mean hitting time of a level and growth rate of maximum process

Question

Let $X_t$ be the absolute value of Brownian motion starting at $0$, let $\tau_x$ be it's first hitting time of the level $x>0$, and let $M_t$ be it's running maximum up to time $t$. Suppose we knew that $E[\tau_x]=x^2$ and wanted to use this fact to get information on the growth rate of $E[M_t]$. Intuitively, since it takes an average of $x^2$ units of time to hit the level $x$, $M_{x^2}$ should be around $x$ on average. Hence, the growth of $E[M_t]$ should be of order $\sqrt{t}$. My question is how do you make this argument rigorous, preferably in a manner that generalizes to other processes beside Brownian motion.

Answer 1

I am assuming our Brownian motions start from $0$. It is a well-known theorem that $$P(M_t >x) = 2P(B_t>x).$$ This can be seen using the reflection principle. Let $\tau$ be a stopping time. Then the Brownian motion $B^*(t)$ reflected at $\tau$ is the Brownian motion up to $\tau$ after which it is reflected along the line $y=B_\tau$. Formally

$$ B^*(t) = B_t1_{t \le \tau} + (2B_{\tau}-B_{t})1_{\tau>t} $$

It is known that $B^*(t)$ also has the distribution of the standard Brownian motion (show this!). Then

$$ P(M_t>x) = P(B_t>x) + P(\tau_x <t , B_t <x) = P(B_t>x) + P(B^*(t) >x) = 2P(B_t>x) $$

You can obtain $E(M_t)$ by integrating the above over $x$ since $B_t$ is just Nomal distribution with variance $t$.