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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}_{{k}}}\right) \mid {\mathcal{E}... | 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\[ \n{P}_{x}\left( {{X}_{m + n} = z}\right) = \mathop{\sum }\limits_{y}{P}_{x}\left( {{X}_{m} = y}\right) {P}_{y}\left( {{X}_{n} = z}\right) \n\] | 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 \( y \) .\n\nProof. 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\le... | 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\\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} & 0 & 0... | (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 th... | 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\\[ \np\\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 a... | We start by setting \( \\varphi \\left( 0\\right) = 0 \) and \( \\varphi \\left( 1\\right) = 1 \) . For the martingale property to hold when \( {X}_{n} = k \\geq 1 \), we must have\n\n\\[ \\varphi \\left( k\\right) = {p}_{k}\\varphi \\left( {k + 1}\\right) + {r}_{k}\\varphi \\left( k\\right) + {q}_{k}\\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, | \[ \varphi \left( n\right) = \mathop{\sum }\limits_{{m = 0}}^{{n - 1}}{\left( \frac{1 - p}{p}\right) }^{m} \] 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 \[ {P}_{x}\left( {{T}_{0} < \infty }\right) = \frac{\varphi \left( \infty \right) - \varphi \left( ... | 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\left( {x, y}\right) = \mu \left( {y + 1}\right) p\left( {y + 1, y}\right) ... | 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 |
Random walks on graphs. A graph is described by giving a countable set of vertices \( S \) and an adjacency matrix \( {a}_{ij} \) that has \( {a}_{ij} = 1 \) if \( i \) and \( j \) are adjacent and 0 otherwise. To have an undirected graph with no loops, we suppose \( {a}_{ij} = {a}_{ji} \) and \( {a}_{ii} = 0 \) . If w... | It is clear from the definition that\n\n\[ \mu \left( i\right) p\left( {i, j}\right) = {a}_{ij} = {a}_{ji} = \mu \left( j\right) p\left( {j, i}\right) \]\n\nso \( \mu \) is a reversible measure for \( p \) . A little thought reveals that if we assume only that\n\n\[ {a}_{ij} = {a}_{ji} \geq 0,\;\mu \left( i\right) = \m... | 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 |
Countable state space. If \( S \) is countable and there is a point \( a \) with \( {\rho }_{xa} > 0 \) for all \( x \) (a condition slightly weaker than irreducibility) then we can take \( A = \{ a\}, B = \{ b\} \), where \( b \) is any state with \( p\left( {a, b}\right) > 0,\mu = {\delta }_{b} \) the point mass at \... | Conversely, if \( S \) is countable and \( \left( {{A}^{\prime },{B}^{\prime }}\right) \) is a pair for which (i) and (ii) hold, then we can without loss of generality reduce \( {B}^{\prime } \) to a single point \( b \) . Having done this, if we set \( A = \{ b\} \), pick \( c \) so that \( p\left( {b, c}\right) > 0 \... | No |
Lemma 6.8.1. \( v\bar{p} = \bar{p} \) and \( \bar{p}v = p \) . | Proof. Before giving the proof, we would like to remind the reader that measures multiply the transition probability on the left, i.e., in the first case we want to show \( {\mu v}\bar{p} = \mu \bar{p} \) . If we first make a transition according to \( v \) and then one according to \( \bar{p} \) , this amounts to one ... | No |
