# Find out expected value of $\xi^2\eta^2$

The task: Find out expected value of $$\xi^2\eta^2$$, where $$(\xi,\eta)$$ has normal distribution with zero mean vector and covariance matrix $$(\begin{matrix} 4 & 1 \\ 1 & 1 \\ \end{matrix})$$

I tryed to find the expected value with new random value as ($$\eta- c\xi$$), where c=const and cov($$\xi, \eta- c\xi$$)=0, but it only complicates the calculations.

• You might prefer to look at $(\xi-\eta,\eta)$ which I suspect has a very simple covariance matrix May 30, 2019 at 15:15

Another is to write $$X=\sqrt 3 Z + Y$$ where $$Z$$ is independent of $$Y$$ and has the same distribution as $$Y$$ and has $$EZ=0$$ and $$V(Z)=1$$ (that is, $$N(0,1)$$). Check: $$V(X) = 3V(Z)+V(Y) = 3 + 1 = 4$$, and $$\text{Cov}(X,Y)= \text{Cov}(Y,Y) =1$$. Then $$X^2Y^2 = 3Z^2Y^2+2\sqrt 3 ZY^3 + Y^4$$, whose expectation is $$EZ^2EY^2+2\sqrt3 EZEY^3+EY^4=3+0+3=6$$. (The 4th moment of $$Y$$ is $$3$$; see the Wikipedia page.)

• Z is a random value? And how we get the coefficient $\sqrt{3}$? And the last question- why $EY^4=3$? May 30, 2019 at 16:13
• I have addressed these issues in an edit. May 30, 2019 at 17:46
• Thank you very much! May 30, 2019 at 17:49

One way is to use the law of total expectation:

$$E(X^2Y^2)=E\left[E(X^2Y^2\mid Y)\right]=E\left[Y^2E(X^2\mid Y)\right]$$

Here $$(X,Y)$$ is jointly normal, so that $$X\mid Y$$ is univariate normal from which you can find $$E(X^2\mid Y)=V(X\mid Y)+(E(X\mid Y))^2$$

• What is X | Y ? May 30, 2019 at 15:18
• @Matthew5335 The random variable $X$ conditioned on the random variable $Y$ ('X given Y'). May 30, 2019 at 15:20
• V(...) is Variance? And how is it possible to calculate $E(X | Y)$? May 30, 2019 at 15:25
• 'V' is variance. And looks like you are not yet acquainted with this theory: en.wikipedia.org/wiki/…. May 30, 2019 at 15:34