Quick way for the expected first hitting time for a 2D Brownian Motion

Question

Let $\{W_t\}_{t\ge 0}$ be a standard 2D Brownian motion starting at $(1,1)\in\Bbb R^2$. What's the probability that $W_t$ hits the positive half of $x$-axis before it hits the negative part?

There are standard "tricks" to solve similar problems in 1D. For example, if $B_t$ is a standard 1D BM starting at 0 and we want to compute $\Bbb P(B_t\,\text{hits 3 before hitting -5})$, then we let $\tau=\inf\{t\ge 0\mid B_t=3\vee B_t=-5\}$ and since $\tau$ is a.s. bounded we have, by the optional stopping theorem, that $$0=B_0=\Bbb E(B_\tau)=\Bbb E(B_\tau I(B_\tau = 3))+\Bbb E(B_\tau I(B_\tau = -5))=3\Bbb P(B_\tau=3)-5(1-\Bbb P(B_\tau = 3)).$$ Simple as above. We don't have to explicitly work out the distribution or expectation of $\tau$ at all.

Now back to the 2D case. Do similar elegant tricks exist? If not, what'd be the quickest possible way to solve it? (For what it's worth, this problem was asked in a Morgan Stanley quant interview, so I expect there to be at least one trick that can solve it in no more than a few minutes.)

Answer 1

Here is a complement to the answer posted by @saz that does not require an explicit use of the distribution of the stopping time. I will just use the notation in that answer. We want to compute $P\{B_{\tau}^1 \geq -1\}$. Since $B^1$ and $\tau$ are independent, we can write

$$P\{B_{\tau}^1 \leq -1\} = \int_{t\geq 0} P\{B_t^1 \leq -1\}f_{\tau}(t)\,dt$$ where $f_{\tau}(t)$ is the density of $\tau$. By the reflection principle,

$$P\{\tau \leq t\} = P\{N^2_t \leq -1\} = P\{M^2_t \geq 1\} = P\{\lvert B^2_t\rvert \geq 1\} = 2P\{B^2_t \leq -1\} = 2P\{B^1_t \leq -1\}$$ where $N^2$ and $M^2$ are the running minimum and the running maximum of $B^2$, respectively.

We can then write $f_{\tau}(t)\,dt = 2 dP\{B^1_t \leq -1\}$. Hence, the integral we want to evaluate becomes

$$2\int_{t\geq 0} P\{B_t^1 \leq -1\}dP\{B^1_t \leq -1\} = \left. P\{B_t^1 \leq -1\}^2\right|_{t=0}^{t=\infty} = \lim_{t\to\infty} P\left\{B_1^1 \leq -\frac{1}{\sqrt{t}}\right\}^2 = P\left\{B_1^1 \leq 0\right\}^2 = \frac{1}{4}$$

The last but one step uses dominated convergence.

Answer 2

The two-dimensional Brownian motion $(W_t)_{t \geq 0}$ has components which are independent Brownian motions, i.e. $$W_t = \begin{pmatrix} 1 \\ 1 \end{pmatrix} +\begin{pmatrix} B_t^1 \\ B_t^2 \end{pmatrix}$$ where $(B_t^i)_{t \geq 0}$, $i=1,2$, are independent one-dimensional Brownian motions started at $0$. Your problem boils down to finding $\mathbb{P}(1+B_{\tau}^1 \geq 0)$ for the stopping time

$$\tau := \inf\{t \geq 0; (1,1)+(B_t^1,B_t^2) \in \mathbb{R} \times (-\infty,0]\}.$$

Clearly,

$$\tau = \inf\{t \geq 0; 1+B_t^2 \leq 0\}$$

which means that $\tau$ is a stopping time with respect to the canonical filtration $\mathcal{F}_t^{2} := \sigma(B_s^2; s \leq t)$. In particular, $\tau$ is independent from $\mathcal{F}_{\infty}^{1} := \sigma(B_s^1; s \geq 0)$. Hence,

$$p:=\mathbb{P}(1+B_{\tau}^1 \geq 0) = \mathbb{E} \bigg[ \mathbb{P}(1+B_t^1 \geq 0) \bigg|_{t=\tau} \bigg].$$

It is known from the reflection principle that $\tau$ has distribution $$\frac{1}{\sqrt{2\pi t^3}} \exp \left(- \frac{1}{2t} \right) 1_{(0,\infty)}(t) \, dt$$ and therefore we get

$$\begin{align*} p &= \int_0^{\infty} \int_{-1}^{\infty} \frac{1}{\sqrt{2\pi t^3}} \frac{1}{\sqrt{2\pi t}} \exp \left(- \frac{1}{2t} - \frac{y^2}{2t} \right) \, dy \, dt \\ &= \frac{1}{2\pi} \int_{-1}^{\infty} \int_0^{\infty} \exp \left(- \frac{1+y^2}{2t} \right) \frac{dt}{t^2} \, dy \\ &= \int_{-1}^{\infty} \frac{1}{\pi} \frac{1}{1+y^2} \, dy \\ &= \frac{1}{2} - \frac{1}{\pi} \arctan(-1). \end{align*}$$

Remark: The above calculation actually shows that $B_{\tau}^1$ is standard Cauchy distributed.

Answer 3

You should be easily able to convert this 2D problem into a 1D one. An intuitive explanation follows.

Converting to polar coordinates, the motion is individually brownian in both $\theta$ and $r$. We don't really care about the $r$ in this case. We see that the current $\theta$ is $\frac{\pi}{4}$. The problem is basically asking which angle among $0$ and $\pi$ will we hit first. Use the exact same formula you quoted for the 1D case and you should get your answer.