Lemma 6.8.3. If \( \mu \) is a probability measure on \( \left( {S,\mathcal{S}}\right) \) then\n\n\[ {E}_{\mu }f\left( {X}_{n}\right) = {E}_{\mu }\bar{f}\left( {\bar{X}}_{n}\right) \] | Proof. Observe that if \( {X}_{n} \) and \( {\bar{X}}_{n} \) are constructed as in Lemma 6.8.2, and \( P\left( {{\bar{X}}_{0} \in }\right. \) \( S) = 1 \) then \( {X}_{0} = {\bar{X}}_{0} \) and \( {X}_{n} \) is obtained from \( {\bar{X}}_{n} \) by making a transition according to \( v \) . | No |
Theorem 6.8.4. Let \( \lambda \left( C\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{2}^{-n}{\bar{p}}^{n}\left( {\alpha, C}\right) \) . In the recurrent case, if \( \lambda \left( C\right) > 0 \) then \( {P}_{\alpha }\left( {{\bar{X}}_{n} \in C\text{i.o.}}\right) = 1 \) . For \( \lambda \) -a.e. \( x,{P}_{x}\left(... | Proof. The first conclusion follows from Lemma 6.3.3. For the second let \( D = \{ x \) : \( \left. {{P}_{x}\left( {R < \infty }\right) < 1}\right\} \) and observe that if \( {p}^{n}\left( {\alpha, D}\right) > 0 \) for some \( n \), then\n\n\[ \n{P}_{\alpha }\left( {{\bar{X}}_{m} = \alpha \text{ i.o. }}\right) \leq \in... | Yes |
Theorem 6.8.5. In the recurrent case, there is a stationary measure. | Proof. Let \( R = \inf \left\{ {n \geq 1 : {\bar{X}}_{n} = \alpha }\right\} \), and let\n\n\[ \bar{\mu }\left( C\right) = {E}_{\alpha }\left( {\mathop{\sum }\limits_{{n = 0}}^{{R - 1}}{1}_{\left\{ {\bar{X}}_{n} \in C\right\} }}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{P}_{\alpha }\left( {{\bar{X}}_{n} \in C, ... | Yes |
Lemma 6.8.6. If \( \nu \) is a \( \sigma \) -finite stationary measure for \( p \), then \( \nu \left( A\right) < \infty \) and \( \bar{\nu } = \nu \bar{p} \) is a stationary measure for \( \bar{p} \) with \( \bar{\nu }\left( \alpha \right) < \infty \) . | Proof. We will first show that \( \nu \left( A\right) < \infty \) . If \( \nu \left( A\right) = \infty \) then part (ii) of the definition implies \( \nu \left( C\right) = \infty \) for all sets \( C \) with \( \rho \left( C\right) > 0 \) . If \( B = { \cup }_{i}{B}_{i} \) with \( \nu \left( {B}_{i}\right) < \infty \) ... | Yes |
Theorem 6.8.7. Suppose \( p \) is recurrent. If \( \nu \) is a \( \sigma \) -finite stationary measure then \( \nu = \bar{\nu }\left( \alpha \right) \mu \), where \( \mu \) is the measure constructed in the proof of Theorem 6.8.5. | Proof. By Lemma 6.8.6, it suffices to prove that if \( \bar{\nu } \) is a stationary measure for \( \bar{p} \) with \( \bar{\nu }\left( \alpha \right) < \infty \) then \( \bar{\nu } = \bar{\nu }\left( \alpha \right) \bar{\mu } \) . Repeating the proof of Theorem 6.5.3 with \( a = \alpha \), it is easy to show that \( \... | Yes |
Theorem 6.8.8. Let \( {X}_{n} \) be an aperiodic recurrent Harris chain with stationary distribution \( \pi \) . If \( {P}_{x}\left( {R < \infty }\right) = 1 \) then as \( n \rightarrow \infty \) , \[ \begin{Vmatrix}{{p}^{n}\left( {x, \cdot }\right) - \pi \left( \cdot \right) }\end{Vmatrix} \rightarrow 0 \] | Proof. In view of Lemma 6.8.3, it suffices to prove the result for \( \bar{p} \) . We begin by observing that the existence of a stationary probability measure and the uniqueness result in Theorem 6.8.7 imply that the measure constructed in Theorem 6.8.5 has \( {E}_{\alpha }R = \bar{\mu }\left( S\right) < \infty \) . A... | Yes |
