \rightarrow \infty \) . This and Theorem 4.2.2 gives the result in one dimension. Our n... | Yes |
Lemma 4.2.4. If \( \mathop{\sum }\limits_{{n = 1}}^{\infty }P\left( {\begin{Vmatrix}{S}_{n}\end{Vmatrix} < \epsilon }\right) < \infty \) then \( P\left( {\begin{Vmatrix}{S}_{n}\end{Vmatrix} < \epsilon \text{i.o.}}\right) = 0 \) . | Proof. The first conclusion follows from the Borel-Cantelli lemma. To prove the second, let \( F = {\left\{ \begin{Vmatrix}{S}_{n}\end{Vmatrix} < \epsilon \text{ i.o. }\right\} }^{c} \) . Breaking things down according to the last time \( \begin{Vmatrix}{S}_{n}\end{Vmatrix} < \epsilon \)\n\n\[ P\left( F\right) = \matho... | Yes |
Lemma 4.2.5. Let \( m \) be an integer \( \geq 2 \) . \[ \mathop{\sum }\limits_{{n = 0}}^{\infty }P\left( {\begin{Vmatrix}{S}_{n}\end{Vmatrix} < {m\epsilon }}\right) \leq {\left( 2m\right) }^{d}\mathop{\sum }\limits_{{n = 0}}^{\infty }P\left( {\begin{Vmatrix}{S}_{n}\end{Vmatrix} < \epsilon }\right) \] | Proof. We begin by observing \[ \mathop{\sum }\limits_{{n = 0}}^{\infty }P\left( {\begin{Vmatrix}{S}_{n}\end{Vmatrix} < {m\epsilon }}\right) \leq \mathop{\sum }\limits_{{n = 0}}^{\infty }\mathop{\sum }\limits_{k}P\left( {{S}_{n} \in {k\epsilon } + \lbrack 0,\epsilon {)}^{d}}\right) \] where the inner sum is over \( k \... | Yes |
Theorem 4.2.7. Chung-Fuchs theorem. Suppose \( d = 1 \) . If the weak law of large numbers holds in the form \( {S}_{n}/n \rightarrow 0 \) in probability, then \( {S}_{n} \) is recurrent. | Proof. Let \( {u}_{n}\left( x\right) = P\left( {\left| {S}_{n}\right| < x}\right) \) for \( x > 0 \) . Lemma 4.2.5 implies\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{u}_{n}\left( 1\right) \geq \frac{1}{2m}\mathop{\sum }\limits_{{n = 0}}^{\infty }{u}_{n}\left( m\right) \geq \frac{1}{2m}\mathop{\sum }\limits_{{n = 0... | Yes |
Theorem 4.2.8. If \( {S}_{n} \) is a random walk in \( {\mathbf{R}}^{2} \) and \( {S}_{n}/{n}^{1/2} \Rightarrow \) a nondegenerate normal distribution then \( {S}_{n} \) is recurrent. | Proof. Let \( u\left( {n, m}\right) = P\left( {\begin{Vmatrix}{S}_{n}\end{Vmatrix} < m}\right) \) . Lemma 4.2.5 implies\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }u\left( {n,1}\right) \geq {\left( 4{m}^{2}\right) }^{-1}\mathop{\sum }\limits_{{n = 0}}^{\infty }u\left( {n, m}\right) \]\n\nIf \( m/\sqrt{n} \rightarrow... | Yes |
Lemma 4.2.11. Parseval relation. Let \( \mu \) and \( \nu \) be probability measures on \( {\mathbf{R}}^{d} \) with ch.f.’s \( \varphi \) and \( \psi \) . | Proof. Since \( {e}^{{it} \cdot x} \) is bounded, Fubini’s theorem implies\n\n\[ \int \psi \left( t\right) \mu \left( {dt}\right) = \iint {e}^{itx}\nu \left( {dx}\right) \mu \left( {dt}\right) = \iint {e}^{itx}\mu \left( {dt}\right) \nu \left( {dx}\right) = \int \varphi \left( x\right) \nu \left( {dx}\right) \] | Yes |
Lemma 4.2.12. If \( \left| x\right| \leq \pi /3 \) then \( 1 - \cos x \geq {x}^{2}/4 \) . | Proof. It suffices to prove the result for \( x > 0 \) . If \( z \leq \pi /3 \) then \( \cos z \geq 1/2 \) ,\n\n\[ \n\sin y = {\int }_{0}^{y}\cos {zdz} \geq \frac{y}{2} \n\]\n\n\[ \n1 - \cos x = {\int }_{0}^{x}\sin {ydy} \geq {\int }_{0}^{x}\frac{y}{2}{dy} = \frac{{x}^{2}}{4} \n\]\n\nwhich proves the desired result. | Yes |
