This commit is contained in:
@@ -1,6 +1,6 @@
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database = "spec.db"
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document = "document.tex"
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siteTitle = "Unnamed Website"
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siteTitle = "Brownian Motion and Stochastic Calculus"
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[compiler]
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compileAll = false
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@@ -1,4 +1,4 @@
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\chapter{Integral Manifolds/Shenanigans}
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\chapter{Integral Manifolds}
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\label{chap:integral}
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\input{./integrable.tex}
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@@ -1,9 +1,14 @@
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\part{Diffusion Processes}
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\part{Stochastic Processes}
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\label{part:diffusion}
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\input{./aws/index.tex}
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\input{./operator/index.tex}
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\part{Stochastic Calculus}
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\label{part:calculus}
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\input{./diffusion/index.tex}
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\input{./calculus/index.tex}
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\input{./sde/index.tex}
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\input{./mal/index.tex}
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\input{./calculus/index.tex}
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40
src/mal/bridge.tex
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40
src/mal/bridge.tex
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\section{The Brownian Bridge}
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\label{section:brownian-bridge}
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\begin{definition}[Brownian Bridge]
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\label{definition:brownian-bridge}
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Let $(\Omega, \cf, \bp)$ be a probability space, $a \in \real^d$, and $X: \Omega \to C([0, 1]; \real^d)$ be a Gaussian process, then $X$ is a \textbf{Brownian bridge} from $0$ to $a$ such that:
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\begin{enumerate}
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\item For each $t \in [0, 1]$, $\ev(X_t) = at$.
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\item For each $s, t \in [0, 1]$, $\text{Cov}(X_s, X_t) = (s \wedge t - st)I$.
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\end{enumerate}
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\end{definition}
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\begin{theorem}[{{\cite[Theorem 40.3]{Rogers}}}]
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\label{theorem:brownian-bridge}
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Let $a \in \real^d$, then the following are equivalent definitions of the distribution of the Brownian bridge from $0$ to $a$.
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\begin{enumerate}
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\item There exists a Brownian motion $B$ such that for each $t \in [0, 1]$,
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\[
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X_t = B_t + t(a - B_1)
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\]
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\item There exists a Brownian motion $B$ such that for each $t \in [0, 1]$,
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\[
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X_t = at + (1 - t)B_{t/(1-t)}
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\]
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\item $X$ is a continuous process such that
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\[
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Y_t = X_t - at + \int_0^1 \frac{X_s - as}{1 - s}ds
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\]
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is a Brownian motion.
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\item The distribution of $X$ is the $h$-transform of the classical Wiener measure on $C([0, 1]; \real^d)$, where
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\[
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h(t, x) = \frac{1}{[2\pi(1 - t)]^{d/2}}\exp\braks{-\frac{\norm{x - a}_{\real^d}^2}{2(1 - t)}}
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\]
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\end{enumerate}
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\end{theorem}
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@@ -1,5 +1,5 @@
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\section{$h$-Brownian Motions}
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\label{section:h-brownian}
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\section{Girsanov Transforms}
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\label{section:girsanov}
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\begin{theorem}[Cameron-Martin-Girsanov, {{\cite[Theorem 38.5]{Rogers}}}]
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\label{theorem:cameron-martin}
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@@ -45,132 +45,3 @@
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\end{enumerate}
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\end{theorem}
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\begin{definition}[Mean Value Property]
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\label{definition:mean-value-property}
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Let $\bp$ be the distribution of a strong Markov process on $D([0, T]; \real^d)$. For each $0 \le s \le t \le T$ and $x \in \real^d$, let $P_{s, t}(x, \cdot)$ be its transition distribution, and $h: (0, \infty) \times \real^d \to [0, \infty)$ be a measurable function, then $h$ satisfies the \textbf{mean-value property} with respect to $\bp$ if for any $0 \le s \le t \le T$ and $x \in \real^d$,
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\[
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h(s, x) = \int_{\real^d}h(t, y)P_{s, t}(x, dy)
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\]
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\end{definition}
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\begin{lemma}
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\label{lemma:h-density-martingale}
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Let $\bp$ be the distribution of a strong Markov process on $D([0, T]; \real^d)$ with transition functions $P_{s, t}(x, \cdot)$, $\bracs{X_t|t \in [0, T]}$ be the canonical process, $\bracs{\cf_t|t \in [0, T]}$ be its natural filtration, $h: (0, \infty) \times \real^d \to [0, \infty)$ be a function satisfying the mean-value property with respect to $\bp$, and $A_0 = \bracs{h(0, \cdot) > 0}$, then the process
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\[
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\one_{\bracs{X_0 \in A_0}} \frac{h(t, X_t)}{h(0, X_0)}
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\]
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is a $\bracs{\mathcal{F}_t}$-martingale with respect to $\bp$.
