Expected Value of X,Y from Covariance Matrix

Question

Let $(X,Y)$ be a bivariate random variable with a Gaussian distribution on $\mathbb{R}^2$, mean zero and variance-covariance matrix:

$$C=\begin{pmatrix} 0.42 & -0.42\\-0.42 & 0.42\end{pmatrix}$$

Find the expected value $\mathbb{E}[|XY|]$.

Not sure how to go about this question. I know that $\mathbb{E}(XY)=\mathbb{E}(X)\mathbb{E}(Y)$ and we can find $Var(X) \ and \ Var(Y)$ from the covariance matrix, as $Cov(X,X) = Var(X) = c_{1,1} = 0.42$. However unsure on a formula to bring everything together?

Answer 1

$Cov(X,Y) = - 0.42$ thus $Cov^2(X,Y) = 0.42^2 = DX \cdot DY$. Hence $$(E XY)^2 = [E (X - EX)(Y-EY)]^2 = [E (X - EX)^2] \cdot [E (Y - EY)^2] = [EX^2][EY^2]$$ and we have equality in Cauchy–Bunyakovsky-Schwarz inequality, hence $C_1 X+ C_2 Y = C_3$ for some constants $C_i$ such that $\sum_{i=1}^3 C_i^2 > 0$. As $EX = EY = 0$ we have $C_3 = E C_3 = E (C_1 X+ C_2 Y) = 0$. Thus $C_1 X = (-C_2)Y$. If $C_1 = 0$ then $Y=0$ and $DY = 0 \ne 0.42$ hence $C_1 \ne 0$. It follows that $X = cY$ where $c = \frac{-C_2}{C_1}$. Further, $0.42 = DX = D(cY) = c^2 DY = c^2 \cdot 0.42$ thus $c^2 = 1$. If $c = 1$ then $-0,42 = Cov(X,Y) = EXY - EX EY = EXY = EX (cX) = EX^2 \ge 0$. Hence $c \ne 1$ and so $c = -1$. Thus $Y = - X$. Finally $E|XY| = E|X(-X)| = EX^2 = DX = 0.42$. The answer is $0.42$.