Theorem 4.3.1. Reflection principle. If \( x, y > 0 \) then the number of paths from \( \left( {0, x}\right) \) to \( \left( {n, y}\right) \) that are 0 at some time is equal to the number of paths from \( \left( {0, - x}\right) \) to \( \left( {n, y}\right) \) . | Proof. Suppose \( \left( {0,{s}_{0}}\right) ,\left( {1,{s}_{1}}\right) ,\ldots ,\left( {n,{s}_{n}}\right) \) is a path from \( \left( {0, x}\right) \) to \( \left( {n, y}\right) \) . Let \( K = \) \( \inf \left\{ {k : {s}_{k} = 0}\right\} \) . Let \( {s}_{k}^{\prime } = - {s}_{k} \) for \( k \leq K,{s}_{k}^{\prime } = ... | Yes |
Theorem 4.3.2. The Ballot Theorem. Suppose that in an election candidate \( A \) gets \( \alpha \) votes, and candidate \( B \) gets \( \beta \) votes where \( \beta < \alpha \) . The probability that throughout the counting \( A \) always leads \( B \) is \( \left( {\alpha - \beta }\right) /\left( {\alpha + \beta }\ri... | Proof. Let \( x = \alpha - \beta, n = \alpha + \beta \) . Clearly, there are as many such outcomes as there are paths from \( \left( {1,1}\right) \) to \( \left( {n, x}\right) \) that are never 0 . The reflection principle implies that the number of paths from \( \left( {1,1}\right) \) to \( \left( {n, x}\right) \) tha... | Yes |
Lemma 4.3.3. \( P\left( {{S}_{1} \neq 0,\ldots ,{S}_{2n} \neq 0}\right) = P\left( {{S}_{2n} = 0}\right) \) . | Proof. \( P\left( {{S}_{1} > 0,\ldots ,{S}_{2n} > 0}\right) = \mathop{\sum }\limits_{{r = 1}}^{\infty }P\left( {{S}_{1} > 0,\ldots ,{S}_{{2n} - 1} > 0,{S}_{2n} = {2r}}\right) \) . From the proof of Theorem 4.3.2, we see that the number of paths from \( \left( {0,0}\right) \) to \( \left( {{2n},{2r}}\right) \) that are ... | Yes |
Lemma 4.3.4. Let \( {u}_{2m} = P\left( {{S}_{2m} = 0}\right) \) . Then \( P\left( {{L}_{2n} = {2k}}\right) = {u}_{2k}{u}_{{2n} - {2k}} \) . | Proof. \( P\left( {{L}_{2n} = {2k}}\right) = P\left( {{S}_{2k} = 0,{S}_{{2k} + 1} \neq 0,\ldots ,{S}_{2n} \neq 0}\right) \), so the desired result follows from Lemma 4.3.3. | No |
Theorem 4.3.5. Arcsine law for the last visit to 0 . For \( 0 < a < b < 1 \) ,\n\n\[ P\left( {a \leq {L}_{2n}/{2n} \leq b}\right) \rightarrow {\int }_{a}^{b}{\pi }^{-1}{\left( x\left( 1 - x\right) \right) }^{-1/2}{dx} \] | Proof of Theorem 4.3.5. From the asymptotic formula for \( {u}_{2n} \), it follows that if \( k/n \rightarrow x \) then\n\n\[ {nP}\left( {{L}_{2n} = {2k}}\right) \rightarrow {\pi }^{-1}{\left( x\left( 1 - x\right) \right) }^{-1/2} \]\n\nTo get from this to the desired result, we let \( {2n}{a}_{n} = \) the smallest eve... | Yes |
Theorem 4.3.6. Arcsine law for time above 0. Let \( {\pi }_{2n} \) be the number of segments \( \left( {k - 1,{S}_{k - 1}}\right) \rightarrow \left( {k,{S}_{k}}\right) \) that lie above the axis (i.e., in \( \{ \left( {x, y}\right) : y \geq 0\} \) ), and let \( {u}_{m} = P\left( {{S}_{m} = 0}\right) . \n\n\[ \nP\left( ... | Proof. Let \( {\beta }_{{2k},{2n}} \) denote the probability of interest. We will prove \( {\beta }_{{2k},{2n}} = {u}_{2k}{u}_{{2n} - {2k}} \) by induction. When \( n = 1 \), it is clear that\n\n\[ \n{\beta }_{0,2} = {\beta }_{2,2} = 1/2 = {u}_{0}{u}_{2} \n\] \n\nFor a general \( n \), first suppose \( k = n \) . From ... | Yes |
