How to find the distribution of $Y = X^2$ for $X$ being $N(\mu,{\sigma}^2)$? I know the results for $N(0,1)$, where $Y = X^2$ has chi-squared distribution with 1 df. But I'm not familiar with that case.
 A: Let $Z \sim N(0,1)$. Then $X\sim \sigma Z + \mu$, and
$$
Y \sim \left(\sigma Z + \mu\right)^2.
$$
In order to determine a random variable's distribution, it suffices to determine its CDF.
Let $t \in \mathbb{R}$. Then
$$
F_Y(t) = P(Z\leq t) = P(\left(\sigma Z + \mu\right)^2 \leq t).
$$
If $t < 0$, the probability on the rhs is zero. So assume $t \geq 0$. Then
$$
\begin{align}
F_Y(t) &= P(-\sqrt{t} \leq \sigma Z + \mu \leq \sqrt{t}) \\
&= P(\frac{-\sqrt{t}-\mu}{\sigma}\leq Z\leq \frac{\sqrt{t}-\mu}{\sigma}) \\
&= P(\frac{-\sqrt{t}-\mu}{\sigma}<Z\leq \frac{\sqrt{t}-\mu}{\sigma}) \\
&= F_Z\left(\frac{\sqrt{t}-\mu}{\sigma}\right) - F_Z\left(\frac{-\sqrt{t}-\mu}{\sigma}\right) \\
&= \Phi\left(\frac{\sqrt{t}-\mu}{\sigma}\right) - \Phi\left(\frac{-\sqrt{t}-\mu}{\sigma}\right) \\
&= \Phi\left(\frac{\sqrt{t}-\mu}{\sigma}\right) - \left(1 - \Phi\left(-\frac{-\sqrt{t}-\mu}{\sigma}\right)\right).
\end{align}
$$
In conclusion,
$$
F_Z(t) = \begin{cases}
\Phi\left(\frac{\sqrt{t}-\mu}{\sigma}\right) + \Phi\left(\frac{\sqrt{t}+\mu}{\sigma}\right) - 1,& t\geq 0\\
0,& t<0
\end{cases}.
$$
