Let $ X \sim N(\mu, \sigma^2) $ and $Z = \frac{X-\mu}{\sigma}$. Suppose I want to prove $f_Z(x) = \frac{1}{\sqrt{2\pi}} \exp(-\frac{1}{2}x^2)$. I tried reasoning:
$$ f_Z(x) = P(Z=x) = P(\frac{X-\mu}{\sigma} = x) = P(X = \sigma x + \mu) = f_X(\sigma x + \mu) \\ = \frac1{\sqrt{2\pi}} \frac1{\sigma} \exp\left(-\frac{1}{2} \left(\frac{ (\sigma x + \mu) - \mu }{\sigma}\right)^2 \right) = \frac1{\sqrt{2\pi}} \frac1{\sigma} \exp\left(-\frac{1}{2} x^2 \right) $$
However, note the pesky $\frac1\sigma$ that still remains in the above expression. That shouldn’t be there. Why is this happening? Is it invalid to equate the PDFs as I did?
Edit: From the comments, I realized my blooper: $f_Z(x)$ is not in fact the same as $P(Z=x)$, because of course the latter is always zero for continuous variables like $X$ and $Z$!
My revised question would be: Is there any way to understand the distribution of a function of an RV without going through the CDF, some method that works using PDFs?