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I want to see if there is known distribution for the below question. Thanks

If X has exponential distribution with parameter $\lambda$, what is the distribution of $Y= \frac{1-e^{cx}}{a-e^{cx}}$, where $c>0$ and $0<a<1$?

Ellie
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  • Haha, I need to show Y which obtained from a map as the above form, follows Beta distribution. I was wondering if there is a way to prove it. For example if X has exponential distribution then $1-e^{-cx}$ follows beta distribution. So my intention is to see if I can make such conclusion with Y – Ellie Sep 20 '17 at 17:57

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It is quite easy to work out although I don't recognize it as a known distribution. The CDF of $Y$ is given by $F_Y(y)=P(Y\le y)=P(\frac{e^{c X}-1}{e^{c X}-a} \le y) = P(X \le \frac{1}{c} Ln(\frac{1-ay}{1-y})) = F_X(\frac{1}{c} Ln(\frac{1-ay}{1-y})) $

So the PDF of $Y$ is given by $f_Y(y)=F_Y'(y) = \frac{(1-a)}{c(1-ay)(1-y)} f_X(\frac{1}{c} Ln(\frac{1-ay}{1-y})) $

Now if $X$ is exponential then $f_X(x)=\lambda e^{-\lambda x}$.

So $f_Y(y)= \frac{(1-a)}{c(1-ay)(1-y)} \lambda e^{-\lambda \frac{1}{c} Ln(\frac{1-ay}{1-y}) }= \frac{(1-a)}{c(1-ay)(1-y)} \lambda (\frac{1-y}{1-ay})^{\frac{\lambda}{c}} = k\frac{(1-a)}{(1-ay)(1-y)} (\frac{1-y}{1-ay})^{k}= k(1-a) \frac{(1-y)^{k-1}}{(1-a y)^{k+1}}$

user121049
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  • Thank you very much. True. I was trying to see if I can find a known distribution for it so I can use some properties of the distribution. But it seems it is not possible. – Ellie Sep 20 '17 at 21:16