# Maximum of standard brownian motion on an interval

I'm trying to find the probability that the maximum of standard Brownian motion on the interval $(t_1, t_2)$ exceeds a value $x$, i.e.,

$$P(max_{t_1 \le s \le t_2}B(s) \gt x)$$

I initially approached this by invoking the reflection principle: Let $M_{t_1, t_2} = max_{t_1 \le s \le t_2}B(s)$. Then $P(B(t_2)>x \mid M_{t_1, t_2} \gt x) = P(B(t_2)<x \mid M_{t_1, t_2} \gt x) = \frac{1}{2}$ by the reflection principle, and

\begin{align} P(B(t_2)>x) & = P(B(t_2)>x \mid M_{t_1, t_2} \gt x) \cdot P(M_{t_1, t_2} \gt x) \ + P(B(t_2)>x \mid M_{t_1, t_2} \lt x) \cdot P(M_{t_1, t_2} \lt x) \\ & = \frac{1}{2} P(M_{t_1, t_2} \gt x) + 0\\ \end{align}

$\Rightarrow P(M_{t_1, t_2} \gt x) = 2P(B(t_2)>x)$.

But this doesn't at all depend on $t_1$. Is this correct? I can't find any flaws in the logic, but the result doesn't seem to make sense. Can anyone explain why my reasoning is wrong?

The only other way I can think of to do this is to condition on $B(t_1) = y$ and integrate over all possible values of $y$, which seems quite messy.

Thanks a bunch!

Think about what the reflection principle says intuitively. It says "if at some time you were at $x$, then at a later time, you are equally likely to be above or below $x$."
Here you're conditioning on $M_{t_1, t_2} > x$. That does not imply that at some time between $t_1$ and $t_2$ you were at $x$. You might have started above $x$ (i.e. $B_{t_1} > x$) and never dropped down to $x$.
You're probably used to applying this argument with $t_1 = 0$ and $x > 0$. Then the argument does work, since you have $B_{t_1} = 0 < x$ unconditionally, and therefore $M_{0, t_2} > x$ does imply that at some time you were at $x$, by the intermediate value theorem. But here it doesn't.
• Thanks Nate!! That helps a lot :) So would the only way to do this be to condition on the value of $B(t_1)$ and integrate over all possible values of $y$? – user3922963 May 7 '16 at 2:44