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Math Help - Quick Brownian Motion question

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

    How do calculate something like...

    \mathbb{E}[W_{\tfrac{1}{2}} W_{\tfrac{3}{4}}] ..?

    Where \{W_t\}_{t \geq 0} is a standard Brownian motion.


    \frac{1}{\sqrt{2 \pi 1/2} \sqrt{2 \pi 3/4}} \int_{-\infty}^{\infty} e^{-x^2}e^{-\frac{2x^2}{3}} dx ..?
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    How do calculate something like...

    \mathbb{E}[W_{\tfrac{1}{2}} W_{\tfrac{3}{4}}] ..?

    Where \{W_t\}_{t \geq 0} is a standard Brownian motion.
    By writing E[W_{1/2}W_{3/4}]=E[W_{1/2}(W_{1/2}+(W_{3/4}-W_{1/2}))]=E[(W_{1/2})^2]+E[W_{1/2}]E[W_{3/4-1/2}] using independence and stationarity of increments.
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  3. #3
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Laurent View Post
    By writing E[W_{1/2}W_{3/4}]=E[W_{1/2}(W_{1/2}+(W_{3/4}-W_{1/2}))]=E[(W_{1/2})^2]+E[W_{1/2}]E[W_{3/4-1/2}] using independence and stationarity of increments.
    Cool, but also grim since I have like, 6 of these to calculate in a question...

    I hope this isn't dumb but...

    \mathbb{E}[(W_{1/2})^2]+\mathbb{E}[W_{1/2}]\mathbb{E}[W_{3/4-1/2}] = \frac{1}{2} + 0?

    So the other ones I have which include...

    \mathbb{E}[W_{1/2}W_1] and \mathbb{E}[W_{3/4}W_1] equal \frac{1}{2} and \frac{3}{4} respectively..?
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  4. #4
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    Quote Originally Posted by Deadstar View Post
    Cool, but also grim since I have like, 6 of these to calculate in a question...

    I hope this isn't dumb but...

    \mathbb{E}[(W_{1/2})^2]+\mathbb{E}[W_{1/2}]\mathbb{E}[W_{3/4-1/2}] = \frac{1}{2} + 0?

    So the other ones I have which include...

    \mathbb{E}[W_{1/2}W_1] and \mathbb{E}[W_{3/4}W_1] equal \frac{1}{2} and \frac{3}{4} respectively..?
    That looks correct, well done.
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  5. #5
    Super Member Deadstar's Avatar
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    Allow me one further 'make or break my reputation with Laurent' conjecture...

    For b > a > 0.

    \mathbb{E}[W_a W_b] = a
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  6. #6
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    Quote Originally Posted by Deadstar View Post
    Allow me one further 'make or break my reputation with Laurent' conjecture...

    For b > a > 0.

    \mathbb{E}[W_a W_b] = a
    No worry for your reputation .

    By the way, one of the definitions of Brownian motion is: a continuous centered Gaussian process ( \forall t, W_t is a centered Gaussian r.v.) with covariance E[W_aW_b]=\min(a,b) for all a,b\geq 0.
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  7. #7
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Laurent View Post
    No worry for your reputation .

    By the way, one of the definitions of Brownian motion is: a continuous centered Gaussian process ( \forall t, W_t is a centered Gaussian r.v.) with covariance E[W_aW_b]=\min(a,b) for all a,b\geq 0.
    Ah cool, the more you know...

    Cheers for the help.
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