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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\[ \n{E}_{\mu }f\left( {X}_{n}\right) = {E}_{\mu }\bar{f}\left( {\bar{X}}_{n}\right) \n\] | 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 and the sequence i... | 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\n\n\[ A = { \cup }_{k = 0}^{n - 1}\left( {B + k/n}\right) \]\n\nthen \( A \) 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}... | No |
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) ,\ldots ,... | 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 \omega }\right) = {S}_{j + 1}\... | 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 |
Since \( \mathcal{I} \) is trivial, the ergodic theorem implies that\n\n\[ \frac{1}{n}\mathop{\sum }\limits_{{m = 0}}^{{n - 1}}{X}_{m} \rightarrow E{X}_{0}\;\text{ a.s. and in }{L}^{1} \]\n\nThe a.s. convergence is the strong law of large numbers. | Remark. We can prove the \( {L}^{1} \) convergence in the law of large numbers without invoking the ergodic theorem. To do this, note that\n\n\[ \frac{1}{n}\mathop{\sum }\limits_{{m = 1}}^{n}{X}_{m}^{ + } \rightarrow E{X}^{ + }\;\text{ a.s. }\;E\left( {\frac{1}{n}\mathop{\sum }\limits_{{m = 1}}^{n}{X}_{m}^{ + }}\right)... | 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 |
Example 7.2.4. 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 o... | \[ \frac{1}{n}\mathop{\sum }\limits_{{m = 0}}^{{n - 1}}{1}_{A}\left( {{\varphi }^{m}0}\right) \rightarrow {\log }_{10}\left( \frac{k + 1}{k}\right) \] 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 |
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 column \( j \) of... | 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 }\li... | 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 8.1.6. 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 |
Theorem 8.2.2. If \( Z \in \mathcal{C} \) is bounded then for all \( s \geq 0 \) and \( x \in {\mathbf{R}}^{d} \) , \[ {E}_{x}\left( {Z \mid {\mathcal{F}}_{s}^{ + }}\right) = {E}_{x}\left( {Z \mid {\mathcal{F}}_{s}^{o}}\right) \] | Proof. As in the proof of Theorem 8.2.1, it suffices to prove the result when \[ Z = \mathop{\prod }\limits_{{m = 1}}^{n}{f}_{m}\left( {B\left( {t}_{m}\right) }\right) \] and the \( {f}_{m} \) are bounded and measurable. In this case, \( Z \) can be written as \( X\left( {Y \circ {\theta }_{s}}\right) \) , where \( X \... | Yes |
Theorem 8.2.3. Blumenthal’s 0-1 law. If \( A \in {\mathcal{F}}_{0}^{ + } \) then for all \( x \in {\mathbf{R}}^{d} \) , \[ {P}_{x}\left( A\right) \in \{ 0,1\} \] | Proof. Using \( A \in {\mathcal{F}}_{0}^{ + } \), Theorem 8.2.2, and \( {\mathcal{F}}_{0}^{o} = \sigma \left( {B}_{0}\right) \) is trivial under \( {P}_{x} \) gives \[ {1}_{A} = {E}_{x}\left( {{1}_{A} \mid {\mathcal{F}}_{0}^{ + }}\right) = {E}_{x}\left( {{1}_{A} \mid {\mathcal{F}}_{0}^{o}}\right) = {P}_{x}\left( A\righ... | Yes |
Theorem 8.2.4. If \( \tau = \inf \left\{ {t \geq 0 : {B}_{t} > 0}\right\} \) then \( {P}_{0}\left( {\tau = 0}\right) = 1 \) . | Proof. \( {P}_{0}\left( {\tau \leq t}\right) \geq {P}_{0}\left( {{B}_{t} > 0}\right) = 1/2 \) since the normal distribution is symmetric about 0 . Letting \( t \downarrow 0 \), we conclude\n\n\[ \n{P}_{0}\left( {\tau = 0}\right) = \mathop{\lim }\limits_{{t \downarrow 0}}{P}_{0}\left( {\tau \leq t}\right) \geq 1/2 \n\]\... | Yes |
Theorem 8.2.6. If \( {B}_{t} \) is a Brownian motion starting at 0, then so is the process defined by \( {X}_{0} = 0 \) and \( {X}_{t} = {tB}\left( {1/t}\right) \) for \( t > 0 \) . | Proof. Here we will check the second definition of Brownian motion. To do this, we note: (i) If \( 0 < {t}_{1} < \ldots < {t}_{n} \), then \( \left( {X\left( {t}_{1}\right) ,\ldots, X\left( {t}_{n}\right) }\right) \) has a multivariate normal distribution with mean 0 . (ii) \( E{X}_{s} = 0 \) and if \( s < t \) then\n\... | Yes |
