Let $$Y =\begin{pmatrix} Y_1 \\ Y_2 \end{pmatrix}
\sim N(0, \Sigma) \quad \Sigma=\begin{pmatrix} \sigma_{11} & \sigma_{12}\\ \sigma_{21} & \sigma_{22} \end{pmatrix}$$
Show that $$\mathbb E(Y_1^2Y_2^2)=\sigma_{11}\sigma_{22} + 2\sigma_{12}^2. $$
I tried to turn this in all the ways but I need $\mathbb {Var}(Y_1Y_2)$ that I don't have. In the solution I was given, it only uses the fact that "the distribution is Gaussian".
Best Answer
Since all Gaussian variables have a kurtosis of $3,$ any Gaussian variable with variance $\sigma^2$ has a central fourth moment of $3\sigma^4.$ In particular, when $Z$ is a zero-mean Gaussian,
$$E[Z^4] = 3\sigma^4.$$
We will apply this below to the Gaussian variables $Z = Y_1\pm Y_2$ by observing (from the symmetry of $\Sigma$) that $$\operatorname{Var}(Y_1\pm Y_2) = \sigma_{11} + \sigma_{22} \pm 2\sigma_{12}.$$
Emulating the method I describe at https://stats.stackexchange.com/a/267021/919 (based on the Polarization Identity for $n=4$ with $x_1 = x_2 = Y_1$ and $x_3 = x_4 = Y_2$), write
$$12Y_1^2Y_2^2 = (Y_1+Y_2)^4 + (Y_1-Y_2)^4 - 2(Y_1^4) - 2(Y_2^4).$$
Use linearity of expectation to compute
$$\begin{aligned} 12 E[Y_1^2 Y_2^2] &= E[(Y_1+Y_2)^4] + E[(Y_1-Y_2)^4] - 2E[Y_1^4] - 2E[Y_2^4]\\ & = 3(\sigma_{11} + \sigma_{22} + 2\sigma_{12})^2 + 3(\sigma_{11} + \sigma_{22} - 2\sigma_{12})^2 - 2(3\sigma_{11}^2) - 2(3\sigma_{22}^2)\\ &= 12\sigma_{11}\sigma_{22} + 24\sigma_{12}^2. \end{aligned}.$$
Dividing both sides by $12$ gives the result in the question.