Let X have the pdf fx xex22 0 x infinity Find the PDF of

Let X have the p.d.f f(x) = xe^-(x^{^2)/2}, 0 < x < infinity. Find the PDF of Y= X².

We have PDF of X = xe^{-x^2 / 2} which is integral of f(x) from 0 < x < infinity.

Now, we have,

Y = X²

Thus, PDF of Y = Integral of f(Y) dy from 0 < y < infinity.

dy = 2x dx

Integral takes the form:

= f(x²) dy where in dy can be replaced as 2x dx

= (2x) x² (e^{-x^4 / 2} ) dx will be the new p.d.f

= 2x³ (e^{-x^4 / 2} )

It can be verified that this p.d.f has the integration value from 0 < x < infinity = 1

P.S : I apologize for the lack of equations format, but the equation editor wasn\'t working.

Ask me if you have doubts.