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I'm confused with this particular problem.

B(t) is a BM. MT is the maximum of B(t) in [0,T].

What is PDF?

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This is worked out in Karatzas and Shreve's book. If $T_a$ is the hitting time of $a > 0$, you compute $P(T_a \ge t)$ using the reflection principle; but this is the same as $P(M_t \le a)$. Then differentiate with respect to $a$ to get the density function. –  Nate Eldredge Nov 3 '12 at 20:01
    
Alvin, this is pretty similar to your last question. Perhaps you can be more specific about what's confusing you? –  amWhy Nov 3 '12 at 20:29
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up vote 2 down vote accepted

Let $M_t := \sup_{s \leq t} B_s$. As Nate Eldredge already mentioned you have to use the reflection principle:

$$\mathbb{P}[M_t \geq b] \stackrel{\text{Def}}{=} \mathbb{P} \left[\sup_{s \leq t} B_s \geq b \right] = 2 \mathbb{P}[B_t \geq b]$$

It follows that $M_t \sim |B_t|$ since

$$\mathbb{P}[M_t \geq b] = 2\mathbb{P}[B_t \geq b] = \mathbb{P}[B_t \geq b]+\underbrace{\mathbb{P}[-B_t \leq -b]}_{\mathbb{P}[B_t \leq -b]} = \mathbb{P}[|B_t| \geq b]$$

(using the fact that $(-B_t)_{t \geq 0}$ is also a Brownian Motion). Thus

$$\mathbb{P}[M_t \leq b] = 2 \mathbb{P}[0 \leq B_t \leq b] = \frac{2}{\sqrt{2\pi \cdot t}} \cdot \int_0^b \exp \left(-\frac{x^2}{2t} \right) \, dx \\ \Rightarrow M_t \sim \frac{\sqrt{2}}{\sqrt{\pi \cdot t}} \cdot \exp \left(-\frac{x^2}{2t} \right) \cdot 1_{(0,\infty)}(x) \, dx$$

Knowing the density function it shouldn't be a problem to calculate the expectation value and variance.

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