Expectation of product of Gaussian random variables Say we have two Gaussian random vectors $p(x_1) = N(0,\Sigma_1), p(x_2) = N(0,\Sigma_2)$, is there a well known result for the expectation of their product $E[x_1x_2^T]$ without assuming independence?
 A: Yes, there is a well-known result.  Based on your edit, we can focus first on individual entries of the array $E[x_1 x_2^T]$.  Such an entry is the product of two variables of zero mean and finite variances, say $\sigma_1^2$ and $\sigma_2^2$.  The Cauchy-Schwarz Inequality implies the absolute value of the expectation of the product cannot exceed $|\sigma_1 \sigma_2|$.  In fact, every value in the interval $[-|\sigma_1 \sigma_2|, |\sigma_1 \sigma_2|]$ is possible because it arises for some binormal distribution.  Therefore, the $i,j$ entry of $E[x_1 x_2^T]$ must be less than or equal to $\sqrt{\Sigma_{1_{i,i}} \Sigma_{2_{j,j}}}$ in absolute value.
If we now assume all variables are normal and that $(x_1; x_2)$ is multinormal, there will be further restrictions because the covariance matrix of $(x_1; x_2)$ must be positive semidefinite.  Rather than belabor the point, I will illustrate.  Suppose $x_1$ has two components $x$ and $y$ and that $x_2$ has one component $z$.  Let $x$ and $y$ have unit variance and correlation $\rho$ (thus specifying $\Sigma_1$) and suppose $z$ has unit variance ($\Sigma_2$).  Let the expectation of $x z$ be $\alpha$ and that of $y z$ be $\beta$.  We have established that $|\alpha| \le 1$ and $|\beta| \le 1$.  However, not all combinations are possible: at a minimum, the determinant of the covariance matrix of $(x_1; x_2)$ cannot be negative.  This imposes the non-trivial condition
$$1-\alpha ^2-\beta ^2+2 \alpha  \beta  \rho -\rho ^2 \ge 0.$$
For any $-1 \lt \rho \lt 1$ this is an ellipse (along with its interior) inscribed within the $\alpha, \beta$ square $[-1, 1] \times [-1, 1]$.
To obtain further restrictions, additional assumptions about the variables are necessary.
Plot of the permissible region $(\rho, \alpha, \beta)$

A: There are no strong results and it does not depend on Gaussianity.  In the case where $x_1$ and $x_2$ are scalars, you are asking if knowing the variance of the variables implies something about their covariance.  whuber’s answer is right.  The Cauchy-Schwarz Inequality and positive semidefiniteness constrain the possible values.  
The simplest example is that the squared covariance of a pair of variables can never exceed the product of their variances.  For covariance matrices there is a generalization.
Consider the block partitioned covariance matrix of $[x_1 \ x_2]$,
$$
\left[
\begin{array}{cc}
\Sigma_{11} & \Sigma_{12} \\
\Sigma_{21} & \Sigma_{22}
\end{array}
\right].
$$
Then
$$\Vert \Sigma_{12} \Vert_q^2 \leq \Vert \Sigma_{11} \Vert_q \Vert \Sigma_{22} \Vert_q$$
for all Schatten q-norms.  Positive (semi)definiteness of the covariance matrix also provides the constraint that 
$$
\Sigma_{11} - \Sigma_{12} \Sigma_{22}^{-1} \Sigma_{21}
$$
must be positive (semi)definite.  $\Sigma_{22}^{-1}$ is the (Moore-Penrose) inverse of $\Sigma_{22}$.
A: suppose $(X,Y)$ is bivariate normal with zero means and correlation $\rho$. then 
${\mathrm E} XY= cov(X,Y)= \rho\sigma_X\sigma_Y$.
all of the entries in the matrix $x_1x_2^T$ are of the form $XY$.
