Expectation of product of two correlated gaussian variables

Question

$\newcommand{\var}{\operatorname{var}}$It seems I can not find the answer anywhere, please point it out how to calculate. Here, I have $X,Y,G,X_D,$ and $Y_D$,both are Gaussian variables, and $X_D \sim N(\mu_x,\var_x)$, $Y_D \sim (\mu_y,\var_y)$, $G\sim (u_g, \var_g).$ And then $X = X_D+G$, $Y = Y_D+G$. $X_D$ , $Y_D$ and $G$ are mutually independent. Now the problem is how to get $\operatorname E(XY),$ and since $X$ and $Y$ both share the contribution of $G,$ I think they are correlated, if I am not wrong

Answer 1

$\newcommand{\E}{\operatorname{E}} \newcommand{\var}{\operatorname{var}}$ It appears to me that what you have in mind is this: \begin{align} X & = G + U \\ Y & = G + V \\ & X,Y,G\text{ are independent}, \\ \text{and} & \text{ $X,Y,G$ are normally distributed (i.e. Gaussian)} \end{align} and you want $\E(XY)$.

\begin{align} \E(XY) & = \E((G+U)(G+V))=\E(G^2+GU+GV+UV) \\ & = \E(G^2)+\E(GU)+\E(GV)+\E(UV) \\ & = (\E(G))^2+\var(G) + \E(G)\E(U)+\E(G)\E(V)+\E(U)\E(V). \end{align} The fact that $\E(GU)=\E(G)\E(U)$ follows from independence of $G$ and $U$, and similarly for the next two terms. The fact that $\E(G^2)=(\E(G))^2+\var(G)$ is a standard identity.

You didn't say anything about independence. Merely saying $X\sim N(a+b_1,c+d_1)$ and $Y\sim N(a+b_2,c+d_2)$ is not enough to imply that there is some $N(a,c)$-distributed random variable (which I called $G$ above) which when subtracted from $X$ and $Y$ yields differences that are actually independent of each other (which I called $U$ and $V$ above).