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Math Help - Brownian Motion

  1. #1
    Super Member Deadstar's Avatar
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    Brownian Motion

    Show that a standard Brownian motion \{W_t\}_{t \geq 0} is a martingale with respect to the filtration \{F_t \}_{t \geq 0}, where F_t = \sigma(W_s : 0 \leq s \leq t). Moreover, show that \{ W_t^2 + t \}_{t \geq 0} is a martingale with respect to the filtration \{F_t \}_{t \geq 0}.

    First part I have done, but for the second part I arrive at a point where I have this...

    \mathbb{E}[(W_t - W_s)^2]

    which is apparently equal to t-s. Why is this?

    (More info if needed)
    \mathbb{E}[(W_t^2 - t |F_s] = \mathbb{E}[(W_t - W_s + W_s)^2 - t|F_s] and continuing I arrive at the above result I'm stuck at.
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    I have this...

    \mathbb{E}[(W_t - W_s)^2]

    which is apparently equal to t-s. Why is this?
    This results possibly from your definition of Brownian motion : for 0\leq s\leq t, W_t-W_s is a Gaussian random variable with mean 0 and variance t-s, that furthermore is independent of W_s. If that doesn't answer your question, please give us your definition of Brownian motion.
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  3. #3
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Laurent View Post
    This results possibly from your definition of Brownian motion : for 0\leq s\leq t, W_t-W_s is a Gaussian random variable with mean 0 and variance t-s, that furthermore is independent of W_s. If that doesn't answer your question, please give us your definition of Brownian motion.
    That seems to be right relative to what we've been told!

    So does that mean...

    \mathbb{E}[(W_t - W_s)^n] = t-s \forall n \geq 2?
    Or perhaps you take t-s to the (n-1)th power..?

    I say 2 since (correct me if I'm wrong which is likely...) \mathbb{E}[W_t - W_s] = 0..?
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  4. #4
    MHF Contributor

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    Quote Originally Posted by Deadstar View Post
    So does that mean...

    \mathbb{E}[(W_t - W_s)^n] = t-s \forall n \geq 2?
    Or perhaps you take t-s to the (n-1)th power..?
    No! It says that E[(W_t-W_s)^n] is the n-th moment of a Gaussian random variable of mean 0 and variance t-s, that's all. For n=2, E[(W_t-W_s)^2] is the variance itself (mean = 0), so it is t-s. We can also write W_t-W_s=\sqrt{t-s}Z, so that Z is a standard Gaussian random variable (if Z\sim \mathcal{N}(0,1) then \lambda Z\sim \mathcal{N}(0,\lambda^2)); then E[(W_t-W_s)^n]=(t-s)^{n/2} E[Z^n] and there are formulas for E[Z^n]. By the way, E[(W_t-W_s)^n]=0 if n is odd (by symmetry).
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