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\end{lemma}
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\begin{proof}
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Let $t \in [0, T]$ and $\mu$ be the initial distribution of $\bp$, then
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\[
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\ev_\bp\braks{\one_{\bracs{X_0 \in A_0}} \frac{h(t, X_t)}{h(0, X_0)}} =
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\int_{A_0} \frac{1}{h(0, x)}\int_{\real^d} h(t, y)P_{0, t}(x, dy) \mu(dx) = \mu(A_0)
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\]
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so the given process is integrable. Now, since $h$ satisfies the mean-value property,
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\begin{align*}
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\ev_\bp\braks{\one_{\bracs{X_0 \in A_0}} \frac{h(T, X_T)}{h(0, X_0)} \bigg | \cf_t} &= \frac{\one_{\bracs{X_0 \in A_0}}}{h(0, X_0)} \cdot \ev_\bp\braks{h(t, X_t)| \cf_t} \\
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&= \one_{\bracs{X_0 \in A_0}} \frac{h(t, X_t)}{h(0, X_0)}
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\end{align*}
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\end{proof}
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\begin{proposition}
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\label{proposition:space-time-regular-measure}
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Let $\bp$ be the distribution of a strong Markov process on $D([0, T]; \real^d)$ with transition functions $P_{s, t}(x, \cdot)$, $\mu$ be its initial distribution, $\bracs{X_t|t \in [0, T]}$ be the canonical process, $\bracs{\cf_t|t \in [0, T]}$ be its natural filtration, $h: (0, \infty) \times \real^d \to [0, \infty)$ be a function satisfying the mean-value property with respect to $\bp$, then there exists a probaiblity measure $\bp^h$ on $D([0, \infty); \real^d)$ such that:
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\begin{enumerate}
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\item For each $t \ge 0$,
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\[
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\frac{d \bp^h}{d \bp} \bigg |_{\cf_{t^+}} = \frac{\one_{\bracs{X_0 \in A_0}}}{\mu(A_0)}\frac{h(t, X_t)}{h(0, X_0)}
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\]
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where $A_0 = \bracs{h(0, \cdot) > 0}$.
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\item Under $\bp^h$, $X_t$ is a Markov process with transition function
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\[
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P_{s, t}^h(x, dy) = \begin{cases}
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\frac{h(t, y)}{h(s, x)}P_{s, t}(x, dy) &h(s, x) > 0 \\
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0 & h(s, x) = 0
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\end{cases}
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\]
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\item If $\bp$ is the classical Wiener measure, then the process
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\[
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\tilde X_t = X_t - \int_0^t \frac{\partial_x h(s, X_s)}{h(s, X_s)}ds
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\]
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is a $\bp^h$-Brownian motion.
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\end{enumerate}
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The distribution $\bp^h$ is the \textbf{$h$-transform} of $\mathbf{P}$.
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\end{proposition}
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\begin{theorem}[Reuter, {{\cite[Theorem 39.66]{Rogers}}}]
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\label{theorem:reuter}
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Let $(\Omega, \cf, \bp)$ be a filtered probability space, $X: \Omega \to C([0, \infty); \real^d)$ be a standard $\bracs{\mathcal{F}_t}$-Brownian motion with drift $\mu$ starting at $0$. If $\tau = \inf\bracs{t \ge 0: |X_t| = 1}$, then $X_\tau$ and $\tau$ are independent.
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\end{theorem}
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\begin{proof}
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If $\mu = 0$, then $X_\tau$ and $\tau$ are independent by rotational invariance.
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Otherwise, let $\wien$ be the classical Wiener measure on $C([0, \infty); \real^d)$, $\mathcal{V}$ be the distribution of $X$, and $Y$ be the canonical process on $C([0, \infty); \real^d)$. For each $t \ge 0$, let $\mathcal{G}_t = \sigma(\bracs{Y_s|0 \le s \le t})$, then by the Cameron-Martin formula,
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\[
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h(t, Y_t) = \frac{d \mathcal{V}}{d \wien} \bigg |_{\mathcal{G}_t} = \exp\braks{\angles{\mu, Y_t}_{\real^d} - \frac{1}{2}\norm{\mu}_{\real^d}^2 t}
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\]
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Since $h(t \wedge \tau, X_{t \wedge \tau})$ is a uniformly integrable martingale,
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\[
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\frac{d \mathcal{V}}{d\wien} \bigg |_{\mathcal{G}_{T^+}} = h(\tau, Y_\tau)
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\]
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In which case, for any measurable functions $f: \mathbb{S}^{d} \to [0, 1]$ and $g: [0, \infty) \to [0, 1]$,
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\begin{align*}
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\ev^{\mathcal{V}}[f(Y_\tau)g(\tau)] &= \ev^{\wien}\braks{f(Y_\tau)g(\tau)\exp(\angles{\mu, Y_\tau}_{\real^d} - \frac{1}{2}\norm{\mu}_{\real^d}^2 \tau)} \\
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&= \ev^\wien[f(Y_\tau)e^{\angles{\mu, Y_\tau}_{\real^d}}]\ev^\wien[g(\tau)e^{- \frac{1}{2}\norm{\mu}_{\real^d}^2 \tau}] \\
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&= \ev^{\mathcal{V}}(f(Y_\tau))\ev^{\mathcal{V}}(g(\tau))
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\end{align*}
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\end{proof}
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\begin{definition}[Brownian Bridge]
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\label{definition:brownian-bridge}
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Let $(\Omega, \cf, \bp)$ be a probability space, $a \in \real^d$, and $X: \Omega \to C([0, 1]; \real^d)$ be a Gaussian process, then $X$ is a \textbf{Brownian bridge} from $0$ to $a$ such that:
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\begin{enumerate}
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\item For each $t \in [0, 1]$, $\ev(X_t) = at$.
