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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}... | Since \( E{X}_{n} = E{X}_{0} = 1 \) for all \( n \) and \( {X}_{\infty } = 0 \), convergence cannot occur in \( {L}^{1} \). | Yes |
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 \( {X}_{k} = \) \( k{X}_{k - 1} \) with probability \( 1/k \) and \( = 0 \) with probability \( 1 - 1/k \) . From the constructi... | 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. \( P\left( {{Z}_{n} = 0\text{for some}n}\right) = \rho \) 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{\varphi }^{\prime \prime }\left( s\right) = \mathop{\sum }\limits_{{k = 2}}^{\infty }k\left( {k ... | 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\[ E\left( {\bar{X}}_{n}^{p}\right) \leq {\left( \frac{p}{p - 1}\right) }^{p}E{\left( {X}_{n}^{ + }\right) }^{p} \]\n\nConsequently, if \( {Y}_{n} \) is a martingale and \( {Y}_{n}^{ * } = \mathop{\max ... | 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 alw... | 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... | \[ E\left( {{X}_{n}^{2} \mid {\mathcal{F}}_{n - 1}}\right) = {X}_{n - 1}^{2} + E\left( {{\left( {X}_{n} - {X}_{n - 1}\right) }^{2} \mid {\mathcal{F}}_{n - 1}}\right) \] To compute the second term, we observe \[ E\left( {{\left( {X}_{n} - {X}_{n - 1}\right) }^{2} \mid {\mathcal{F}}_{n - 1}}\right) = E\left( {{\left( {Z}... | 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 | 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 \( n \rightarrow \infty \) ,\n\n\[ E\left( {X \mid {\mathcal{F}}_... | 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 \[ {\mathcal{F}}_{-n} = \sigma \left( {{S}_{n},{S}_{n + 1},{S}_{n + 2},\ldots }\right) = \sigm... | 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 \[ E\left( {{\xi }_{n + 1} \mid {\mathcal{F}}_{-n - 1}}\right) = ... | 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 |
Lemma 5.6.4. Suppose \( {X}_{1},{X}_{2},\ldots \) are i.i.d. and let\n\n\[ \n{A}_{n}\left( \varphi \right) = \frac{1}{{\left( n\right) }_{k}}\mathop{\sum }\limits_{i}\varphi \left( {{X}_{{i}_{1}},\ldots ,{X}_{{i}_{k}}}\right)\n\]\n\nwhere the sum is over all sequences of distinct integers \( 1 \leq {i}_{1},\ldots ,{i}_... | Proof. \( {A}_{n}\left( \varphi \right) \in {\mathcal{E}}_{n} \), so\n\n\[ \n{A}_{n}\left( \varphi \right) = E\left( {{A}_{n}\left( \varphi \right) \mid {\mathcal{E}}_{n}}\right) = \frac{1}{{\left( n\right) }_{k}}\mathop{\sum }\limits_{i}E\left( {\varphi \left( {{X}_{{i}_{1}},\ldots ,{X}_{{i}_{k}}}\right) \mid {\mathca... | Yes |
Theorem 5.6.6. If \( {X}_{1},{X}_{2},\ldots \) are exchangeable and take values in \( \{ 0,1\} \) then there is a probability distribution on \( \left\lbrack {0,1}\right\rbrack \) so that\n\n\[ P\left( {{X}_{1} = 1,\ldots ,{X}_{k} = 1,{X}_{k + 1} = 0,\ldots ,{X}_{n} = 0}\right) = {\int }_{0}^{1}{\theta }^{k}{\left( 1 -... | This result is useful for people concerned about the foundations of statistics (see Section 3.7 of Savage (1972)), since from the palatable assumption of symmetry one gets the powerful conclusion that the sequence is a mixture of i.i.d. sequences. Theorem 5.6.6 has been proved in a variety of different ways. See Feller... | No |
Theorem 5.7.1. If \( {X}_{n} \) is a uniformly integrable submartingale then for any stopping time \( N,{X}_{N \land n} \) is uniformly integrable. | Proof. \( {X}_{n}^{ + } \) is a submartingale, so Theorem 5.4.1 implies \( E{X}_{N \land n}^{ + } \leq E{X}_{n}^{ + } \) . Since \( {X}_{n}^{ + } \) is uniformly integrable, it follows from the remark after the definition that\n\n\[ \mathop{\sup }\limits_{n}E{X}_{N \land n}^{ + } \leq \mathop{\sup }\limits_{n}E{X}_{n}^... | Yes |
Theorem 5.7.3. If \( {X}_{n} \) is a uniformly integrable submartingale then for any stopping time \( N \leq \infty \), we have \( E{X}_{0} \leq E{X}_{N} \leq E{X}_{\infty } \), where \( {X}_{\infty } = \lim {X}_{n} \). | Proof. Theorem 5.4.1 implies \( E{X}_{0} \leq E{X}_{N \land n} \leq E{X}_{n} \). Letting \( n \rightarrow \infty \) and observing that Theorem 5.7.1 and 5.5.3 imply \( {X}_{N \land n} \rightarrow {X}_{N} \) and \( {X}_{n} \rightarrow {X}_{\infty } \) in \( {L}^{1} \) gives the desired result. | Yes |