Theorem 8.2.7. If \( A \in \mathcal{T} \) then either \( {P}_{x}\left( A\right) \equiv 0 \) or \( {P}_{x}\left( A\right) \equiv 1 \) . | Proof. Since the tail \( \sigma \) -field of \( B \) is the same as the germ \( \sigma \) -field for \( X \), it follows that \( {P}_{0}\left( A\right) \in \{ 0,1\} \) . To improve this to the conclusion given, observe that \( A \in {\mathcal{F}}_{1}^{\prime } \), so \( {1}_{A} \) can be written as \( {1}_{D} \circ {\t... | Yes |
Theorem 8.2.8. Let \( {B}_{t} \) be a one-dimensional Brownian motion starting at 0 then with probability 1 ,\n\n\[ \mathop{\limsup }\limits_{{t \rightarrow \infty }}{B}_{t}/\sqrt{t} = \infty \;\mathop{\liminf }\limits_{{t \rightarrow \infty }}{B}_{t}/\sqrt{t} = - \infty \] | Proof. Let \( K < \infty \) . By Exercise 2.3.1 and scaling\n\n\[ {P}_{0}\left( {{B}_{n}/\sqrt{n} \geq K\text{ i.o. }}\right) \geq \mathop{\limsup }\limits_{{n \rightarrow \infty }}{P}_{0}\left( {{B}_{n} \geq K\sqrt{n}}\right) = {P}_{0}\left( {{B}_{1} \geq K}\right) > 0 \]\n\nso the 0-1 law in Theorem 8.2.7 implies the... | No |
Theorem 8.3.1. If \( G \) is an open set and \( T = \inf \left\{ {t \geq 0 : {B}_{t} \in G}\right\} \) then \( T \) is a stopping time. | Proof. Since \( G \) is open and \( t \rightarrow {B}_{t} \) is continuous, \( \{ T < t\} = { \cup }_{q < t}\left\{ {{B}_{q} \in G}\right\} \), where the union is over all rational \( q \), so \( \{ T < t\} \in {\mathcal{F}}_{t} \) . Here we need to use the rationals to get a countable union, and hence a measurable set... | Yes |
Theorem 8.3.2. If \( {T}_{n} \) is a sequence of stopping times and \( {T}_{n} \downarrow T \) then \( T \) is a stopping time. | \[ \text{Proof.}\{ T < t\} = { \cup }_{n}\left\{ {{T}_{n} < t}\right\} \text{.} \] | No |
Theorem 8.3.3. If \( {T}_{n} \) is a sequence of stopping times and \( {T}_{n} \uparrow T \) then \( T \) is a stopping time. | \[ \text{Proof.}\{ T \leq t\} = { \cap }_{n}\left\{ {{T}_{n} \leq t}\right\} \text{.} \] | No |
Theorem 8.3.4. If \( K \) is a closed set and \( T = \inf \left\{ {t \geq 0 : {B}_{t} \in K}\right\} \) then \( T \) is a stopping time. | Proof. Let \( B\left( {x, r}\right) = \{ y : \left| {y - x}\right| < r\} \), let \( {G}_{n} = { \cup }_{x \in K}B\left( {x,1/n}\right) \) and let \( {T}_{n} = \inf \{ t \geq \) \( \left. {0 : {B}_{t} \in {G}_{n}}\right\} \) . Since \( {G}_{n} \) is open, it follows from Theorem 8.3.1 that \( {T}_{n} \) is a stopping ti... | Yes |
Theorem 8.3.5. If \( S \leq T \) are stopping times then \( {\mathcal{F}}_{S} \subset {\mathcal{F}}_{T} \) . | Proof. If \( A \in {\mathcal{F}}_{S} \) then \( A \cap \{ T \leq t\} = \left( {A\cap \{ S \leq t\} }\right) \cap \{ T \leq t\} \in {\mathcal{F}}_{t} \) . | Yes |
Theorem 8.3.6. If \( {T}_{n} \downarrow T \) are stopping times then \( {\mathcal{F}}_{T} = \cap \mathcal{F}\left( {T}_{n}\right) \) . | Proof. Theorem 8.3.5 implies \( \mathcal{F}\left( {T}_{n}\right) \supset {\mathcal{F}}_{T} \) for all \( n \) . To prove the other inclusion, let \( A \in \cap \mathcal{F}\left( {T}_{n}\right) \) . Since \( A \cap \left\{ {{T}_{n} < t}\right\} \in {\mathcal{F}}_{t} \) and \( {T}_{n} \downarrow T \), it follows that \( ... | Yes |
Theorem 8.4.1. Under \( {P}_{0},\left\{ {{T}_{a}, a \geq 0}\right\} \) has stationary independent increments. | Proof. The first step is to notice that if \( 0 < a < b \) then\n\n\[ \n{T}_{b} \circ {\theta }_{{T}_{a}} = {T}_{b} - {T}_{a} \n\] \n\nso if \( f \) is bounded and measurable, the strong Markov property,8.3.7 and translation invariance imply\n\n\[ \n{E}_{0}\left( {f\left( {{T}_{b} - {T}_{a}}\right) \mid {\mathcal{F}}_{... | Yes |
Reflection principle. Let \( a > 0 \) and let \( {T}_{a} = \inf \left\{ {t : {B}_{t} = a}\right\} \) . Then | \[ {P}_{0}\left( {{T}_{a} < t}\right) = 2{P}_{0}\left( {{B}_{t} \geq a}\right) \] (8.4.4) Intuitive proof. We observe that if \( {B}_{s} \) hits \( a \) at some time \( s < t \), then the strong Markov property implies that \( {B}_{t} - B\left( {T}_{a}\right) \) is independent of what happened before time \( {T}_{a} \)... | Yes |
