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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\[ \nE{S}_{N} = \int {S}_{N}{dP} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\int {S}_{n}{1}_{\{ N = n\} }{dP} = \mathop{\sum }\limits_{{n = 1}}^{\infty }\mathop{\sum }\limits_{{m = 1}}^{n}\int {X}_{m}{1}_{\{ N = n\} }{dP} \n\] \n\nSince the \( {X}_{i} \geq 0 \), we can... | Yes |
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 check that \( {EN... | 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 |
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 \( a = - L,{EN} = {L}^{2} \). | No |
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... | No |
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 }\right) \... | 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 |
Theorem 4.4.4. 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\ri... | 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 2.1... | 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 |
(a) Conditional expectation is linear:\n\n\[ E\left( {{aX} + Y \mid \mathcal{F}}\right) = {aE}\left( {X \mid \mathcal{F}}\right) + E\left( {Y \mid \mathcal{F}}\right) \] | Proof. 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 \mi... | 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\[ \n{\int }_{A}{XdP} = {\int }_{A}E\left( {X \mid \mathcal{G}}\right) {dP} \n\] | 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 |
Theorem 5.2.9. If \( {X}_{n} \geq 0 \) is a supermartingale then as \( n \rightarrow \infty ,{X}_{n} \rightarrow X \) a.s. and \( {EX} \leq E{X}_{0} \) . | Proof. \( {Y}_{n} = - {X}_{n} \leq 0 \) is a submartingale with \( E{Y}_{n}^{ + } = 0 \) . Since \( E{X}_{0} \geq E{X}_{n} \), the inequality follows from Fatou's lemma. | No |
The first shows that the assumptions of Theorem 5.2.9 (or 5.2.8) do not guarantee convergence in \( {L}^{1} \) . Let \( {S}_{n} \) be a symmetric simple random walk with \( {S}_{0} = 1 \), i.e., \( {S}_{n} = {S}_{n - 1} + {\xi }_{n} \) where \( {\xi }_{1},{\xi }_{2},\ldots \) are i.i.d. with \( P\left( {{\xi }_{i} = 1}... | Theorem 5.2.9 implies \( {X}_{n} \) converges to a limit \( {X}_{\infty } < \infty \) that must be \( \equiv 0 \), since convergence to \( k > 0 \) is impossible. (If \( {X}_{n} = k > 0 \) then \( \left. {{X}_{n + 1} = k \pm 1\text{.}}\right) \) Since \( E{X}_{n} = E{X}_{0} = 1 \) for all \( n \) and \( {X}_{\infty } =... | Yes |
Example 5.2.4. We will now give an example of a martingale with \( {X}_{k} \rightarrow 0 \) in probability but not a.s. Let \( {X}_{0} = 0 \) . When \( {X}_{k - 1} = 0 \), let \( {X}_{k} = 1 \) or -1 with probability \( 1/{2k} \) and \( = 0 \) with probability \( 1 - 1/k \) . When \( {X}_{k - 1} \neq 0 \), let \( {X}_{... | From the construction, \( P\left( {{X}_{k} = 0}\right) = 1 - 1/k \) so \( {X}_{k} \rightarrow 0 \) in probability. On the other hand, the second Borel-Cantelli lemma implies \( P\left( {{X}_{k} = 0}\right. \) for \( \left. {k \geq K}\right) = 0 \), and values in \( \left( {-1,1}\right) - \{ 0\} \) are impossible, so \(... | Yes |
