Sum of squared differences between standard normal random variables

Question

I'm struggling with the following problem:

Let $Z_1, Z_2, Z_3$ be independent, identically distributed standard normal random variables. Show that $$\frac{1}{3}((Z_1 - Z_2)^2 + (Z_2 - Z_3)^2 + (Z_3 - Z_1)^2) \sim \chi_2^2.$$

Now, considering for example $Z_1 - Z_2$, I see that it is a normal random variable with mean $0$ and variance $2$, and so $\frac{1}{\sqrt{2}}(Z_1-Z_2)$ is a standard normal random variable. Thus, $\frac{1}{2}(Z_1-Z_2)^2$ is a $\chi_1^2$ random variable, and of course the same holds for $\frac{1}{2}(Z_2 - Z_3)^2$ and $\frac{1}{2}(Z_3 - Z_1)^2$. I don't see how to continue from there, though.

Answer 1

Hint: write the sum as a quadratic form with $z'=(z_1,z_2,z_3)$ and your sum is $Q=z'A z$. It follows that A is a matrix of the form $\left[ \begin{array}{ccc} \dfrac{2}{3} & \dfrac{-1}{3} & \dfrac{-1}{3} \\ \dfrac{-1}{3} & \dfrac{2}{3} & \dfrac{-1}{3} \\ \dfrac{-1}{3} & \dfrac{-1}{3} & \dfrac{2}{3} \end{array} \right]$

A has a one zero eigenvalue and two unitary eigenvalues and from here you mimic the proof of the Cochran's theorem for the sum squares of demeaned standard normals.