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I am working on this problem for class, where the setup is the following:

Let X be a single observation from the $beta(\theta,1)$ pdf.

(a) Let $Y=-(logX)^{-1}$. Evaluate the confidence coefficient of the set $[y/2,y]$.

The part I am having trouble with is computing the pdf of Y. Since Y is a function of a random variable X with a known distribution. I did the following, $X = e^{-1/y}$, but this doesn't seem to be a one-one transformation since for a beta distribution $0 \le x \le 1$. The book we are using in class shows that I should get $f_y = \frac{\theta}{y^2} * e^{-\frac{\theta}{y}}$. I am a little confused as to how to get there.

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  • $\begingroup$ This is standard enough that GPT can give you a reasonable answer. $\endgroup$
    – user225256
    Commented Apr 8 at 17:26

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As this is a strictly monotonic transformation ($Y$ increases as $X$ increases), you can obtain the answer immediately using the change-of-variable formula.

To find the PDF of $Y$ from first principles, start with the CDF. We have $$ F_Y(y)=\Pr(Y \leq y) = \Pr(-(\log X)^{-1} \leq y) = \Pr (X \leq e^{-1/y})=F_X(e^{-1/y}) $$ You can now differentiate this using the chain rule to obtain $f_Y(y)$ in terms of $f_X(x)$.

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  • $\begingroup$ Ahh that makes so much sense. Thank you! $\endgroup$
    – Harry Lofi
    Commented Apr 8 at 21:19

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