Exponential service time. Suppose \( P\left( {{\eta }_{n} > x}\right) = {e}^{-{\beta x}} \) and \( E{\zeta }_{n} > E{\eta }_{n} \) . Let \( T = \inf \left\{ {n : {S}_{n} > 0}\right\} \) and \( L = {S}_{T} \), setting \( L = - \infty \) if \( T = \infty \) . The lack of memory property of the exponential distribution im... | \[ P\left( {M = x}\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }{r}^{k}\left( {1 - r}\right) {e}^{-{\beta x}}{\beta }^{k}{x}^{k - 1}/\left( {k - 1}\right) ! = {\beta r}\left( {1 - r}\right) {e}^{-{\beta x}\left( {1 - r}\right) } \] | Yes |
Poisson arrivals. Suppose \( P\left( {{\zeta }_{n} > x}\right) = {e}^{-{\alpha x}} \) and \( E{\zeta }_{n} > E{\eta }_{n} \) . Let \( {\bar{S}}_{n} = - {S}_{n} \) . Reversing time as in (ii) of Exercise 6.8.14, we see (for \( n \geq 1 \) ) | \[ P\left( {\mathop{\max }\limits_{{0 \leq k < n}}{\bar{S}}_{k} < {\bar{S}}_{n} \in A}\right) = P\left( {\mathop{\min }\limits_{{1 \leq k \leq n}}{\bar{S}}_{k} > 0,{\bar{S}}_{n} \in A}\right) \] Let \( {\psi }_{n}\left( A\right) \) be the common value of the last two expression and let \( \psi \left( A\right) = \mathop... | Yes |
Theorem 7.1.1. If \( {X}_{0},{X}_{1},\ldots \) is a stationary sequence and \( g : {\mathbf{R}}^{\{ 0,1,\ldots \} } \rightarrow \mathbf{R} \) is measurable then \( {Y}_{k} = g\left( {{X}_{k},{X}_{k + 1},\ldots }\right) \) is a stationary sequence. | Proof. If \( x \in {\mathbf{R}}^{\{ 0,1,\ldots \} } \), let \( {g}_{k}\left( x\right) = g\left( {{x}_{k},{x}_{k + 1},\ldots }\right) \), and if \( B \in {\mathcal{R}}^{\{ 0,1,\ldots \} } \) let\n\n\[ A = \left\{ {x : \left( {{g}_{0}\left( x\right) ,{g}_{1}\left( x\right) ,\ldots }\right) \in B}\right\} \]\n\nTo check s... | Yes |
Let \( \left( {\Omega ,\mathcal{F}, P}\right) \) be a probability space. A measurable map \( \varphi : \Omega \rightarrow \Omega \) is said to be measure preserving if \( P\left( {{\varphi }^{-1}A}\right) = P\left( A\right) \) for all \( A \in \mathcal{F} \). Let \( {\varphi }^{n} \) be the \( n \) th iterate of \( \va... | To check this, let \( B \in {\mathcal{R}}^{n + 1} \) and \( A = \left\{ {\omega : \left( {{X}_{0}\left( \omega \right) ,\ldots ,{X}_{n}\left( \omega \right) }\right) \in B}\right\} \). Then\n\n\[ P\left( {\left( {{X}_{k},\ldots ,{X}_{k + n}}\right) \in B}\right) = P\left( {{\varphi }^{k}\omega \in A}\right) = P\left( {... | Yes |
Theorem 7.1.2. Any stationary sequence \( \left\{ {{X}_{n}, n \geq 0}\right\} \) can be embedded in a two-sided stationary sequence \( \left\{ {{Y}_{n} : n \in \mathbf{Z}}\right\} \) . | Proof. We observe that\n\n\[ P\left( {{Y}_{-m} \in {A}_{0},\ldots ,{Y}_{n} \in {A}_{m + n}}\right) = P\left( {{X}_{0} \in {A}_{0},\ldots ,{X}_{m + n} \in {A}_{m + n}}\right) \]\n\nis a consistent set of finite dimensional distributions, so a trivial generalization of the Kolmogorov extension theorem implies there is a ... | Yes |
