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Math Help - Product of Random variables - Density

  1. #1
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    Product of Random variables - Density

    A random variable \xi has Gaussian distribution with mean zero and variance one, while a random variable \eta has the distribution with the density:
    p_{\eta}(t)= \{ \begin{array}{cc}t e^{-t^2/2} & if\quad t\geq 0 \\ 0 & otherwise \end{array}
    What is the distribution of \zeta = \xi.\eta assuming that \xi and \eta are independent?

    One approach is to use the formula of change of variables (e.g. using Jacobian) for a product of random variables X,Y (see for example Grimmett and Stirzaker, p. 109) through the map: u = xy, v=x which gives:
    f_{U,V}(u,v)=f_{X,Y}(v,u/v)|v|^{-1} , use independence and integrate over v to obtain the result.
    But the formula is somewhat cumbersome as it gives 2 different expressions whether you assign \xi or \eta to X (no sign problem with \eta). I assume you get to the same result although the calculation doesn't seem tractable.

    Another (simpler?) approach is to use the formula of total probability and use independence. But again it doesn't seem you can go beyond an integral of the form:

    f_{\xi\eta}(a)=\frac{1}{\sqrt{2\pi}}\int_0^{\infty  } e^{-\frac{1}{2}((\frac{a}{x})^2+x^2)}dx

    Is this correct? Is it possible to do better?
    Thanks for any help.
    Last edited by akbar; November 19th 2009 at 11:52 AM.
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  2. #2
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    Quote Originally Posted by akbar View Post
    A random variable \xi has Gaussian distribution with mean zero and variance one, while a random variable \eta has the distribution with the density:
    p_{\eta}(t)= \{ \begin{array}{cc}t\exp^{-t^2/2} & if\quad t\geq 0 \\ 0 & otherwise \end{array}
    What is the distribution of \zeta = \xi.\eta assuming that \xi and \eta are independent?

    One approach is to use the formula of change of variables (e.g. using Jacobian) for a product of random variables X,Y (see for example Grimmett and Stirzaker, p. 109) through the map: u = xy, v=x which gives:
    f_{U,V}(u,v)=f_{X,Y}(v,u/v)|v|^{-1} , use independence and integrate over v to obtain the result.
    But the formula is somewhat cumbersome as it gives 2 different expressions whether you assign \xi or \eta to X (no sign problem with \eta). I assume you get to the same result although the calculation doesn't seem tractable.

    Another (simpler?) approach is to use the formula of total probability and use independence. But again it doesn't seem you can go beyond an integral of the form:

    f_{\xi\eta}(a)=\frac{1}{\sqrt{2\pi}}\int_0^{\infty  }\exp^{-\frac{1}{2}((\frac{a}{x})^2+x^2)}dx

    Is this correct? Is it possible to do better?
    Thanks for any help.
    Use characteristic functions: cf. here. In this case, if my quick computation is correct, you should get a Laplace distribution: e^{-|x|}dx.
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  3. #3
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    Although the problem wasn't related (in the book) to the use of characteristic functions, I agree the result is almost immediate that way. I still wonder how you can get it without the transform...

    One question though: you get \Phi_{XY}(\lambda)=\frac{1}{1+\lambda^2} when you first use the Gaussian random variable, then evaluate the final expectation using the other distribution. Have you tried the other way round? e.g. first calculate the CF of the distribution (let me know your result), then evaluate the final expectation with the Gaussian distribution.

    In short: do you get back on your feet?

    Thanks for your lights.
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