The distribution of \( L = \sup \left\{ {t \leq 1 : {B}_{t} = 0}\right\} \) . | By (8.2.4),\n\n\[ \n{P}_{0}\left( {L \leq s}\right) = {\int }_{-\infty }^{\infty }{p}_{s}\left( {0, x}\right) {P}_{x}\left( {{T}_{0} > 1 - s}\right) {dx} \]\n\n\[ \n= 2{\int }_{0}^{\infty }{\left( 2\pi s\right) }^{-1/2}\exp \left( {-{x}^{2}/{2s}}\right) {\int }_{1 - s}^{\infty }{\left( 2\pi {r}^{3}\right) }^{-1/2}x\exp... | Yes |
Theorem 8.4.2. With probability 1,\n\n\\[ \n\\mathop{\\limsup }\\limits_{{\\delta \\rightarrow 0}}\\operatorname{osc}\\left( \\delta \\right) /{\\left( \\delta \\log \\left( 1/\\delta \\right) \\right) }^{1/2} \\leq 6 \n\\]\n\nRemark. The constant 6 is not the best possible because the end of the proof is sloppy. Lévy ... | Proof. Let \\( {I}_{m, n} = \\left\\lbrack {m{2}^{-n},\\left( {m + 1}\\right) {2}^{-n}}\\right\\rbrack \\), and \\( {\\Delta }_{m, n} = \\sup \\left\\{ {\\left| {{B}_{t} - B\\left( {m{2}^{-n}}\\right) }\\right| : t \\in {I}_{m, n}}\\right\\} \\) . From (8.4.4) and the scaling relation, it follows that\n\n\\[ \nP\\left(... | Yes |
Theorem 8.5.1. Let \( {X}_{t} \) be a right continuous martingale adapted to a right continuous filtration. If \( T \) is a bounded stopping time, then \( E{X}_{T} = E{X}_{0} \) . | Proof. Let \( n \) be an integer so that \( P\left( {T \leq n - 1}\right) = 1 \) . As in the proof of the strong Markov property, let \( {T}_{m} = \left( {\left\lbrack {{2}^{m}T}\right\rbrack + 1}\right) /{2}^{m}.{Y}_{k}^{m} = X\left( {k{2}^{-m}}\right) \) is a martingale with respect to \( {\mathcal{F}}_{k}^{m} = \mat... | No |
Theorem 8.5.2. \( {B}_{t} \) is a martingale w.r.t. the \( \sigma \) -fields \( {\mathcal{F}}_{t} \) defined in Section 8.2. | Proof. The Markov property implies that\n\n\[ \n{E}_{x}\left( {{B}_{t} \mid {\mathcal{F}}_{s}}\right) = {E}_{{B}_{s}}\left( {B}_{t - s}\right) = {B}_{s} \n\]\n\nsince symmetry implies \( {E}_{y}{B}_{u} = y \) for all \( u \geq 0 \) . | Yes |
Theorem 8.5.3. If \( a < x < b \) then \( {P}_{x}\left( {{T}_{a} < {T}_{b}}\right) = \left( {b - x}\right) /\left( {b - a}\right) \) . | Proof. Let \( T = {T}_{a} \land {T}_{b} \) . Theorem 8.2.8 implies that \( T < \infty \) a.s. Using Theorems 8.5.1 and 8.5.2, it follows that \( x = {E}_{x}B\left( {T \land t}\right) \) . Letting \( t \rightarrow \infty \) and using the bounded convergence theorem, it follows that\n\n\[ x = a{P}_{x}\left( {{T}_{a} < {T... | Yes |
Theorem 8.5.4. \( {B}_{t}^{2} - t \) is a martingale. | Proof. Writing \( {B}_{t}^{2} = {\left( {B}_{s} + {B}_{t} - {B}_{s}\right) }^{2} \) we have\n\n\[ \n{E}_{x}\left( {{B}_{t}^{2} \mid {\mathcal{F}}_{s}}\right) = {E}_{x}\left( {{B}_{s}^{2} + 2{B}_{s}\left( {{B}_{t} - {B}_{s}}\right) + {\left( {B}_{t} - {B}_{s}\right) }^{2} \mid {\mathcal{F}}_{s}}\right) \n\]\n\n\[ \n= {B... | Yes |
Theorem 8.5.5. Let \( T = \inf \left\{ {t : {B}_{t} \notin \left( {a, b}\right) }\right\} \), where \( a < 0 < b \) . \[ {E}_{0}T = - {ab} \] | Proof Theorem 8.5.1 and 8.5.4 imply \( {E}_{0}\left( {{B}^{2}\left( {T \land t}\right) }\right) = {E}_{0}\left( {T \land t}\right) ) \) . Letting \( t \rightarrow \infty \) and using the monotone convergence theorem gives \( {E}_{0}\left( {T \land t}\right) \uparrow {E}_{0}T \) . Using the bounded convergence theorem a... | Yes |
Theorem 8.5.6. \( \\exp \\left( {\\theta {B}_{t} - \\left( {{\\theta }^{2}t/2}\\right) }\\right) \) is a martingale. | Proof. Bringing \( \\exp \\left( {\\theta {B}_{s}}\\right) \) outside\n\n\[ \n{E}_{x}\\left( {\\exp \\left( {\\theta {B}_{t}}\\right) \\mid {\\mathcal{F}}_{s}}\\right) = \\exp \\left( {\\theta {B}_{s}}\\right) E\\left( {\\exp \\left( {\\theta \\left( {{B}_{t} - {B}_{s}}\\right) }\\right) \\mid {\\mathcal{F}}_{s}}\\righ... | Yes |
