Let $Y=-\log{X}$ and $X$ has pdf given by:
$f_X(x)=\frac{(n+m+1)!}{n!m!}x^{n}(1-x)^m; 0<x<1$. Find the pdf of $Y$.
MY WORKING
Since pdf of $Y$ equals the derivative of cdf of $Y$ first we find cdf then go for pdf.
Now the cdf is given by: $P(Y\leq y)=P(-\log{X}\leq y)=P(y\geq \log{X})=P(X\leq e^{y})=\int_{0}^{e^{y}} \frac{(n+m+1)!}{n!m!}x^{n}(1-x)^m \,dx$. I am not sure whether I have gone okay so far, if yes then I am not able to simply the integral expression further. Kindly guide me
I would suggest we find pdf directly using change of variable.
$f_X(x)=\frac{(n+m+1)!}{n!m!}x^{n}(1-x)^m; 0 \lt x \lt 1$
$Y = g(X) = - \ln X $
As $g(X)$ is differentiable and a strictly decreasing function in $0 \lt X \lt 1$, we can use transformation technique,
$f_Y(y) = f_X(g^{-1}(y)) \left|\frac{dx}{dy}\right|$
$ = f_X(e^{-y}) \ e^{-y}$
$ = \displaystyle \frac{(n+m+1)!}{n! \ m!} e^{-(n+1)y}(1-e^{-y})^m \ , \ 0 \lt y \lt \infty$