Theorem 5.2.10. Doob’s decomposition. Any submartingale \( {X}_{n}, n \geq 0 \), can be written in a unique way as \( {X}_{n} = {M}_{n} + {A}_{n} \), where \( {M}_{n} \) is a martingale and \( {A}_{n} \) is a predictable increasing sequence with \( {A}_{0} = 0 \) . | Proof. We want \( {X}_{n} = {M}_{n} + {A}_{n}, E\left( {{M}_{n} \mid {\mathcal{F}}_{n - 1}}\right) = {M}_{n - 1} \), and \( {A}_{n} \in {\mathcal{F}}_{n - 1} \) . So we must have\n\n\[ E\left( {{X}_{n} \mid {\mathcal{F}}_{n - 1}}\right) = E\left( {{M}_{n} \mid {\mathcal{F}}_{n - 1}}\right) + E\left( {{A}_{n} \mid {\mat... | Yes |
Theorem 5.3.1. Let \( {X}_{1},{X}_{2},\ldots \) be a martingale with \( \left| {{X}_{n + 1} - {X}_{n}}\right| \leq M < \infty \) . Let\n\n\[ C = \left\{ {\lim {X}_{n}}\right. \text{exists and is finite}\} \]\n\n\[ D = \left\{ {\lim \sup {X}_{n} = + \infty }\right. \text{and}\left. {\lim \inf {X}_{n} = - \infty }\right\... | Proof. Since \( {X}_{n} - {X}_{0} \) is a martingale, we can without loss of generality suppose that \( {X}_{0} = 0 \) . Let \( 0 < K < \infty \) and let \( N = \inf \left\{ {n : {X}_{n} \leq - K}\right\} .{X}_{n \land N} \) is a martingale with \( {X}_{n \land N} \geq - K - M \) a.s. so applying Theorem 5.2.9 to \( {X... | Yes |
Theorem 5.3.2. Second Borel-Cantelli lemma, II. Let \( {\mathcal{F}}_{n}, n \geq 0 \) be a filtration with \( {\mathcal{F}}_{0} = \{ \varnothing ,\Omega \} \) and \( {A}_{n}, n \geq 1 \) a sequence of events with \( {A}_{n} \in {\mathcal{F}}_{n} \) . Then\n\n\[ \n\left\{ {{A}_{n}\text{ i.o. }}\right\} = \left\{ {\matho... | Proof. If we let \( {X}_{0} = 0 \) and \( {X}_{n} = \mathop{\sum }\limits_{{m = 1}}^{n}{1}_{{A}_{m}} - P\left( {{A}_{m} \mid {\mathcal{F}}_{m - 1}}\right) \) for \( n \geq 1 \) then \( {X}_{n} \) is a martingale with \( \left| {{X}_{n} - {X}_{n - 1}}\right| \leq 1 \) . Using the notation of Theorem 5.3.1 we have:\n\n\[... | Yes |
Theorem 5.3.3. Suppose \( {\mu }_{n} \ll {\nu }_{n} \) for all \( n \) . Let \( {X}_{n} = d{\mu }_{n}/d{\nu }_{n} \) and let \( X = \) \( \lim \sup {X}_{n} \) . Then\n\n\[ \mu \left( A\right) = {\int }_{A}{Xd\nu } + \mu \left( {A\cap \{ X = \infty \} }\right) \] | Proof. As the reader can probably anticipate: | No |
Lemma 5.3.4. \( {X}_{n} \) (defined on \( \left( {\Omega ,\mathcal{F},\nu }\right) \) ) is a martingale w.r.t. \( {\mathcal{F}}_{n} \) . | Proof. We observe that, by definition, \( {X}_{n} \in {\mathcal{F}}_{n} \) . Let \( A \in {\mathcal{F}}_{n} \) . Since \( {X}_{n} \in {\mathcal{F}}_{n} \) and \( {\nu }_{n} \) is the restriction of \( \nu \) to \( {\mathcal{F}}_{n} \)\n\n\[ \n{\int }_{A}{X}_{n}{d\nu } = {\int }_{A}{X}_{n}d{\nu }_{n} \n\]\n\nUsing the d... | No |
Theorem 5.3.5. \( \mu \ll \nu \) or \( \mu \bot \nu \), according as \( \mathop{\prod }\limits_{{m = 1}}^{\infty }\int \sqrt{{q}_{m}}d{G}_{m} > 0 \) or \( = 0 \) . | Proof. Jensen's inequality and Exercise A.4.7 imply\n\n\[ \n{\left( \int \sqrt{{q}_{m}}d{G}_{m}\right) }^{2} \leq \int {q}_{m}d{G}_{m} = \int d{F}_{m} = 1 \n\]\n\nso the infinite product of the integrals is well defined and \( \leq 1 \) . Let\n\n\[ \n{X}_{n} = \mathop{\prod }\limits_{{m \leq n}}{q}_{m}\left( {\omega }_... | Yes |
