Brownian motion and expectation

Question

I'm having trouble solving the following exercise:

Let $T_{[-a,a]} = \inf \{t: B_t \notin \{-a, a\} \}.$ Show that $E[T_{[-a,a]}]$ $=$ $a^{2} \times E[T_{[-1,1]}]$.

I don't see how I can solve this. If someone could help me it would be awesome.

Thanks in advance!

Answer 1

This can be shown directly. As shown in this post Expectation of Exit Time of Brownian Motion from Interval, we have $\mathbb{E}[T_{[-a,a]}] = a^2$. Hence $\mathbb{E}[T_{[-1,1]}] = 1$ and the result follows.