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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 imp... | 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\[
{T}_{b} \circ {\theta }_{{T}_{a}} = {T}_{b} - {T}_{a}
\]\n\nso if \( f \) is bounded and measurable, the strong Markov property,8.3.7 and translation invariance imply\n\n\[
{E}_{0}\left( {f\left( {{T}_{b} - {T}_{a}}\right) \mid {\mathcal{F}}_{{T}_{a}... | 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\[ \n{E}_{x}\left( {\exp \left( {\theta {B}_{t}}\right) \mid {\mathcal{F}}_{s}}\right) = \exp \left( {\theta {B}_{s}}\right) E\left( {\exp \left( {\theta \left( {{B}_{t} - {B}_{s}}\right) }\right) \mid {\mathcal{F}}_{s}}\right) \]\n\n\[ \n= \exp \left... | 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 converge... | 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 |
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 |
Taking \( f = {x}^{2} \) in (8.6.1) we have\n\n\[ \n{B}_{t}^{2} - {B}_{0}^{2} = {\int }_{0}^{t}2{B}_{s}d{B}_{s} + t \]\n\nso if \( {B}_{0} = 0 \), we have\n\n\[ \n{\int }_{0}^{t}2{B}_{s}d{B}_{s} = {B}_{t}^{2} - t \]\n\nIn contrast to the calculus formula \( {\int }_{0}^{a}{2xdx} = {a}^{2} \) . | To give one reason for the difference, we note that | No |
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 probabil... | 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... | No |
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 |
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\n\n\[{\int }_{0}^{1}{\left( S\left( nt\right) /\sqrt{n}\right) }^{k}{dt} \Rightarrow {\int }_{0}^{1}{B}_{t}^{k}{dt}\... | 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\n\n\[ \left| {{x}^{k} - {y}^{k}}\right| \leq {\int }_{x}^{y}k{\left| z\right| }^{k - 1}{dz} \leq {k\epsilon }{M}^{k ... | 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 \( M \) . | Yes |
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 \( \delta \in {\mathcal{G}}_{0} \) , | Proof. We proceed by induction. When \( n = 1 \), the conclusion says\n\n\[ P\left( {{A}_{1} \mid {\mathcal{G}}_{0}}\right) \leq \delta + P\left( {P\left( {{A}_{1} \mid {\mathcal{G}}_{0}}\right) > \delta \mid {\mathcal{G}}_{0}}\right) \]\n\nThis is obviously true on \( {\Omega }_{ - } \equiv \left\{ {P\left( {{A}_{1} \... | 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. | To identify the distribution of the limit in Theorem 8.9.2, we will first prove\n\n\[ \left\{ {{B}_{t} - t{B}_{1},0 \leq t \leq 1}\right\} \overset{d}{ = }\left\{ {{B}_{t},0 \leq t \leq 1 \mid {B}_{1} = 0}\right\} \]\n\n(8.9.3)\n\na process we will denote by \( {B}_{t}^{0} \) and call the Brownian bridge. The event \( ... | 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 |
Lemma 8.10.4. If each subsequence of \( {\mu }_{n} \) has a further subsequence that converges to \( \nu \) then \( {\mu }_{n} \Rightarrow \nu \) . | Proof. Note that if \( f \) is a bounded continuous function, the sequence of real numbers \( \int f\left( \omega \right) {\mu }_{n}\left( {d\omega }\right) \) have the property that every subsequence has a further subsequence that converges to \( \int f\left( \omega \right) \nu \left( {d\omega }\right) \) . Exercise 8... | No |