Theorem 4.3.7. Let \( {\nu }_{n} = \left| \left\{ {k : 1 \leq k \leq n,{S}_{k} > 0}\right\} \right| \) . Then\n\n(i) \( P\left( {{\nu }_{n} = k}\right) = P\left( {{\nu }_{k} = k}\right) P\left( {{\nu }_{n - k} = 0}\right) \) | Proof. Taking things in reverse order, (iii) is an immediate consequence of (ii) and the proof of Theorem 4.3.5. Our next step is to show (ii) follows from (i) by induction. When \( n = 1 \), our assumptions imply \( P\left( {{\nu }_{1} = 0}\right) = 1/2 = {u}_{0}{u}_{2} \) . If \( n > 1 \) and \( 1 \leq k < n \) , the... | Yes |
Lemma 4.3.8. \( \\left( {{\\ell }_{n},{S}_{n}}\\right) \) and \( \\left( {n - {\\ell }_{n}^{\\prime },{S}_{n}}\\right) \) have the same distribution. | Proof. If we let \( {T}_{k} = {S}_{n} - {S}_{n - k} = {X}_{n} + \\cdots + {X}_{n - k + 1} \), then \( {T}_{k}0 \\leq k \\leq n \) has the same distribution as \( {S}_{k},0 \\leq k \\leq n \) . Clearly,\n\n\[ \n\\mathop{\\max }\\limits_{{0 \\leq k \\leq n}}{T}_{k} = {S}_{n} - \\mathop{\\min }\\limits_{{0 \\leq k \\leq n... | No |
Lemma 4.3.9. \( \left( {{\ell }_{n},{S}_{n}}\right) \) and \( \left( {{\nu }_{n},{S}_{n}}\right) \) have the same distribution. | Proof. When \( n = 1,\left\{ {{\ell }_{1} = 0}\right\} = \left\{ {{S}_{1} \leq 0}\right\} = \left\{ {{\nu }_{1} = 0}\right\} \), and \( \left\{ {{\ell }_{1}^{\prime } = 0}\right\} = \left\{ {{S}_{1} > }\right. \) \( 0\} = \left\{ {{\nu }_{1}^{\prime } = 0}\right\} \) . We shall prove the general case by induction, supp... | Yes |
Theorem 4.4.2. As \( t \rightarrow \infty, U\left( t\right) /t \rightarrow 1/\mu \) . | Proof. We will apply Wald’s equation to the stopping time \( {N}_{t} \) . The first step is to show that \( E{N}_{t} < \infty \) . To do this, pick \( \delta > 0 \) so that \( P\left( {{\xi }_{i} > \delta }\right) = \epsilon > 0 \) and pick \( K \) so that \( {K\delta } \geq t \) . Since \( K \) consecutive \( {\xi }_{... | Yes |
Theorem 4.4.3. Blackwell’s renewal theorem. If \( F \) is nonarithmetic then\n\n\[ U\left( \left\lbrack {t, t + h}\right\rbrack \right) \rightarrow h/\mu \;\text{ as }t \rightarrow \infty . \]\n | We will prove the result in the case \( \mu < \infty \) by \ | No |
Example 4.4.2. \( h\left( t\right) = G\left( t\right) : V\left( t\right) = G\left( t\right) + {\int }_{0}^{t}V\left( {t - s}\right) {dF}\left( s\right) \) | The last equation is valid for an arbitrary delay distribution. If we let \( G \) be the distribution in (4.4.4) and subtract the last two equations, we get | No |
Let \( x > 0 \) be fixed, and let \( H\left( t\right) = P\left( {{T}_{N\left( t\right) } - t > x}\right) \). By considering the value of \( {T}_{1} \), we get | \[ H\left( t\right) = \left( {1 - F\left( {t + x}\right) }\right) + {\int }_{0}^{t}H\left( {t - s}\right) {dF}\left( s\right) \] | Yes |
