Expectation of a stopping time of a Wiener process

Question

How can we calculate $\mathbb{E}(\tau)$ when $\tau=\inf\{t\geq0:B^2_t=1-t\}$?

If we can prove that $\tau$ is bounded a.s. (i.e. $\mathbb{E}[\tau]<\infty$), then we can use the fact that $\mathbb{E}[B^2_\tau]=\mathbb{E}[\tau]$. Hence, $1-\mathbb{E}[\tau]=\mathbb{E}[\tau]$ and so $\mathbb{E}[\tau]=\dfrac{1}{2 }$.

Is that right? If so, how can we show that $\mathbb{E}[\tau]<\infty$?

Answer 1

The random function $u:t\mapsto B_t^2-1+t$ is almost surely continuous. Since $u(0)=-1$ and $u(1)=B_1^2$, $u(0)\lt0\leqslant u(1)$. By the intermediate value theorem, there exists almost surely $t$ in $[0,1]$ such that $u(t)=0$, that is, $\tau\leqslant1$ almost surely.

Answer 2

Let $f_{\pm}(t) := \pm \sqrt{1-t}$. Then $\tau=\inf\{t \geq 0; B_t = f_+(t) \vee B_t = f_-(t)\}$. Let $w \in \Omega$. There are 3 cases:

• $B_1(w)=0$: Then clearly $\tau(w) \leq 1$ holds (since $B_1(w)=0=f_+(1)$).
• $B_1(w)>0$: Define $g(t) := f_+(t)-B_t(w)$. Then we have $$g(0)=1-0 =1>0 \qquad \qquad g(1)=0-B_1(w)<0$$ since $g$ is continuous we conclude that there exists $t_0 \in (0,1)$ such that $g(t_0)=0$ (intermediate value theorem), i.e. $f_+(t_0)=B_{t_0}(w)$. Hence $\tau(w) \leq t_0 <1$.
• $B_1(w)<0$: Similar proof as for $B_1(w)>0$, replace $f_+$ by $f_-$.

Thus $\tau \leq 1$ almost surely.

(Your conclusions regarding $\mathbb{E}\tau = \frac{1}{2}$ are correct.)