Q: $\displaystyle M_t$ is an $\displaystyle L^2$ martingale. Show that if $\displaystyle \langle M,M {\rangle}_t = \langle M,M {\rangle}_a$ then $\displaystyle M_t = M_a$ a.s.

Here is the answer:

$\displaystyle E{[M_t - M_a]}^2 = E[M_t^2] - E[M_a^2] = E \langle M {\rangle}_t - E \langle M {\rangle}_a = 0 \Rightarrow M_t = M_a $ a.s.

Couple of questions : 1. How do you justify the second inequality (I know how to show the first). $\displaystyle \langle M {\rangle}_t = \lim_{\Delta_n \to 0} \sum_{i=0}^{n-1} {|M_{t_{i+1}}-M_{t_i}|}^2$ how is this equal to $\displaystyle M_t^2$ ??

2. Is there an "easy" way of computing the quadratic variation of stochastic processes. For example the q.v. of a standard BM is t, and I understand the derivation, but it's using the definition! For that matter martingales, semi-marts or any other useful stochastic process......always using the definition? Thanks