Theorem 5.7.4. Optional Stopping Theorem. If \( L \leq M \) are stopping times and \( {Y}_{M \land n} \) is a uniformly integrable submartingale, then \( E{Y}_{L} \leq E{Y}_{M} \) and\n\n\[{Y}_{L} \leq E\left( {{Y}_{M} \mid {\mathcal{F}}_{L}}\right)\] | Proof. Use the inequality \( E{X}_{N} \leq E{X}_{\infty } \) in Theorem 5.7.3 with \( {X}_{n} = {Y}_{M \land n} \) and \( N = L \) . To prove the second result, let \( A \in {\mathcal{F}}_{L} \) and\n\n\[N = \left\{ \begin{array}{ll} L & \text{ on }A \\ M & \text{ on }{A}^{c} \end{array}\right.\]\n\nis a stopping time ... | Yes |
Theorem 5.7.5. Suppose \( {X}_{n} \) is a submartingale and \( E\left( {\left| {{X}_{n + 1} - {X}_{n}}\right| \mid {\mathcal{F}}_{n}}\right) \leq B \) a.s. If \( N \) is a stopping time with \( {EN} < \infty \) then \( {X}_{N \land n} \) is uniformly integrable and hence \( E{X}_{N} \geq E{X}_{0} \) | Proof. We begin by observing that\n\n\[ \left| {X}_{N \land n}\right| \leq \left| {X}_{0}\right| + \mathop{\sum }\limits_{{m = 0}}^{\infty }\left| {{X}_{m + 1} - {X}_{m}}\right| {1}_{\left( N > m\right) } \]\n\nTo prove uniform integrability, it suffices to show that the right-hand side has finite expectation for then ... | Yes |
Theorem 5.7.6. If \( {X}_{n} \) is a nonnegative supermartingale and \( N \leq \infty \) is a stopping time, then \( E{X}_{0} \geq E{X}_{N} \) where \( {X}_{\infty } = \lim {X}_{n} \), which exists by Theorem 5.2.9. | Proof. Using Theorem 5.4.1 and Fatou's Lemma,\n\n\[ E{X}_{0} \geq \mathop{\liminf }\limits_{{n \rightarrow \infty }}E{X}_{N \land n} \geq E{X}_{N} \] | Yes |
Theorem 5.7.7. Asymmetric simple random walk refers to the special case in which \( P\left( {{\xi }_{i} = 1}\right) = p \) and \( P\left( {{\xi }_{i} = - 1}\right) = q \equiv 1 - p \) with \( p \neq q \) . Without loss of generality we assume \( 1/2 < p < 1 \) .\n\n(a) If \( \varphi \left( x\right) = \{ \left( {1 - p}\... | Proof. Since \( {S}_{n} \) and \( {\xi }_{n + 1} \) are independent, Example 5.1.5 implies that on \( \left\{ {{S}_{n} = m}\right\} \) ,\n\n\[ E\left( {\varphi \left( {S}_{n + 1}\right) \mid {\mathcal{F}}_{n}}\right) = p \cdot {\left( \frac{1 - p}{p}\right) }^{m + 1} + \left( {1 - p}\right) {\left( \frac{1 - p}{p}\righ... | Yes |
Theorem 6.1.1. \( {X}_{n} \) is a Markov chain (with respect to \( {\mathcal{F}}_{n} = \sigma \left( {{X}_{0},{X}_{1},\ldots ,{X}_{n}}\right) \) ) with transition probability \( p \) . | Proof. To prove this, we let \( A = \left\{ {{X}_{0} \in {B}_{0},{X}_{1} \in {B}_{1},\ldots ,{X}_{n} \in {B}_{n}}\right\} ,{B}_{n + 1} = B \), and observe that using the definition of the integral, the definition of \( A \), and the definition of \( {P}_{\mu } \)\n\n\[ \n{\int }_{A}{1}_{\left( {X}_{n + 1} \in B\right) ... | Yes |
Theorem 6.1.2. If \( {X}_{n} \) is a Markov chain with transition probabilities \( p \) and initial distribution \( \mu \), then the finite dimensional distributions are given by (6.1.1). | Proof. Our first step is to show that if \( {X}_{n} \) has transition probability \( p \) then for any bounded measurable \( f \)\n\n\[ E\left( {f\left( {X}_{n + 1}\right) \mid {\mathcal{F}}_{n}}\right) = \int p\left( {{X}_{n},{dy}}\right) f\left( y\right) \]\n\n(6.1.2)\n\nThe desired conclusion is a consequence of the... | No |
Theorem 6.1.3. Monotone class theorem. Let \( \mathcal{A} \) be a \( \pi \) -system that contains \( \Omega \) and let \( \mathcal{H} \) be a collection of real-valued functions that satisfies:\n\n(i) If \( A \in \mathcal{A} \), then \( {1}_{A} \in \mathcal{H} \).\n\n(ii) If \( f, g \in \mathcal{H} \), then \( f + g \)... | Proof. The assumption \( \Omega \in \mathcal{A} \) ,(ii), and (iii) imply that \( \mathcal{G} = \left\{ {A : {1}_{A} \in \mathcal{H}}\right\} \) is a \( \lambda \) -system so by (i) and the \( \pi - \lambda \) theorem, Theorem 2.1.2, \( \mathcal{G} \supset \sigma \left( \mathcal{A}\right) \) . (ii) implies \( \mathcal{... | Yes |
Example 6.2.1. Random walk. Let \( {\xi }_{1},{\xi }_{2},\ldots \in {\mathbf{R}}^{d} \) be independent with distribution \( \mu \) . Let \( {X}_{0} = x \in {\mathbf{R}}^{d} \) and let \( {X}_{n} = {X}_{0} + {\xi }_{1} + \cdots + {\xi }_{n} \) . Then \( {X}_{n} \) is a Markov chain with transition probability. | \[ p\left( {x, A}\right) = \mu \left( {A - x}\right) \] where \( A - x = \{ y - x : y \in A\} \). | Yes |