If \( h \) is bounded then the function\n\n\[ H\left( t\right) = {\int }_{0}^{t}h\left( {t - s}\right) {dU}\left( s\right) \]\n\nis the unique solution of the renewal equation that is bounded on bounded intervals. | Proof. Let \( {U}_{n}\left( A\right) = \mathop{\sum }\limits_{{m = 0}}^{n}P\left( {{T}_{m} \in A}\right) \) and\n\n\[ {H}_{n}\left( t\right) = {\int }_{0}^{t}h\left( {t - s}\right) d{U}_{n}\left( s\right) = \mathop{\sum }\limits_{{m = 0}}^{n}\left( {h * {F}^{m * }}\right) \left( t\right) \]\n\nHere, \( {F}^{m * } \) is... | Yes |
A chicken wants to cross a road (we won't ask why) on which the traffic is a Poisson process with rate \( \lambda \) . She needs one unit of time with no arrival to safely cross the road. Let \( M = \inf \{ t \geq 0 \) : there are no arrivals in \( t, t + 1\rbrack \} \) be the waiting time until she starts to cross the... | \[ H\left( t\right) = {e}^{-\lambda } + {\int }_{0}^{1}H\left( {t - y}\right) \lambda {e}^{-{\lambda y}}{dy} \] Comparing with Example 4.4.1 and using Theorem 4.4.4, we see that \[ H\left( t\right) = {e}^{-\lambda }\mathop{\sum }\limits_{{n = 0}}^{\infty }{F}^{n * }\left( t\right) \] We could have gotten this answer wi... | Yes |
Consider an insurance company that collects money at rate \( c \) and experiences i.i.d. claims at the arrival times of a Poisson process \( {N}_{t} \) with rate 1 . If its initial capital is \( x \), its wealth at time \( t \) is\n\n\[ \n{W}_{x}\left( t\right) = x + {ct} - \mathop{\sum }\limits_{{m = 1}}^{{Nt}}{Y}_{i}... | This does not look much like a renewal equation, but with some ingenuity it can be transformed into one. Changing variables \( t = x + {cs} \)\n\n\[ \nR\left( x\right) {e}^{-x/c} = {\int }_{x}^{\infty }{e}^{-t/c}{\int }_{0}^{t}R\left( {t - y}\right) {dG}\left( y\right) \frac{dt}{c}\n\]\n\nDifferentiating w.r.t. \( x \)... | Yes |
Theorem 4.4.5. The renewal theorem. If \( F \) is nonarithmetic and \( h \) is directly Riemann integrable then as \( t \rightarrow \infty \)\n\n\[ H\left( t\right) \rightarrow \frac{1}{\mu }{\int }_{0}^{\infty }h\left( s\right) {ds} \] | Proof. Suppose\n\n\[ h\left( s\right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }{a}_{k}{1}_{\lbrack {k\delta },\left( {k + 1}\right) \delta )}\left( s\right) \]\n\nwhere \( \mathop{\sum }\limits_{{k = 0}}^{\infty }\left| {a}_{k}\right| < \infty \) . Since \( U\left( \left\lbrack {t, t + \delta }\right\rbrack \right) \... | Yes |
Lemma 4.4.6. If \( h\left( x\right) \geq 0 \) is decreasing with \( h\left( 0\right) < \infty \) and \( {\int }_{0}^{\infty }h\left( x\right) {dx} < \infty \), then \( h \) is directly Riemann integrable. | Proof. Because \( h \) is decreasing, \( {I}^{\delta } = \mathop{\sum }\limits_{{k = 0}}^{\infty }{\delta h}\left( {k\delta }\right) \) and \( {I}_{\delta } = \mathop{\sum }\limits_{{k = 0}}^{\infty }{\delta h}\left( {\left( {k + 1}\right) \delta }\right) \) . So\n\n\[ \n{I}^{\delta } \geq {\int }_{0}^{\infty }h\left( ... | Yes |
