Probability Theory – Probability on First Hitting Time of Brownian Motion with Drift

brownian motionmartingalesprobability theorystopping-times

I am struggling with the following problem:

Let $B$ be a one dimensional Brownian motion and $a,b>0$. Show that $$P[B_t=a + bt \text{ for some } t\geq 0] = e^{-2ab}.$$

The following hint is given: Consider the martingale $(X_t)_{t\geq 0} = (\exp(2bB_t -2b^2 t))_{t \geq 0}$.

I already showed that $(X_t)$ is a martingale but I do not have any idea how I can use this to prove the statement.

Could somebody help me?
Thanks in advance!

Best Answer

Hints:

Define a stopping time $\tau$ by $$\tau := \inf\{t \geq 0; X_t = e^{2ab}\}.$$ Show that $$\mathbb{P}(\tau<\infty) = \mathbb{P}(\exists t \geq 0: B_t = a+bt). \tag{1}$$
Apply the optional stopping theorem to show that $$\mathbb{E}(X_{t \wedge \tau}) = \mathbb{E}(X_0)=1 \tag{2}$$ for all $t \geq 0$.
Show that $$\lim_{t \to \infty} X_{t \wedge \tau}(\omega)=e^{2ab} \quad \text{for $\omega \in \{\tau<\infty\}$}$$ and $$\lim_{t \to \infty} X_{t \wedge \tau}(\omega)=0 \quad \text{for $\omega \in \{\tau=\infty\}$}$$ (use $\lim_{t \to \infty} B_t/t=0$ almost surely).
By Step 3, we have $$\lim_{t \to \infty} X_{t \wedge \tau} = e^{2ab} \cdot 1_{\{\tau<\infty\}}.$$ Use the fact that $|X_{t \wedge \tau}| \leq e^{2ab}$ to conclude from $(2)$ and the dominated convergence theorem that $$\mathbb{P}(\tau<\infty) = e^{-2ab}.$$

Related Solutions

Probability Theory – Density of First Hitting Time of Brownian Motion with Drift

For $c=0$ this result is knows as reflection principle (see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 6) and follows from the Markov property and symmetry of Brownian motion. However, for $c>0$ the proof is more involved since we have to get rid of the drift term.

Since by definition

$$[H_a \leq t] = \left[ \sup_{s \leq t} X_s \geq a \right] \tag{1}$$

determining the distribution of $H_a$ is equivalent to finding the distribution of $\sup_{s \leq t} X_s$. In order to find the distribution of the latter, we need two ingredients: Girsanov's theorem and the joint distribution $(X_t,\sup_{s \leq t} X_s)$ for a Brownian motion $(X_t)_{t \ge 0}$.

Girsanov theorem: Let $c \in \mathbb{R}$ and $(B_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal{A},\mathbb{P})$. Then $$X_t := B_t+ct, \qquad t \leq T,$$ is a Brownian motion with respect to the probability measure $$d\mathbb{Q} := d\mathbb{Q}_T := \exp \left( -c B_T - \frac{c^2}{2} T \right) d\mathbb{P}.$$

For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 18.

Joint distribution of $(X_t, \sup_{s \leq t} X_s)$: Let $(X_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal{A},\mathbb{Q})$. Then the joint distribution $(X_t,\sup_{s \leq t} X_s)$ equals $$\mathbb{Q} \left[ X_t \in dx, \sup_{s \leq t} X_s \in dy \right] = \frac{2 (2y-x)}{\sqrt{2\pi t^3}} \exp \left(- \frac{(2y-x)^2}{2t} \right) 1_{[-\infty,y]}(x) \, dx \, dy. \tag{2}$$

For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Exercise 6.8 (there are full solutions available on the web).

So let's finally put it all together: It follows from $(1)$ and the definition of the probability measure $\mathbb{Q}_T$ that

$$\begin{align*} \mathbb{P}[H_a \leq T] &= \mathbb{P} \left[ \sup_{s \leq T} X_s \geq a \right] = \int 1_{[a,\infty)} \left( \sup_{s \leq T} X_s \right) \, d\mathbb{P} \\ &= \int 1_{[a,\infty)}\left( \sup_{s \leq T} X_s \right) \exp \left( c B_T + \frac{c^2}{2} T \right) \, d\mathbb{Q}_T \\ &= \int 1_{[a,\infty)}\left( \sup_{s \leq T} X_s \right)\exp\left(c X_T- \frac{c^2}{2} T \right) \, d\mathbb{Q}_T. \end{align*}$$