Lemma 6.2.1. Let \( X \) and \( Y \) take values in \( \left( {S,\mathcal{S}}\right) \) . Suppose \( \mathcal{F} \) and \( Y \) are independent. Let \( X \in \mathcal{F},\varphi \) be a function with \( E\left| {\varphi \left( {X, Y}\right) }\right| < \infty \) and let \( g\left( x\right) = E\left( {\varphi \left( {x, ... | Proof. Suppose first that \( \phi \left( {x, y}\right) = {1}_{A}\left( x\right) {1}_{B}\left( y\right) \) and let \( C \in \mathcal{F} \). \[ E\left( {\varphi \left( {X, Y}\right) ;C}\right) = P\left( {\{ X \in A\} \cap C\cap \{ Y \in B\} }\right) \] \[ = P\left( {\{ X \in A\} \cap C}\right) P\left( {\{ Y \in B\} }\rig... | Yes |
Example 6.2.3. Renewal chain. \( S = \{ 0,1,2,\ldots \} ,{f}_{k} \geq 0 \), and \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{f}_{k} = 1 \) . | \[ p\left( {0, j}\right) = {f}_{j + 1}\;\text{ for }j \geq 0 \] \[ p\left( {i, i - 1}\right) = 1\;\text{ for }i \geq 1 \] \[ p\left( {i, j}\right) = 0\;\text{ otherwise } \] To explain the definition, let \( {\xi }_{1},{\xi }_{2},\ldots \) be i.i.d. with \( P\left( {{\xi }_{m} = j}\right) = {f}_{j} \), let \( {T}_{0} =... | Yes |
Example 6.2.4. \( \mathrm{M}/\mathrm{G}/1 \) queue. In this model, customers arrive according to a Poisson process with rate \( \lambda \) . (M is for Markov and refers to the fact that in a Poisson process the number of arrivals in disjoint time intervals is independent.) Each customer requires an independent amount o... | To define our Markov chain \( {X}_{n} \), let\n\n\[ \n{a}_{k} = {\int }_{0}^{\infty }{e}^{-{\lambda t}}\frac{{\left( \lambda t\right) }^{k}}{k!}{dF}\left( t\right) \]\n\nbe the probability that \( k \) customers arrive during a service time. Let \( {\xi }_{1},{\xi }_{2},\ldots \) be i.i.d. with \( P\left( {{\xi }_{i} =... | Yes |
Theorem 6.3.1. The Markov property. Let \( Y : {\Omega }_{o} \rightarrow \mathbf{R} \) be bounded and measurable.\n\n\[ \n{E}_{\mu }\left( {Y \circ {\theta }_{m} \mid {\mathcal{F}}_{m}}\right) = {E}_{{X}_{m}}Y \n\] | Proof. We begin by proving the result in a special case and then use the \( \pi - \lambda \) and monotone class theorems to get the general result. Let \( A = \left\{ {\omega : {\omega }_{0} \in {A}_{0},\ldots ,{\omega }_{m} \in }\right. \) \( \left. {A}_{m}\right\} \) and \( {g}_{0},\ldots {g}_{n} \) be bounded and me... | Yes |
Theorem 6.3.2. Chapman-Kolmogorov equation.\n\n\[ \n{P}_{x}\left( {{X}_{m + n} = z}\right) = \mathop{\sum }\limits_{y}{P}_{x}\left( {{X}_{m} = y}\right) {P}_{y}\left( {{X}_{n} = z}\right) \n\]\n\nIntuitively, in order to go from \( x \) to \( z \) in \( m + n \) steps we have to be at some \( y \) at time \( m \) and t... | Proof. \( {P}_{x}\left( {{X}_{n + m} = z}\right) = {E}_{x}\left( {{P}_{x}\left( {{X}_{n + m} = z \mid {\mathcal{F}}_{m}}\right) }\right) = {E}_{x}\left( {{P}_{{X}_{m}}\left( {{X}_{n} = z}\right) }\right) \) by the Markov property, Theorem 6.3.1 since \( {1}_{\left( {X}_{n} = z\right) } \circ {\theta }_{m} = {1}_{\left(... | Yes |
Theorem 6.3.3. Let \( {X}_{n} \) be a Markov chain and suppose\n\n\[ P\left( {\left. {{ \cup }_{m = n + 1}^{\infty }\left\{ {{X}_{m} \in {B}_{m}}\right\} }\right| \;{X}_{n}}\right) \geq \delta > 0\;\text{ on }\left\{ {{X}_{n} \in {A}_{n}}\right\} \]\n\nThen \( P\left( {\left\{ {{X}_{n} \in {A}_{n}\text{ i.o. }}\right\}... | Proof. Let \( {\Lambda }_{n} = { \cup }_{m = n + 1}^{\infty }\left\{ {{X}_{m} \in {B}_{m}}\right\} \), let \( \Lambda = \cap {\Lambda }_{n} = \left\{ {{X}_{n} \in {B}_{n}}\right. \) i.o. \( \} \), and let \( \Gamma = \left\{ {{X}_{n} \in {A}_{n}}\right. \) i.o. \( \} \). Let \( {\mathcal{F}}_{n} = \sigma \left( {{X}_{0... | Yes |
Theorem 6.3.4. Strong Markov property. Suppose that for each \( n,{Y}_{n} : {\Omega }_{0} \rightarrow \mathbf{R} \) is measurable and \( \left| {Y}_{n}\right| \leq M \) for all \( n \) . Then\n\n\[ \n{E}_{\mu }\left( {{Y}_{N} \circ {\theta }_{N} \mid {\mathcal{F}}_{N}}\right) = {E}_{{X}_{N}}{Y}_{N}\text{ on }\{ N < \in... | Proof. Let \( A \in {\mathcal{F}}_{N} \) . Breaking things down according to the value of \( N \) .\n\n\[ \n{E}_{\mu }\left( {{Y}_{N} \circ {\theta }_{N};A\cap \{ N < \infty \} }\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{E}_{\mu }\left( {{Y}_{n} \circ {\theta }_{n};A\cap \{ N = n\} }\right)\n\]\n\nSince \( A \... | Yes |