We begin by observing that if \( \Omega = {\mathbf{R}}^{\{ 0,1,\ldots \} } \) and \( \varphi \) is the shift operator, then an invariant set \( A \) has \( \{ \omega : \omega \in A\} = \{ \omega : {\varphi \omega } \in A\} \in \) \( \sigma \left( {{X}_{1},{X}_{2},\ldots }\right) \) . | Iterating gives \[ A \in { \cap }_{n = 1}^{\infty }\sigma \left( {{X}_{n},{X}_{n + 1},\ldots }\right) = \mathcal{T},\;\text{ the tail }\sigma \text{-field } \] so \( \mathcal{I} \subset \mathcal{I} \) . For an i.i.d. sequence, Kolmogorov’s 0-1 law implies \( \mathcal{T} \) is trivial, so \( \mathcal{I} \) is trivial an... | Yes |
Suppose the state space \( S \) is countable and the stationary distribution has \( \pi \left( x\right) > 0 \) for all \( x \in S \) . By Theorems 6.5.4 and 6.4.5, all states are recurrent, and we can write \( S = \cup {R}_{i} \), where the \( {R}_{i} \) are disjoint irreducible closed sets. If \( {X}_{0} \in {R}_{i} \... | \[ {E}_{\pi }\left( {{1}_{A} \mid {\mathcal{F}}_{n}}\right) = {E}_{\pi }\left( {{1}_{A} \circ {\theta }_{n} \mid {\mathcal{F}}_{n}}\right) = h\left( {X}_{n}\right) \] where \( h\left( x\right) = {E}_{x}{1}_{A} \) . Lévy’s 0-1 law implies that the left-hand side converges to \( {1}_{A} \) as \( n \rightarrow \infty \) .... | Yes |
Rotation of the circle is not ergodic if \( \theta = m/n \) where \( m < n \) are positive integers. If \( B \) is a Borel subset of \( \lbrack 0,1/n) \) and \[ A = { \cup }_{k = 0}^{n - 1}\left( {B + k/n}\right) \] then \( A \) is invariant. | Conversely, if \( \theta \) is irrational, then \( \varphi \) is ergodic. To prove this, we need a fact from Fourier analysis. If \( f \) is a measurable function on \( \lbrack 0,1) \) with \( \int {f}^{2}\left( x\right) {dx} < \infty \), then \( f \) can be written as \( f\left( x\right) = \mathop{\sum }\limits_{k}{c}... | Yes |
Bernoulli shift is ergodic. | To prove this, we recall that the stationary sequence \( {Y}_{n}\left( \omega \right) = {\varphi }^{n}\left( \omega \right) \) can be represented as\n\n\[ \n{Y}_{n} = \mathop{\sum }\limits_{{m = 0}}^{\infty }{2}^{-\left( {m + 1}\right) }{X}_{n + m} \]\n\nwhere \( {X}_{0},{X}_{1},\ldots \) are i.i.d. with \( P\left( {{X... | No |
Theorem 7.1.3. Let \( g : {\mathbf{R}}^{\{ 0,1,\ldots \} } \rightarrow \mathbf{R} \) be measurable. If \( {X}_{0},{X}_{1},\ldots \) is an ergodic stationary sequence, then \( {Y}_{k} = g\left( {{X}_{k},{X}_{k + 1},\ldots }\right) \) is ergodic. | Proof. Suppose \( {X}_{0},{X}_{1},\ldots \) is defined on sequence space with \( {X}_{n}\left( \omega \right) = {\omega }_{n} \) . If \( B \) has \( \left\{ {\omega : \left( {{Y}_{0},{Y}_{1},\ldots }\right) \in B}\right\} = \left\{ {\omega : \left( {{Y}_{1},{Y}_{2},\ldots }\right) \in B}\right\} \) then \( A = \left\{ ... | Yes |
Lemma 7.2.2. Maximal ergodic lemma. Let \( {X}_{j}\left( \omega \right) = X\left( {{\varphi }^{j}\omega }\right) ,{S}_{k}\left( \omega \right) = {X}_{0}\left( \omega \right) + \\) \( \\ldots + {X}_{k - 1}\left( \omega \right) \), and \( {M}_{k}\left( \omega \right) = \\max \\left( {0,{S}_{1}\left( \omega \right) ,\\ldo... | Proof. If \( j \\leq k \) then \( {M}_{k}\\left( {\\varphi \\omega }\\right) \\geq {S}_{j}\\left( {\\varphi \\omega }\\right) \), so adding \( X\\left( \\omega \\right) \) gives\n\n\[ \nX\\left( \\omega \\right) + {M}_{k}\\left( {\\varphi \\omega }\\right) \\geq X\\left( \\omega \\right) + {S}_{j}\\left( {\\varphi \\om... | Yes |
