I am looking for a proof of the statement "If the variance of the prior distribution is greater, the posterior is more affected by the data".
More specifically, if X, X' are priors such that E(X)=E(X') and Var(X)>Var(X'), and S is data such that E(X|S)=Z and E(X'|S)=Z', then |E(Z) - E(S)| =< |E(Z') - E(S)|, where E(S) is the mean of the data, and E(X|S) is the mean of X given the data.
I know how to prove this for a few specific cases (like normal with known variance, where the mean is unknown but the prior on it is also normal).
Is there a reference to a more general proof?
My current application involves a three sided die with possibly unequal probabilities (Dirichlet), but I might like to extend it to other, possibly continuous, cases.
This statement is found, among other places, in the highly rated answer by COOLSerdash (under the heading Rules of Thumb) to the question: Help me understand Bayesian prior and posterior distributions