Theorem 6.3.5. Reflection principle. Let \( {\xi }_{1},{\xi }_{2},\ldots \) be independent and identically distributed with a distribution that is symmetric about 0 . Let \( {S}_{n} = {\xi }_{1} + \cdots + {\xi }_{n} \) . If \( a > 0 \) then \[ P\left( {\mathop{\sup }\limits_{{m \leq n}}{S}_{m} > a}\right) \leq {2P}\le... | Proof. Let \( {Y}_{m}\left( \omega \right) = 1 \) if \( m \leq n \) and \( {\omega }_{n - m} > a,{Y}_{m}\left( \omega \right) = 0 \) otherwise. The definition of \( {Y}_{m} \) is chosen so that \( \left( {{Y}_{N} \circ {\theta }_{N}}\right) \left( \omega \right) = 1 \) if \( {\omega }_{n} > a \) (and hence \( N \leq n ... | Yes |
Theorem 6.4.1. \( {P}_{x}\left( {{T}_{y}^{k} < \infty }\right) = {\rho }_{xy}{\rho }_{yy}^{k - 1} \) | Intuitively, in order to make \( k \) visits to \( y \), we first have to go from \( x \) to \( y \) and then return \( k - 1 \) times to \( y \) . Proof. When \( k = 1 \), the result is trivial, so we suppose \( k \geq 2 \) . Let \( Y\left( \omega \right) = 1 \) if \( {\omega }_{n} = y \) for some \( n \geq 1, Y\left(... | Yes |
Theorem 6.4.3. If \( x \) is recurrent and \( {\rho }_{xy} > 0 \) then \( y \) is recurrent and \( {\rho }_{yx} = 1 \) . | Proof. We will first show \( {\rho }_{yx} = 1 \) by showing that if \( {\rho }_{xy} > 0 \) and \( {\rho }_{yx} < 1 \) then \( {\rho }_{xx} < 1 \) . Let \( K = \inf \left\{ {k : {p}^{k}\left( {x, y}\right) > 0}\right\} \) . There is a sequence \( {y}_{1},\ldots ,{y}_{K - 1} \) so that\n\n\[ p\left( {x,{y}_{1}}\right) p\... | Yes |
Theorem 6.4.4. Let \( C \) be a finite closed set. Then \( C \) contains a recurrent state. If \( C \) is irreducible then all states in \( C \) are recurrent. | Proof. In view of Theorem 6.4.3, it suffices to prove the first claim. Suppose it is false. Then for all \( y \in C,{\rho }_{yy} < 1 \) and \( {E}_{x}N\left( y\right) = {\rho }_{xy}/\left( {1 - {\rho }_{yy}}\right) \), but this is ridiculous since it implies\n\n\[ \infty > \mathop{\sum }\limits_{{y \in C}}{E}_{x}N\left... | Yes |
Example 6.4.1. A Seven-state chain. Consider the transition probability:\n\n\[ \n\\begin{matrix} & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 1 & {.3} & 0 & 0 & 0 & {.7} & 0 & 0 \\ 2 & {.1} & {.2} & {.3} & {.4} & 0 & 0 & 0 \\ 3 & 0 & 0 & {.5} & {.5} & 0 & 0 & 0 \\ 4 & 0 & 0 & 0 & {.5} & 0 & {.5} & 0 \\ 5 & {.6} & 0 & 0 & 0 & {.4} & ... | (i) \( {\\rho }_{21} > 0 \) and \( {\\rho }_{12} = 0 \) so 2 must be transient, or we would contradict Theorem 6.4.3. Similarly, \( {\\rho }_{34} > 0 \) and \( {\\rho }_{43} = 0 \) so 3 must be transient\n\n(ii) \( \\{ 1,5\\} \) and \( \\{ 4,6,7\\} \) are irreducible closed sets, so Theorem 6.4.4 implies these states a... | Yes |
Theorem 6.4.5. Decomposition theorem. Let \( R = \left\{ {x : {\rho }_{xx} = 1}\right\} \) be the recurrent states of a Markov chain. \( R \) can be written as \( { \cup }_{i}{R}_{i} \), where each \( {R}_{i} \) is closed and irreducible. | Proof. If \( x \in R \) let \( {C}_{x} = \left\{ {y : {\rho }_{xy} > 0}\right\} \) . By Theorem 6.4.3, \( {C}_{x} \subset R \), and if \( y \in {C}_{x} \) then \( {\rho }_{yx} > 0 \) . From this it follows easily that either \( {C}_{x} \cap {C}_{y} = \varnothing \) or \( {C}_{x} = {C}_{y} \) . To prove the last claim, ... | Yes |
Example 6.4.4. Birth and death chains on \( \\{ 0,1,2,\\ldots \\} \) . Let\n\n\[ p\\left( {i, i + 1}\\right) = {p}_{i}\\;p\\left( {i, i - 1}\\right) = {q}_{i}\\;p\\left( {i, i}\\right) = {r}_{i} \]\n\nwhere \( {q}_{0} = 0 \) . Let \( N = \\inf \\left\\{ {n : {X}_{n} = 0}\\right\\} \) . To analyze this example, we are g... | Using \( {r}_{k} = 1 - \\left( {{p}_{k} + {q}_{k}}\\right) \), we can rewrite the last equation as\n\n\[ {q}_{k}\\left( {\\varphi \\left( k\\right) - \\varphi \\left( {k - 1}\\right) }\\right) = {p}_{k}\\left( {\\varphi \\left( {k + 1}\\right) - \\varphi \\left( k\\right) }\\right) \]\n\n\[ \\text{or}\\varphi \\left( {... | Yes |