Example 4.4.7. Continuation of Example 4.4.3. \( h\left( t\right) = \frac{1}{\mu }{\int }_{\lbrack t,\infty )}1 - F\left( s\right) {ds} \) . \( h \) is decreasing, \( h\left( 0\right) = 1 \), and | \[ \mu {\int }_{0}^{\infty }h\left( t\right) {dt} = {\int }_{0}^{\infty }{\int }_{t}^{\infty }1 - F\left( s\right) {dsdt} \] \[ = {\int }_{0}^{\infty }{\int }_{0}^{s}1 - F\left( s\right) {dtds} = {\int }_{0}^{\infty }s\left( {1 - F\left( s\right) }\right) {ds} = E\left( {{\xi }_{i}^{2}/2}\right) \] So, if \( \nu \equiv... | Yes |
Example 4.4.8. Continuation of Example 4.4.4. \( h\left( t\right) = 1 - F\left( {t + x}\right) \) . Again, \( h \) is decreasing, but this time \( h\left( 0\right) \leq 1 \) and the integral of \( h \) is finite when \( \mu = E\left( {\xi }_{i}\right) < \infty \) . Applying Lemma 4.4.6 and Theorem 4.4.5 now gives | \[ P\left( {{T}_{N\left( t\right) } - t > x}\right) \rightarrow \frac{1}{\mu }{\int }_{0}^{\infty }h\left( s\right) {ds} = \frac{1}{\mu }{\int }_{x}^{\infty }1 - F\left( t\right) {dt} \] so (when \( \mu < \infty \) ) the distribution of the residual waiting time \( {T}_{N\left( t\right) } - t \) converges to the delay ... | Yes |
Exercise 4.4.11. (i) Show that for any pattern of length \( k, E{t}_{j} = {2}^{k} \) for \( j \geq 2 \) . (ii) Compute \( E{t}_{1} \) when the pattern is \( \mathrm{{HH}} \), and when it is HT. Hint: For \( \mathrm{{HH}} \), observe | \[ E{t}_{1} = P\left( {HH}\right) + P\left( {HT}\right) E\left( {{t}_{1} + 2}\right) + P\left( T\right) E\left( {{t}_{1} + 1}\right) \] | No |
Lemma 5.1.1. If \( Y \) satisfies (i) and (ii), then it is integrable. | Proof. Letting \( A = \{ Y > 0\} \in \mathcal{F} \), using (ii) twice, and then adding\n\n\[{\int }_{A}{YdP} = {\int }_{A}{XdP} \leq {\int }_{A}\left| X\right| {dP}\]\n\n\[{\int }_{{A}^{c}} - {YdP} = {\int }_{{A}^{c}} - {XdP} \leq {\int }_{{A}^{c}}\left| X\right| {dP}\]\n\nSo we have \( E\left| Y\right| \leq E\left| X\... | Yes |
At the other extreme from perfect information is no information. Suppose \( X \) is independent of \( \mathcal{F} \), i.e., for all \( B \in \mathcal{R} \) and \( A \in \mathcal{F} \)\n\n\[ P\left( {\{ X \in B\} \cap A}\right) = P\left( {X \in B}\right) P\left( A\right) \]\n\nWe claim that, in this case, \( E\left( {X ... | To check the definition, note that \( {EX} \in \mathcal{F} \) so (i). To verify (ii), we observe that if \( A \in \mathcal{F} \) then since \( X \) and \( {1}_{A} \in \mathcal{F} \) are independent, Theorem 2.1.9 implies\n\n\[ {\int }_{A}{XdP} = E\left( {X{1}_{A}}\right) = {EXE}{1}_{A} = {\int }_{A}{EXdP} \] | Yes |
In this example, we relate the new definition of conditional expectation to the first one taught in an undergraduate probability course. Suppose \( {\Omega }_{1},{\Omega }_{2},\ldots \) is a finite or infinite partition of \( \Omega \) into disjoint sets, each of which has positive probability, and let \( \mathcal{F} =... | To prove our guess is correct, observe that the proposed formula is constant on each \( {\Omega }_{i} \), so it is measurable with respect to \( \mathcal{F} \). To verify (ii), it is enough to check the equality for \( A = {\Omega }_{i} \), but this is trivial: \[ {\int }_{{\Omega }_{i}}\frac{E\left( {X;{\Omega }_{i}}\... | Yes |