Theorem 8.10.5. The sequence \( {\mu }_{n} \) is tight if and only if for each \( \epsilon > 0 \) there are \( {n}_{0}, M \) and \( \delta \) so that\n\n\[ \text{(i)}{\mu }_{n}\left( {\left| {\omega \left( 0\right) }\right| > M}\right) \leq \epsilon \text{for all}n \geq {n}_{0} \]\n\n\[ \text{(ii)}{\mu }_{n}\left( {{\o... | Proof. We begin by recalling (see e.g., Royden (1988), page 169) | No |
Theorem 8.10.6. Arzela-Ascoli Theorem. A subset A of \( C \) has compact closure if and only if \( \mathop{\sup }\limits_{{\omega \in A}}\left| {\omega \left( 0\right) }\right| < \infty \) and \( \mathop{\lim }\limits_{{\delta \rightarrow 0}}\mathop{\sup }\limits_{{\omega \in A}}{\operatorname{osc}}_{\delta }\left( \om... | To prove the necessity of (i) and (ii), we note that if \( {\mu }_{n} \) is tight and \( \epsilon > 0 \) we can choose a compact set \( K \) so that \( {\mu }_{n}\left( K\right) \geq 1 - \epsilon \) for all \( n \) . By Theorem 8.10.6, \( K \subset \{ X\left( 0\right) \leq M\} \) for large \( M \) and if \( \epsilon > ... | Yes |
For \( 1 \leq n \leq \infty \) let\n\n\[ \n{f}_{n}\left( t\right) = \left\{ \begin{array}{ll} 0 & t \in \lbrack 0,\left( {n + 1}\right) /{2n}) \\ 1 & t \in \left\lbrack {\left( {n + 1}\right) /{2n},1}\right\rbrack \end{array}\right.\n\]\n\nwhere \( \left( {n + 1}\right) /{2n} = 1/2 \) for \( n = \infty \) . We certainl... | Let \( \Lambda \) be the class of strictly increasing continuous mappings of \( \left\lbrack {0,1}\right\rbrack \) onto itself. Such functions necessarily have \( \lambda \left( 0\right) = 0 \) and \( \lambda \left( 1\right) = 1 \) . For \( f, g \in D \) define \( d\left( {f, g}\right) \) to be the infimum of those pos... | Yes |
For \( 1 \leq n < \infty \) let\n\n\[ \n{g}_{n}\left( t\right) = \left\{ \begin{array}{ll} 0 & t \in \lbrack 0,1/2) \\ 1 & t \in \lbrack 1/2,\left( {n + 1}\right) /{2n}) \\ 0 & t \in \left\lbrack {\left( {n + 1}\right) /{2n},1}\right\rbrack \end{array}\right. \n\]\n\nIn order to have \( \epsilon < 1 \) in the definitio... | To fix the problem with completeness we require that \( \lambda \) be close to the identity in a more stringent sense: the slopes of all of its chords are close to 1 . If \( \lambda \in \Lambda \) let\n\n\[ \n\parallel \lambda \parallel = \mathop{\sup }\limits_{{s \neq t}}\left| {\log \left( \frac{\lambda \left( t\righ... | Yes |
Theorem 8.11.2. Kolmogorov’s test. If \( h\left( t\right) \uparrow \) and \( {t}^{-1/2}h\left( t\right) \downarrow \) then \( h \) is upper or lower class according as | Approximating \( h \) from above by piecewise constant functions, it is easy to show that if the integral in Theorem 8.11.2 converges, \( h\left( t\right) \) is an upper class function. The proof of the other direction is much more difficult; see Motoo (1959) or Section 4.12 of Itô and McKean (1965). | No |
Theorem 8.11.3. If \( {X}_{1},{X}_{2},\ldots \) are i.i.d. with \( E{X}_{i} = 0 \) and \( E{X}_{i}^{2} = 1 \) then\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}{S}_{n}/{\left( 2n\log \log n\right) }^{1/2} = 1 \] | Proof. By Theorem 8.7.2, we can write \( {S}_{n} = B\left( {T}_{n}\right) \) with \( {T}_{n}/n \rightarrow 1 \) a.s. As in the proof of Donsker's theorem, this is all we will use in the argument below. Theorem 8.11.3 will follow from Theorem 8.11.1 once we show\n\n\[ \left( {{S}_{\left\lbrack t\right\rbrack } - {B}_{t}... | Yes |
