Properties of Brownian Motion (Expected value)

Question

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!

Thanks in advance!

Answer 1

Yes, and there is also an alternative proof via explicit computation. See: Proof of Expected value of Brownian Motion.