Expectation of product of two correlated gaussian variables $\newcommand{\var}{\operatorname{var}}$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 
 A: $\newcommand{\E}{\operatorname{E}}
\newcommand{\var}{\operatorname{var}}$
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).