Theorem 8.11.4. Strassen’s (1964) invariance principle. Let \( {X}_{1},{X}_{2},\ldots {be} \) i.i.d. with \( E{X}_{i} = 0 \) and \( E{X}_{i}^{2} = 1 \), let \( {S}_{n} = {X}_{1} + \cdots + {X}_{n} \), and let \( {S}_{\left( n \cdot \right) } \) be the usual linear interpolation. The limit set (i.e., the collection of l... | Jensen’s inequality implies \( f{\left( 1\right) }^{2} \leq {\int }_{0}^{1}g{\left( y\right) }^{2}{dy} \leq 1 \) with equality if and only if \( f\left( t\right) = t \), so Theorem 8.11.4 contains Theorem 8.11.3 as a special case and provides some information about how the large value of \( {S}_{n} \) came about. | No |
Proposition 2.1. Let \( P \) be a probability on \( \left( {S,\mathcal{B}\left( S\right) }\right) \). Then, for every \( B \in \mathcal{B}\left( S\right) \)\n\n\[ P\left( B\right) = \mathop{\sup }\limits_{{F \subset B, F : \text{ closed }}}P\left( F\right) = \mathop{\inf }\limits_{{B \subset G, G : \text{ open }}}P\lef... | Proof. Set \( \mathcal{C} = \{ B \in \mathcal{B}\left( S\right) ; (2.1) holds \} \). If \( B \in \mathcal{C} \), then clearly \( {B}^{c} \) (the complement of \( B \) ) \( \in \mathcal{C} \). If \( {B}_{n} \in \mathcal{C}, n = 1,2,\ldots \), then \( \cup {B}_{n} \in \mathcal{C} \). Indeed, for given \( \varepsilon > 0 ... | Yes |
Proposition 2.2. Let \( P \) and \( Q \) be probabilities on \( \left( {S,\mathcal{B}\left( S\right) }\right) \) . If \( {\int }_{S}f\left( x\right) P\left( {dx}\right) = {\int }_{S}f\left( x\right) Q\left( {dx}\right) \) for every \( f \in {C}_{b}\left( S\right) \), then \( P = Q. \) | Proof. By Proposition 2.1, it is sufficient to show that \( P\left( F\right) = Q\left( F\right) \) for every closed set \( F \) . If we set \( {f}_{n}\left( x\right) = \phi \left( {{n\rho }\left( {x, F}\right) }\right) \) where \( \phi \left( t\right) \) is the function defined by\n\n\[ \phi \left( t\right) = \left\{ \... | Yes |
Proposition 2.3. Suppose that \( S \) is complete with respect to the metric \( \rho \), that is, every \( \rho \) -Cauchy sequence is a convergent sequence. Then any probability \( P \) on \( \left( {S,\mathcal{B}\left( S\right) }\right) \) is inner regular in the sense that for every \( B \in \) \( \mathcal{B}\left( ... | Proof. We first prove that for every \( \varepsilon > 0 \) there exists a compact set\n\n* \( \rho \left( {x, A}\right) = \mathop{\inf }\limits_{{y \in A}}\rho \left( {x, y}\right) \) . \( K \subset S \) such that \( P\left( K\right) > 1 - \varepsilon \) . Since \( S \) is separable, \( S \) can be covered by a countab... | Yes |
Proposition 2.4. The following five conditions are equivalent:\n\n(i) \( {P}_{n}\overset{w}{ \rightarrow }P \) .\n\n(ii) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{\int }_{S}f\left( x\right) {P}_{n}\left( {dx}\right) = {\int }_{S}f\left( x\right) P\left( {dx}\right) \) for every uniformly continuous \( f \in {C... | Proof. \ | Yes |
Proposition 2.5. Weak convergence of probabilities is a metric concept. To be precise, we can define a metric \( d \) on the totality \( \mathcal{P}\left( S\right) \) of probabilities on \( \left( {S,\mathcal{B}\left( S\right) }\right) \) such that\n\n\[ \n{P}_{n}\overset{w}{ \rightarrow }P\;\text{ is equivalent to }\;... | Proof. The well known Prohorov metric, which is a generalization of that of Lévy in the case \( S = \mathbf{R} \), is such a metric (cf. [142]). Here we give an equivalent metric in the following way (cf. [165]). If \( S \) is a separable metric space, we can choose an equivalent metric so that \( S \) is totally bound... | Yes |