To continue making connection with definitions of conditional expectation from undergraduate probability, suppose \( X \) and \( Y \) have joint density \( f\left( {x, y}\right) \) , i.e., \[ P\left( {\left( {X, Y}\right) \in B}\right) = {\int }_{B}f\left( {x, y}\right) {dxdy}\;\text{ for }B \in {\mathcal{R}}^{2} \] an... | To \ | No |
Suppose \( X \) and \( Y \) are independent. Let \( \varphi \) be a function with \( E\left| {\varphi \left( {X, Y}\right) }\right| < \infty \) and let \( g\left( x\right) = E\left( {\varphi \left( {x, Y}\right) }\right) \) . We will now show that\n\n\[ E\left( {\varphi \left( {X, Y}\right) \mid X}\right) = g\left( X\r... | Proof. It is clear that \( g\left( X\right) \in \sigma \left( X\right) \) . To check (ii), note that if \( A \in \sigma \left( X\right) \) then \( A = \{ X \in C\} \), so using the change of variables formula (Theorem 1.6.9) and the fact that the distribution of \( \left( {X, Y}\right) \) is product measure (Theorem 2.... | Yes |
Example 5.1.6. Borel’s paradox. Let \( X \) be a randomly chosen point on the earth, let \( \theta \) be its longitude, and \( \varphi \) be its latitude. It is customary to take \( \theta \in \lbrack 0,{2\pi }) \) and \( \varphi \in ( - \pi /2,\pi /2\rbrack \) but we can equally well take \( \theta \in \lbrack 0,\pi )... | At first glance it might seem that if \( X \) is uniform on the globe then \( \theta \) and the angle \( \varphi \) on the great circle should both be uniform over their possible values. \( \theta \) is uniform but \( \varphi \) is not. The paradox completely evaporates once we realize that in the new or in the traditi... | Yes |
Conditional expectation is linear: | To prove (a), we need to check that the right-hand side is a version of the left. It clearly is \( \mathcal{F} \) -measurable. To check (ii), we observe that if \( A \in \mathcal{F} \) then by linearity of the integral and the defining properties of \( E\left( {X \mid \mathcal{F}}\right) \) and \( E\left( {Y \mid \math... | Yes |
Theorem 5.1.3. If \( \varphi \) is convex and \( E\left| X\right|, E\left| {\varphi \left( X\right) }\right| < \infty \) then\n\n\[ \varphi \left( {E\left( {X \mid \mathcal{F}}\right) }\right) \leq E\left( {\varphi \left( X\right) \mid \mathcal{F}}\right) \] | Proof. If \( \varphi \) is linear, the result is trivial, so we will suppose \( \varphi \) is not linear. We do this so that if we let \( S = \{ \left( {a, b}\right) : a, b \in \mathbf{Q},{ax} + b \leq \varphi \left( x\right) \) for all \( x\} \), then \( \varphi \left( x\right) = \sup \{ {ax} + b : \left( {a, b}\right... | Yes |
Theorem 5.1.4. Conditional expectation is a contraction in \( {L}^{p}, p \geq 1 \) . | Proof. (5.1.4) implies \( {\left| E\left( X \mid \mathcal{F}\right) \right| }^{p} \leq E\left( {{\left| X\right| }^{p} \mid \mathcal{F}}\right) \) . Taking expected values gives\n\n\[ E\left( {\left| E\left( X \mid \mathcal{F}\right) \right| }^{p}\right) \leq E\left( {E\left( {{\left| X\right| }^{p} \mid \mathcal{F}}\r... | Yes |
