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Math Help - stochastic processes

  1. #1
    Junior Member elenaas's Avatar
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    stochastic processes

    let be Brownian motion on R, . Put
    a) Prove
    for all

    b) Use power series expansion of the exponential function on both sides, compare the terms with the same power of u and deduse that


    c) Prove that

    for all functions f such that the integral of the right converges.
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  2. #2
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    Quote Originally Posted by elenaas View Post
    let be Brownian motion on R, . Put
    a) Prove
    for all

    b) Use power series expansion of the exponential function on both sides, compare the terms with the same power of u and deduse that


    c) Prove that

    for all functions f such that the integral of the right converges.
    a) and c) are easy consequences of the fact that B_t is a Gaussian random variable of mean 0 and variance t. As for b), you already have a hint given, you should specify what causes you difficulty.
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  3. #3
    Junior Member elenaas's Avatar
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    I have a problem with a)
    if is a random variable with density then for any (well behaved) function :


    From that:

    So I have a problem to calculate this integral
    I tried with integration by parts and got more complicated integral.
    So what do you think?
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  4. #4
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    Quote Originally Posted by elenaas View Post
    I have a problem with a)
    if is a random variable with density then for any (well behaved) function :


    From that:

    So I have a problem to calculate this integral
    I tried with integration by parts and got more complicated integral.
    So what do you think?
    I think your teacher assumes you already know the characteristic function of a Gaussian random variable: if X\sim \mathcal{N}(\mu,\sigma^2), then \Phi_X(t)=E[e^{itX}]=e^{it\mu-t^2\sigma^2/2}. This is well-known but not quite easy to prove, you can't find it by direct computation. One way is to see that, if X\sim \mathcal{N}(0,\sigma^2) and f(t)=E[e^{itX}], then f'(t)=-\sigma^2 t f(t) (you need to differentiate and then integrate by part), and f(0)=1, hence f(t)=e^{-\sigma^2 t^2/2}. The case \mathcal{N}(\mu,\sigma^2) follows. But, once again: you can't be expected to invent that, this is the kind of thing one is supposed to know (not the proof, but the formula).
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  5. #5
    Junior Member elenaas's Avatar
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    Thanks for your help,
    I solved a) and c) but now I have difficulty to prove that
    which implies b).
    What do you think?
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  6. #6
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    Quote Originally Posted by elenaas View Post
    Thanks for your help,
    I solved a) and c) but now I have difficulty to prove that
    which implies b).
    What do you think?
    From a), you have E[e^{iuB_t}]=e^{-\frac{u^2 t}{2}}, hence E[\sum_{n=0}^\infty \frac{(iuB_t)^n}{n!}]=\sum_{k=0}^\infty \frac{(-u^2 t)^k}{2^k k!} by using the power series expansion of the exponentials. Using the bounded convergence theorem, you can justify that E[\sum_{n=0}^\infty \frac{(iuB_t)^n}{n!}]=\sum_{n=0}^\infty E [\frac{(iuB_t)^n}{n!}]. So that you get, for all real u,

    \sum_{n=0}^\infty \frac{i^n E[B_t^n]}{n!}u^n = \sum_{k=0}^\infty \frac{(-1)^k t^k}{2^k k!}u^{2k}.

    Identifying coefficients gives E[B_t^n]=0 if n is odd, and your formula if n=2k.
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  7. #7
    Junior Member elenaas's Avatar
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    Thank you!
    I really appreciate your help.
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