Finding the PDF using the CDF Method I am currently working through a series of lecture notes on functions of random variables and I have been given the following as an exercise. 
Find the PDF of $X=Z^2$ given $Z$~$N(0,1)$ using the CDF method. 
I was under the impression that the normal distribution was not easily integrable, which would make  this question a bad candidate for the cdf method.
Is this thinking correct or am I missing something ?
 A: Yes, your thinking is wrong since we don't actually need to evaluate the CDF. We can simply  notice that $\Phi'(z) = F_Z'(z) = f_Z(z) = \phi(z)$, where $\Phi$ and $\phi$ are the standard normal CDF and PDF respectively. 
Proceed directly,
$$P(X\leq x) = P(Z^2\leq x) = P(|Z|\leq \sqrt{x}).$$
What can you say about $|Z|\leq \sqrt x$? Can you rewrite $P(|Z|\leq \sqrt x)$ in terms of $\Phi$?
A: Denoting the CDF of $Z$ by $\Phi$ and the CDF of $X$ by $F_X$ we find for $x>0$: $$F_X(x)=\Phi(x^{\frac12})-\Phi(-x^{\frac12})\tag1$$
To find the PDF just take the derivative of $F_X$ to end up with $$f_X(x)=\frac12x^{-\frac12}\phi(x^{\frac12})-\left[-\frac12x^{-\frac12}\phi(-x^{\frac12})\right]=x^{-\frac12}\phi(x^{\frac12})\tag2$$
where $\phi$ - as derivative of $\Phi$ - denotes the PDF of $X$.
It is well known that $\phi(x)=\frac1{\sqrt{2\pi}}e^{-\frac12x^2}$ and the second equality of $(2)$ is consequence of $\phi(y)=\phi(-y)$ for every $y$.
Final result for $x>0$:$$f_X(x)=\frac{x^{-\frac12}e^{-\frac12x}}{\sqrt{2\pi}}$$
Evidently we have $f(x)=0$ for $x\leq0$.