Theorem 7.2.3. Wiener’s maximal inequality. Let \( {X}_{j}\left( \omega \right) = X\left( {{\varphi }^{j}\omega }\right) ,{S}_{k}\left( \omega \right) = \) \( {X}_{0}\left( \omega \right) + \cdots + {X}_{k - 1}\left( \omega \right) ,{A}_{k}\left( \omega \right) = {S}_{k}\left( \omega \right) /k \), and \( {D}_{k} = \ma... | Proof. Let \( B = \left\{ {{D}_{k} > \alpha }\right\} \) . Applying Lemma 7.2.2 to \( {X}^{\prime } = X - \alpha \), with \( {X}_{j}^{\prime }\left( \omega \right) = \) \( {X}^{\prime }\left( {{\varphi }^{j}\omega }\right) ,{S}_{k}^{\prime } = {X}_{0}^{\prime }\left( \omega \right) + \cdots + {X}_{k - 1}^{\prime } \), ... | Yes |
Let \( {X}_{n} \) be an irreducible Markov chain on a countable state space that has a stationary distribution \( \pi \) . Let \( f \) be a function with\n\n\[ \mathop{\sum }\limits_{x}\left| {f\left( x\right) }\right| \pi \left( x\right) < \infty \]\n\nIn Example 7.1.7, we showed that \( \mathcal{I} \) is trivial, so ... | \[ \frac{1}{n}\mathop{\sum }\limits_{{m = 0}}^{{n - 1}}f\left( {X}_{m}\right) \rightarrow \mathop{\sum }\limits_{x}f\left( x\right) \pi \left( x\right) \;\text{ a.s. and in }{L}^{1} \] | Yes |
Example 7.2.3. Rotation of the circle. \( \Omega = \lbrack 0,1)\varphi \left( \omega \right) = \left( {\omega + \theta }\right) {\;\operatorname{mod}\;1} \) . Suppose that \( \theta \in \left( {0,1}\right) \) is irrational, so that by a result in Section \( {7.1}\mathcal{I} \) is trivial. If we set \( X\left( \omega \r... | \[ \frac{1}{n}\mathop{\sum }\limits_{{m = 0}}^{{n - 1}}{1}_{\left( {\varphi }^{m}\omega \in A\right) } \rightarrow \left| A\right| \;\text{ a.s. } \] where \( \left| A\right| \) denotes the Lebesgue measure of \( A \) . The last result for \( \omega = 0 \) is usually called Weyl's equidistribution theorem, although Boh... | Yes |
Theorem 7.2.4. If \( A = \lbrack a, b) \) then the exceptional set is \( \varnothing \) . | Proof. Let \( {A}_{k} = \lbrack a + 1/k, b - 1/k) \) . If \( b - a > 2/k \), the ergodic theorem implies\n\n\[\n\frac{1}{n}\mathop{\sum }\limits_{{m = 0}}^{{n - 1}}{1}_{{A}_{k}}\left( {{\varphi }^{m}\omega }\right) \rightarrow b - a - \frac{2}{k}\n\]\n\nfor \( \omega \in {\Omega }_{k} \) with \( P\left( {\Omega }_{k}\r... | Yes |
Benford's law. As Gelfand first observed, the equidistribution theorem says something interesting about \( {2}^{m} \) . Let \( \theta = {\log }_{10}2,1 \leq k \leq 9 \), and \( {A}_{k} = \left\lbrack {{\log }_{10}k,{\log }_{10}\left( {k + 1}\right) }\right) \) where \( {\log }_{10}y \) is the logarithm of \( y \) to th... | \[ \frac{1}{n}\mathop{\sum }\limits_{{m = 0}}^{{n - 1}}{1}_{A}\left( {{\varphi }^{m}0}\right) \rightarrow {\log }_{10}\left( \frac{k + 1}{k}\right) \] A little thought reveals that the first digit of \( {2}^{m} \) is \( k \) if and only if \( {m\theta }{\;\operatorname{mod}\;1} \in {A}_{k} \) . The numerical values of ... | Yes |
Example 7.2.5. Bernoulli shift. \( \Omega = \lbrack 0,1),\varphi \left( \omega \right) = \left( {2\omega }\right) {\;\operatorname{mod}\;1} \) . Let \( {i}_{1},\ldots ,{i}_{k} \in \) \( \{ 0,1\} \), let \( r = {i}_{1}{2}^{-1} + \cdots + {i}_{k}{2}^{-k} \), and let \( X\left( \omega \right) = 1 \) if \( r \leq \omega < ... | \[ \frac{1}{n}\mathop{\sum }\limits_{{m = 0}}^{{n - 1}}X\left( {{\varphi }^{m}\omega }\right) \rightarrow {2}^{-k}\;\text{ a.s. } \] i.e., in almost every \( \omega \in \lbrack 0,1) \) the pattern \( {i}_{1},\ldots ,{i}_{k} \) occurs with its expected frequency. Since there are only a countable number of patterns of fi... | Yes |
