# Sum of squared differences between standard normal random variables

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.

• Hint: write this variable as $Z\cdot SZ$ for symmetric $S\in\Bbb R^{3\times3}$, with $Z$ the vector of the $Z_i$. Now find $S$'s eigenvalues.
– J.G.
Commented Jun 8, 2023 at 21:52
• Thank you for your hint, but sadly we haven't worked with matrices at all in this course. So while I'm comfortable with linear algebra, I'm completely unfamiliar with the application of linear algebra to statistics. Something we worked out earlier was that $\frac{1}{2}(Z_1-Z_2)^2 + \frac{1}{2}(Z_1+Z_2)^2 \sim \chi_2^2$. I get the feeling that I probably have to use that somehow, but I can't figure out how to make it work. Commented Jun 8, 2023 at 21:59
• Answered at math.stackexchange.com/q/3975602/321264 Commented Jun 9, 2023 at 7:10

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 $$\left[ \begin{array}{ccc} \dfrac{2}{3} & \dfrac{-1}{3} & \dfrac{-1}{3} \\ \dfrac{-1}{3} & \dfrac{2}{3} & \dfrac{-1}{3} \\ \dfrac{-1}{3} & \dfrac{-1}{3} & \dfrac{2}{3} \end{array} \right]$$