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Math Help - MGF of a multiplication of RV

  1. #1
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    MGF of a multiplication of RV

    Hi, I am at a loss for a problem...

    If Y1 and Y2 are independent random variables, each having a normal distribution with mean 0 and variance 1, find the moment-generating function of U=Y1Y2. Use this moment-generating function to find E(U) and V(U). Check the result by evaluating E(U) and V(U) directly from the density functions for Y1 and Y2.


    I try to do Mu(t) = E(e^tu) the normal way, but then I get: E(e^ty1y2) which doesn't help me in the least...
    I know that E(U) = E(Y1)*E(Y2) = 0 and V(U) = 1 similarly, but I cannot find the MGF. Any pointers? Am I missing something obvious, or not so obvious?
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  2. #2
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    Quote Originally Posted by Volcanicrain View Post
    Hi, I am at a loss for a problem...

    If Y1 and Y2 are independent random variables, each having a normal distribution with mean 0 and variance 1, find the moment-generating function of U=Y1Y2. Use this moment-generating function to find E(U) and V(U). Check the result by evaluating E(U) and V(U) directly from the density functions for Y1 and Y2.


    I try to do Mu(t) = E(e^tu) the normal way, but then I get: E(e^ty1y2) which doesn't help me in the least...
    I know that E(U) = E(Y1)*E(Y2) = 0 and V(U) = 1 similarly, but I cannot find the MGF. Any pointers? Am I missing something obvious, or not so obvious?
    You're meant to know that if U = X+Y and X and Y are independent then m_U(t) = m_X(t) \cdot m_Y(t).
    Last edited by mr fantastic; January 30th 2010 at 12:50 AM. Reason: Missing a lot of pluses today
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  3. #3
    Moo
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    Hello,
    Quote Originally Posted by mr fantastic View Post
    You're meant to know that if U = XY and X and Y are independent then m_U(t) = m_X(t) \cdot m_Y(t).
    Isn't it for when U=X+Y, but not XY ?


    But I admit that I don't understand the order of the questions since you're first asked to find the mgf and then the expectations and all the stuff. And finding that mgf is more difficult than finding the expectation if it's indeed a product of normal distributions
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  4. #4
    MHF Contributor matheagle's Avatar
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    I hate to say, but katie is right.
    http://comics.com/pearls_before_swine/2010-01-19/
    However, my guess is that there should be a plus sign in there.
    http://comics.com/pearls_before_swine/2010-01-29/
    Last edited by matheagle; January 29th 2010 at 08:32 PM.
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  5. #5
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    Quote Originally Posted by Moo View Post
    Hello,

    Isn't it for when U=X+Y, but not XY ?


    But I admit that I don't understand the order of the questions since you're first asked to find the mgf and then the expectations and all the stuff. And finding that mgf is more difficult than finding the expectation if it's indeed a product of normal distributions
    *sigh* I imagined a plus sign that isn't there (but probably is meant to be there).
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  6. #6
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    I e-mailed my teacher and he said there was no mistake... :S that it had to do with conditional probability. Oye :S
    I appreciate your insight though ^^
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  7. #7
    MHF Contributor matheagle's Avatar
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    I would let W=Y1 and make a 2-2 transformation and then integrate out the W.
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  8. #8
    MHF Contributor matheagle's Avatar
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    Maybe this will work.
    I'll use X and Z as standard normal rvs.
    Let U=XZ. You want ...

    M_U(t)=E( e^{XZt} )=E(E(e^{XZt}|X))=E(M_Z(Xt)|X))

    What I'm doing is fixing the X which makes the argument now Xt.
    So plug in Xt into the MGF of Z and see what you get.

    Using the MGF of Z and plugging in Xt for t, I get

    E( e^{.5X^2t^2} )

    and now using the density of X...

    ={1\over \sqrt{2\pi}}\int_{-\infty}^{\infty}e^{.5x^2t^2-.5x^2}dx

    ={1\over \sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-.5x^2(1-t^2)}dx

    which you can compute without integrating by observing the st deviation here as (1-t^2)^{-.5}
    Last edited by matheagle; January 30th 2010 at 07:14 PM.
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  9. #9
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    Wow, thank you very much.
    I've been flirting around with the "given X" part of your answer, I just hadn't pinpointed that yet haha.

    I'm just a tad confused at the end - how did you solve that final integral? The integral is equal to the MGF of U, right?
    Thanks a million for your persisted help in this problem
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  10. #10
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    Quote Originally Posted by Volcanicrain View Post
    Wow, thank you very much.
    I've been flirting around with the "given X" part of your answer, I just hadn't pinpointed that yet haha.

    I'm just a tad confused at the end - how did you solve that final integral? The integral is equal to the MGF of U, right?
    Thanks a million for your persisted help in this problem
    Compare the final integral to the integral of a Gaussian (note that t is treated as a constant ....)
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  11. #11
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    Thank you both so very much!
    I've learned a lot in this thread - a ton actually! and now I can finally solve this problem I'd been thinking about since last Monday.

    Thank you thank you thank you!
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  12. #12
    MHF Contributor matheagle's Avatar
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    both?
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  13. #13
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    Well to you and to mr fantastic for helping me with the integral
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