Lemma 5.3.6. Let \( {\mathcal{F}}_{n} = \sigma \left( {{\xi }_{i}^{m} : i \geq 1,1 \leq m \leq n}\right) \) and \( \mu = E{\xi }_{i}^{m} \in \left( {0,\infty }\right) \) . Then \( {Z}_{n}/{\mu }^{n} \) is a martingale w.r.t. \( {\mathcal{F}}_{n} \) . | Proof. Clearly, \( {Z}_{n} \in {\mathcal{F}}_{n} \). \n\n\[ \nE\left( {{Z}_{n + 1} \mid {\mathcal{F}}_{n}}\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }E\left( {{Z}_{n + 1}{1}_{\left\{ {Z}_{n} = k\right\} } \mid {\mathcal{F}}_{n}}\right) \n\] \n\nby the linearity of conditional expectation, (5.1.1), and the monoto... | Yes |
Theorem 5.3.7. If \( \mu < 1 \) then \( {Z}_{n} = 0 \) for all \( n \) sufficiently large, so \( {Z}_{n}/{\mu }^{n} \rightarrow 0 \) . | Proof. \( E\left( {{Z}_{n}/{\mu }^{n}}\right) = E\left( {Z}_{0}\right) = 1 \), so \( E\left( {Z}_{n}\right) = {\mu }^{n} \) . Now \( {Z}_{n} \geq 1 \) on \( \left\{ {{Z}_{n} > 0}\right\} \) so\n\n\[ P\left( {{Z}_{n} > 0}\right) \leq E\left( {{Z}_{n};{Z}_{n} > 0}\right) = E\left( {Z}_{n}\right) = {\mu }^{n} \rightarrow ... | Yes |
Theorem 5.3.8. If \( \mu = 1 \) and \( P\left( {{\xi }_{i}^{m} = 1}\right) < 1 \) then \( {Z}_{n} = 0 \) for all \( n \) sufficiently large. | Proof. When \( \mu = 1,{Z}_{n} \) is itself a nonnegative martingale. Since \( {Z}_{n} \) is integer valued and by Theorem 5.2.9 converges to an a.s. finite limit \( {Z}_{\infty } \), we must have \( {Z}_{n} = {Z}_{\infty } \) for large \( n \) . If \( P\left( {{\xi }_{i}^{m} = 1}\right) < 1 \) and \( k > 0 \) then \( ... | Yes |
Theorem 5.3.9. \( P\left( {{Z}_{n} = 0}\right. \) for some \( \left. n\right) = \rho \) the unique fixed point of \( \varphi \) in \( \lbrack 0,1) \) . | Proof. Differentiating and referring to Theorem A.5.2 for the justification gives for \( s < 1 \)\n\n\[ \n{\varphi }^{\prime }\left( s\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }k{p}_{k}{s}^{k - 1} \geq 0 \n\]\n\n\[ \n{\varphi }^{\prime \prime }\left( s\right) = \mathop{\sum }\limits_{{k = 2}}^{\infty }k\left( {... | Yes |
Theorem 5.3.10. \( W = \lim {Z}_{n}/{\mu }^{n} \) is not \( \equiv 0 \) if and only if \( \sum {p}_{k}k\log k < \infty \) . | For a proof, see Athreya and Ney (1972), p. 24-29. | No |
Theorem 5.4.1. If \( {X}_{n} \) is a submartingale and \( N \) is a stopping time with \( P(N \leq \) \( k) = 1 \) then\n\n\[ E{X}_{0} \leq E{X}_{N} \leq E{X}_{k} \] | Proof. Theorem 5.2.6 implies \( {X}_{N \land n} \) is a submartingale, so it follows that\n\n\[ E{X}_{0} = E{X}_{N \land 0} \leq E{X}_{N \land k} = E{X}_{N} \]\n\nTo prove the other inequality, let \( {K}_{n} = {1}_{\{ N < n\} } = {1}_{\{ N \leq n - 1\} }.{K}_{n} \) is predictable, so Theorem 5.2.5 implies \( {\left( K... | Yes |