Theorem 7.3.1. As \( n \rightarrow \infty ,{R}_{n}/n \rightarrow E\left( {{1}_{A} \mid \mathcal{I}}\right) \) a.s. | Proof. Suppose \( {X}_{1},{X}_{2},\ldots \) are constructed on \( {\left( {\mathbf{R}}^{d}\right) }^{\{ 0,1,\ldots \} } \) with \( {X}_{n}\left( \omega \right) = {\omega }_{n} \), and let \( \varphi \) be the shift operator. It is clear that\n\n\[ \n{R}_{n} \geq \mathop{\sum }\limits_{{m = 1}}^{n}{1}_{A}\left( {{\varph... | Yes |
Theorem 7.3.2. Let \( {X}_{1},{X}_{2},\ldots \) be a stationary sequence taking values in \( \mathbf{Z} \) with \( E\left| {X}_{i}\right| < \infty \) . Let \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \), and let \( A = \left\{ {{S}_{1} \neq 0,{S}_{2} \neq 0,\ldots }\right\} \) . (i) If \( E\left( {{X}_{1} \mid \mathcal{I}}... | Proof. If \( E\left( {{X}_{1} \mid \mathcal{I}}\right) = 0 \) then the ergodic theorem implies \( {S}_{n}/n \rightarrow 0 \) a.s. Now\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}\left( {\mathop{\max }\limits_{{1 \leq k \leq n}}\left| {S}_{k}\right| /n}\right) = \mathop{\limsup }\limits_{{n \rightarrow \inft... | Yes |
Theorem 7.3.3. If \( P\left( {{X}_{n} \in A}\right. \) at least once \( ) = 1 \), then under \( P\left( {\cdot \mid {X}_{0} \in A}\right) ,{t}_{n} = \) \( {T}_{n} - {T}_{n - 1} \) is a stationary sequence with \( E\left( {{T}_{1} \mid {X}_{0} \in A}\right) = 1/P\left( {{X}_{0} \in A}\right) \) . | Proof. We first show that under \( P\left( {\cdot \mid {X}_{0} \in A}\right) ,{t}_{1},{t}_{2},\ldots \) is stationary. To cut down on ...'s, we will only show that\n\n\[ P\left( {{t}_{1} = m,{t}_{2} = n \mid {X}_{0} \in A}\right) = P\left( {{t}_{2} = m,{t}_{3} = n \mid {X}_{0} \in A}\right) \]\n\nIt will be clear that ... | Yes |
Theorem 7.3.4. Suppose \( \varphi : \Omega \rightarrow \Omega \) preserves \( P \), that is, \( P \circ {\varphi }^{-1} = P \) . (i) \( {T}_{A} < \infty \) a.s. on \( A \), that is, \( P\left( {\omega \in A,{T}_{A} = \infty }\right) = 0 \) . (ii) \( \left\{ {{\varphi }^{n}\left( \omega \right) \in A\text{i.o.}}\right\}... | Proof. Let \( B = \left\{ {\omega \in A,{T}_{A} = \infty }\right\} \) . A little thought shows that if \( \omega \in {\varphi }^{-m}B \) then \( {\varphi }^{m}\left( \omega \right) \in A \), but \( {\varphi }^{n}\left( \omega \right) \notin A \) for \( n > m \), so the \( {\varphi }^{-m}B \) are pairwise disjoint. The ... | Yes |
Example 7.4.3. Longest common subsequences. Given are ergodic stationary sequences \( {X}_{1},{X}_{2},{X}_{3},\ldots \) and \( {Y}_{1},{Y}_{2},{Y}_{3},\ldots \) be Let \( {L}_{m, n} = \max \left\{ {K : {X}_{{i}_{k}} = {Y}_{{j}_{k}}}\right. \) for \( 1 \leq k \leq K \), where \( \left. {m < {i}_{1} < {i}_{2}\ldots < {i}... | \[ {L}_{0, m} + {L}_{m, n} \geq {L}_{0, n} \] so \( {X}_{m, n} = - {L}_{m, n} \) is subadditive. \( 0 \leq {L}_{0, n} \leq n \) so (iv) holds. Applying Theorem 7.4.1 now, we conclude that \[ {L}_{0, n}/n \rightarrow \gamma = \mathop{\sup }\limits_{{m \geq 1}}E\left( {{L}_{0, m}/m}\right) \] | Yes |
