What is the probability distribution of one chi-square variable with zero mean but arbitrary standard deviation? I keep seeing chi-squared distributions or generalized chi-squared distributions online defined for unit variance, but I'm just trying to find the regular chi-squared distribution (one degree of freedom!) of a Gaussian random variable with zero mean but some non-unit standard deviation (or variance). Can someone help me please?
 A: Does this help?

First let us compute the CDF. Let $Y \sim (N(0,\sigma^2))^2$. Let $X\sim N(0,\sigma^2)$
Case 1: if $t<0,\quad \mathbb P(Y<t) = 0$
Case 2: if $t\geq0$
Let the CDF of Y upto $t\in \mathbb R$ be $F_Y(t)$
\begin{align}
F_Y(t) &= \mathbb P(Y<t)\\
&= \mathbb P(X^2<t)\\
&= \mathbb P(-\sqrt{t}<X<\sqrt{t})\\
&= \mathbb P(X<\sqrt{t}) - \mathbb P(X<-\sqrt{t})\\
&= F_X(\sqrt{t}) - F_X(-\sqrt{t})
\end{align}
But the normal distribution is symmetric about the origin. So $F_X(-\sqrt{t}) = 1-F_X(\sqrt{t})$
\begin{align}
F_Y(t) 
&= F_X(\sqrt{t}) - F_X(-\sqrt{t})\\
&= 2F_X(\sqrt{t}) - 1\\
&= 2\int\limits_{x=-\infty}^{\sqrt{t}} \frac{1}{\sqrt{2\pi}\sigma}\exp(-\frac{t^2}{2\sigma^2})- 1\\
&= erf(\frac{\sqrt{t}}{\sigma\sqrt{2}})
\end{align}
This is the cumulative distribution you want. To get the pdf, you can differentiate the above with respect to t.
\begin{align*}
f_y(t) &= \frac{d}{dt} F_Y(t)\\
&= \frac{d}{dt}\left(2\int\limits_{x=-\infty}^{\sqrt{t}} \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{t^2}{2\sigma^2}\right)- 1\right)\\
&= 2\frac{d}{dt}\int\limits_{x=-\infty}^{\sqrt{t}} \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{t^2}{2\sigma^2}\right)\\
&= 2\frac{1}{\sqrt{2\pi}\sigma} \exp\left(-\frac{(\sqrt{t})^2}{2\sigma^2}\right) \frac{d}{dt}\sqrt{t}\\
&= \frac{1}{\sigma\sqrt{2\pi t}} \exp\left(-\frac{t}{2\sigma^2}\right) \\
\end{align*}
This is the pdf you are looking for.