Theorem 5.4.2. Doob’s inequality. Let \( {X}_{m} \) be a submartingale,\n\n\[ \n{\bar{X}}_{n} = \mathop{\max }\limits_{{0 \leq m \leq n}}{X}_{m}^{ + }\n\]\n\n\( \lambda > 0 \), and \( A = \left\{ {{\bar{X}}_{n} \geq \lambda }\right\} \) . Then\n\n\[ \n{\lambda P}\left( A\right) \leq E{X}_{n}{1}_{A} \leq E{X}_{n}^{ + }\... | Proof. Let \( N = \inf \left\{ {m : {X}_{m} \geq \lambda }\right. \) or \( \left. {m = n}\right\} \) . Since \( {X}_{N} \geq \lambda \) on \( A \) ,\n\n\[ \n{\lambda P}\left( A\right) \leq E{X}_{N}{1}_{A} \leq E{X}_{n}{1}_{A}\n\]\n\nThe second inequality follows from the fact that Theorem 5.4.1 implies \( E{X}_{N} \leq... | Yes |
Example 5.4.1. Random walks. If we let \( {S}_{n} = {\xi }_{1} + \cdots + {\xi }_{n} \) where the \( {\xi }_{m} \) are independent and have \( E{\xi }_{m} = 0,{\sigma }_{m}^{2} = E{\xi }_{m}^{2} < \infty \), then Theorem 5.2.3 implies \( {X}_{n} = {S}_{n}^{2} \) is a submartingale. If we let \( \lambda = {x}^{2} \) and... | \[ P\left( {\mathop{\max }\limits_{{1 \leq m \leq n}}\left| {S}_{m}\right| \geq x}\right) \leq {x}^{-2}\operatorname{var}\left( {S}_{n}\right) \] | Yes |
Theorem 5.4.3. \( {L}^{p} \) maximum inequality. If \( {X}_{n} \) is a submartingale then for \( 1 < p < \infty \) ,\n\n\[ E\left( {\bar{X}}_{n}^{p}\right) \leq {\left( \frac{p}{p - 1}\right) }^{p}E{\left( {X}_{n}^{ + }\right) }^{p} \] | Proof. The second inequality follows by applying the first to \( {X}_{n} = \left| {Y}_{n}\right| \) . To prove the first we will, for reasons that will become clear in a moment, work with \( {\bar{X}}_{n} \land M \) rather than \( {\bar{X}}_{n} \) . Since \( \left\{ {{\bar{X}}_{n} \land M \geq \lambda }\right\} \) is a... | Yes |
Theorem 5.4.3 is false when \( \mathrm{p} = 1 \) | Again, the counterexample is provided by Example 5.2.3. Let \( {S}_{n} \) be a simple random walk starting from \( {S}_{0} = 1 \) , \( N = \inf \left\{ {n : {S}_{n} = 0}\right\} \), and \( {X}_{n} = {S}_{N \land n} \) . Theorem 5.4.1 implies \( E{X}_{n} = E{S}_{N \land n} = \) \( E{S}_{0} = 1 \) for all \( n \) . Using... | Yes |
Theorem 5.4.4. Let \( {X}_{n} \) be a submartingale and \( {\log }^{ + }x = \max \left( {\log x,0}\right) \). \[ E{\bar{X}}_{n} \leq {\left( 1 - {e}^{-1}\right) }^{-1}\left\{ {1 + E\left( {{X}_{n}^{ + }{\log }^{ + }\left( {X}_{n}^{ + }\right) }\right) }\right\} \] | Exercise 5.4.6. Prove Theorem 5.4.4 by carrying out the following steps: (i) Imitate the proof of 5.4.2 but use the trivial bound \( P\left( A\right) \leq 1 \) for \( \lambda \leq 1 \) to show \[ E\left( {{\bar{X}}_{n} \land M}\right) \leq 1 + \int {X}_{n}^{ + }\log \left( {{\bar{X}}_{n} \land M}\right) {dP} \] (ii) Us... | No |
Theorem 5.4.5. \( {L}^{p} \) convergence theorem. If \( {X}_{n} \) is a martingale with \( \sup E{\left| {X}_{n}\right| }^{p} < \) \( \infty \) where \( p > 1 \), then \( {X}_{n} \rightarrow X \) a.s. and in \( {L}^{p} \) . | Proof. \( {\left( E{X}_{n}^{ + }\right) }^{p} \leq {\left( E\left| {X}_{n}\right| \right) }^{p} \leq E{\left| {X}_{n}\right| }^{p} \), so it follows from the martingale convergence theorem (5.2.8) that \( {X}_{n} \rightarrow X \) a.s. The second conclusion in Theorem 5.4.3 implies\n\n\[ E{\left( \mathop{\sup }\limits_{... | Yes |
