Take B$_t$ as a standard Brownian motion such that B$_0$ = 0. And M$_t$ is the corresponding running maximum. i.e. M$_t$ = max$_{0\leq s \leq t}$ B$_s$. My goal is to compute:
(i) Quadratic Variation of M$_t$ on interval [0,T]
The quadratic variation of B$_t$ over [0,T] is T, but how can we compute that for running maximum
M$_t$?
(ii) The probability density function of M$_t$
I calculate it by using the joint density function of M$_t$ and B$_t$, which is
f$_{M(t),B(t)}$(m,w) = $\frac{2(2m-w)}{t\sqrt{2 \pi t}}$e$^{-\frac{(2m-w)^2}{2t}}$. It implies that f$_{M_t}$(m) = $\int_{-\infty}^{+\infty}$ f$_{M(t),B(t)}$(m,w) dw, but what I get is zero after some calculation. A bit weird. Any comments?
You forgot that $M_t \geq B_t$. The joint density you have obtained is only for $m \geq w$ so the integral for computing $f_{M_t}(m)$ is from $-\infty$ to $m$.