Theorem 6.4.6. If \( a < x < b \) then\n\n\[ \n{P}_{x}\left( {{T}_{a} < {T}_{b}}\right) = \frac{\varphi \left( b\right) - \varphi \left( x\right) }{\varphi \left( b\right) - \varphi \left( a\right) }\;{P}_{x}\left( {{T}_{b} < {T}_{a}}\right) = \frac{\varphi \left( x\right) - \varphi \left( a\right) }{\varphi \left( b\r... | Proof. If we let \( T = {T}_{a} \land {T}_{b} \) then \( \varphi \left( {X}_{n \land T}\right) \) is a bounded martingale and \( T < \infty \) a.s. by Theorem 6.3.3, so \( \varphi \left( x\right) = {E}_{x}\varphi \left( {X}_{T}\right) \) by Theorem 5.7.4. Since \( {X}_{T} \in \{ a, b\} \) a.s.,\n\n\[ \n\varphi \left( x... | Yes |
Theorem 6.4.7. 0 is recurrent if and only if \( \varphi \left( M\right) \rightarrow \infty \) as \( M \rightarrow \infty \), i.e., | \[ \varphi \left( \infty \right) \equiv \mathop{\sum }\limits_{{m = 0}}^{\infty }\mathop{\prod }\limits_{{j = 1}}^{m}\frac{{q}_{j}}{{p}_{j}} = \infty \] If \( \varphi \left( \infty \right) < \infty \) then \( {P}_{x}\left( {{T}_{0} = \infty }\right) = \varphi \left( x\right) /\varphi \left( \infty \right) \) . | Yes |
Example 6.4.5. Asymmetric simple random walk. Suppose \( {p}_{j} = p \) and \( {q}_{j} = \) \( 1 - p \) for \( j \geq 1 \) . In this case,\n\n\[ \varphi \left( n\right) = \mathop{\sum }\limits_{{m = 0}}^{{n - 1}}{\left( \frac{1 - p}{p}\right) }^{m} \]\n\nFrom Theorem 6.4.7, it follows that 0 is recurrent if and only if... | From Theorem 6.4.7, it follows that 0 is recurrent if and only if \( p \leq 1/2 \), and if \( p > 1/2 \), then\n\n\[ {P}_{x}\left( {{T}_{0} < \infty }\right) = \frac{\varphi \left( \infty \right) - \varphi \left( x\right) }{\varphi \left( \infty \right) } = {\left( \frac{1 - p}{p}\right) }^{x} \] | Yes |
To probe the boundary between recurrence and transience, suppose \( {p}_{j} = 1/2 + {\epsilon }_{j} \) where \( {\epsilon }_{j} \sim C{j}^{-\alpha } \) as \( j \rightarrow \infty \), and \( {q}_{j} = 1 - {p}_{j} \) . | A little arithmetic shows\n\n\[ \frac{{q}_{j}}{{p}_{j}} = \frac{1/2 - {\epsilon }_{j}}{1/2 + {\epsilon }_{j}} = 1 - \frac{2{\epsilon }_{j}}{1/2 + {\epsilon }_{j}} \approx 1 - {4C}{j}^{-\alpha }\;\text{ for large }j \]\n\nCase 1: \( \alpha > 1 \) . It is easy to show that if \( 0 < {\delta }_{j} < 1 \), then \( \mathop{... | No |
Let \( \mu = \sum k{a}_{k} \) be the mean number of customers that arrive during one service time. We will now show that if \( \mu > 1 \), the chain is transient (i.e., all states are), but if \( \mu \leq 1 \), it is recurrent. | For the case \( \mu > 1 \) , we observe that if \( {\xi }_{1},{\xi }_{2},\ldots \) are i.i.d. with \( P\left( {{\xi }_{m} = j}\right) = {a}_{j + 1} \) for \( j \geq - 1 \) and \( {S}_{n} = \) \( {\xi }_{1} + \cdots + {\xi }_{n} \), then \( {X}_{0} + {S}_{n} \) and \( {X}_{n} \) behave the same until time \( N = \inf \l... | Yes |
Theorem 6.4.8. Suppose \( S \) is irreducible, and \( \varphi \geq 0 \) with \( {E}_{x}\varphi \left( {X}_{1}\right) \leq \varphi \left( x\right) \) for \( x \notin F \), a finite set, and \( \varphi \left( x\right) \rightarrow \infty \) as \( x \rightarrow \infty \), i.e., \( \{ x : \varphi \left( x\right) \leq M\} \)... | Proof. Let \( \tau = \inf \left\{ {n > 0 : {X}_{n} \in F}\right\} \) . Our assumptions imply that \( {Y}_{n} = \varphi \left( {X}_{n \land \tau }\right) \) is a supermartingale. Let \( {T}_{M} = \inf \left\{ {n > 0 : {X}_{n} \in F\text{or}\varphi \left( {X}_{n}\right) > M}\right\} \) . Since \( \{ x : \varphi \left( x\... | Yes |
Example 6.5.1. Random walk. \( S = {\mathbf{Z}}^{d}.p\left( {x, y}\right) = f\left( {y - x}\right) \), where \( f\left( z\right) \geq 0 \) and \( \sum f\left( z\right) = 1 \) . In this case, \( \mu \left( x\right) \equiv 1 \) is a stationary measure since | \[ \mathop{\sum }\limits_{x}p\left( {x, y}\right) = \mathop{\sum }\limits_{x}f\left( {y - x}\right) = 1 \] | Yes |
Asymmetric simple random walk. \( S = \mathbf{Z} \). | \[ p\left( {x, x + 1}\right) = p\;p\left( {x, x - 1}\right) = q = 1 - p \] By the last example, \( \mu \left( x\right) \equiv 1 \) is a stationary measure. When \( p \neq q,\mu \left( x\right) = {\left( p/q\right) }^{x} \) is a second one. To check this, we observe that \[ \mathop{\sum }\limits_{x}\mu \left( x\right) p... | Yes |