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\item For each $s, t \in [0, 1]$, $\text{Cov}(X_s, X_t) = (s \wedge t - st)I$.
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\end{enumerate}
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\end{definition}
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\begin{theorem}[{{\cite[Theorem 40.3]{Rogers}}}]
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\label{theorem:brownian-bridge}
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Let $a \in \real^d$, then the following are equivalent definitions of the distribution of the Brownian bridge from $0$ to $a$.
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\begin{enumerate}
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\item There exists a Brownian motion $B$ such that for each $t \in [0, 1]$,
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\[
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X_t = B_t + t(a - B_1)
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\]
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\item There exists a Brownian motion $B$ such that for each $t \in [0, 1]$,
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\[
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X_t = at + (1 - t)B_{t/(1-t)}
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\]
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\item $X$ is a continuous process such that
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\[
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Y_t = X_t - at + \int_0^1 \frac{X_s - as}{1 - s}ds
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\]
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is a Brownian motion.
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\item The distribution of $X$ is the $h$-transform of the classical Wiener measure on $C([0, 1]; \real^d)$, where
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\[
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h(t, x) = \frac{1}{[2\pi(1 - t)]^{d/2}}\exp\braks{-\frac{\norm{x - a}_{\real^d}^2}{2(1 - t)}}
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\]
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\end{enumerate}
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\end{theorem}
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93
src/mal/h.tex
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93
src/mal/h.tex
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@@ -0,0 +1,93 @@
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\section{$h$-Transforms}
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\label{section:h-transform}
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\begin{definition}[Mean Value Property]
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\label{definition:mean-value-property}
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Let $\bp$ be the distribution of a strong Markov process on $D([0, T]; \real^d)$. For each $0 \le s \le t \le T$ and $x \in \real^d$, let $P_{s, t}(x, \cdot)$ be its transition distribution, and $h: (0, \infty) \times \real^d \to [0, \infty)$ be a measurable function, then $h$ satisfies the \textbf{mean-value property} with respect to $\bp$ if for any $0 \le s \le t \le T$ and $x \in \real^d$,
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\[
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h(s, x) = \int_{\real^d}h(t, y)P_{s, t}(x, dy)
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\]
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\end{definition}
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\begin{lemma}
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\label{lemma:h-density-martingale}
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Let $\bp$ be the distribution of a strong Markov process on $D([0, T]; \real^d)$ with transition functions $P_{s, t}(x, \cdot)$, $\bracs{X_t|t \in [0, T]}$ be the canonical process, $\bracs{\cf_t|t \in [0, T]}$ be its natural filtration, $h: (0, \infty) \times \real^d \to [0, \infty)$ be a function satisfying the mean-value property with respect to $\bp$, and $A_0 = \bracs{h(0, \cdot) > 0}$, then the process
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\[
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\one_{\bracs{X_0 \in A_0}} \frac{h(t, X_t)}{h(0, X_0)}
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\]
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is a $\bracs{\mathcal{F}_t}$-martingale with respect to $\bp$.