Theorem 8.5.7. If \( {T}_{a} = \inf \left\{ {t : {B}_{t} = a}\right\} \) then \( {E}_{0}\exp \left( {-\lambda {T}_{a}}\right) = \exp \left( {-a\sqrt{2\lambda }}\right) \). | Proof. Theorem 8.5.1 and 8.5.6 imply that \( 1 = {E}_{0}\exp \left( {{\theta B}\left( {T \land t}\right) - {\theta }^{2}\left( {{T}_{a} \land t}\right) /2}\right) \). Taking \( \theta = \sqrt{2\lambda } \), letting \( t \rightarrow \infty \) and using the bounded convergence theorem gives \( 1 = {E}_{0}\exp \left( {a\s... | Yes |
Theorem 8.5.8. If \( u\left( {t, x}\right) \) is a polynomial in \( t \) and \( x \) with\n\n\[ \frac{\partial u}{\partial t} + \frac{1}{2}\frac{{\partial }^{2}u}{\partial {x}^{2}} = 0 \]\n\nthen \( u\left( {t,{B}_{t}}\right) \) is a martingale. | Proof. Let \( {p}_{t}\left( {x, y}\right) = {\left( 2\pi \right) }^{-1/2}{t}^{-1/2}\exp \left( {-{\left( y - x\right) }^{2}/{2t}}\right) \) . The first step is to check that \( {p}_{t} \) satisfies the heat equation: \( \partial {p}_{t}/\partial t = \left( {1/2}\right) {\partial }^{2}{p}_{t}/\partial {y}^{2} \) .\n\n\[... | Yes |
Theorem 8.5.9. If \( T = \inf \left\{ {t : {B}_{t} \notin \left( {-a, a}\right) }\right\} \) then \( E{T}^{2} = 5{a}^{4}/3 \) . | Proof. Theorem 8.5.1 implies\n\n\[ E\left( {B{\left( T \land t\right) }^{4} - 6\left( {T \land t}\right) B{\left( T \land t\right) }^{2}}\right) = - {3E}{\left( T \land t\right) }^{2}. \]\n\nFrom Theorem 8.5.5, we know that \( {ET} = {a}^{2} < \infty \). Letting \( t \rightarrow \infty \), using the dominated convergen... | Yes |
Theorem 8.5.10. Suppose \( v \in {C}^{2} \), i.e., all first and second order partial derivatives exist and are continuous, and \( v \) has compact support. Then\n\n\[ v\left( {B}_{t}\right) - {\int }_{0}^{t}\frac{1}{2}{\Delta v}\left( {B}_{s}\right) {ds}\;\text{ is a martingale. } \] | Proof. Repeating the proof of Theorem 8.5.8\n\n\[ \frac{\partial }{\partial t}{E}_{x}v\left( {B}_{t}\right) = \int v\left( y\right) \frac{\partial }{\partial t}{p}_{t}\left( {x, y}\right) {dy} \]\n\n\[ = \int \frac{1}{2}v\left( y\right) \left( {{\Delta }_{y}{p}_{t}\left( {x, y}\right) }\right) {dy} \]\n\n\[ = \int \fra... | Yes |
Theorem 8.5.11. If \( \left| x\right| < R \) then \( {E}_{x}{S}_{R} = \left( {{R}^{2} - {\left| x\right| }^{2}}\right) /d \) . | Proof. It follows from Theorem 8.5.4 that \( {\left| {B}_{t}\right| }^{2} - {dt} = \mathop{\sum }\limits_{{i = 1}}^{d}{\left( {B}_{t}^{i}\right) }^{2} - t \) is a martingale. Theorem 8.5.1 implies \( {\left| x\right| }^{2} = E{\left| {B}_{{S}_{R} \land t}\right| }^{2} - {dE}\left( {{S}_{R} \land t}\right) \) . Letting ... | Yes |
Lemma 8.5.12. \( \varphi \left( x\right) = {E}_{x}\varphi \left( {B}_{\tau }\right) \) | Proof. Define \( \psi \left( x\right) = g\left( \left| x\right| \right) \) to be \( {C}^{2} \) and have compact support, and have \( \psi \left( x\right) = \) \( \phi \left( x\right) \) when \( r < \left| x\right| < R \) . Theorem 8.5.10 implies that \( \psi \left( x\right) = {E}_{x}\psi \left( {B}_{t \land \tau }\righ... | Yes |
Theorem 8.5.13. As \( t \rightarrow \infty ,\left| {B}_{t}\right| \rightarrow \infty \) a.s. | Proof. Let \( {A}_{n} = \left\{ {\left| {B}_{t}\right| > {n}^{1 - \epsilon }}\right. \) for all \( \left. {t \geq {S}_{n}}\right\} \) . The strong Markov property implies\n\n\[ \n{P}_{x}\left( {A}_{n}^{c}\right) = {E}_{x}\left( {{P}_{B\left( {S}_{n}\right) }\left( {{S}_{{n}^{1 - \epsilon }} < \infty }\right) }\right) =... | Yes |
Theorem 8.5.14. Suppose \( g\left( t\right) \) is positive and decreasing. Then\n\n\[ \n{P}_{0}\left( {\left| {B}_{t}\right| \leq g\left( t\right) \sqrt{t}\text{ i.o. as }t \uparrow \infty }\right) = 1\text{ or }0 \]\n\naccording as \( {\int }^{\infty }g{\left( t\right) }^{d - 2}/{tdt} = \infty \) or \( < \infty \) . | Here the absence of the lower limit implies that we are only concerned with the behavior of the integral \ | No |
