M asked me a question which essentially looks like this: In a mediation model a and b are regression coefficient through the mediation path, and the final quantity of interest is therefore the product $ab$. In a multilevel model, for each group $j$, we model both a[j] and b[j] varying within group, where we could model in Stan as a multivariate normal

\[(a_j,b_j)^T \sim MVN ((a_0, b_0)^T, \Sigma),\]

According to literature xxx, we could estimate the expectation of ab in a typical group by (the sample mean of)

\[a_0 b_0 + \sigma_{12}^2,\]

for $\sigma_{12}$ the off-diagonal element in $\Sigma$. Does it makes sense to summarize the uncertainty by the sample sd of the draws above?

The answer is No. The formula really comes from point estimation context with

\[E[ab]= E[a]E[b] + Cov(a,b) = a_0 b_0 + \sigma_{12}^2\]

The law of total variance says

\[Var[ab]= E Var [ab| a_0 b_0 + \sigma_{12}^2 ] + Var E[ab| a_0 b_0 + \sigma_{12}^2 ]\]

The sample deviation of $a_0 b_0 + \sigma_{12}^2$ draws only amounts to the second term and I don’t think it means anything. The easiest way to solve the problem is to obtain posterior draws of group-level indirect effects $a_jb_j$ directly.