I have the following question:
Let $Z$ be a random variable whose density function is:
$$ f(t) = \begin{cases} \frac{1}{t^2} & \text{if } t \ge 1 \\ 0 & \text{if } t < 1 \end{cases}$$ What is the density function of the random variable $\ln(Z)$?
The solution is as follows:
Let $Y = \ln(Z)$, then we can write $Z = e^X = g(X)$ with $g(x) = e^x$. Now we have the density function of Z:
$$h(x) = f(g(x))\frac{dg}{dx}(x) = \begin{cases}e^{-x} & \text{if } x \ge 0 \\ 0 & \text{if not}\end{cases}$$
So my question here is, why do we let this random variable $Y$ be $\ln(Z)$? It looks like we are doing something weird from the CDF to the PDF, but I am confused of what is going on here. Also, are we taking the derivative of the original density function? It looks like we have because of the $-x$ in the power of $e$.
When $g(x)\ge1$: $$f(g(x))=\frac{1}{(g(x))^2}=\frac{1}{(e^x)^2}=e^{-2x}$$And: $$\frac{\mathrm{d}g}{\mathrm{d}x}=e^x$$So: $$h(x)=e^{-2x}\cdot e^x=e^{-x}$$Whenever $g(x)\ge1$, i.e. whenever $e^x\ge1,\,x\ge0$. When $g(x)<1,\,x<0$, $f(g(x))=0$.