Compute the probability that a brownian motion starting at $0$ hits the line $t+1$ before the line $t-1$.

Here is what I did:

I figured it has to do with optional stopping theorem. The probability is the same as the probability that $X(t) = B(t) - t$ hits $+1$ before $-1$.

So consider the exponential martingale. $M_{\lambda, t} = e^{\lambda B(t)-t\lambda^2/2 }$

Now we can consider the stopping time $\alpha$ that says $B(t) - t$ hits $+1$ before $-1$.

So now since $M_{\lambda, t}$ is a martingale and $\alpha$ is bounded, then by optional stopping theorem the stopped process $e^{2(B(t) - t)\wedge \alpha}$ (setting $\lambda = 2$) is also a martingale.

Now how do I go from here to compute the actual probability using optional stopping?

  • 1
    $\begingroup$ $\alpha$ is not bounded. But you don't need boundedness to conclude that $M_{\lambda, t \wedge \alpha} = e^{\lambda B(t \wedge \alpha) - (t \wedge \alpha) \lambda^2/2}$ (note that this is not what you wrote) is a martingale. $\endgroup$ – Nate Eldredge Feb 18 '15 at 1:38

As @NateEldredge pointed out, the stopping time $\alpha$ is not bounded. But obviously the stopping time $t \wedge \alpha$ is bounded for each $t \geq 0$ and therefore it follows from the optional stopping theorem that

$$\mathbb{E}\exp(2X_{t \wedge \alpha}) = \mathbb{E}\exp\big(2 B_{t \wedge \alpha}- 2(t \wedge \alpha)\big) = 1.$$

As $|X_{t \wedge \alpha}| \leq 1$, the dominated convergence theorem gives

$$\mathbb{E}\exp(2X_{\alpha}) = 1.$$

Now use

$$\exp(2X_{\alpha}) = e^2 1_{\{X_{\alpha} = 1\}}+ e^{-2} 1_{\{X_{\alpha}=-1\}}$$

to calculate $\mathbb{P}(X_{\alpha}=1) = 1-\mathbb{P}(X_{\alpha}=-1)$.

  • $\begingroup$ Worked like a charm, thanks! $\endgroup$ – Alex Feb 18 '15 at 17:09

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