By Girsanov's theorem, $(X_t)_{t \leq T}$ is a Brownian motion with respect to $\mathbb{Q}_T$ and therefore $(2)$ gives

$$\begin{align*} \mathbb{P}[H_a \leq T] &= \exp\left(- \frac{c^2}{2} T \right) \int_{y \geq a} \int_{x \leq y} e^{cx} \frac{2 (2y-x)}{\sqrt{2\pi T^3}} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dx \, dy. \end{align*}$$

It remains to calculate the integral expression. First of all, by Fubini's theorem,

$$\begin{align*} \mathbb{P}[H_a \leq T] &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \left(\int_{x \geq a} e^{cx} I_1(x) \, dx + \int_{x \leq a} e^{-cx} I_2(x) \, dx \right) \\ &:=J_1+J_2 \tag{3} \end{align*}$$

where

$$\begin{align*} I_1(x):= \int_{y \geq x} \frac{2(2y-x)}{T} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dy &= \left[ - \exp \left(- \frac{(2y-x)^2}{2T} \right) \right]_{y=x}^{\infty} \\ &= \exp \left(- \frac{x^2}{2T} \right) \\ I_2(x) := \int_{y \geq a} \frac{2(2y-x)}{T} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dy &=\exp \left(- \frac{(2a-x)^2}{2T} \right). \end{align*}$$

Hence, $$\begin{align*} J_1 &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{x \geq a} e^{cx} I_1(x) \, dx \\ &= \frac{1}{\sqrt{2\pi T}} \int_{x \geq a} \exp \left(- \frac{(x-cT)^2}{2T} \right) \, dx \\ &= \frac{1}{\sqrt{\pi}} \int_{z \geq \frac{a-cT}{\sqrt{2T}}} \exp(-z^2) \, dz \tag{4} \end{align*}$$

and

$$\begin{align*} J_2 &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{x \leq a} e^{cx} I_2(x) \, dx \\ &\stackrel{u:=2a-x}{=} \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{u \geq a} e^{c(2a-u)} \exp\left(-\frac{u^2}{2T} \right) \, du \\ &=\ldots = \frac{e^{2ac}}{\sqrt{\pi}} \int_{z \geq \frac{a+CT}{\sqrt{2T}}} \exp(-z^2) \, dz. \tag{5} \end{align*}$$

Now if we differentiate $(3)$ with respect to $T$, using $(4)$ and $(5)$, we get

$$\begin{align*} \frac{d}{dT} \mathbb{P}(H_a \leq T) &= \frac{-1}{\sqrt{\pi}} \exp \left( - \frac{(a-cT)^2}{2T} \right) \left( \frac{-c}{\sqrt{2T}} - \frac{a-cT}{2 \sqrt{2} T^{3/2}} \right) \\ &\quad + \frac{-1}{\sqrt{\pi}} \underbrace{e^{2ac} \exp \left( - \frac{(a+cT)^2}{2T} \right)}_{\exp(-(a-cT)^2/2T)} \left( \frac{c}{\sqrt{2T}} - \frac{a+cT}{2 \sqrt{2} T^{3/2}} \right) \\ &= \frac{a}{\sqrt{2\pi T^3}} \exp \left(- \frac{(a-cT)^2}{2T} \right). \end{align*}$$

Brownian Motion – Expectation Stopped Brownian Motion with Drift

Hints:

Check that for fixed $\alpha>0$ the process $$M_t := \exp \left(\alpha X_t-\alpha \mu t- \frac{1}{2} \alpha^2 t \right)$$ is a martingale (with respect to the canonical filtration of the Brownian motion).
By the optional stopping theorem, $$\mathbb{E}(M_{\tau \wedge t}) = \mathbb{E}(M_0) = 1, \qquad t \geq 0.$$
Show that $|M_{t \wedge \tau}| \leq e^{\alpha a}$. Deduce from the dominated convergence theorem that $$\mathbb{E}(M_{\tau}) = 1.$$
Since $(X_t)_{t \geq 0}$ has continuous sample paths, we have $X_{\tau}=a$. Hence, $$M_{\tau} = e^{\alpha a} \exp \left( - \left[ \mu \alpha + \frac{1}{2} \alpha^2 \right] \tau \right).$$
It follows from step 3 and 4 that $$\mathbb{E} \exp\left( - \left[ \mu \alpha + \frac{1}{2} \alpha^2 \right] \tau \right) = e^{-\alpha a}.$$ Setting $\lambda := \mu \alpha + \frac{1}{2} \alpha^2$ proves the assertion.

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