# Expected Value of X,Y from Covariance Matrix

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?

• The equality $\mathbb{E}(XY)=\mathbb{E}(X)\mathbb{E}(Y)$ is not true, because we $X$ and $Y$ are dependent. Nov 2, 2021 at 13:10
• Thanks, must have taken this down wrong in my notes Nov 2, 2021 at 13:28

## 1 Answer

$$Cov(X,Y) = - 0.42$$ 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 $$0.42$$.