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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.

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

2 Answers 2

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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.)

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  • $\begingroup$ Z is a random value? And how we get the coefficient $\sqrt{3}$? And the last question- why $EY^4=3$? $\endgroup$ Commented May 30, 2019 at 16:13
  • $\begingroup$ I have addressed these issues in an edit. $\endgroup$ Commented May 30, 2019 at 17:46
  • $\begingroup$ Thank you very much! $\endgroup$ Commented May 30, 2019 at 17:49
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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 $$E(X^2\mid Y)=V(X\mid Y)+(E(X\mid Y))^2$$

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

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