Theorem 3.2. Let \( \\left( {\\Omega ,\\mathcal{F}}\\right) \) be a standard measurable space and \( P \) be a probability on \( \\left( {\\Omega ,\\mathcal{F}}\\right) \). Let \( \\mathcal{G} \) be a sub \( \\sigma \) -field of \( \\mathcal{F} \) and \( p\\left( {\\omega, d{\\omega }^{\\prime }}\\right) \) be a regula... | Proof. Let \( {\\mathcal{H}}_{0} \\subset \\mathcal{H} \) be a countable set in Definition 3.4 for \( \\mathcal{H} \). Clearly if \( A \\in {\\mathcal{H}}_{0} \), then there exists \( {N}_{A} \\in \\mathcal{G} \) of \( P \) -measure 0 such that \( p\\left( {\\omega, A}\\right) = {I}_{A}\\left( \\omega \\right) \) if \(... | Yes |
Theorem 3.3. Let \( \\left( {\\Omega ,\\mathcal{F}}\\right) \) be a standard measurable space and \( P \) be a probability on \( \\left( {\\Omega ,\\mathcal{F}}\\right) \). Let \( \\xi \\left( \\omega \\right) \) be a mapping from \( \\Omega \) into a measurable space \( \\left( {S,\\mathcal{B}}\\right) \) such that it... | Thus, for every integrable random variable \( X,{\\int }_{\\Omega }X\\left( \\omega \\right) p\\left( {x,{d\\omega }}\\right) \) coincides with \( E\\left( {X \\mid \\xi = x}\\right) ,{P}^{\\xi } \) -a.e. \( x.p\\left( {x, A}\\right) \) is called the regular conditional probability given \( \\xi = x \). In the same way... | No |
Proposition 4.1. \( \sigma \left\lbrack \mathcal{C}\right\rbrack = \mathcal{B}\left( {W}^{d}\right) \) . | Proof. It is only necessary to show that \( \sigma \left\lbrack \mathcal{C}\right\rbrack \supset \mathcal{B}\left( {W}^{d}\right) \) . The totality of sets of the form \( \left\{ {w;\mathop{\max }\limits_{{0 \leq t \leq n}}\left| {w\left( t\right) - {w}_{0}\left( t\right) }\right| \leq \varepsilon }\right\} ,{w}_{0} \i... | Yes |
If \( X = \left( {X}_{t}\right) \) is a right-continuous, \( \left( {\mathcal{F}}_{t}\right) \) -adapted process and \( E \subset {\mathbf{R}}^{d} \) is an open set in \( {\mathbf{R}}^{d} \), then the first hitting time \( {\sigma }_{E} \) to the set \( E \) , defined by\n\n(5.3)\n\n\[ \n{\sigma }_{E}\left( \omega \rig... | Indeed,\n\n\[ \n\left\{ {{\sigma }_{E}\left( \omega \right) \leq t}\right\} = \mathop{\bigcap }\limits_{n}\left\{ {{\sigma }_{E}\left( \omega \right) < t + 1/n}\right\} \n\]\n\n\[ \n= \mathop{\bigcap }\limits_{n}\mathop{\bigcup }\limits_{\substack{{r \in Q} \\ {r < t + 1/n} }}\left\{ {{X}_{r}\left( \omega \right) \in E... | Yes |
Proposition 5.2. Let \( \sigma ,\tau ,{\sigma }_{n}, n = 1,2,\ldots \) be stopping times. Then\n\n(i) \( \sigma \vee \tau ,\;\sigma \land \tau \) ,\n\n(ii) \( \sigma = \mathop{\lim }\limits_{n}{\sigma }_{n} \), when \( {\sigma }_{n} \uparrow \) or \( {\sigma }_{n} \downarrow \) ,\n\nare all stopping times. | Proof. Since \( \{ \sigma \vee \tau \leq t{\} }^{*2} = \{ \sigma \leq t\} \cap \{ \tau \leq t\} ,\sigma \vee \tau \) is a stopping time. Similarly, \( \sigma \land \tau \) is also a stopping time. If \( {\sigma }_{n} \uparrow \sigma \) then \( \{ \sigma \leq t\} = \mathop{\bigcap }\limits_{n}\left\{ {{\sigma }_{n} \leq... | Yes |
Lemma 5.2. \( \sigma \) is a stopping time if and only if \( \{ \sigma < t\} \in {\mathcal{F}}_{t} \) for every \( t \) . | Proof. If \( \sigma \) is a stopping time then \( \{ \sigma < t\} = \mathop{\bigcup }\limits_{n}\{ \sigma \left( \omega \right) \leq t - 1/n\} \) \( \in {\mathcal{F}}_{t} \) . Conversely, if \( \{ \sigma < t\} \in {\mathcal{F}}_{t} \) for every \( t \), then \( \{ \sigma \left( \omega \right) \leq t\} = \) \( \mathop{\... | Yes |