Example 7.5.1. Products of random matrices. Suppose \( {A}_{1},{A}_{2},\ldots \) is a stationary sequence of \( k \times k \) matrices with positive entries and let\n\n\[ \n{\alpha }_{m, n}\left( {i, j}\right) = \left( {{A}_{m + 1}\cdots {A}_{n}}\right) \left( {i, j}\right) \n\] \n\ni.e., the entry in row \( i \) of co... | To check (iv), we observe that\n\n\[ \n\mathop{\prod }\limits_{{m = 1}}^{n}{A}_{m}\left( {1,1}\right) \leq {\alpha }_{0, n}\left( {1,1}\right) \leq {k}^{n - 1}\mathop{\prod }\limits_{{m = 1}}^{n}\left( {\mathop{\sup }\limits_{{i, j}}{A}_{m}\left( {i, j}\right) }\right) \n\] \n\nor taking logs\n\n\[ \n- \mathop{\sum }\l... | Yes |
Increasing sequences in random permutations. Let \( \pi \) be a permutation of \( \{ 1,2,\ldots, n\} \) and let \( \ell \left( \pi \right) \) be the length of the longest increasing sequence in \( \pi \) . That is, the largest \( k \) for which there are integers \( {i}_{1} < {i}_{2}\ldots < {i}_{k} \) so that \( \pi \... | Hammersley (1970) attacked this problem by putting a rate one Poisson process in the plane, and for \( s < t \in \lbrack 0,\infty ) \), letting \( {Y}_{s, t} \) denote the length of the longest increasing path lying in the square \( {R}_{s, t} \) with vertices \( \left( {s, s}\right) ,\left( {s, t}\right) ,\left( {t, t... | No |
Lemma 7.5.1. \( \tau \left( n\right) /\sqrt{n} \rightarrow 1 \) a.s. | Proof. Let \( {S}_{n} \) be the number of points in \( {R}_{0,\sqrt{n}}.{S}_{n} - {S}_{n - 1} \) are independent Poisson r.v.’s with mean 1, so the strong law of large numbers implies \( {S}_{n}/n \rightarrow 1 \) a.s. If \( \epsilon > 0 \) then for large \( n,{S}_{n\left( {1 - \epsilon }\right) } < n < {S}_{n\left( {1... | Yes |
Suppose \( {p}_{0} = 0 \), let \( {X}_{0, m} \) be the birth time of the first member of generation \( m \) , and let \( {X}_{m, n} \) be the time lag necessary for that individual to have an offspring in generation \( n \) . In case of ties, pick an individual at random from those in generation \( m \) born at time \(... | \[ {X}_{0, n}/n \rightarrow \gamma \;\text{ a.s. } \] The limit is constant because the sequences \( \left\{ {{X}_{{nk},\left( {n + 1}\right) k}, n \geq 0}\right\} \) are i.i.d. | Yes |
Consider \( {\mathbf{Z}}^{d} \) as a graph with edges connecting each \( x, y \in {\mathbf{Z}}^{d} \) with \( \left| {x - y}\right| = 1 \) . Assign an independent nonnegative random variable \( \tau \left( e\right) \) to each edge that represents the time required to traverse the edge going in either direction. If \( e... | Clearly \( {X}_{0, m} + {X}_{m, n} \geq {X}_{0, n}.\;{X}_{0, n} \geq 0 \) so if \( {E\tau }\left( {x, y}\right) < \infty \) then (iv) holds, and Theorem 7.4.1 implies that \( {X}_{0, n}/n \rightarrow X \) a.s. To see that the limit is constant, enumerate the edges in some order \( {e}_{1},{e}_{2},\ldots \) and observe ... | Yes |
Theorem 7.5.2. For any passage time distribution \( F \) with \( F\left( 0\right) = 0 \), there is a convex set \( A \) so that for any \( \epsilon > 0 \) we have with probability one\n\n\[{\xi }_{t} \subset \left( {1 + \epsilon }\right) {tA}\text{for all}t\text{sufficiently large}\]\n\nand \( \left| {{\xi }_{t}^{\epsi... | Ignoring the boring details of how to state things precisely, the last result says \( {\xi }_{t}/t \rightarrow \) \( A \) a.s. It implies that \( {a}_{n}/n \rightarrow \gamma \) a.s., where \( \gamma = 1/\sup \left\{ {{x}_{1} : x \in A}\right\} \) . (Use the convexity and reflection symmetry of \( A \) .) When the dist... | No |
