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I want find the PDF of $Y$ when $Y=-\log X$ and $X$ has a beta distribution.
I found the below formula as the answer, but I think there should be $(1-e^{-y})^{b-1}$ part should added to this.
Is that correct ?
$$f_Y(y)=e^{-y}f_X(e^{−y})=e^{−y}\cdot\dfrac {(e^{−y})^{a−1}}{B(a,1)}=ae^{−ay}, 0≤y<∞$$
The PDF of $X \sim \operatorname{Beta}(a,b)$ is given by $$f_X(x) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1} (1-x)^{b-1}, \quad 0 < x < 1.$$ Then for a monotone transformation $Y = g(X) = -\log X$, we have $$f_Y(y) = f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right| = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} (e^{-y})^{a-1} (1-e^{-y})^{b-1} e^{-y}, \quad y > 0.$$ We can simplify this slightly by writing $$f_Y(y) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} e^{-ay} (1-e^{-y})^{b-1}.$$ In the special case where $b = 1$, we obtain $$f_Y(y) = a e^{-ay}, \quad y > 0,$$ which of course is an exponential distribution with rate parameter $a$.
That's correct if $X$ has a beta distribution with shape parameter of $1$.
You are also correct that if the parameter is otherwise, $b$, then an additional term needs to be included. $(1-e^{-y})^{b-1}$, which equals $1$ when $b=1$.
$$\begin{align}f_Y(y)~=~&e^{-y}f_X(e^{−y})\\~=~&e^{−y}\cdot\dfrac {(e^{−y})^{a−1}(1-e^{-y})^{b-1}}{B(a,b)}\\~=~&\dfrac{(e^{−y})^{a}(1-e^{-y})^{b-1}\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\end{align}\tag{$0≤y<∞$}$$
