$$ 
\begin{aligned}
\mathrm{Var}_{S'}(x_0, y_0) = \frac{1}{n} \sum_i \left[ \sum_{x,y} K^2(x_0 - x, y_0 - y) [S_i(x, y) - \bar{S}(x, y)]^2 + \right. \\
\left. \sum_{\substack{x,x',y,y' \\ x \neq x' \wedge y \neq y'}} K(x_0 - x, y_0 - y) K(x_0 - x', y_0 - y') [S_i(x, y) - \bar{S}(x, y)] [S_i(x', y') - \bar{S}(x', y')] \right].
\end{aligned} 
\tag{D.4} $$

If we invert the order of the first and the two inner sums, we finally get

$$ 
\begin{aligned}
\mathrm{Var}_{S'}(x_0, y_0) = \sum_{x,y} K^2(x_0 - x, y_0 - y) \mathrm{Var}_S(x, y) + \\
\sum_{\substack{x,x',y,y' \\ x \neq x' \wedge y \neq y'}} K(x_0 - x, y_0 - y) K(x_0 - x', y_0 - y') \mathrm{Cov}_S((x, y), (x', y')),
\end{aligned} 
\tag{D.5} $$

where $\mathrm{Cov}_S((x, y), (x', y'))$ is the covariance between two pixels at position $(x, y)$ and $(x', y')$. If the covariance is null or negligible, then the variance of the convolved signal is given by the variance map convolved by the square of the kernel.