The joint probability distribution function of X and Y:
2xe^(-y), x > 0, y > x^2
Obtain the pdf of V = (X^2)/Y
The Attempt at a Solution
The interval of V is (0,1) because Y is always greater than X^2.
Fv(v) = P(V <= v)
= P(X^2/Y <= v)
= P(Y <= X^2/v)
= double integral from (0 to ?) and (x^2 to x^2/v) of (2xe^(-y)) dy dx
I can't figure out what the limits of integration would be the first integral, and I'm not sure if the limits for the second integral are correct either. I'm not really sure if I'm doing this right... Any help?