Theorem 5.4.6. Orthogonality of martingale increments. Let \( {X}_{n} \) be a martingale with \( E{X}_{n}^{2} < \infty \) for all \( n \) . If \( m \leq n \) and \( Y \in {\mathcal{F}}_{m} \) has \( E{Y}^{2} < \infty \) then\n\n\[ E\left( {\left( {{X}_{n} - {X}_{m}}\right) Y}\right) = 0 \] | Proof. The Cauchy-Schwarz inequality implies \( E\left| {\left( {{X}_{n} - {X}_{m}}\right) Y}\right| < \infty \) . Using (5.1.5), Theorem 5.1.7, and the definition of a martingale,\n\n\[ E\left( {\left( {{X}_{n} - {X}_{m}}\right) Y}\right) = E\left\lbrack {E\left( {\left( {{X}_{n} - {X}_{m}}\right) Y \mid {\mathcal{F}}... | Yes |
Theorem 5.4.7. Conditional variance formula. If \( {X}_{n} \) is a martingale with \( E{X}_{n}^{2} < \infty \) for all \( n \) , \[ E\left( {{\left( {X}_{n} - {X}_{m}\right) }^{2} \mid {\mathcal{F}}_{m}}\right) = E\left( {{X}_{n}^{2} \mid {\mathcal{F}}_{m}}\right) - {X}_{m}^{2}. \] | Proof. Using the linearity of conditional expectation and then Theorem 5.1.7, we have \[ E\left( {{X}_{n}^{2} - 2{X}_{n}{X}_{m} + {X}_{m}^{2} \mid {\mathcal{F}}_{m}}\right) = E\left( {{X}_{n}^{2} \mid {\mathcal{F}}_{m}}\right) - 2{X}_{m}E\left( {{X}_{n} \mid {\mathcal{F}}_{m}}\right) + {X}_{m}^{2} \] \[ = E\left( {{X}_... | Yes |
Example 5.4.3. Branching processes. We continue the study begun at the end of the last section. Using the notation introduced there, we suppose \( \mu = E\left( {\xi }_{i}^{m}\right) > 1 \) and \( \operatorname{var}\left( {\xi }_{i}^{m}\right) = {\sigma }^{2} < \infty \) . Let \( {X}_{n} = {Z}_{n}/{\mu }^{n} \) . Takin... | \[ E\left( {{X}_{n}^{2} \mid {\mathcal{F}}_{n - 1}}\right) = {X}_{n - 1}^{2} + E\left( {{\left( {X}_{n} - {X}_{n - 1}\right) }^{2} \mid {\mathcal{F}}_{n - 1}}\right) \] \n\nTo compute the second term, we observe \n\n\[ E\left( {{\left( {X}_{n} - {X}_{n - 1}\right) }^{2} \mid {\mathcal{F}}_{n - 1}}\right) = E\left( {{\l... | Yes |
Theorem 5.4.8. \( E\left( {\mathop{\sup }\limits_{m}{\left| {X}_{m}\right| }^{2}}\right) \leq {4E}{A}_{\infty } \) . | Proof. Applying the \( {L}^{2} \) maximum inequality (Theorem 5.4.3) to \( {X}_{n} \) gives\n\n\[ E\left( {\mathop{\sup }\limits_{{0 \leq m \leq n}}{\left| {X}_{m}\right| }^{2}}\right) \leq {4E}{X}_{n}^{2} = {4E}{A}_{n} \]\n\nsince \( E{X}_{n}^{2} = E{M}_{n} + E{A}_{n} \) and \( E{M}_{n} = E{M}_{0} = E{X}_{0}^{2} = 0 \... | Yes |
Theorem 5.4.9. \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{X}_{n} \) exists and is finite a.s. on \( \left\{ {{A}_{\infty } < \infty }\right\} \) . | Proof. Let \( a > 0 \) . Since \( {A}_{n + 1} \in {\mathcal{F}}_{n}, N = \inf \left\{ {n : {A}_{n + 1} > {a}^{2}}\right\} \) is a stopping time. Applying Theorem 5.4.8 to \( {X}_{N \land n} \) and noticing \( {A}_{N \land n} \leq {a}^{2} \) gives\n\n\[ E\left( {\mathop{\sup }\limits_{n}{\left| {X}_{N \land n}\right| }^... | Yes |
