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I have a question which says the random variable X has a pdf of

$f_{X}(x)= \frac{x}{8},\ 0<x<4 $

$f_{X}(x)= 0, \ $ otherwise

I have been asked to find the pdf for $Z=log_{e}(X/4)$

Can anyone help me on how to go about this please? I'm guessing that intergration needs to happen but i'm not really sure on which part and the $log_{e}$ has confused me a bit.

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Hint: Consider the CDF of $Z$, which I'll write as $F_Z(t)$. $$ F_Z(t) = P(Z \leq t) = P( \log(X/4) \leq t) = P( X \leq 4 e^{t}) = F_X(4e^t). $$ Here $F_X(t)$ is the CDF of $X$. Now, to find the PDF of $Z$, it comes down to remembering that $f_Z(t) = \frac{d}{dt} F_Z(t)$.

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To complete Tom's answer, if $f_{X}(x)= \frac{x}{8},\ 0\lt x\lt 4$ then $F_{X}(x)= P(X \le x)= \int_{-\infty}^x f_{X}(y)\,dy=\int_0^x f_{X}(y)\,dy = \frac{x^2}{16},\ 0\lt x\lt 4$.

So $F_{Z}(z) = P(Z\le z) =P(X \le 4e^z) = \frac{(4e^z)^2}{16} = e^{2z},\ -\infty\lt z\lt 0$ leading to $f_{Z}(z)=\frac{dF_z(z)}{dz}=2e^{2z},\ -\infty\lt z\lt 0$.

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