Theorem 8.1.1. Let \( {\Omega }_{o} = \{ \) functions \( \omega : \lbrack 0,\infty ) \rightarrow \mathbf{R}\} \) and \( {\mathcal{F}}_{o} \) be the \( \sigma \) -field generated by the finite dimensional sets \( \left\{ {\omega : \omega \left( {t}_{i}\right) \in {A}_{i}}\right. \) for \( \left. {1 \leq i \leq n}\right\... | This follows from a generalization of Kolmogorov's extension theorem, (7.1) in the Appendix. We will not bother with the details since at this point we are at the dead end referred to above. | No |
Theorem 8.1.2. Let \( T < \infty \) and \( x \in \mathbf{R}.{\nu }_{x} \) assigns probability one to paths \( \omega \) : \( {\mathbf{Q}}_{2} \rightarrow \mathbf{R} \) that are uniformly continuous on \( {\mathbf{Q}}_{2} \cap \left\lbrack {0, T}\right\rbrack \) . | Proof. By translation invariance and scaling (8.1.1), we can without loss of generality suppose \( {B}_{0} = 0 \) and prove the result for \( T = 1 \) . In this case, part (b) of the definition and the scaling relation imply\n\n\[ \n{E}_{0}{\left( \left| {B}_{t} - {B}_{s}\right| \right) }^{4} = {E}_{0}{\left| {B}_{t - ... | Yes |
Theorem 8.1.3. Suppose \( E{\left| {X}_{s} - {X}_{t}\right| }^{\beta } \leq K{\left| t - s\right| }^{1 + \alpha } \) where \( \alpha ,\beta > 0 \) . If \( \gamma < \alpha /\beta \) then with probability one there is a constant \( C\left( \omega \right) \) so that | Proof. Let \( {G}_{n} = \left\{ {\left| {X\left( {i/{2}^{n}}\right) - X\left( {\left( {i - 1}\right) /{2}^{n}}\right) }\right| \leq {2}^{-{\gamma n}}}\right. \) for all \( \left. {0 < i \leq {2}^{n}}\right\} \) . Chebyshev’s inequality implies \( P\left( {\left| Y\right| > a}\right) \leq {a}^{-\beta }E{\left| Y\right| ... | No |
Lemma 8.1.4. On \( {H}_{N} = { \cap }_{n = N}^{\infty }{G}_{n} \) we have\n\n\[ \left| {X\left( q\right) - X\left( r\right) }\right| \leq \frac{3}{1 - {2}^{-\gamma }}{\left| q - r\right| }^{\gamma } \]\n\nfor \( q, r \in {\mathbf{Q}}_{2} \cap \left\lbrack {0,1}\right\rbrack \) with \( \left| {q - r}\right| < {2}^{-N} \... | Proof of Lemma 8.1.4. Let \( q, r \in {\mathbf{Q}}_{2} \cap \left\lbrack {0,1}\right\rbrack \) with \( 0 < r - q < {2}^{-N} \) . For some \( m \geq N \) we can write\n\n\[ r = i{2}^{-m} + {2}^{-r\left( 1\right) } + \cdots + {2}^{-r\left( \ell \right) } \]\n\n\[ q = \left( {i - 1}\right) {2}^{-m} - {2}^{-q\left( 1\right... | Yes |
Theorem 8.1.5. Brownian paths are Hölder continuous for any exponent \( \gamma < 1/2 \) . | It is easy to show: | No |
Theorem 8.1.6. With probability one, Brownian paths are not Lipschitz continuous (and hence not differentiable) at any point. | Proof. Fix a constant \( C < \infty \) and let \( {A}_{n} = \{ \omega \) : there is an \( s \in \left\lbrack {0,1}\right\rbrack \) so that \( \left| {{B}_{t} - {B}_{s}}\right| \leq C\left| {t - s}\right| \) when \( \left| {t - s}\right| \leq 3/n\} \) . For \( 1 \leq k \leq n - 2 \), let\n\n\[ \n{Y}_{k, n} = \max \left\... | Yes |
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