Proposition 5.3. Let \( \sigma ,\tau ,{\sigma }_{n}, n = 1,2,\ldots \), be stopping times.\n\n(1) If \( \sigma \left( \omega \right) \leq \tau \left( \omega \right) \) for all \( \omega \), then \( {\mathcal{F}}_{\sigma } \subset {\mathcal{F}}_{\tau } \) . | Proof. (1) follows from the definition. | No |
Proposition 5.4. Let \( \sigma \) be a stopping time. Then the following hold.\n\n(1) \( \sigma : \Omega \ni \omega \mapsto \sigma \left( \omega \right) \in \left\lbrack {0,\infty }\right\rbrack \) is \( {\mathcal{F}}_{\sigma }/\mathcal{B}\left( \left\lbrack {0,\infty }\right\rbrack \right) \) -measurable.\n\n(2) If \(... | Proof. (1) is easy and hence omitted. To prove (2), we may assume \( X \) is right-continuous by Proposition 5.1. Let \( {\sigma }_{n}\left( \omega \right) = k/{2}^{n} \) if \( \sigma \left( \omega \right) \in \lbrack (k - \) 1) \( /{2}^{n}, k/{2}^{n} \) ). Then \( {\sigma }_{n} \) is a stopping time and \( {\sigma }_{... | No |
Theorem 6.1. (Optional sampling theorem). \( {}^{*1} \) Let \( X = {\left( {X}_{n}\right) }_{n \in T} \) be a martingale (supermartingale, submartingale) relative to \( \left( {\mathcal{F}}_{n}\right) \) and let \( \sigma \) and \( \tau \) be bounded*2 stopping times such that \( \sigma \left( \omega \right) \leq \tau ... | Proof. First we show (6.3). Let \( m \in T \) be such that \( \tau \left( \omega \right) \leq m \) for all \( \omega \) . Set \( {f}_{n} = {I}_{\left( \sigma < n \leq \tau \right) } = {I}_{\left( n \leq \tau \right) } - {I}_{\left( n \leq \sigma \right) } \), for \( n = 1,2,\ldots \) . Then \( f = \) \( \left( {f}_{n}\... | Yes |
Theorem 6.3. Let \( X = {\left( {X}_{n}\right) }_{n \in T} \) be a submartingale. Then for every \( N \in T \) and real numbers \( a \) and \( b \) such that \( a < b \), \[ E\left( {{U}_{N}^{X}\left( {a, b}\right) }\right) \leq \frac{1}{b - a}\left( {E\left\{ {{\left( {X}_{N} - a\right) }^{ + } - {\left( {X}_{0} - a\r... | Proof. By Jensen’s inequality, \( Y = \left( {Y}_{n}\right) \) where \( {Y}_{n} = {\left( {X}_{n} - a\right) }^{ + } \), \( n = 0,1,2,\ldots \), is also a submartingale and \( {U}_{N}^{x}\left( {a, b}\right) = {U}_{N}^{y}\left( {0, b - a}\right) \). Let \( {\tau }_{1},{\tau }_{2},\ldots \) be defined as in (6.8) with \... | Yes |
Theorem 6.4. If \( X = {\left( {X}_{n}\right) }_{n \in T} \) is a submartingale such that\n\n(6.12)\n\n\[ \mathop{\sup }\limits_{n}E\left( {X}_{n}^{ + }\right) < \infty \]\n\nthen \( {X}_{\infty } = \mathop{\lim }\limits_{{n \rightarrow \infty }}{X}_{n} \) exists almost surely and \( {X}_{\infty } \) is integrable. \( ... | Proof. Since \( E\left( \left| {X}_{n}\right| \right) = {2E}\left( {X}_{n}^{ + }\right) - E\left( {X}_{n}\right) \leq {2E}\left( {X}_{n}^{ + }\right) - E\left( {X}_{0}\right) \), we have \( \mathop{\sup }\limits_{n}E\left( \left| {X}_{n}\right| \right) < \infty \) by (6.12). Thus, if \( {X}_{\infty } = \mathop{\lim }\l... | Yes |
Theorem 6.5. Let \( X = {\left( {X}_{n}\right) }_{n \in T} \) be a submartingale satisfying the condition (6.12) and let \( {X}_{\infty } = \mathop{\lim }\limits_{{n \rightarrow \infty }}{X}_{n} \). In order that \( \bar{X} = {\left( {X}_{n}\right) }_{n \in T} \) be a submar-tingale, i.e., \( {X}_{n} \leq E\left( {{X}_... | Proof. If \( {X}_{n} \leq E\left( {{X}_{\infty } \mid {\mathcal{F}}_{n}}\right), n = 0,1,2,\ldots \), then, by Jensen’s inequality, \( {X}_{n}^{ + } \leq E\left( {{X}_{\infty }^{ + } \mid {\mathcal{F}}_{n}}\right) \) and hence \( E\left( {{X}_{n}^{ + } : {X}_{n}^{ + } > \lambda }\right) \leq E\left( {{X}_{\infty }^{ + ... | Yes |
