Question concerning maximum increment of Brownian motion

Question

Suppose $B$ is the standard Brownian motion, how to calculate $$\mathbb{P}\left((\max_{0\le s\le t}B(s))-B(t)<a\right)$$ I tried $B(t)-B(s)=B(t-s)$, $$P(B(t-s)>-a)=\int_{-a}^{\infty}\frac{1}{\sqrt{2\pi(t-s)}}e^{-\frac{y^2}{2(t-s)}}dy$$ And then I integrate it via $s$, that is $$\mathbb{P}\left((\max_{0\le s\le t}B(s))-B(t)<a\right)=\int_{0}^{t}\int_{-a}^{\infty}\frac{1}{\sqrt{2\pi(t-s)}}e^{-\frac{y^2}{2(t-s)}}dy ds$$ Am I right in this process? I am novice to stochastic process, and I don't know if these manipulations are legal. Also, I don't know how to integrate this stuff. Thanks for your attention!

Answer 1

You need a couple of tricks for this one.

First use that fact that if $s\mapsto B(s)$ is a Brownian motion then $s\mapsto B(t-s)-B(t)$ is also a Brownian motion. Hence

$$\mathbb{P}\left((\max_{0\le s\le t}B(s))-B(t)<a\right)= \mathbb{P}\left(\max_{0\le s\le t}B(s)<a\right)$$

Now we can use the reflection principle.

Let $T_a$ be the first hitting time of $a$ and set $$B_a(s) = \begin{cases}B(s) &\mathop{ if } s\leq T_a \\ a-B(s) &\mathop{ if } s\geq T_a \end{cases}$$

Now $B_a(s)$ is also a Brownian motion and $\max_{0\le s\le t}B(s)<a$ if and only if both $B(t) < a$ and $B_a(t) < a$.

As the events $B(t) \geq a$ and $B_a(t) \geq a$ are mutually exclusive (up to an event of probabililty $0$) and have identical probabilities we have that

$$\mathbb{P}\left((\max_{0\le s\le t}B(s))-B(t)<a\right)= 1 - 2\mathbb P\left( B(t)\geq a\right).$$

Which is easy as $B(t)$ is normally distributed.