Theorem 8.6.1. Suppose that \( f \in {C}^{2} \), i.e., it has two continuous derivatives. The with probability one, for all \( t \geq 0 \) , \[ f\left( {B}_{t}\right) - f\left( {B}_{0}\right) = {\int }_{0}^{t}{f}^{\prime }\left( {B}_{s}\right) d{B}_{s} + \frac{1}{2}{\int }_{0}^{t}{f}^{\prime \prime }\left( {B}_{s}\righ... | Proof. To derive (8.6.1) we let \( {t}_{i}^{n} = {ti}/{2}^{n} \) for \( 0 \leq i \leq k\left( n\right) = {2}^{n} \) . From calculus we know that for any \( a \) and \( b \), there is a \( c\left( {a, b}\right) \) in between \( a \) and \( b \) such that \[ f\left( b\right) - f\left( a\right) = \left( {b - a}\right) {f}... | Yes |
Lemma 8.6.2. If (i) measures \( {\mu }_{n} \) on \( \left\lbrack {0, t}\right\rbrack \) converge weakly to \( {\mu }_{\infty } \), a finite measure, and (ii) \( {g}_{n} \) is a sequence of functions with \( \left| {g}_{n}\right| \leq K \) that have the property that whenever \( {s}_{n} \in \left\lbrack {0, t}\right\rbr... | Proof. By letting \( {\mu }_{n}^{\prime }\left( A\right) = {\mu }_{n}\left( A\right) /{\mu }_{n}\left( \left\lbrack {0, t}\right\rbrack \right) \), we can assume that all the \( {\mu }_{n} \) are probability measures. A standard construction (see Theorem 3.2.2) shows that there is a sequence of random variables \( {X}_... | Yes |
Theorem 8.6.3. If \( f \in {C}^{2} \) and \( E{\int }_{0}^{t}{\left| {f}^{\prime }\left( {B}_{s}\right) \right| }^{2}{ds} < \infty \) then \( {\int }_{0}^{t}{f}^{\prime }\left( {B}_{s}\right) d{B}_{s} \) is a continuous martingale. | Proof. We first prove the result assuming \( \left| {f}^{\prime }\right| ,\left| {f}^{\prime \prime }\right| \leq K \) . Let\n\n\[ \n{I}_{n}^{1}\left( s\right) = \mathop{\sum }\limits_{{i : {t}_{i + 1}^{n} \leq s}}{f}^{\prime }\left( {B}_{{t}_{i}^{n}}\right) \left( {{B}_{{t}_{i + 1}^{n}} - {B}_{{t}_{i}^{n}}}\right) + {... | Yes |
Using (8.6.8) we can prove Theorem 8.5.8: if \( u\left( {t, x}\right) \) is a polynomial in \( t \) and \( x \) with \( \partial u/\partial t + \left( {1/2}\right) {\partial }^{2}u/\partial {x}^{2} = 0 \) then Ito’s formula implies | \[ u\left( {t,{B}_{t}}\right) - u\left( {0,{B}_{0}}\right) = {\int }_{0}^{t}\frac{\partial u}{\partial x}\left( {s,{B}_{s}}\right) d{B}_{s} \] Since \( \partial u/\partial x \) is a polynomial it satisfies the integrability condition in Theorem 8.6.3, \( u\left( {t,{B}_{t}}\right) \) is a martingale. To get a new concl... | Yes |
Theorem 8.7.2. Let \( {X}_{1},{X}_{2},\ldots \) be i.i.d. with a distribution \( F \), which has mean 0 and variance 1, and let \( {S}_{n} = {X}_{1} + \ldots + {X}_{n} \) . There is a sequence of stopping times \( {T}_{0} = \) \( 0,{T}_{1},{T}_{2},\ldots \) such that \( {S}_{n}{ = }_{d}B\left( {T}_{n}\right) \) and \( ... | Proof. Let \( \left( {{U}_{1},{V}_{1}}\right) ,\left( {{U}_{2},{V}_{2}}\right) ,\ldots \) be i.i.d. and have distribution given in (8.7.1) and let \( {B}_{t} \) be an independent Brownian motion. Let \( {T}_{0} = 0 \), and for \( n \geq 1 \), let\n\n\[ \n{T}_{n} = \inf \left\{ {t \geq {T}_{n - 1} : {B}_{t} - B\left( {T... | Yes |
Theorem 8.7.3. Central limit theorem. Under the hypotheses of Theorem 8.7.2, \( {S}_{n}/\sqrt{n} \Rightarrow \chi \), where \( \chi \) has the standard normal distribution. | Proof. If we let \( {W}_{n}\left( t\right) = B\left( {nt}\right) /\sqrt{n} = {}_{d}{B}_{t} \) by Brownian scaling, then\n\n\[ \n{S}_{n}/\sqrt{n}\overset{d}{ = }B\left( {T}_{n}\right) /\sqrt{n} = {W}_{n}\left( {{T}_{n}/n}\right) \n\] \n\nThe weak law of large numbers implies that \( {T}_{n}/n \rightarrow 1 \) in probabi... | Yes |
Lemma 8.7.4. \( \mathcal{B} \) is the same as \( \mathcal{C} \) the \( \sigma \) -field generated by the finite dimensional sets \( \left\{ {\omega : \omega \left( {t}_{i}\right) \in {A}_{i}}\right\} \) | Proof. Observe that if \( \xi \) is a given continuous function\n\n\[
\{ \omega : \parallel \omega - \xi \parallel \leq r - 1/n\} = { \cap }_{q}\{ \omega : \left| {\omega \left( q\right) - \xi \left( q\right) }\right| \leq r - 1/n\}
\]\n\nwhere the intersection is over all rationals in \( \left\lbrack {0,1}\right\rbrac... | Yes |