Theorem 6.6. For \( Y \in {\mathcal{L}}_{1}\left( {\Omega ,\mathcal{F}, P}\right) \) set \( {X}_{n} = E\left( {Y \mid {\mathcal{F}}_{n}}\right), n \in T \) . Then \( X = \left( {X}_{n}\right) \) is an equi-integrable martingale and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{X}_{n} = {X}_{\infty } \) exists almo... | Proof. Since \( \left| {X}_{n}\right| \leq E\left( {\left| Y\right| \mid {\mathcal{F}}_{n}}\right), n \in T,\left\{ {X}_{n}\right\} \) is equi-integrable as in the proof of Theorem 6.5. Also \( \left( {X}_{n}\right) \) is a martingale since \( E\left( {{X}_{m} \mid {\mathcal{F}}_{n}}\right) = \) \( E\left( {E\left( {Y ... | Yes |
Theorem 6.7. Let \( X = {\left( {X}_{n}\right) }_{n = 0, - 1, - 2,\ldots } \) be a submartingale such that\n\n(6.14)\n\n\[ \mathop{\inf }\limits_{n}E\left( {X}_{n}\right) > - \infty \text{.} \]\n\nThen \( X \) is equi-integrable and \( \mathop{\lim }\limits_{{n \rightarrow - \infty }}{X}_{n} = {X}_{-\infty } \) exists ... | Proof. Since \( E\left( {X}_{n}\right) \) is decreasing as \( n \downarrow - \infty \) ,(6.14) implies that lim \( E\left( {X}_{n}\right) \) exists finitely as \( n \downarrow - \infty \) . Let \( \varepsilon > 0 \) and take \( k \) such that \( E\left( {X}_{k}\right) - \mathop{\lim }\limits_{{n \rightarrow - \infty }}... | Yes |
Theorem 6.8. With probability one, \( t \in T \cap Q \mapsto {X}_{t} \) is bounded and possesses\n\n\[ \mathop{\lim }\limits_{{Q \cap T \ni s \downarrow t}}{X}_{s}\text{ and }\mathop{\lim }\limits_{{Q \cap T \ni s \uparrow t}}{X}_{s} \]\n\nfor every \( t \geq 0 \) . | Proof. Let \( T > 0 \) be given and \( \left\{ {{r}_{1},{r}_{2},\ldots }\right\} \) be an enumeration of the set \( Q \cap \left\lbrack {0, T}\right\rbrack \) . For every \( n \), if \( \left\lbrack {{s}_{1},{s}_{2},\ldots ,{s}_{n}}\right\rbrack \) is the set \( \left\lbrack {{r}_{1},{r}_{2},\ldots }\right. \) , \( \le... | Yes |
Theorem 6.12. If \( X = \left( {X}_{t}\right) \) is a submartingale of class (DL), then it is expressible as the sum of a martingale \( M = \left( {M}_{t}\right) \) and an integrable increasing process \( A = \left( {A}_{t}\right) \) . Furthermore, \( A \) can be chosen to be natural and, under this condition, the deco... | Proof.* First we prove that the decomposition (6.21) with \( A \) natural is unique. Indeed, if \( {X}_{t} = {M}_{t} + {A}_{t} = {M}_{t}^{\prime } + {A}_{t}^{\prime } \) are two such decompositions, then since \( {A}_{t} - {A}_{t}^{\prime } = {M}_{t}^{\prime } - {M}_{t} \) is a martingale, we have, for any bounded mart... | Yes |
Lemma 6.1. \( \left\{ {{A}_{a}^{\left( n\right) }, n = 1,2,\ldots }\right\} \) is equi-integrable. | Proof. Let \( c > 0 \) be fixed and set\n\n\[ \n{\sigma }_{c}^{\left( n\right) } = \left\{ \begin{matrix} \inf \left\{ {{t}_{k - 1}^{\left( n\right) };{A}_{{t}_{k}^{\left( n\right) }}^{\left( n\right) } > c}\right\} , \\ a,\text{ if }\{ \text{ 最小 }\} = \phi . \end{matrix}\right. \]\n\nThen \( {\sigma }_{c}^{\left( n\ri... | Yes |