Theorem 5.1.5. If \( \mathcal{F} \subset \mathcal{G} \) and \( E\left( {X \mid \mathcal{G}}\right) \in \mathcal{F} \) then \( E\left( {X \mid \mathcal{F}}\right) = E\left( {X \mid \mathcal{G}}\right) \) . | Proof. By assumption \( E\left( {X \mid \mathcal{G}}\right) \in \mathcal{F} \). To check the other part of the definition we note that if \( A \in \mathcal{F} \subset \mathcal{G} \) then\n\n\[{\int }_{A}{XdP} = {\int }_{A}E\left( {X \mid \mathcal{G}}\right) {dP}\] | Yes |
Theorem 5.1.6. If \( {\mathcal{F}}_{1} \subset {\mathcal{F}}_{2} \) then (i) \( E\left( {E\left( {X \mid {\mathcal{F}}_{1}}\right) \mid {\mathcal{F}}_{2}}\right) = E\left( {X \mid {\mathcal{F}}_{1}}\right) \)\n\n(ii) \( E\left( {E\left( {X \mid {\mathcal{F}}_{2}}\right) \mid {\mathcal{F}}_{1}}\right) = E\left( {X \mid ... | Proof. Once we notice that \( E\left( {X \mid {\mathcal{F}}_{1}}\right) \in {\mathcal{F}}_{2} \) ,(i) follows from Example 5.1.1. To prove (ii), notice that \( E\left( {X \mid {\mathcal{F}}_{1}}\right) \in {\mathcal{F}}_{1} \), and if \( A \in {\mathcal{F}}_{1} \subset {\mathcal{F}}_{2} \) then\n\n\[ \n{\int }_{A}E\lef... | Yes |
Theorem 5.1.7. If \( X \in \mathcal{F} \) and \( E\left| Y\right|, E\left| {XY}\right| < \infty \) then\n\n\[ E\left( {{XY} \mid \mathcal{F}}\right) = {XE}\left( {Y \mid \mathcal{F}}\right) . | Proof. The right-hand side \( \in \mathcal{F} \), so we have to check (ii). To do this, we use the usual four-step procedure. First, suppose \( X = {1}_{B} \) with \( B \in \mathcal{F} \) . In this case, if \( A \in \mathcal{F} \)\n\n\[ {\int }_{A}{1}_{B}E\left( {Y \mid \mathcal{F}}\right) {dP} = {\int }_{A \cap B}E\le... | Yes |
Consider the successive tosses of a fair coin and let \( {\xi }_{n} = 1 \) if the \( n \) th tossis heads and \( {\xi }_{n} = - 1 \) if the \( n \) th toss is tails. Let \( {X}_{n} = {\xi }_{1} + \cdots + {\xi }_{n} \) and \( {\mathcal{F}}_{n} = \sigma \left( {{\xi }_{1},\ldots ,{\xi }_{n}}\right) \) for \( n \geq 1,{X... | To prove this, we observe that \( {X}_{n} \in {\mathcal{F}}_{n}, E\left| {X}_{n}\right| < \infty \), and \( {\xi }_{n + 1} \) is independent of \( {\mathcal{F}}_{n} \), so using the linearity of conditional expectation, (5.1.1), and Example 5.1.2,\n\n\[ E\left( {{X}_{n + 1} \mid {\mathcal{F}}_{n}}\right) = E\left( {{X}... | Yes |
Theorem 5.2.1. If \( {X}_{n} \) is a supermartingale then for \( n > m, E\left( {{X}_{n} \mid {\mathcal{F}}_{m}}\right) \leq {X}_{m} \) . | Proof. The definition gives the result for \( n = m + 1 \) . Suppose \( n = m + k \) with \( k \geq 2 \) . By Theorem 5.1.2,\n\n\[ E\left( {{X}_{m + k} \mid {\mathcal{F}}_{m}}\right) = E\left( {E\left( {{X}_{m + k} \mid {\mathcal{F}}_{m + k - 1}}\right) \mid {\mathcal{F}}_{m}}\right) \leq E\left( {{X}_{m + k - 1} \mid ... | Yes |
Theorem 5.2.2. (i) If \( {X}_{n} \) is a submartingale then for \( n > m, E\left( {{X}_{n} \mid {\mathcal{F}}_{m}}\right) \geq {X}_{m} \) . (ii) If \( {X}_{n} \) is a martingale then for \( n > m, E\left( {{X}_{n} \mid {\mathcal{F}}_{m}}\right) = {X}_{m} \) . | Proof. To prove (i), note that \( - {X}_{n} \) is a supermartingale and use (5.1.1). For (ii), observe that \( {X}_{n} \) is a supermartingale and a submartingale. | No |
