Question

From the pdf f(x)=2xe^-x^2, where x>0, find the distribution of Y=X^2

Answer 1

this is \(f(x)=2xe^{-x^2}\) yes?

Answer 2

start with
\(X=\sqrt{Y}\)

Answer 3

yes

Answer 4

then if i remember correctly you are supposed to compute
\[\int_0^xf(\sqrt{y})dx=\int_0^x2\sqrt{y}e^{-y}\frac{1}{2\sqrt{y}}dy\]
it has been a while so my notation is probably off, but i think that this is correct because miracle occurs and everything cancels, leaving
\[\int_0^xe^{-y}dy\]

Answer 5

Many thanks for your assistance

Answer 6

yw, i only hope it is right, and i wouldn't bet the ranch on it, but since it worked out so nicely i think it is probably on the right track