The Ehrenfest chain. \( S = \{ 0,1,\ldots, r\} \) . \[ p\left( {k, k + 1}\right) = \left( {r - k}\right) /r\;p\left( {k, k - 1}\right) = k/r \] | In this case, \( \mu \left( x\right) = {2}^{-r}\left( \begin{array}{l} r \\ x \end{array}\right) \) is a stationary distribution. One can check this without pencil and paper by observing that \( \mu \) corresponds to flipping \( r \) coins to determine which urn each ball is to be placed in, and the transitions of the ... | Yes |
Example 6.5.4. Birth and death chains. \( S = \{ 0,1,2,\ldots \} \)\n\n\[ p\left( {x, x + 1}\right) = {p}_{x}\;p\left( {x, x}\right) = {r}_{x}\;p\left( {x, x - 1}\right) = {q}_{x} \]\n\nwith \( {q}_{0} = 0 \) and \( p\left( {i, j}\right) = 0 \) otherwise. In this case, there is the measure\n\n\[ \mu \left( x\right) = \... | Since \( p\left( {x, y}\right) = 0 \) when \( \left| {x - y}\right| > 1 \), it follows that\n\n\[ \mu \left( x\right) p\left( {x, y}\right) = \mu \left( y\right) p\left( {y, x}\right) \;\text{ for all }x, y \]\n\n(6.5.1)\n\nSumming over \( x \) gives\n\n\[ \mathop{\sum }\limits_{x}\mu \left( x\right) p\left( {x, y}\rig... | Yes |
Theorem 6.5.1. Suppose \( p \) is irreducible. A necessary and sufficient condition for the existence of a reversible measure is that (i) \( p\left( {x, y}\right) > 0 \) implies \( p\left( {y, x}\right) > 0 \), and (ii) for any loop \( {x}_{0},{x}_{1},\ldots ,{x}_{n} = {x}_{0} \) with \( \mathop{\prod }\limits_{{1 \leq... | Proof. To prove the necessity of this cycle condition, due to Kolmogorov, we note that irreducibility implies that any stationary measure has \( \mu \left( x\right) > 0 \) for all \( x \), so (6.5.1) implies (i) holds. To check (ii), note that (6.5.1) implies that for the sequences considered above\n\n\[ \mathop{\prod ... | Yes |
Theorem 6.5.2. Let \( x \) be a recurrent state, and let \( T = \inf \left\{ {n \geq 1 : {X}_{n} = x}\right\} \) . Then\n\n\[ \n{\mu }_{x}\left( y\right) = {E}_{x}\left( {\mathop{\sum }\limits_{{n = 0}}^{{T - 1}}{1}_{\left\{ {X}_{n} = y\right\} }}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{P}_{x}\left( {{X}_{n}... | Proof. This is called the \ | No |
Theorem 6.5.3. If \( p \) is irreducible and recurrent (i.e., all states are) then the stationary measure is unique up to constant multiples. | Proof. Let \( \nu \) be a stationary measure and let \( a \in S \) .\n\n\[ \nu \left( z\right) = \mathop{\sum }\limits_{y}\nu \left( y\right) p\left( {y, z}\right) = \nu \left( a\right) p\left( {a, z}\right) + \mathop{\sum }\limits_{{y \neq a}}\nu \left( y\right) p\left( {y, z}\right) \]\n\nUsing the last identity to r... | Yes |
Theorem 6.5.4. If there is a stationary distribution then all states \( y \) that have \( \pi \left( y\right) > \) 0 are recurrent. | Proof. Since \( \pi {p}^{n} = \pi \), Fubini’s theorem implies\n\n\[\n\mathop{\sum }\limits_{x}\pi \left( x\right) \mathop{\sum }\limits_{{n = 1}}^{\infty }{p}^{n}\left( {x, y}\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\pi \left( y\right) = \infty\n\]\n\nwhen \( \pi \left( y\right) > 0 \) . Using Theorem 6.4.2 ... | Yes |
Theorem 6.5.5. If \( p \) is irreducible and has stationary distribution \( \pi \), then\n\n\[ \pi \left( x\right) = 1/{E}_{x}{T}_{x} \] | Proof. Irreducibility implies \( \pi \left( x\right) > 0 \) so all states are recurrent by Theorem 6.5.4. From Theorem 6.5.2,\n\n\[ {\mu }_{x}\left( y\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{P}_{x}\left( {{X}_{n} = y,{T}_{x} > n}\right) \]\n\ndefines a stationary measure with \( {\mu }_{x}\left( x\right) = 1... | Yes |
Theorem 6.5.6. If \( p \) is irreducible then the following are equivalent:\n\n(i) Some \( x \) is positive recurrent.\n\n(ii) There is a stationary distribution.\n\n(iii) All states are positive recurrent. | Proof. (i) implies (ii). If \( x \) is positive recurrent then\n\n\[ \pi \left( y\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{P}_{x}\left( {{X}_{n} = y,{T}_{x} > n}\right) /{E}_{x}{T}_{x} \]\n\ndefines a stationary distribution.\n\n(ii) implies (iii). Theorem 6.5.5 implies \( \pi \left( y\right) = 1/{E}_{y}{T}_{... | Yes |