Theorem 5.2.3. If \( {X}_{n} \) is a martingale w.r.t. \( {\mathcal{F}}_{n} \) and \( \varphi \) is a convex function with \( E\left| {\varphi \left( {X}_{n}\right) }\right| < \infty \) for all \( n \) then \( \varphi \left( {X}_{n}\right) \) is a submartingale w.r.t. \( {\mathcal{F}}_{n} \) . Consequently, if \( p \ge... | Proof By Jensen's inequality and the definition\n\n\[ E\left( {\varphi \left( {X}_{n + 1}\right) \mid {\mathcal{F}}_{n}}\right) \geq \varphi \left( {E\left( {{X}_{n + 1} \mid {\mathcal{F}}_{n}}\right) }\right) = \varphi \left( {X}_{n}\right) \] | Yes |
Theorem 5.2.4. If \( {X}_{n} \) is a submartingale w.r.t. \( {\mathcal{F}}_{n} \) and \( \varphi \) is an increasing convex function with \( E\left| {\varphi \left( {X}_{n}\right) }\right| < \infty \) for all \( n \), then \( \varphi \left( {X}_{n}\right) \) is a submartingale w.r.t. \( {\mathcal{F}}_{n} \) . Consequen... | Proof By Jensen's inequality and the assumptions\n\n\[ E\left( {\varphi \left( {X}_{n + 1}\right) \mid {\mathcal{F}}_{n}}\right) \geq \varphi \left( {E\left( {{X}_{n + 1} \mid {\mathcal{F}}_{n}}\right) }\right) \geq \varphi \left( {X}_{n}\right) \] | Yes |
Theorem 5.2.5. Let \( {X}_{n}, n \geq 0 \), be a supermartingale. If \( {H}_{n} \geq 0 \) is predictable and each \( {H}_{n} \) is bounded then \( {\left( H \cdot X\right) }_{n} \) is a supermartingale. | Proof. Using the fact that conditional expectation is linear, \( {\left( H \cdot X\right) }_{n} \in {\mathcal{F}}_{n},{H}_{n} \in \) \( {\mathcal{F}}_{n - 1} \), and (5.1.7), we have\n\n\[ E\left( {{\left( H \cdot X\right) }_{n + 1} \mid {\mathcal{F}}_{n}}\right) = {\left( H \cdot X\right) }_{n} + E\left( {{H}_{n + 1}\... | Yes |
Theorem 5.2.7. Upcrossing inequality. If \( {X}_{m}, m \geq 0 \), is a submartingale then\n\n\[ \left( {b - a}\right) E{U}_{n} \leq E{\left( {X}_{n} - a\right) }^{ + } - E{\left( {X}_{0} - a\right) }^{ + } \] | Proof. Let \( {Y}_{m} = a + {\left( {X}_{m} - a\right) }^{ + } \) . By Theorem 5.2.4, \( {Y}_{m} \) is a submartingale. Clearly, it upcrosses \( \left\lbrack {a, b}\right\rbrack \) the same number of times that \( {X}_{m} \) does, and we have \( \left( {b - a}\right) {U}_{n} \leq \) \( {\left( H \cdot Y\right) }_{n} \)... | Yes |
Theorem 5.2.8. Martingale convergence theorem. If \( {X}_{n} \) is a submartingale with \( \sup E{X}_{n}^{ + } < \infty \) then as \( n \rightarrow \infty ,{X}_{n} \) converges a.s. to a limit \( X \) with \( E\left| X\right| < \infty \) . | Proof. Since \( {\left( X - a\right) }^{ + } \leq {X}^{ + } + \left| a\right| \), Theorem 5.2.7 implies that\n\n\[ E{U}_{n} \leq \left( {\left| a\right| + E{X}_{n}^{ + }}\right) /\left( {b - a}\right) \]\n\nAs \( n \uparrow \infty ,{U}_{n} \uparrow U \) the number of upcrossings of \( \left\lbrack {a, b}\right\rbrack \... | Yes |
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