Probability subject. The question is:
If fx(x)=xe^(-x^2/2) for x>0 and Y=lnX find the density function for Y
The solution is:(e)^(2y-1/2e^2y)
I'm stuck on the part of the solution that uses this formula:
fy(y)=fx(v(y)) times |v'(y)|
Thanks to anyone who's willing to help.
@Randomgirl is correct. Translating the bad formatting from the link:
"Let X be a continuous random variable with generic probability density function $f(x)$ defined over the support $c_1 < x < c_2$. And, let $Y = u(X)$ be an invertible function of X with inverse function $X = v(Y)$.
Then, using the change-of-variable technique, the probability density function of Y is:
$$f_Y(y)=f_X(v(y)) |v'(y)|$$
defined over the support $u(c_1) < y < u(c_2)$. "