Consider the following exercise:
Let $T_{[-a,a]} = \inf \{t: B_t \notin [-a, a] \}.$ Show that $E[T_{[-a,a]}]$ $=$ $a^{2} \times E[T_{[-1,1]}]$.
Please tell me if this reasoning is correct:
$T_{[-a,a]} = \inf \{t: B_t \notin [-a, a] \}$ $ = \inf \{t: \frac{1}{a}B_t \notin [-1, 1] \} $
Then since the Brownian Motion is 0.5-self-similar we have:
$ \frac{1}{a} B_{t} = (\frac{1}{a^2})^{\frac{1}{2}} B_{t} =^{d} B_{t \times \frac{1}{a^2}} $
From here it follows:
$E[T_{[-a,a]}] = E[\inf \{t: \frac{1}{a}B_t \notin [-1, 1] \}]$ $ = E[\inf \{t: B_{t \times \frac{1}{a^2}} \notin [-1, 1] \}] $ $ = E[ {a^2} \times \inf \{t: B_{t} \notin [-1, 1] \}] $ $ = {a^2} \times E[T_{[-1,1]}] $
Can this be a possible resolution to this problem! Please give me some feedback!
What you wrote seems entirely correct. An alternative way to do this is to actually explicitly compute (takes even less effort perhaps). Note that $M_t = B_{t}^{2}-t$ is a martingale and set $\tau_n = \min(T_{[-a,a],}, n).$ Then this is easily seen to be a stopping time, as the minimum of a deterministic stopping time and the hitting time of an open set.
Defining $N_t = M_{t \wedge \tau_n}$, we see that this is a uniformly integrable martingale, since it is bounded by $a^2+n$. Therefore, we may apply the Optional Stopping Theorem to it. \begin{equation} \mathbb{E}[M_{\tau_n}] = \mathbb{E}[M_{0}] = 0 \end{equation} Then, by monotone and dominated convergence, you may carefully take the limit as $n \to \infty$ and see $\mathbb{E}[T_{[-a,a]}]=\mathbb{E}\left[B_{T_{[-a,a]}}^{2}\right]=a^2$. Your result follows immediately.