Assume that we have a random variable $Z = X + Y$ for $X$ and $Y$ independent. Then if w use two independent data-sets $D_1$ and $D_2$ to try and approximate the distribution of $Z$, i.e. $$p(Z|D_1,D_2)$$
If we use the fact that $X$ and $Y$ are independent then we have apply bayes theorem and rearrange stuff to get:
$$ p(Z\mid D_1,D_2)=p(X + Y\mid D_1,D_2) = p(X\mid D_1,D_2) \ast p(Y\mid D_1,D_2) $$ $$ \propto [p(X\mid D_1)p(X\mid D_2))]\ast [p(Y\mid D_1)p(Y\mid D_2)] $$
However, if we ignore the fact that $X$ and $Y$ are independent (or we don't know) then we can similarly derive the expression:
$$ p(Z\mid D_1,D_2)\propto p(Z\mid D_1)p(Z\mid D_2) $$ the independence of $X$ and $Y$ then mean that this last line would be equal to $$ [p(X\mid D_1) \ast p(Y\mid D_1)][p(X\mid D_2) \ast p(Y\mid D_2)] $$
But if for example we are dealing with normal distributions the convolutions just adds the means and variances of the various distributions. Hence in the first case we would be essentially taking a weighted average of the predictions for $X$ form $D_1$ and $D_2$ and similarly for $Y$ and then adding these two new predictions together. But in the second case we take a weighted average of the full predictions for $Z$ and the $X$ and $Y$ parts essentially get lumped together. As far as I can tell these two formulae would give different results in practise.
So my questions are:
1) Are the above formulations correct? and is either 'right' or 'wrong' as an approach to the analysis
2) If we don't know whether $X$ and $Y$ are independent but they are, will the second formulation give the same thing as the first where we assume independence?