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\end{lemma}
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\begin{proof}
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Let $t \in [0, T]$ and $\mu$ be the initial distribution of $\bp$, then
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\[
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\ev_\bp\braks{\one_{\bracs{X_0 \in A_0}} \frac{h(t, X_t)}{h(0, X_0)}} =
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\int_{A_0} \frac{1}{h(0, x)}\int_{\real^d} h(t, y)P_{0, t}(x, dy) \mu(dx) = \mu(A_0)
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\]
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so the given process is integrable. Now, since $h$ satisfies the mean-value property,
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\begin{align*}
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\ev_\bp\braks{\one_{\bracs{X_0 \in A_0}} \frac{h(T, X_T)}{h(0, X_0)} \bigg | \cf_t} &= \frac{\one_{\bracs{X_0 \in A_0}}}{h(0, X_0)} \cdot \ev_\bp\braks{h(t, X_t)| \cf_t} \\
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&= \one_{\bracs{X_0 \in A_0}} \frac{h(t, X_t)}{h(0, X_0)}
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\end{align*}
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\end{proof}
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\begin{proposition}
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\label{proposition:space-time-regular-measure}
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Let $\bp$ be the distribution of a strong Markov process on $D([0, T]; \real^d)$ with transition functions $P_{s, t}(x, \cdot)$, $\mu$ be its initial distribution, $\bracs{X_t|t \in [0, T]}$ be the canonical process, $\bracs{\cf_t|t \in [0, T]}$ be its natural filtration, $h: (0, \infty) \times \real^d \to [0, \infty)$ be a function satisfying the mean-value property with respect to $\bp$, then there exists a probaiblity measure $\bp^h$ on $D([0, \infty); \real^d)$ such that:
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\begin{enumerate}
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\item For each $t \ge 0$,
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\[
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\frac{d \bp^h}{d \bp} \bigg |_{\cf_{t^+}} = \frac{\one_{\bracs{X_0 \in A_0}}}{\mu(A_0)}\frac{h(t, X_t)}{h(0, X_0)}
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\]
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where $A_0 = \bracs{h(0, \cdot) > 0}$.
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\item Under $\bp^h$, $X_t$ is a Markov process with transition function
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\[
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P_{s, t}^h(x, dy) = \begin{cases}
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\frac{h(t, y)}{h(s, x)}P_{s, t}(x, dy) &h(s, x) > 0 \\
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0 & h(s, x) = 0
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\end{cases}
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\]
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\item If $\bp$ is the classical Wiener measure, then the process
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\[
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\tilde X_t = X_t - \int_0^t \frac{\partial_x h(s, X_s)}{h(s, X_s)}ds
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\]
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is a $\bp^h$-Brownian motion.
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\end{enumerate}
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The distribution $\bp^h$ is the \textbf{$h$-transform} of $\mathbf{P}$.
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\end{proposition}
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\begin{theorem}[Reuter, {{\cite[Theorem 39.66]{Rogers}}}]
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\label{theorem:reuter}
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Let $(\Omega, \cf, \bp)$ be a filtered probability space, $X: \Omega \to C([0, \infty); \real^d)$ be a standard $\bracs{\mathcal{F}_t}$-Brownian motion with drift $\mu$ starting at $0$. If $\tau = \inf\bracs{t \ge 0: |X_t| = 1}$, then $X_\tau$ and $\tau$ are independent.
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||||
\end{theorem}
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\begin{proof}
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||||
If $\mu = 0$, then $X_\tau$ and $\tau$ are independent by rotational invariance.
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Otherwise, let $\wien$ be the classical Wiener measure on $C([0, \infty); \real^d)$, $\mathcal{V}$ be the distribution of $X$, and $Y$ be the canonical process on $C([0, \infty); \real^d)$. For each $t \ge 0$, let $\mathcal{G}_t = \sigma(\bracs{Y_s|0 \le s \le t})$, then by the Cameron-Martin formula,
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\[
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h(t, Y_t) = \frac{d \mathcal{V}}{d \wien} \bigg |_{\mathcal{G}_t} = \exp\braks{\angles{\mu, Y_t}_{\real^d} - \frac{1}{2}\norm{\mu}_{\real^d}^2 t}
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\]
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Since $h(t \wedge \tau, X_{t \wedge \tau})$ is a uniformly integrable martingale,
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\[
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\frac{d \mathcal{V}}{d\wien} \bigg |_{\mathcal{G}_{T^+}} = h(\tau, Y_\tau)
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\]
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In which case, for any measurable functions $f: \mathbb{S}^{d} \to [0, 1]$ and $g: [0, \infty) \to [0, 1]$,
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\begin{align*}
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\ev^{\mathcal{V}}[f(Y_\tau)g(\tau)] &= \ev^{\wien}\braks{f(Y_\tau)g(\tau)\exp(\angles{\mu, Y_\tau}_{\real^d} - \frac{1}{2}\norm{\mu}_{\real^d}^2 \tau)} \\
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&= \ev^\wien[f(Y_\tau)e^{\angles{\mu, Y_\tau}_{\real^d}}]\ev^\wien[g(\tau)e^{- \frac{1}{2}\norm{\mu}_{\real^d}^2 \tau}] \\
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&= \ev^{\mathcal{V}}(f(Y_\tau))\ev^{\mathcal{V}}(g(\tau))
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\end{align*}
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\end{proof}
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@@ -1,4 +1,6 @@
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\chapter{Unknown Chapter}
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\chapter{$h$-Transforms}
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\label{chap:mal}
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\input{./brownian.tex}
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\input{./h.tex}
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\input{./bridge.tex}
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Reference in New Issue
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