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Math Help - characteristic function

  1. #1
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    characteristic function

    Hi, can someone please help me answer this question:
    Let  W_t and  B_t be independent standard Brownian Motions,  t \in [0,T], T< \infty

    find the characteristic function of
     \phi(u) = Eexp[iuW_tB_T]
    where u is a real variable,  i = \sqrt{-1}

    well i know that im supposed to integrate this, but i have no idea how to do it.. should i just:

     \int_0^\infty exp[iuW_tB_T] \times \frac{1}{\sigma\sqrt{2\pi}} exp[\frac{(y-\mu)^2}{2\sigma^2}] dW_tB_T ?

    is this the right way to answer it?
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  2. #2
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    Quote Originally Posted by Pengu View Post
    Hi, can someone please help me answer this question:
    Let  W_t and  B_t be independent standard Brownian Motions,  t \in [0,T], T< \infty

    find the characteristic function of
     \phi(u) = Eexp[iuW_tB_T]
    where u is a real variable,  i = \sqrt{-1}

    well i know that im supposed to integrate this, but i have no idea how to do it.. should i just:

     \int_0^\infty exp[iuW_tB_T] \times \frac{1}{\sigma\sqrt{2\pi}} exp[\frac{(y-\mu)^2}{2\sigma^2}] dW_tB_T ?

    is this the right way to answer it?
    You should have a look here.
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  3. #3
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    so is this right...

    \int_0^\infty exp[-u^2x^2/2] \times \frac{1}{\sqrt{2\pi T}} exp[\frac{(x)^2}{2T}] dx

    is that right now??
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  4. #4
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    lol

    is that you henry?? LOL

    are you sure variance of a standard brownian motion is 1....thats the correct answer if the s.b.m~N(0,1)
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  5. #5
    MHF Contributor

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    Quote Originally Posted by Pengu View Post
    so is this right...

    \int_0^\infty exp[-u^2x^2/2] \times \frac{1}{\sqrt{2\pi T}} exp[\frac{(x)^2}{2T}] dx

    is that right now??
    Yes, except that you forgot t, a minus sign, and you should integrate on \mathbb{R}. I.e.: E[e^{iuB_t W_T}]=\int e^{-\frac{u^2 x^2 t}{2}} \frac{e^{-\frac{x^2}{2T}}}{\sqrt{2\pi T}} dx. I let you find where you did a mistake.

    Now you can compute this integral. Remember that the density of a Gaussian integrates to 1: \int e^{-\frac{x^2}{2v}}dx = \sqrt{2\pi v} for any v>0. Apply this here for the suitable v and conclude.
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