The random variable X has probability density function
\begin{cases} \frac{x}{8} & 0<x<4 \\ 0 & otherwise \end{cases}
Find the probability density function of $Z = \log_e (X/4)$
I tried and got this:
\begin{cases} \frac{4e^z}{8} & 0<Z<e^z \\ 0 & otherwise \end{cases} but the limits don't seem right to me, is this correct?
if $f(x)$ is the profability density function for $x$ and $z = z(x)$, then the pdf for $z$ can be calculated as
$$ f(z) = f(x)\left|\frac{dx}{dz}\right| \tag{1} $$
In this case you already know the expression $z=z(x)$:
$$ z = \ln x/4 \quad\Rightarrow\quad x = 4e^{z} $$
so that
$$ \frac{dx}{dz} = 4e^z $$
Before replacing in (1) note that when $z(x\to 0^+) = -\infty$, and $z(x=1) = 0$, so the $z$ will range in $(\infty, 0)$. So in the this range
$$ f(z) = f(x)|4e^z| = 2e^{2z} \quad\mbox{for}\quad z<0 $$
or in other words
$$ f(z) = \left\{\begin{array}{lcl} 2e^{2z} &,& z < 0 \\ 0 &,& {\rm otherwise}\end{array}\right. $$
Here's a simulation, the dashed line is the function $2e^{2z}$