Theorem 7.1. For any probability \( \mu \) on \( \left( {{\mathbf{R}}^{d},\mathcal{B}\left( {\mathbf{R}}^{d}\right) }\right) \) the \( d \) -dimensional Wiener measure \( {P}_{\mu } \) with the initial distribution \( \mu \) exists uniquely. | Proof. The uniqueness is obvious. We will show its existence. First, we consider the case \( d = 1 \) and \( \mu = {\delta }_{0} \) . On a probability space, we construct a family of real random variables \( \{ X\left( t\right), t \in \lbrack 0,\infty )\} \) such that \( X\left( 0\right) = 0 \) and for every \( 0 < {t}... | Yes |
Theorem 7.2. If \( X = \left\{ {{X}_{t} = \left( {{X}_{t}^{1},{X}_{t}^{2},\ldots ,{X}_{t}^{d}}\right) }\right\} \) is a \( d \) -dimensional \( \left( {\mathcal{F}}_{t}\right) \) -Brownian motion, then for every \( t > s \geq 0 \) , (7.5) \[ E\left( {{X}_{t}^{t} - {X}_{s}^{t} \mid {\mathcal{F}}_{s}}\right) = 0,\text{ a... | Thus, if \( E\left( {\left| {X}_{0}\right| }^{2}\right) < \infty \), then each \( \left( {X}_{i}^{t}\right), i = 1,2,\ldots, d \), is a square-integrable martingale relative to \( \left( {\mathcal{F}}_{t}\right) \) such that \( {X}_{t}^{t}{X}_{t}^{j} - {\delta }_{ij}t \) is a martingale relative to \( \left( {\mathcal{... | No |
Theorem 8.1. Given a \( \sigma \) -finite measure \( \lambda \) on \( \left( {\mathbf{X},{\mathcal{B}}_{\mathbf{X}}}\right) \), there exists a Poisson random measure \( \mu \) with \( E\left( {\mu \left( B\right) }\right) = \lambda \left( B\right) \) for every \( B \in {\mathcal{B}}_{X} \) . | Proof. Let \( {U}_{n} \in {\mathcal{B}}_{X} \) be disjoint, \( 0 < \lambda \left( {U}_{n}\right) < \infty \) and \( \mathop{\cup }\limits_{n}{U}_{n} = X \) . On a probability space we construct the following:\n\n(i) for each \( n = 1,2,\ldots \) and \( i = 1,2,\ldots ,{\xi }_{i}^{\left( n\right) } \) is a \( {U}_{n} \)... | No |
Theorem 9.1. Given a \( \sigma \) -finite measure \( n \) on \( \left( {X,{\mathcal{B}}_{X}}\right) \) there exists a stationary Poisson point process on \( \mathbf{X} \) with the characteristic measure \( n \) . | The following construction is essentially the same as in Theorem 8.1 ; indeed, \( p \) may be identified with a Poisson random measure on \( \left( {0,\infty }\right) \times \mathbf{X} \) having the intensity measure \( \operatorname{dtn}\left( {dx}\right) \). Let \( {U}_{k} \in {\mathcal{B}}_{X}, k = 1,2,\ldots \), be... | Yes |
Lemma 1.1. \( {\mathcal{L}}_{0} \) is dense in \( {\mathcal{L}}_{2} \) with respect to the metric \( \parallel \cdot {\parallel }_{2} \) . | Proof. Take \( \Phi \in {\mathcal{L}}_{2}\;\mathrm{{and}}\;\mathrm{{set}}\;{\Phi }^{M}\left( {t,\omega }\right) = \Phi \left( {t,\omega }\right) {I}_{\left\lbrack -M, M\right\rbrack }\left( {\Phi \left( {t,\omega }\right) }\right) . \) Then \( {\Phi }^{M} \in {\mathcal{L}}_{2} \) and \( {\begin{Vmatrix}\Phi - {\Phi }^{... | Yes |
Lemma 1.2. \( {\mathcal{M}}_{2} \) is a complete metric space in the metric \( \parallel X - Y\parallel \) , \( X, Y \in {\mathcal{M}}_{2} \), and \( {\mathcal{M}}_{2}^{c} \) is a closed subspace of \( {\mathcal{M}}_{2} \) . | Proof. First we note that if \( \parallel X - Y\parallel = 0 \), then \( X = Y \) ; indeed, \( \parallel X - Y\parallel = 0 \) implies \( {X}_{n} = {Y}_{n} \) a.s., \( n = 1,2,\ldots \), and so \( {X}_{t} = \) \( E\left\lbrack {{X}_{n} \mid {\mathcal{F}}_{t}}\right\rbrack = E\left\lbrack {{Y}_{n} \mid {\mathcal{F}}_{t}... | Yes |
