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Math Help - Quadratic variation of Martingales

  1. #1
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    Quadratic variation of Martingales

    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
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  2. #2
    Moo
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    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 ?
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  3. #3
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    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.
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  4. #4
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    Sorry, talking about \Delta n, not t
    And I don't know the answer (yet ?), so can't help just pointing out stuff
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  5. #5
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    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.
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  6. #6
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    Quote Originally Posted by davidmccormick View Post
    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.
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