Theorem 5.4.10. Let \( f \geq 1 \) be increasing with \( {\int }_{0}^{\infty }f{\left( t\right) }^{-2}{dt} < \infty \) . Then \( {X}_{n}/f\left( {A}_{n}\right) \rightarrow \) 0 a.s. on \( \left\{ {{A}_{\infty } = \infty }\right\} \) . | Proof. \( {H}_{m} = f{\left( {A}_{m}\right) }^{-1} \) is bounded and predictable, so Theorem 5.2.5 implies\n\n\[ \n{Y}_{n} \equiv {\left( H \cdot X\right) }_{n} = \mathop{\sum }\limits_{{m = 1}}^{n}\frac{{X}_{m} - {X}_{m - 1}}{f\left( {A}_{m}\right) }\;\text{ is a martingale }\n\]\n\nIf \( {B}_{n} \) is the increasing ... | Yes |
Theorem 5.4.11. Second Borel-Cantelli Lemma, III. Suppose \( {B}_{n} \) is adapted to \( {\mathcal{F}}_{n} \) and let \( {p}_{n} = P\left( {{B}_{n} \mid {\mathcal{F}}_{n - 1}}\right) \) . Then\n\n\[ \mathop{\sum }\limits_{{m = 1}}^{n}{1}_{B\left( m\right) }/\mathop{\sum }\limits_{{m = 1}}^{n}{p}_{m} \rightarrow 1\;\tex... | Proof. Define a martingale by \( {X}_{0} = 0 \) and \( {X}_{n} - {X}_{n - 1} = {1}_{{B}_{n}} - P\left( {{B}_{n} \mid {\mathcal{F}}_{n - 1}}\right) \) for \( n \geq 1 \) so that we have\n\n\[ \left( {\mathop{\sum }\limits_{{m = 1}}^{n}{1}_{B\left( m\right) }/\mathop{\sum }\limits_{{m = 1}}^{n}{p}_{m}}\right) - 1 = {X}_{... | Yes |
Theorem 5.4.12. \( E\left( {\mathop{\sup }\limits_{n}\left| {X}_{n}\right| }\right) \leq {3E}{A}_{\infty }^{1/2} \) . | Proof. As in the proof of Theorem 5.4.9 we let \( a > 0 \) and let \( N = \inf \left\{ {n : {A}_{n + 1} > {a}^{2}}\right\} \) . This time, however, our starting point is\n\n\[ P\left( {\mathop{\sup }\limits_{m}\left| {X}_{m}\right| > a}\right) \leq P\left( {N < \infty }\right) + P\left( {\mathop{\sup }\limits_{m}\left|... | Yes |
Theorem 5.5.1. Given a probability space \( \left( {\Omega ,{\mathcal{F}}_{o}, P}\right) \) and an \( X \in {L}^{1} \), then \( \{ E\left( {X \mid \mathcal{F}}\right) \) : \( \mathcal{F} \) is a \( \sigma \) -field \( \left. { \subset {\mathcal{F}}_{o}}\right\} \) is uniformly integrable. | Proof. If \( {A}_{n} \) is a sequence of sets with \( P\left( {A}_{n}\right) \rightarrow 0 \) then the dominated convergence theorem implies \( E\left( {\left| X\right| ;{A}_{n}}\right) \rightarrow 0 \) . From the last result, it follows that if \( \epsilon > 0 \), we can pick \( \delta > 0 \) so that if \( P\left( A\r... | Yes |
Theorem 5.5.3. For a submartingale, the following are equivalent:\n\n(i) It is uniformly integrable.\n\n(ii) It converges a.s. and in \( {L}^{1} \) .\n\n(iii) It converges in \( {L}^{1} \) . | Proof. (i) implies (ii). Uniform integrability implies sup \( E\left| {X}_{n}\right| < \infty \) so the martingale convergence theorem implies \( {X}_{n} \rightarrow X \) a.s., and Theorem 5.5.2 implies \( {X}_{n} \rightarrow X \) in \( {L}^{1} \) . (ii) implies (iii). Trivial. (iii) implies (i). \( {X}_{n} \rightarrow... | Yes |
Lemma 5.5.4. If integrable random variables \( {X}_{n} \rightarrow X \) in \( {L}^{1} \) then\n\n\[ E\left( {{X}_{n};A}\right) \rightarrow E\left( {X;A}\right) \] | Proof. \( \left| {E{X}_{m}{1}_{A} - {EX}{1}_{A}}\right| \leq E\left| {{X}_{m}{1}_{A} - X{1}_{A}}\right| \leq E\left| {{X}_{m} - X}\right| \rightarrow 0 \) | Yes |
