Expectation of brownian motion at hitting time

Question

Am i correct in my derivation? I want to calculate $\mathbb{E}B_{\tau_a}$. From the definition of the hitting time i get $B_{\tau_a}=a$, so $$\mathbb{E}B_{\tau_a}=\mathbb{E}a=a$$ I am new to the subject and it seems too simple to be true...

Answer 1

Yes - given that the Brownian motion hits $a$, $\mathbb{E}[B_{\tau_a}]=a$.

But note that a Brownian motion with drift away from $a$ will have a positive probability of never hitting $a$.

Even if you have a standard Brownian motion (a Wiener process with no drift, $B_0=0$ and $B_t−B_s$~$ N(0, t−s)$ for $0 \le s \le t$, $\mathbb{E}[{\tau_a}]=\infty$. If on the other hand $\tau_a$ were the first time the process hits either $a$ or $-a$ (a common use of the notation), then $\mathbb{E}[B_{\tau_a}]=0$ and $\mathbb{E}[{\tau_a}]=a^2$.