• May 1st 2011, 04:40 AM
davidmccormick
Q: $M_t$ is an $L^2$ martingale. Show that if $\langle M,M {\rangle}_t = \langle M,M {\rangle}_a$ then $M_t = M_a$ a.s.

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

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

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
• May 1st 2011, 08:25 AM
Moo
Hello,

Not clear, what is $\Delta t$ in your example ?

Also, it's not that <M>_t that is equal to M_t^2, it's their expectations ! How can you make such a mistake while you're already studying BM's ?
• May 1st 2011, 09:27 AM
davidmccormick
Hi, not sure what you mean by $\Delta t$ in my example. Also, how do you show that their expectations are equal as you said? I really don't know.....I find stochastic analysis quite difficult to understand.
• May 1st 2011, 09:43 AM
Moo
Sorry, talking about $\Delta n$, not t :)
And I don't know the answer (yet ?), so can't help just pointing out stuff :p
• May 1st 2011, 09:58 AM
davidmccormick
By \Delta_n I just mean a partition of the interval [0,t] - you know, just from the definition of q.v. - nothing to do with the question.
• May 8th 2011, 02:51 PM
Focus
Quote:

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

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