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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 $\mathbb{E}[\tau]<\infty$?

saz
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Nick Papadopoulos
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2 Answers2

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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.

Did
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  • $B_t$ is continuous almost surely. Can we apply the intermediate value theorem in this case? It doesn't seem correct... Also, what does "there exists almost surely $t$ in $[0,1]$" mean? – Nick Papadopoulos Dec 10 '12 at 18:21
  • Can we apply the intermediate value theorem in this case? Yes we can, to the function $t\mapsto B_t^2(\omega)-1+t$ for every $\omega$ such that $t\mapsto B_t(\omega)$ is continuous. // what does "there exists almost surely t in [0,1]" mean? That the set of $\omega$ such that there exists $t$ such that $u(t)(\omega)=0$ has probability $1$. – Did Dec 10 '12 at 20:31
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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.)

saz
  • 120,083