Lemma 5.5.5. If a martingale \( {X}_{n} \rightarrow X \) in \( {L}^{1} \) then \( {X}_{n} = E\left( {X \mid {\mathcal{F}}_{n}}\right) \) . | Proof. The martingale property implies that if \( m > n, E\left( {{X}_{m} \mid {\mathcal{F}}_{n}}\right) = {X}_{n} \), so if \( A \in \) \( {\mathcal{F}}_{n}, E\left( {{X}_{n};A}\right) = E\left( {{X}_{m};A}\right) \) . Lemma 5.5.4 implies \( E\left( {{X}_{m};A}\right) \rightarrow E\left( {X;A}\right) \), so we have \(... | Yes |
Theorem 5.5.6. For a martingale, the following are equivalent:\n\n(i) It is uniformly integrable.\n\n(ii) It converges a.s. and in \( {L}^{1} \).\n\n(iii) It converges in \( {L}^{1} \).\n\n(iv) There is an integrable random variable \( X \) so that \( {X}_{n} = E\left( {X \mid {\mathcal{F}}_{n}}\right) \). | Proof. (i) implies (ii). Since martingales are also submartingales, this follows from Theorem 5.5.3. (ii) implies (iii). Trivial. (iii) implies (iv). Follows from Lemma 5.5.5. (iv) implies (i). This follows from Theorem 5.5.1. | Yes |
Theorem 5.5.7. Suppose \( {\mathcal{F}}_{n} \uparrow {\mathcal{F}}_{\infty } \), i.e., \( {\mathcal{F}}_{n} \) is an increasing sequence of \( \sigma \) -fields and \( {\mathcal{F}}_{\infty } = \sigma \left( {{ \cup }_{n}{\mathcal{F}}_{n}}\right) \) . As \( n \rightarrow \infty \) | Proof. The first step is to note that if \( m > n \) then Theorem 5.1.6 implies\n\n\[ E\left( {E\left( {X \mid {\mathcal{F}}_{m}}\right) \mid {\mathcal{F}}_{n}}\right) = E\left( {X \mid {\mathcal{F}}_{n}}\right) \]\n\nso \( {Y}_{n} = E\left( {X \mid {\mathcal{F}}_{n}}\right) \) is a martingale. Theorem 5.5.1 implies th... | Yes |
Theorem 5.5.8. Lévy’s 0-1 law. If \( {\mathcal{F}}_{n} \uparrow {\mathcal{F}}_{\infty } \) and \( A \in {\mathcal{F}}_{\infty } \) then \( E\left( {{1}_{A} \mid {\mathcal{F}}_{n}}\right) \rightarrow {1}_{A} \) a.s. | To steal a line from Chung: \ | No |
Theorem 5.5.9. Dominated convergence theorem for conditional expectations. Suppose \( {Y}_{n} \rightarrow Y \) a.s. and \( \left| {Y}_{n}\right| \leq Z \) for all \( n \) where \( {EZ} < \infty \) . If \( {\mathcal{F}}_{n} \uparrow {\mathcal{F}}_{\infty } \) then\n\n\[ E\left( {{Y}_{n} \mid {\mathcal{F}}_{n}}\right) \r... | Proof. Let \( {W}_{N} = \sup \left\{ {\left| {{Y}_{n} - {Y}_{m}}\right| : n, m \geq N}\right\} .{W}_{N} \leq {2Z} \), so \( E{W}_{N} < \infty \) . Using monotonicity (5.1.2) and applying Theorem 5.5.7 to \( {W}_{N} \) gives\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}E\left( {\left| {{Y}_{n} - Y}\right| \mi... | Yes |
We will now show that \( E\left( {{X}_{n} \mid \mathcal{F}}\right) \) need not converge a.s. | Let \( {Y}_{1},{Y}_{2},\ldots \) and \( {Z}_{1},{Z}_{2},\ldots \) be independent r.v.’s with\n\n\[ P\left( {{Y}_{n} = 1}\right) = 1/n\;P\left( {{Y}_{n} = 0}\right) = 1 - 1/n \]\n\n\[ P\left( {{Z}_{n} = n}\right) = 1/n\;P\left( {{Z}_{n} = 0}\right) = 1 - 1/n \]\n\nLet \( {X}_{n} = {Y}_{n}{Z}_{n}.P\left( {{X}_{n} > 0}\ri... | Yes |
