LaGatta's answer nails it, and may be useful for drawing simulations, etc.

This is just a note to remind that if one is only interested in the mean of the product of normally-distributed (possibly correlated) random variables, then the answer is straightforward, using the identity $\mathrm{E}XY = \mathrm{Cov}(X,Y) + \mathrm{E}X\,\mathrm{E}Y$.

LaGatta's answer nails it, and may be useful for drawing simulations, etc.

This is just a note to remind that if one is only interested in the mean of the product of normally-distributed (possibly correlated) random variables, then the answer is straightforward, using the identity $\operatorname{E}XY= \operatorname{Cov}(X,Y) + (\operatorname{E}X)(\operatorname{E}Y)$.

LaGatta's answer nails it, and may be useful for drawing simulations, etc.

This is just a note to remind that if one is only interested in the mean of the product of normally-distributed (possibly correlated) random variables, then the answer is straightforward, using the identity E[x.y] = Cov(x,y) + E[x].E[y].

LaGatta's answer nails it, and may be useful for drawing simulations, etc.

This is just a note to remind that if one is only interested in the mean of the product of normally-distributed (possibly correlated) random variables, then the answer is straightforward, using the identity $\mathrm{E}XY = \mathrm{Cov}(X,Y) + \mathrm{E}X\,\mathrm{E}Y$.