Expected distance between two vectors that belong to two different Gaussian distributions Let $X$, $Y$ be two random variables that follow the Gaussian distributions with mean vectors $\mu_x$, $\mu_y$, and covariance matrices $\Sigma_x$, $\Sigma_y$, respectively. The probability density functions of $X$, $Y$ are given, respectively, by
$$
f_{X}(\mathbf{x})=\frac{1}{(2\pi)^{\frac{n}{2}}\lvert \Sigma_x \rvert^{\frac{1}{2}}}
\exp\Big\{-\frac{1}{2}(\mathbf{x}-\mu_x)^\top\Sigma_x^{-1}(\mathbf{x}-\mu_x)\Big\},
$$
and
$$
f_{Y}(\mathbf{y})=\frac{1}{(2\pi)^{\frac{n}{2}}\lvert \Sigma_y \rvert^{\frac{1}{2}}}
\exp\Big\{-\frac{1}{2}(\mathbf{y}-\mu_y)^\top\Sigma_x^{-1}(\mathbf{y}-\mu_y)\Big\},
$$
where $\mathbf{x},\mathbf{y}\in\Bbb{R}^n$. We will be thinking of $\mathbf{x}$, $\mathbf{y}$ as "members" of the distributions $X$, $Y$, respectively.
If we have two fixed vectors, say $\mathbf{x}$, $\mathbf{y}$, then the squared Euclidean distance between them would be equal to
$$
\big\lVert \mathbf{x} - \mathbf{y} \big\rVert^2.
$$
If we think about $\mathbf{x}$, $\mathbf{y}$ as above, i.e., as members of $X$, $Y$, respectively, then what would be the expected value of this distance?
Thank you very much for your help!

If $X$ and $Y$ are independent and normal $(\mu_X,\Sigma_X)$ and $(\mu_Y,\Sigma_Y)$ respectively, then: 

$$E(\|X-Y\|^2)=\|\mu_X-\mu_Y\|^2+\mathrm{tr}(\Sigma_X+\Sigma_Y)$$

To show this, note that, by independence, $X-Y$ is normal $(\mu_X-\mu_Y,\Sigma_X+\Sigma_Y)$ and that every random variable $Z$ normal $(\mu,\Sigma)$ can be written as $Z=\mu+LU$ where $LL^\top=\Sigma$ and $U$ is standard normal, hence a little bit of matrix calculus should yield the result. 
To wit, note that the decomposition $$\|Z\|^2=Z^\top Z=\mu^\top\mu+\mu^\top LU+U^\top L^\top\mu+U^\top L^\top LU,$$ and the fact that $E(U)=0$ and $E(U^\top)=0^\top$ yield $$E(\|Z\|^2)=\mu^\top\mu+E(U^\top L^\top LU).$$
Now, $\mu^\top\mu=\|\mu\|^2$ and $$U^\top L^\top LU=\sum_{k,\ell}(L^\top L)_{k,\ell}U_kU_\ell,\quad E(U_k^2)=1,\quad E(U_kU_\ell)=0\ (k\ne\ell),$$ hence $$E(U^\top L^\top LU)=\sum_{k}(L^\top L)_{k,k}=\mathrm{tr}(L^\top L)=\mathrm{tr}(LL^\top)=\mathrm{tr}(\Sigma).$$ Finally, as desired, $$E(\|Z\|^2)=\|\mu\|^2+\mathrm{tr}(\Sigma).$$
Remarks: 


*

*This nowhere uses the explicit forms of the densities. As a matter of fact,  when solving problems about normal random variables, a useful principle is to avoid as much as possible to manipulate the gaussian densities themselves. Instead, write each $(\mu,\Sigma)$ normal random variable as $\mu+LU$ with $LL^\top=\Sigma$, as we did, and proceed with the standard normal $U$.

*The mapping $(x,y)\mapsto\|x-y\|^2$ is not a metrics, only $(x,y)\mapsto\|x-y\|$ is.