Proposition 1.1. The stochastic integral with respect to an \( \left( {\mathcal{F}}_{t}\right) \) - Brownian motion has the following properties:\n\n(i) \( I\left( \Phi \right) \left( 0\right) = 0\; \) a.s.\n\n(ii) For each \( \;t > s \geqq 0 \), \n\n(1.11)\n\n\[ E\left\lbrack {\left( {I\left( \Phi \right) \left( t\rig... | Proof. (i) is obvious. (1.11) is clear since \( I\left( \Phi \right) \) is a martingale. (1.12) is easily proved first for \( \Phi \in {\mathcal{L}}_{0} \) and then by taking limits. (1.13) and (1.14) are consequences of Doob's optional sampling theorem. Therefore, we need only prove (iv). \( {}^{*2} \) First consider ... | Yes |
Proposition 1.2. For \( t > s \geq 0 \) , \[ E\left\lbrack {{\int }_{s}^{t}{\Phi }_{l}\left( u\right) d{B}^{l}\left( u\right) {\int }_{s}^{t}{\Phi }_{j}\left( u\right) d{B}^{j}\left( u\right) \mid {\mathcal{F}}_{s}}\right\rbrack \] \[ = {\delta }_{ij}E\left\lbrack {{\int }_{s}^{t}{\Phi }_{i}\left( u\right) {\Phi }_{j}\... | Proof. It is easy to prove this when \( {\Phi }_{l} \in {\mathcal{L}}_{0}, i = 1,2,\ldots, r \) . The general case follows by taking limits. | No |
Proposition 2.1. (i) Let \( M = \left( {M}_{t}\right) \in {\mathcal{M}}_{2} \) . Then there exists a natural integrable increasing process \( {}^{*1}A = \left( {A}_{t}\right) \) such that \( {M}_{t}^{2} - {A}_{t} \) is an \( \left( {\mathcal{F}}_{t}\right) \) - martingale. Furthermore, \( A \) is uniquely determined. \... | Proof. Let \( M = \left( {M}_{t}\right) \in {\mathcal{M}}_{2} \) . Then \( t \mapsto {M}_{t}^{2} \) is a non-negative sub-martingale of the class (DL) and hence by Doob-Meyer's decomposition theorem* \( {}^{*3} \), there exists a unique natural integrable increasing process \( A = \left( {A}_{t}\right) \) such that \( ... | Yes |
Proposition 2.3. For \( t > s \geq 0 \) ,\n\n\[ E\left\lbrack {\left( {{I}^{M}\left( \Phi \right) \left( t\right) - {I}^{M}\left( \Phi \right) \left( s\right) }\right) \left( {{I}^{N}\left( \Psi \right) \left( t\right) - {I}^{N}\left( \Psi \right) \left( s\right) }\right) \mid \mathcal{F}}\right\rbrack \]\n\n(2.10)\n\n... | The proof is easily obtained first for \( \Phi ,\Psi \in {\mathcal{L}}_{0} \) and then by a limiting procedure. (2.10) is another way of saying that\n\n(2.11)\n\n\[ \left\langle {{I}^{M}\left( \Phi \right) ,{I}^{N}\left( \Psi \right) }\right\rangle \left( t\right) = {\int }_{0}^{t}\left( {\Phi \Psi }\right) \left( u\ri... | No |
Proposition 2.4. Let \( M \in {\mathcal{M}}_{2} \) and \( \Phi \in {\mathcal{L}}_{2}\left( M\right) \) (or \( M \in {\mathcal{M}}_{2}^{loc} \) , \( \Phi \in {\mathcal{L}}_{2}^{loc}\left( M\right) \) ). Then \( X = {I}^{M}\left( \Phi \right) \) is characterized as the unique \( X \in {\mathcal{M}}_{2} \) \( \left( {X \i... | Proof. We need only show uniqueness. If \( {X}^{\prime } \in {\mathcal{M}}_{2} \) also satisfies (2.15), then \( \left\langle {X - {X}^{\prime }, N}\right\rangle = 0 \) for every \( N \in {\mathcal{M}}_{2} \) and so, by taking \( N = \) \( X - {X}^{\prime },\left\langle {X - {X}^{\prime }}\right\rangle = 0 \) . Hence \... | Yes |
Proposition 2.5. (i) Let \( M, N \in {\mathcal{M}}_{2}^{loc} \) and \( \Phi \in {\mathcal{L}}_{2}^{loc}\left( M\right) \cap {\mathcal{L}}_{2}^{loc}\left( N\right) \). Then \( \Phi \in {\mathcal{L}}_{2}^{loc}\left( {M + N}\right) \) and | \[ {\int }_{0}^{t}\Phi \left( u\right) d\left( {M + N}\right) \left( u\right) = {\int }_{0}^{t}\Phi \left( u\right) {dM}\left( u\right) + {\int }_{0}^{t}\Phi \left( u\right) {dN}\left( u\right) . \] | Yes |
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