Theorem 8.7.6. If \( \psi : C\left\lbrack {0,1}\right\rbrack \rightarrow \mathbf{R} \) has the property that it is continuous \( {P}_{0} \) -a.s. then | \[ \psi \left( {S\left( {n \cdot }\right) /\sqrt{n}}\right) \Rightarrow \psi \left( {B\left( \cdot \right) }\right) \] | No |
Example 8.7.2. Maxima. Let \( \psi \left( \omega \right) = \max \{ \omega \left( t\right) : 0 \leq t \leq 1\} \) . Again, \( \psi : C\left\lbrack {0,1}\right\rbrack \rightarrow \) \( \mathbf{R} \) is continuous. This time Theorem 8.7.6 implies | \[ \mathop{\max }\limits_{{0 \leq m \leq n}}{S}_{m}/\sqrt{n} \Rightarrow {M}_{1} \equiv \mathop{\max }\limits_{{0 \leq t \leq 1}}{B}_{t} \] To complete the picture, we observe that by (8.4.4) the distribution of the right-hand side is \[ {P}_{0}\left( {{M}_{1} \geq a}\right) = {P}_{0}\left( {{T}_{a} \leq 1}\right) = 2{... | No |
Example 8.7.3. Last 0 before time \( n \) . Let \( \psi \left( \omega \right) = \sup \{ t \leq 1 : \omega \left( t\right) = 0\} \) . This time, \( \psi \) is not continuous, for if \( {\omega }_{\epsilon } \) with \( {\omega }_{\epsilon }\left( 0\right) = 0 \) is piecewise linear with slope 1 on \( \left\lbrack {0,1/3 ... | It is easy to see that if \( \psi \left( \omega \right) < 1 \) and \( \omega \left( t\right) \) has positive and negative values in each interval \( \left( {\psi \left( \omega \right) - \delta ,\psi \left( \omega \right) }\right) \), then \( \psi \) is continuous at \( \omega \) . By arguments in Subection 8.4.1, the l... | No |
As we will now show, with a little work, one can convert this into the more natural result\n\n\[ \left| \left\{ {m \leq n : {S}_{m} > a\sqrt{n}}\right\} \right| /n \Rightarrow \left| \left\{ {t \in \left\lbrack {0,1}\right\rbrack : {B}_{t} > a}\right\} \right| \] | Proof. Application of Theorem 8.7.6 gives that for any \( a \) ,\n\n\[ \left| {\{ t \in \left\lbrack {0,1}\right\rbrack : S\left( {nt}\right) > a\sqrt{n}\} }\right| \Rightarrow \left| \left\{ {t \in \left\lbrack {0,1}\right\rbrack : {B}_{t} > a}\right\} \right| \]\n\nTo convert this into a result about \( \left| \left\... | Yes |
Example 8.7.5. Let \( \psi \left( \omega \right) = {\int }_{\left\lbrack 0,1\right\rbrack }\omega {\left( t\right) }^{k}{dt} \) where \( k > 0 \) is an integer. \( \psi \) is continuous, so applying Theorem 8.7.6 gives \[ {\int }_{0}^{1}{\left( S\left( nt\right) /\sqrt{n}\right) }^{k}{dt} \Rightarrow {\int }_{0}^{1}{B}... | To convert this into a result about the original sequence, we begin by observing that if \( x < y \) with \( \left| {x - y}\right| \leq \epsilon \) and \( \left| x\right| ,\left| y\right| \leq M \), then \[ \left| {{x}^{k} - {y}^{k}}\right| \leq {\int }_{x}^{y}k{\left| z\right| }^{k - 1}{dz} \leq {k\epsilon }{M}^{k - 1... | Yes |
Lemma 8.7.7. If \( {\tau }_{\left\lbrack ns\right\rbrack }^{n} \rightarrow s \) in probability for each \( s \in \left\lbrack {0,1}\right\rbrack \) then\n\n\[ \begin{Vmatrix}{{S}_{n,\left( {n \cdot }\right) } - B\left( \cdot \right) }\end{Vmatrix} \rightarrow 0\;\text{ in probability } \] | Proof. The fact that \( B \) has continuous paths (and hence uniformly continuous on \( \left\lbrack {0,1}\right\rbrack ) \) implies that if \( \epsilon > 0 \) then there is a \( \delta > 0 \) so that \( 1/\delta \) is an integer and\n\n(a)\n\n\[ P\left( {\left| {{B}_{t} - {B}_{s}}\right| < \epsilon \text{ for all }0 \... | Yes |
Lemma 8.7.8. If \( \varphi \) is bounded and continuous then \( {E\varphi }\left( {S}_{n,\left( {n \cdot }\right) }\right) \rightarrow {E\varphi }\left( {B\left( \cdot \right) }\right) \) . | Proof. For fixed \( \epsilon > 0 \), let \( {G}_{\delta } = \left\{ {\omega : }\right. \) if \( \left. {\begin{Vmatrix}{\omega - {\omega }^{\prime }}\end{Vmatrix} < \delta \text{then}\left| {\varphi \left( \omega \right) - \varphi \left( {\omega }^{\prime }\right) }\right| < \epsilon }\right\} \) . Since \( \varphi \) ... | Yes |
