Finding the chi-squared distribution of a squared difference of two independent normal variables Given two independent random variables $X\sim N(0, \sigma^2), Y\sim N(0, \sigma^2)$, what is the distribution of a variable $Q = (X-Y)^2/4$ ? What would be the expected value and variance of $Q$?
 A: Below, I presume that you actually want X and Y to have variance $\sigma^2$, and not $\sigma$.
Since $N(0, \sigma^2)$ is symmetric over zero, the random variable $Y$ is equal to the random variable $-Y$. Thus $X-Y$ is just the sum of two independent normal distributions $N(0, \sigma^2)$. Using the general formula for the sum of two independent normal distributions $A\sim N(\mu_A, \sigma_A^2), B\sim N(\mu_B, \sigma_B^2)$:
$$
A+B \sim N(\mu_A + \mu_B, \; \sigma^2_A + \sigma_B^2)
$$
we get for $X-Y$:
$$
\begin{align}
X-Y &\sim N(0, 2\sigma^2),\\
\frac{X-Y}{\sqrt{2}\sigma} &\sim N(0, 1).
\end{align}
$$
Next, the square of a $N(0, 1)$ variable has a $\chi^2$ distribution of degree one, i.e. $\left(\frac{X-Y}{\sqrt 2\sigma}\right)^2$ is $\chi^2_1$-distributed. Furthermore
$$
\begin{align}
\frac{(X-Y)^2}{4} &= \frac{1}{4}\left(\frac{\sqrt 2 \sigma (X-Y)}{\sqrt 2 \sigma}\right)^2\\
    &= \frac{\sigma^2}{2}\left(\frac{X-Y}{\sqrt 2 \sigma}\right)^2.\\
\end{align}
$$
Thus, it is a scaled $\chi^2_1$-distribution, and the scaling factor is $\frac{\sigma^2}{2}$. Now, the $\chi^2_1$ distribution has mean equal to one, and variance equal to two. Since the expectation is linear, we have:
$$
\begin{align}
E\left[\frac{(X-Y)^2}{4}\right] &= \frac{\sigma^2}{2}E\left[\left(\frac{X-Y}{\sqrt 2 \sigma}\right)^2\right]\\
    &=\frac{\sigma^2}{2}.
\end{align}
$$
The variance is quadratic, thus
$$
\begin{align}
Var\left[\frac{(X-Y)^2}{4}\right] &= \left(\frac{\sigma^2}{2}\right)^2 Var\left[\left(\frac{X-Y}{\sqrt 2 \sigma}\right)^2\right]\\
    &=\frac{\sigma^4}{4}\cdot 2\\
    &= \frac{\sigma^4}{2}.
\end{align}
$$
