Brownian Motion and Probability Density Function 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?

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.
