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Math Help - Marginal Distribution - Double exponential

  1. #1
    Junior Member
    Joined
    Jan 2009
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    Unhappy Marginal Distribution - Double exponential

    Mx,y(s,t)=1/((1-s)^2-t^2);

    As Mx,y(0,0)=1;

    I found that Mx=Mx,y(s,0)= (1-s)^-2
    Therefore X has gamma distribution with parameters (1,2)
    E(X)=2 and Var(x)=2.

    then I found that My=Mx,y(0,t)=1/(1-t^2)
    which can be considered as a double exponential distribution

    but what are the parameters for this one

    am i right if say they are Y~Laplace(0,1)

    Does it mean that E(Y)=0 and Var(y)= 2 and the by differentiating Mx,y(s,t) with repect to t and s I got E[XY]=0 and at the end I got cov(x,y)=0 aswell.

    Then at the end I have to find the coefficient of correlation p(X,Y) which I got to be 0.

    This means X and Y should be independent isn't.

    But when I tried to do Mx(s)*My(t) to find Mx,y(s,t) I get different results.

    PLEASE HELP ME AM LOST NOW COMPLETELY

    Please help me with this

    thankyou
    Last edited by rajr; December 9th 2009 at 08:46 AM.
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