Theorem 8.7.9. \( S\left( {n \cdot }\right) /\sqrt{n} \Rightarrow B\left( \cdot \right) \), i.e., the associated measures on \( C\lbrack 0,\infty ) \) converge weakly. | Proof. By definition, all we have to show is that weak convergence occurs on \( C\left\lbrack {0, M}\right\rbrack \) for all \( M < \infty \) . The proof of Theorem 8.7.5 works in the same way when 1 is replaced by \( M \) . | No |
Let \( {N}_{n} = \inf \left\{ {m : {S}_{m} \geq \sqrt{n}}\right\} \) and \( {T}_{1} = \inf \left\{ {t : {B}_{t} \geq 1}\right\} \). Since \( \psi \left( \omega \right) = {T}_{1}\left( \omega \right) \land 1 \) is continuous \( {P}_{0} \) a.s. on \( C\left\lbrack {0,1}\right\rbrack \) and the distribution of \( {T}_{1} ... | \[ P\left( {{N}_{n} \leq {nt}}\right) \rightarrow P\left( {{T}_{1} \leq t}\right) \] | Yes |
Theorem 8.8.1. If \( {S}_{n} \) is a square integrable martingale with \( {S}_{0} = 0 \), and \( {B}_{t} \) is a Brownian motion, then there is a sequence of stopping times \( 0 = {T}_{0} \leq {T}_{1} \leq {T}_{2}\ldots \) for the Brownian motion so that \[ \left( {{S}_{0},{S}_{1},\ldots ,{S}_{k}}\right) \overset{d}{ =... | Proof. We include \( {S}_{0} = 0 = B\left( {T}_{0}\right) \) only for the sake of starting the induction argument. Suppose we have \( \left( {{S}_{0},\ldots ,{S}_{k - 1}}\right) { = }_{d}\left( {B\left( {T}_{0}\right) ,\ldots, B\left( {T}_{k - 1}\right) }\right) \) for some \( k \geq 1 \) . The strong Markov property i... | Yes |
Theorem 8.8.2. Let \( {\mathcal{F}}_{m} = \sigma \left( {{S}_{0},{S}_{1},\ldots {S}_{m}}\right) .\mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{n} \) exists and is finite on \( \mathop{\sum }\limits_{{m = 1}}^{\infty }E\left( {{\left( {S}_{m} - {S}_{m - 1}\right) }^{2} \mid {\mathcal{F}}_{m - 1}}\right) < \infty .... | Proof. Let \( {\mathcal{B}}_{t} \) be the filtration generated by Brownian motion, and let \( {t}_{m} = {T}_{m} - \) \( {T}_{m - 1} \) . By construction we have\n\n\[ E\left( {{\left( {S}_{m} - {S}_{m - 1}\right) }^{2} \mid {\mathcal{F}}_{m - 1}}\right) = E\left( {{t}_{m} \mid \mathcal{B}\left( {T}_{m - 1}\right) }\rig... | Yes |
Theorem 8.8.3. Suppose \( \left\{ {{X}_{n, m},{\mathcal{F}}_{n, m}}\right\} \) is a martingale difference array.\n\nIf (i) for each \( t,{V}_{n,\left\lbrack {nt}\right\rbrack } \rightarrow t \) in probability,\n\n(i) \( \left| {X}_{n, m}\right| \leq {\epsilon }_{n} \) for all \( m \) with \( {\epsilon }_{n} \rightarrow... | Proof. (i) implies \( {V}_{n, n} \rightarrow 1 \) in probability. By stopping each sequence at the first time \( {V}_{n, k} > 2 \) and setting the later \( {X}_{n, m} = 0 \), we can suppose without loss of generality that \( {V}_{n, n} \leq 2 + {\epsilon }_{n}^{2} \) for all \( n \) . By Theorem 8.8.1, we can find stop... | Yes |
Theorem 8.8.4. Lindeberg-Feller theorem for martingales. Suppose \( {X}_{n, m} \) , \( {\mathcal{F}}_{n, m},1 \leq m \leq n \) is a martingale difference array. If (i) \( {V}_{n,\left\lbrack {nt}\right\rbrack } \rightarrow t \) in probability for all \( t \in \left\lbrack {0,1}\right\rbrack \) and (ii) for all \( \epsi... | Proof. The first step is to truncate so that we can apply Theorem 8.8.3. Let \[ {\widehat{V}}_{n}\left( \epsilon \right) = \mathop{\sum }\limits_{{m = 1}}^{n}E\left( {{X}_{n, m}^{2}{1}_{\left( \left| {X}_{n, m}\right| > {\epsilon }_{n}\right) } \mid {\mathcal{F}}_{n, m - 1}}\right) \] | No |
Lemma 8.8.5. If \( {\epsilon }_{n} \rightarrow 0 \) slowly enough then \( {\epsilon }_{n}^{-2}{\widehat{V}}_{n}\left( {\epsilon }_{n}\right) \rightarrow 0 \) in probability. | Proof. Let \( {N}_{m} \) be chosen so that \( P\left( {{m}^{2}{\widehat{V}}_{n}\left( {1/m}\right) > 1/m}\right) \leq 1/m \) for \( n \geq {N}_{m} \) . Let \( {\epsilon }_{n} = 1/m \) for \( n \in \left\lbrack {{N}_{m},{N}_{m + 1}}\right) \) and \( {\epsilon }_{n} = 1 \) if \( n < {N}_{1} \) . If \( \delta > 0 \) and \... | No |
