Product of two variables: variance of the glm coefficients If we have two stochastic variables X and Y, and we assume independence (they are actually not). We would like to know the product of X and Y, and we get:
$E[XY]=E[X]E[Y]$
We proceed with GLM analysis of both X and Y, both regressed on a covariate $z$ and obtain a set of coefficients (using for example R):
For X:  $\beta_{11}$ and $\beta_{12}$
For Y:  $\beta_{21}$ and $\beta_{22}$
R also provides us with standard deviations for $\beta_{12}$ and $\beta_{22}$, namely $\sigma_{12}$ and $\sigma_{22}$, respectively.
We thus get: $E[XY]=E[X]E[Y]=\exp{[\beta_{11}+\beta_{21} + (\beta_{12}+\beta_{22})z]}$
In my practice of finding confidence intervals for the sum of the coefficients (beta's) of equal covariates in X and Y I have done the following:
$\sigma=\sqrt{\sigma_{12}^2+\sigma_{22}^2}$.
But now I wonder if this is correct?
EDIT: clarified notation.
 A: Your approach to obtaining confidence intervals for sums of coefficients is wrong. You have not taken account of covariance. It is a basic math fact that 
$$\mbox{Var}(X + Y) = \mbox{Var}(X) + \mbox{Var}(Y) + 2\mbox{Cov}(X, y).$$
Obtaining covariance of two parameters in R requires use of the vcov function which returns a variance-covariance matrix for regression coefficients.
The variance of the product is more complicated since expectation is only a linear operator. You can approximate variance of products using the $\delta$ method. Since $f(X, Y) = XY$, $\nabla f (X, Y) = [Y, X]$, and so:
$\mbox{Var}(XY) \approx [E[Y], E[X]] \left[ \begin{array}{cc} \sigma^2_x & \sigma_{xy} \\ 
\sigma_{xy}  & \sigma^2_y \\ \end{array} \right] [E[Y], E[X]]^T
$
However, computationally an exact fit is obtained by simply using the bootstrap. It's easy and fast on virtually any computer.
Under the rare assumption of independence, it is true as you said, the product of the expectation is the expectation of the product. It also applies to higher moments:
$$ E[Y^2X^2] = E[Y^2]E[X^2]$$
and $$\mbox{Var}(YX) = E[Y^2X^2]  - E[YX]^2 = E[Y]^2 \mbox{Var}(X) + E[X]^2 \mbox{Var}(Y)$$
which is exactly the same as the expression from Delta method setting $\sigma_{xy}$ to 0.
