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Math Help - Brownian Motion

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

    I don't understand this one small point about Brownian motion...

    Here's the question, my answer and the part I always miss out...

    Let \{W_t\}_{t \geq 0} be a standard Brownian motion. Show that.
    (a) \{X_t\}_{t \geq0} is a standard Brownian motion, where X_t = -W_t

    Question is easy enough. Can show that X_0 = 0 almost surely and get that for 0 \leq s \leq t, X_t - X_s = -(W_t - W_s) is distributed with mean 0 and variance t-s and is independent of F_s = \sigma (W_u: 0 \leq u \leq s). (Not ENTIRELY sure why this is though)

    Then comes this small statement that I don't know how to show (and hence don't know if it's true).

    X_t has continuous paths.


    How do I show that? It's only ever stated in answers so I don't need some in depth proof, just a simple, it has continuous paths because...
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Deadstar View Post
    it has continuous paths because...
    because if a function f is continuous, then so is -f. That's all! Now you get why it's not extensively proved in the answers...

    If t\mapsto W_t is continuous, then so is t\mapsto -W_t.
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  3. #3
    Member Focus's Avatar
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    Quote Originally Posted by Deadstar View Post
    Can show that X_0 = 0 almost surely and get that for 0 \leq s \leq t, X_t - X_s = -(W_t - W_s) is distributed with mean 0 and variance t-s and is independent of F_s = \sigma (W_u: 0 \leq u \leq s). (Not ENTIRELY sure why this is though)
    The reason for the law is because the normal distribution is symmetric. The independence follows rather easily if you use the weak definition (see below), that is for any continuous bounded function f and  A \in \mathcal{F}_s you have that (by independent increments of BM)
    \mathbb{E}[f(W_t-W_s)\mathbf{1}_{A}]=\mathbb{E}[f(W_t-W_s)]\mathbb{P}(A).

    Now if you consider any f that is bounded and continuous then g(x)=f(-x) is bounded and continuous hence

    \mathbb{E}[f(-(W_t-W_s))\mathbf{1}_A]=\mathbb{E}[g(W_t-W_s)\mathbf{1}_A]=\mathbb{E}[g(W_t-W_s)]\mathbb{P}(A)
    =\mathbb{E}[f(-(W_t-W_s))]\mathbb{P}(A).

    A rather long and boring way of proving what your intuition tells you.
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