# Thread: Marginal Distribution - Double exponential

1. ## 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.

thankyou