Let \( \mu = \sum k{a}_{k} \) be the mean number of customers that arrive during one service time. We will now show that the chain is positive recurrent if and only if \( \mu < 1 \). | First, suppose that \( \mu < 1 \). When \( {X}_{n} > 0 \), the chain behaves like a random walk that has jumps with mean \( \mu - 1 \), so if \( N = \inf \left\{ {n \geq 0 : {X}_{n} = 0}\right\} \) then \( {X}_{N \land n} - \left( {\mu - 1}\right) \left( {N \land n}\right) \) is a martingale. If \( {X}_{0} = x > 0 \) t... | Yes |
In the \( M/M/\infty \) queue, where \( {X}_{n + 1} = \mathop{\sum }\limits_{{m = 1}}^{{Xn}}{\xi }_{n, m} + {Y}_{n + 1} \), with \( {\xi }_{n, m} \) being i.i.d. Bernoulli with mean \( p \) and \( {Y}_{n + 1} \) being an independent Poisson with mean \( \lambda \), show that if \( {X}_{n} \) is Poisson with mean \( \mu... | It follows from properties of the Poisson distribution that if \( {X}_{n} \) is Poisson with mean \( \mu \), then \( {X}_{n + 1} \) is Poisson with mean \( {\mu p} + \lambda \). Setting \( \mu = {\mu p} + \lambda \), we find that a Poisson distribution with mean \( \mu = \lambda /\left( {1 - p}\right) \) is a stationar... | No |
Theorem 6.5.7. If \( p \) is irreducible and has a stationary distribution \( \pi \) then any other stationary measure is a multiple of \( \pi \) . | Proof. Since \( p \) is irreducible, \( \pi \left( x\right) > 0 \) for all \( x \) . Let \( \varphi \) be a concave function that is bounded on \( \left( {0,\infty }\right) \), e.g., \( \varphi \left( x\right) = x/\left( {x + 1}\right) \) . Define the entropy of \( \mu \) by\n\n\[ \mathcal{E}\left( \mu \right) = \matho... | Yes |
Theorem 6.6.1. Suppose \( y \) is recurrent. For any \( x \in S \), as \( n \rightarrow \infty \)\n\n\[ \frac{{N}_{n}\left( y\right) }{n} \rightarrow \frac{1}{{E}_{y}{T}_{y}}{1}_{\left\{ {T}_{y} < \infty \right\} }\;{P}_{x}\text{-a.s. } \] | Proof. Suppose first that we start at \( y \) . Let \( R\left( k\right) = \min \left\{ {n \geq 1 : {N}_{n}\left( y\right) = k}\right\} = \) the time of the \( k \) th return to \( y \) . Let \( {t}_{k} = R\left( k\right) - R\left( {k - 1}\right) \), where \( R\left( 0\right) = 0 \) . Since we have assumed \( {X}_{0} = ... | Yes |
Lemma 6.6.2. If \( {\rho }_{xy} > 0 \) then \( {d}_{y} = {d}_{x} \) . | Proof. Let \( K \) and \( L \) be such that \( {p}^{K}\left( {x, y}\right) > 0 \) and \( {p}^{L}\left( {y, x}\right) > 0 \) . ( \( x \) is recurrent, so \( \left. {{\rho }_{yx} > 0\text{.}}\right) \)\n\n\[ \n{p}^{K + L}\left( {y, y}\right) \geq {p}^{L}\left( {y, x}\right) {p}^{K}\left( {x, y}\right) > 0 \n\] \n\nso \( ... | Yes |
Lemma 6.6.3. If \( {d}_{x} = 1 \) then \( {p}^{m}\left( {x, x}\right) > 0 \) for \( m \geq {m}_{0} \) . | Proof by example. Suppose \( 4,7 \in {I}_{x}.{p}^{m + n}\left( {x, x}\right) \geq {p}^{m}\left( {x, x}\right) {p}^{n}\left( {x, x}\right) \) so \( {I}_{x} \) is closed under addition, i.e., if \( m, n \in {I}_{x} \) then \( m + n \in {I}_{x} \) . A little calculation shows that in the example\n\n\[ \n{I}_{x} \supset \{... | No |
The state of a deck of \( n \) cards can be represented by a permutation, \( \pi \left( i\right) \) giving the location of the \( i \) th card. Consider the following method of mixing the deck up. The top card is removed and inserted under one of the \( n - 1 \) cards that remain. I claim that by following the bottom c... | This card stays at the bottom until the first time \( \left( {T}_{1}\right) \) a card is inserted below it. It is easy to see that when the \( k \) th card is inserted below the original bottom card (at time \( {T}_{k} \) ), all \( k \) ! arrangements of the cards below are equally likely, so at time \( {\tau }_{n} = {... | Yes |
Lemma 6.7.1. Suppose \( p \) is irreducible, recurrent, and all states have period \( d \) . Fix \( x \in S \), and for each \( y \in S \), let \( {K}_{y} = \left\{ {n \geq 1 : {p}^{n}\left( {x, y}\right) > 0}\right\} \) . (i) There is an \( {r}_{y} \in \{ 0,1,\ldots, d - 1\} \) so that if \( n \in {K}_{y} \) then \( n... | Proof. (i) Let \( m\left( y\right) \) be such that \( {p}^{m\left( y\right) }\left( {y, x}\right) > 0 \) . If \( n \in {K}_{y} \) then \( {p}^{n + m\left( y\right) }\left( {x, x}\right) \) is positive so \( d \mid \left( {n + m}\right) \) . Let \( {r}_{y} = \left( {d - m\left( y\right) }\right) {\;\operatorname{mod}\;d... | Yes |
