Why do $\underset{t\leq u \leq T}{\max}W_u-W_t$ and $\underset{0\leq u \leq T-t}{\max}W_u-W_0$ have the same distribution?

Question

Let $W$ a Brownian motion.

Why do $$\max_{t\leq u \leq T} W_u-W_t$$

and $$\max_{0\leq u \leq T-t}W_u-W_0$$

( with $ T >t$ ) have the same distribution?

How can I prove it? Thanks

Answer 1

Here is a try. To conceptionally know why, brownian motion is an integral of white noice or if you like picture it as a sum of iid symmetrik random variables that scaled correctly, so you have

Assume $W_0 = 0$, $$ W_t = \int_0^t dW_t $$ and for $u > t$ you have $$ W_u = \int_0^u dW_t = \int_0^t dW_t + \int_t^u dW_t = W_t + \int_t^u dW_t $$ so $$ W_u - W_t = \int_t^u dW_t $$ Now the integral is over i.i.d and hence where the intervall is located in the integral does not matter so, $$ \{W_u - W_t\}_{t\leq u \leq T} =^d \left \{\int_0^{u} dW_\tau \right \} _{0 \leq u \leq T - t} = \{W_{u-t}\}_{0 \leq u-t \leq T - t} = $$ This proves the question for $W_0=0$.

So, $$ \max_{t\leq u \leq T} (W_u-W_t) =^d \max_{0\leq u \leq T-t} W_u $$ Because the max is the a max over the same range of processes that are equal in distribution.

It's uncertain if the max is for both term or just the one that varies, this is equivalent because of the identity

$$ \max_{u\in I} (W_u + c) = (\max_{u\in I} W_u) + c. $$

The case $W_0=a$ is left as an exercise.

Answer 2

Recall the renewal property of Brownian motion.

Lemma Let $(W_s)_{s \geq 0}$ be a Brownian motion. If $t>0$ is a fixed number, then $$B_s := W_{t+s}-W_t, \qquad s \geq 0$$ is a Brownian motion.

From the Lemma it follows, in particular, the processes $(W_s)_{s \geq 0}$ and $(B_s)_{s \geq 0}$ are equal in distribution. Since the processes have continuous sample paths this, in turn, implies that the random variables

$$X:=\sup_{0 \leq u \leq M} W_u$$

and

$$Y:=\sup_{0 \leq u \leq M} B_u$$

have the same distribution; here $M>0$ is some fixed number. By the very definition of $(B_s)_{s \geq 0}$, we have

$$Y = \sup_{0 \leq u \leq M} W_{t+u} - W_t = \sup_{t \leq v \leq M+t} W_v-W_t.$$

Choosing $M:=T-t$ this proves the assertion.