Lemma 8.8.6. If we define \( {\widetilde{S}}_{n,\left( {n \cdot }\right) } \) in the obvious way then Theorem 8.8.3 implies \( {\widetilde{S}}_{n,\left( {n \cdot }\right) } \Rightarrow B\left( \cdot \right) . | Proof. Since \( \left| {\widetilde{X}}_{n, m}\right| \leq 2{\epsilon }_{n} \), we only have to check (ii) in Theorem 8.8.3. To do this, we observe that the conditional variance formula, Theorem 5.4.7, implies\n\n\[ E\left( {{\widetilde{X}}_{n, m}^{2} \mid {\mathcal{F}}_{n, m - 1}}\right) = E\left( {{\bar{X}}_{n, m}^{2}... | Yes |
Lemma 8.8.7. If \( {A}_{n} \) is adapted to \( {\mathcal{G}}_{n} \) then for any nonnegative \( \delta \in {\mathcal{G}}_{0} \), \[ P\left( {{ \cup }_{m = 1}^{n}{A}_{m} \mid {\mathcal{G}}_{0}}\right) \leq \delta + P\left( {\mathop{\sum }\limits_{{m = 1}}^{n}P\left( {{A}_{m} \mid {\mathcal{G}}_{m - 1}}\right) > \delta \... | Proof. We proceed by induction. When \( n = 1 \), the conclusion says \[ P\left( {{A}_{1} \mid {\mathcal{G}}_{0}}\right) \leq \delta + P\left( {P\left( {{A}_{1} \mid {\mathcal{G}}_{0}}\right) > \delta \mid {\mathcal{G}}_{0}}\right) \] This is obviously true on \( {\Omega }_{ - } \equiv \left\{ {P\left( {{A}_{1} \mid {\... | Yes |
Theorem 8.8.8. Martingale central limit theorem. Suppose \( {X}_{n},{\mathcal{F}}_{n}, n \geq 1 \), is a martingale difference sequence and let \( {V}_{k} = \mathop{\sum }\limits_{{1 \leq n \leq k}}E\left( {{X}_{n}^{2} \mid {\mathcal{F}}_{n - 1}}\right) \) . If (i) \( {V}_{k}/k \rightarrow {\sigma }^{2} > 0 \) in proba... | Proof. Let \( {X}_{n, m} = {X}_{m}/\sigma \sqrt{n},{\mathcal{F}}_{n, m} = {\mathcal{F}}_{m} \) . Changing notation and letting \( k = {nt} \) , our first assumption becomes (i) of Theorem 8.8.4. To check (ii), observe that\n\n\[ E\mathop{\sum }\limits_{{m = 1}}^{n}E\left( {{X}_{n, m}^{2}{1}_{\left( \left| {X}_{n, m}\ri... | Yes |
Lemma 8.9.1. \( \\left\\{ {{U}_{k}^{n} : 1 \\leq k \\leq n}\\right\\} \\overset{d}{ = }\\left\\{ {{Z}_{k}/{Z}_{n + 1} : 1 \\leq k \\leq n}\\right\\} \) | Proof. We change variables \( v = r\\left( t\\right) \), where \( {v}_{i} = {t}_{i}/{t}_{n + 1} \) for \( i \\leq n,{v}_{n + 1} = {t}_{n + 1} \). The inverse function is\n\n\[ s\\left( v\\right) = \\left( {{v}_{1}{v}_{n + 1},\\ldots ,{v}_{n}{v}_{n + 1},{v}_{n + 1}}\\right) \]\n\nwhich has matrix of partial derivatives ... | Yes |
Theorem 8.9.2. \( {D}_{n} \Rightarrow \mathop{\max }\limits_{{0 \leq t \leq 1}}\left| {{B}_{t} - t{B}_{1}}\right| \), where \( {B}_{t} \) is a Brownian motion starting at 0. | Proof of (8.9.3). Formula (8.9.4) shows that the f.d.d.’s of \( {B}_{t}^{0} \) are multivariate normal and have mean 0 . Since \( {B}_{t} - t{B}_{1} \) also has this property, it suffices to show that the covariances are equal. We begin with the easier computation. If \( s < t \) then\n\n\[ E\left( {\left( {{B}_{s} - s... | Yes |
Let \( {a}_{n} = 1/2 - 1/{2n},{b}_{n} = 1/2 - 1/{4n} \), and let \( {\mu }_{n} \) be the point mass on the function that is 1 at \( {b}_{n} \), is 0 at \( 0,{a}_{n},1/2 \), and 1, and linear in between these points. As \( n \rightarrow \infty ,{f}_{n}\left( t\right) \rightarrow {f}_{\infty } \equiv 0 \) but not uniform... | \[ \int h\left( \omega \right) {\mu }_{n}\left( {d\omega }\right) = 1 \nrightarrow 0 = \int h\left( \omega \right) {\mu }_{\infty }\left( {d\omega }\right) \] | Yes |
Theorem 8.10.3. Let \( {\mu }_{n},1 \leq n \leq \infty \) be probability measures on \( \mathcal{C} \) . If the finite dimensional distributions of \( {\mu }_{n} \) converge to those of \( {\mu }_{\infty } \) and if the \( {\mu }_{n} \) are tight then \( {\mu }_{n} \Rightarrow {\mu }_{\infty } \) | Proof. If \( {\mu }_{n} \) is tight then by Theorem 8.10.1 it is relatively compact and hence each subsequence \( {\mu }_{{n}_{m}} \) has a further subsequence \( {\mu }_{{n}_{m}^{\prime }} \) that converges to a limit \( \nu \) . If \( f \) : \( {\mathbf{R}}^{k} \rightarrow \mathbf{R} \) is bounded and continuous then... | No |
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