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Thread: Brownian Motion

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

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

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

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

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

    (More info if needed)
    $\displaystyle \mathbb{E}[(W_t^2 - t |F_s] = \mathbb{E}[(W_t - W_s + W_s)^2 - t|F_s]$ 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...

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

    which is apparently equal to $\displaystyle t-s$. Why is this?
    This results possibly from your definition of Brownian motion : for $\displaystyle 0\leq s\leq t$, $\displaystyle W_t-W_s$ is a Gaussian random variable with mean 0 and variance $\displaystyle t-s$, that furthermore is independent of $\displaystyle W_s$. 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 $\displaystyle 0\leq s\leq t$, $\displaystyle W_t-W_s$ is a Gaussian random variable with mean 0 and variance $\displaystyle t-s$, that furthermore is independent of $\displaystyle W_s$. 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...

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

    I say 2 since (correct me if I'm wrong which is likely...) $\displaystyle \mathbb{E}[W_t - W_s] = 0$..?
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  4. #4
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    Quote Originally Posted by Deadstar View Post
    So does that mean...

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