Quadratic variation of Martingales

**davidmccormick** · May 1st 2011, 04:40 AM

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.

Here is the answer:

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

**Moo** · May 1st 2011, 08:25 AM

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 ?

**davidmccormick** · May 1st 2011, 09:27 AM

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.

**Moo** · May 1st 2011, 09:43 AM

Sorry, talking about $\Delta n$ , not t

And I don't know the answer (yet ?), so can't help just pointing out stuff

**davidmccormick** · May 1st 2011, 09:58 AM

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.

**Focus** · May 8th 2011, 02:51 PM

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.

Here is the answer:

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

The second equality follows from the fact that [M] is the unique increasing process for which M^2-[M] is a (local) martingale, which in your case as M is square integrable is a true martingale.

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

There is no easy way of explicitly working out the quadratic variation, but if you can write it as an SDE in terms of a BM then your task is easy.

Thread: Quadratic variation of Martingales

LinkBack

Thread Tools

Display

Quadratic variation of Martingales

Similar Math Help Forum Discussions

Martingales

Martingales - Explaination.

basic question on martingales definition

Square Integrable Martingales

Problems understanding Riemann integral with Martingales

Search Tags