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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 ?

$$f_Y(y)=e^{-y}f_X(e^{−y})=e^{−y}\cdot\dfrac {(e^{−y})^{a−1}}{B(a,1)}=ae^{−ay}, 0≤y<∞$$

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2 Answers 2

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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 $a$.

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  • $\begingroup$ Hi Thank you . When i want to calculate the Expected value of Y , when Y =-log x , how to do that ? i think some substitution should be made inorder to do that $\endgroup$
    – Sam88
    Oct 24, 2016 at 0:31
  • $\begingroup$ @dhanushkaSampath That was not part of your original question, which was answered in good faith. If you wish to have your new question answered, you will need to demonstrate your own efforts first. $\endgroup$
    – heropup
    Oct 24, 2016 at 2:27
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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 $b=1$.

$$\begin{align}f_Y(y)~=~&e^{-y}f_X(e^{−y})\\~=~&e^{−y}\cdot\dfrac {(e^{−y})^{a−1}(1-e^{-y})^{b-1}}{B(a,b)}\\~=~&\dfrac{(e^{−y})^{a}(1-e^{-y})^{b-1}\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\end{align}\tag{$0≤y<∞$}$$

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  • $\begingroup$ So if they asked to calculate E(Y), how do we do that ? $\endgroup$
    – Sam88
    Oct 24, 2016 at 0:37

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