Theorem 5.6.1. \( {X}_{-\infty } = \mathop{\lim }\limits_{{n \rightarrow - \infty }}{X}_{n} \) exists a.s. and in \( {L}^{1} \) . | Proof. Let \( {U}_{n} \) be the number of upcrossings of \( \left\lbrack {a, b}\right\rbrack \) by \( {X}_{-n},\ldots ,{X}_{0} \) . The upcrossing inequality, Theorem 5.2.7 implies \( \left( {b - a}\right) E{U}_{n} \leq E{\left( {X}_{0} - a\right) }^{ + } \) . Letting \( n \rightarrow \infty \) and using the monotone c... | Yes |
Theorem 5.6.2. If \( {X}_{-\infty } = \mathop{\lim }\limits_{{n \rightarrow - \infty }}{X}_{n} \) and \( {\mathcal{F}}_{-\infty } = { \cap }_{n}{\mathcal{F}}_{n} \), then \( {X}_{-\infty } = E\left( {{X}_{0} \mid {\mathcal{F}}_{-\infty }}\right) \). | Proof. Clearly, \( {X}_{-\infty } \in {\mathcal{F}}_{-\infty }.{X}_{n} = E\left( {{X}_{0} \mid {\mathcal{F}}_{n}}\right) \), so if \( A \in {\mathcal{F}}_{-\infty } \subset {\mathcal{F}}_{n} \) then\n\n\[ \n{\int }_{A}{X}_{n}{dP} = {\int }_{A}{X}_{0}{dP} \n\]\n\nTheorem 5.6.1 and Lemma 5.5.4 imply \( E\left( {{X}_{n};A... | Yes |
Theorem 5.6.3. If \( {\mathcal{F}}_{n} \downarrow {\mathcal{F}}_{-\infty } \) as \( n \downarrow - \infty \) (i.e., \( {\mathcal{F}}_{-\infty } = { \cap }_{n}{\mathcal{F}}_{n} \) ), then\n\n\[ E\left( {Y \mid {\mathcal{F}}_{n}}\right) \rightarrow E\left( {Y \mid {\mathcal{F}}_{-\infty }}\right) \;\text{ a.s. and in }{L... | Proof. \( {X}_{n} = E\left( {Y \mid {\mathcal{F}}_{n}}\right) \) is a backwards martingale, so Theorem 5.6.1 and 5.6.2 imply that as \( n \downarrow - \infty ,{X}_{n} \rightarrow {X}_{-\infty } \) a.s. and in \( {L}^{1} \), where\n\n\[ {X}_{-\infty } = E\left( {{X}_{0} \mid {\mathcal{F}}_{-\infty }}\right) = E\left( {E... | Yes |
Strong law of large numbers. Let \( {\xi }_{1},{\xi }_{2},\ldots \) be i.i.d. with \( E\left| {\xi }_{i}\right| < \) \( \infty \) . Let \( {S}_{n} = {\xi }_{1} + \cdots + {\xi }_{n} \), let \( {X}_{-n} = {S}_{n}/n \), and let\n\n\[ \n{\mathcal{F}}_{-n} = \sigma \left( {{S}_{n},{S}_{n + 1},{S}_{n + 2},\ldots }\right) = ... | To compute \( E\left( {{X}_{-n} \mid {\mathcal{F}}_{-n - 1}}\right) \), we observe that if \( j, k \leq n + 1 \), symmetry implies \( E\left( {{\xi }_{j} \mid {\mathcal{F}}_{-n - 1}}\right) = E\left( {{\xi }_{k} \mid {\mathcal{F}}_{-n - 1}}\right) \), so\n\n\[ \nE\left( {{\xi }_{n + 1} \mid {\mathcal{F}}_{-n - 1}}\righ... | Yes |
Ballot theorem. Let \( \left\{ {{\xi }_{j},1 \leq j \leq n}\right\} \) be i.i.d. nonnegative integer-valued r.v.’s, let \( {S}_{k} = {\xi }_{1} + \cdots + {\xi }_{k} \), and let \( G = \left\{ {{S}_{j} < j}\right. \) for \( \left. {1 \leq j \leq n}\right\} \) . Then \[ P\left( {G \mid {S}_{n}}\right) = {\left( 1 - {S}_... | Proof. The result is trivial when \( {S}_{n} \geq n \), so suppose \( {S}_{n} < n \) . Computations in Example 5.6.1 show that \( {X}_{-j} = {S}_{j}/j \) is a martingale w.r.t. \( {\mathcal{F}}_{-j} = \sigma \left( {{S}_{j},\ldots ,{S}_{n}}\right) \) . Let \( T = \inf \left\{ {k \geq - n : {X}_{k} \geq 1}\right\} \) an... | Yes |
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