Suppose we have a function, f(x,y) = e^-x * e^-y , 0<=x< ∞, 0<=y<∞,
where X and Y are exponential random variables with mean = 1. (For those who may not know, all this means is ∫(x*e^(-x) dx) from 0 to ∞ = 1, and the same for y)
Suppose we want to transform f(x,y) into f(z), where the transformation is Z = X-Y. Find f(z)
f(x,y) = e^-x * e^-y , 0<=x< ∞, 0<=y<∞
Z = X-Y
The Attempt at a Solution
So I decided to transform -Y into W. So we have -Y<=W, which implies Y>=-W
∫e^-y dy from -w to ∞...
-e^-y from -w to ∞
0 + e^-(-w)
Differentiate wrt w
f(w) = e^w -∞ < w <= 0
So now we have Z = X+W
Z-W = X
We'll just let W stay as is for this problem.
The jacobian of this transformation is 1.
So we have f(z) = ∫e^-(z-w)*e^w dw from -∞ to z,
This becomes ∫e^-z*e^2w dw from -∞ to z
This becomes e^-z * (e^2w)/2 from -∞ to z
This becomes e^-z * ((e^2z)-0)/2
Which becomes (e^z)/2
The domain of z is -∞ to ∞, however, this integral does not evaluate to 1. As a matter of fact, it does not even converge.
Any help would be greatly appreciated!!