If $f_X(x) = x/8$, find the pdf of $Z = \log (x/4)$.

Question

Given the function $f_{X}(x) = \left\{ \begin{array}{lr} \ \frac{x}{8} \quad & : 0 <x <4 \\ 0 & :\text{Otherwise} \end{array} \right. $
Find the PDF of $Z = \log_{e}(\frac{x}{4})$

By using method of transformations I got my equation to be $f_{Z}(z) = \left\{ \begin{array}{lr} \ 2e^{2z} \\ 0 & :\text{Otherwise} \end{array} \right. $

But I get my limits for $2e^{2z}$ as $-{\infty} < z < 0$ which I don't think looks right. Just wondering how I find my limits for this method?

Answer 1

The cdf of $X$ is $$F_X(x)=\begin{cases} 0, &\text{if }& -\infty<x<0,\\ \int_0^x \frac y8 dy=\frac 1{16}x^2,&\text{if }&0\le x\le4,\\ 1,&\text{otherwise} \end{cases}$$

With this, the cdf of $Z$:

$$F_Z(x)=P\left(\ln \left(\frac X4\right)<x\right)=P\left(X<4e^x\right)=F_X(4e^x)=$$ $$=\begin{cases}e^{2x}, \text{ if } &-\infty<x\le0\\ 1, &\text{otherwise}\end{cases}$$

because $4e^x$ is between $0$ and $4$ if $-\infty<x\le 0$.

So, the pdf of $Z$, $$f_Z(x)=\frac{dF_Z(x)}{dx}=\begin{cases}2e^{2x},\text{ if } &-\infty<x\le0\\ 0,& \text{otherwise.}\end{cases} $$