Theorem 6.7.2. Convergence theorem, periodic case. Suppose p is irreducible, has a stationary distribution \( \pi \), and all states have period \( d \) . Let \( x \in S \), and let \( {S}_{0},{S}_{1},\ldots ,{S}_{d - 1} \) be the cyclic decomposition of the state space with \( x \in {S}_{0} \) . If \( y \in {S}_{r} \)... | Proof. If \( y \in {S}_{0} \) then using (iii) in Lemma 6.7.1 and applying Theorem 6.6.4 to \( {p}^{d} \) shows\n\n\[ \mathop{\lim }\limits_{{m \rightarrow \infty }}{p}^{md}\left( {x, y}\right) \text{ exists } \]\n\nTo identify the limit, we note that (6.6.1) implies\n\n\[ \frac{1}{n}\mathop{\sum }\limits_{{m = 1}}^{n}... | Yes |
Theorem 6.7.3. Suppose \( p \) is irreducible, recurrent, and all states have period \( d \) , \( \mathcal{T} = \sigma \left( {\left\{ {{X}_{0} \in {S}_{r}}\right\} : 0 \leq r < d}\right) . | Proof. We build up to the general result in three steps.\n\nCase 1. Suppose \( P\left( {{X}_{0} = x}\right) = 1 \) . Let \( {T}_{0} = 0 \), and for \( n \geq 1 \), let \( {T}_{n} = \inf \left\{ {m > {T}_{n - 1}}\right. \) : \( \left. {{X}_{m} = x}\right\} \) be the time of the \( n \) th return to \( x \) . Let\n\n\[ \... | Yes |
Theorem 6.7.4. Suppose \( {X}_{0} \) has initial distribution \( \mu \) . The equations\n\n\[ h\left( {{X}_{n}, n}\right) = {E}_{\mu }\left( {Z \mid {\mathcal{F}}_{n}}\right) \;\text{ and }\;Z = \mathop{\lim }\limits_{{n \rightarrow \infty }}h\left( {{X}_{n}, n}\right) \]\n\nset up a 1-1 correspondence between bounded ... | Proof. Let \( Z \in \mathcal{T} \), write \( Z = {Y}_{n} \circ {\theta }_{n} \), and let \( h\left( {x, n}\right) = {E}_{x}{Y}_{n} \).\n\n\[ {E}_{\mu }\left( {Z \mid {\mathcal{F}}_{n}}\right) = {E}_{\mu }\left( {{Y}_{n} \circ {\theta }_{n} \mid {\mathcal{F}}_{n}}\right) = h\left( {{X}_{n}, n}\right) \]\n\nby the Markov... | Yes |
Example 6.7.1. Simple random walk in d dimensions. We begin by constructing a coupling for this process. Let \( {i}_{1},{i}_{2},\ldots \) be i.i.d. uniform on \( \{ 1,\ldots, d\} \) . Let \( {\xi }_{1},{\xi }_{2},\ldots \) and \( {\eta }_{1},{\eta }_{2},\ldots \) be i.i.d. uniform on \( \{ - 1,1\} \) . Let \( {e}_{j} \... | Let \( {L}_{0} = \left\{ {z \in {\mathbf{Z}}^{d} : {z}^{1} + \cdots + {z}^{d}}\right. \) is even \( \} \) and \( {L}_{1} = {\mathbf{Z}}^{d} - {L}_{0} \) . Although we have only defined the notion for the recurrent case, it should be clear that \( {L}_{0},{L}_{1} \) is the cyclic decomposition of the state space for sim... | No |
Theorem 6.7.5. For d-dimensional simple random walk, \n\n\[ \n\mathcal{T} = \sigma \left( {\left\{ {{X}_{0} \in {L}_{i}}\right\}, i = 0,1}\right) \n\] | Proof. Let \( x, y \in {L}_{i} \), and let \( {X}_{n},{Y}_{n} \) be a realization of the coupling defined above for \( {X}_{0} = x \) and \( {Y}_{0} = y \) . Let \( h\left( {x, n}\right) \) be a bounded space-time harmonic function. The martingale property implies \( h\left( {x,0}\right) = {E}_{x}h\left( {{X}_{n}, n}\r... | Yes |
Example 6.7.2. Ornstein’s coupling. Let \( p\left( {x, y}\right) = f\left( {y - x}\right) \) be the transition probability for an irreducible aperiodic random walk on \( \mathbf{Z} \) . To prove that the tail \( \sigma \) -field is trivial, pick \( M \) large enough so that the random walk generated by the probability ... | \[ {Y}_{n} = \left\{ \begin{array}{ll} {Y}_{n - 1} + {Z}_{n} & \text{ if }\left| {Z}_{n}\right| > m \\ {Y}_{n - 1} + {W}_{n} & \text{ if }\left| {Z}_{n}\right| \leq m \end{array}\right. \] In words, the big jumps are taken in parallel and the small jumps are independent. The recurrence of one-dimensional random walks w... | Yes |
Example 6.7.3. Random walk on a tree. To facilitate definitions, we will consider the system as a random walk on a group with 3 generators \( a, b, c \) that have \( {a}^{2} = {b}^{2} = \) \( {c}^{2} = e \), the identity element. To form the random walk, let \( {\xi }_{1},{\xi }_{2},\ldots \) be i.i.d. with \( P\left( ... | \[ p\left( {j, j - 1}\right) = 1/3\;p\left( {j, j + 1}\right) = 2/3\;\text{ for }j \geq 1 \] As \( n \rightarrow \infty ,{L}_{n} \rightarrow \infty \) . From this, it follows easily that the word \( {X}_{n} \) has a limit in the sense that the \( i \) th letter \( {X}_{n}^{i} \) stays the same for large \( n \) . Let \... | No |
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