Brownian Motion and Stochastic Calculus

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/Part 2: Stochastic Calculus/Chapter 5: $h$-Transforms

5.2 $h$-Transforms

Definition 5.2.1 (Mean Value Property).label 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 mean-value property with respect to $\bp$ if for any $0 \le s \le t \le T$ and $x \in \real^{d}$,

\[h(s, x) = \int_{\real^d}h(t, y)P_{s, t}(x, dy)\]

Lemma 5.2.2.label 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

\[\one_{\bracs{X_0 \in A_0}}\frac{h(t, X_{t})}{h(0, X_{0})}\]

is a $\bracs{\mathcal{F}_t}$-martingale with respect to $\bp$.

Proof. Let $t \in [0, T]$ and $\mu$ be the initial distribution of $\bp$, then

\[\ev_{\bp}\braks{\one_{\bracs{X_0 \in A_0}} \frac{h(t, X_{t})}{h(0, X_{0})}}= \int_{A_0}\frac{1}{h(0, x)}\int_{\real^d}h(t, y)P_{0, t}(x, dy) \mu(dx) = \mu(A_{0})\]

so the given process is integrable. Now, since $h$ satisfies the mean-value property,

\begin{align*}\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}\\&= \one_{\bracs{X_0 \in A_0}}\frac{h(t, X_{t})}{h(0, X_{0})}\end{align*}

$\square$

Proposition 5.2.3.label 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:

  1. (1)

    For each $t \ge 0$,

    \[\frac{d \bp^{h}}{d \bp}\bigg |_{\cf_{t^+}}= \frac{\one_{\bracs{X_0 \in A_0}}}{\mu(A_{0})}\frac{h(t, X_{t})}{h(0, X_{0})}\]

    where $A_{0} = \bracs{h(0, \cdot) > 0}$.

  2. (2)

    Under $\bp^{h}$, $X_{t}$ is a Markov process with transition function

    \[P_{s, t}^{h}(x, dy) = \begin{cases}\frac{h(t, y)}{h(s, x)}P_{s, t}(x, dy)&h(s, x) > 0 \\ 0&h(s, x) = 0\end{cases}\]

  3. (3)

    If $\bp$ is the classical Wiener measure, then the process

    \[\tilde X_{t} = X_{t} - \int_{0}^{t} \frac{\partial_{x} h(s, X_{s})}{h(s, X_{s})}ds\]

    is a $\bp^{h}$-Brownian motion.

The distribution $\bp^{h}$ is the $h$-transform of $\mathbf{P}$.

Theorem 5.2.4 (Reuter, [Theorem 39.66, RW89]).label 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.

Proof. If $\mu = 0$, then $X_{\tau}$ and $\tau$ are independent by rotational invariance.

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,

\[h(t, Y_{t}) = \frac{d \mathcal{V}}{d \wien}\bigg |_{\mathcal{G}_t}= \exp\braks{\angles{\mu, Y_t}_{\real^d} - \frac{1}{2}\norm{\mu}_{\real^d}^2 t}\]

Since $h(t \wedge \tau, X_{t \wedge \tau})$ is a uniformly integrable martingale,

\[\frac{d \mathcal{V}}{d\wien}\bigg |_{\mathcal{G}_{T^+}}= h(\tau, Y_{\tau})\]

In which case, for any measurable functions $f: \mathbb{S}^{d}\to [0, 1]$ and $g: [0, \infty) \to [0, 1]$,

\begin{align*}\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)}\\&= \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}] \\&= \ev^{\mathcal{V}}(f(Y_{\tau}))\ev^{\mathcal{V}}(g(\tau))\end{align*}

$\square$

Brownian Motion and Stochastic Calculus

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