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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

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    $\begingroup$ While not a duplicate, there are common points with this question. $\endgroup$
    – Xi'an
    Feb 14, 2019 at 15:45
  • $\begingroup$ What is the meaning of $E(S)$ and how does it relate with $E[X|S]$? $\endgroup$
    – Xi'an
    Feb 14, 2019 at 15:47
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    $\begingroup$ By E[S], do you mean the empirical average of the data S? Am I right in assuming that your parameter of interest is the mean (or position) of the distribution of S? $\endgroup$
    – Robin Ryder
    Feb 14, 2019 at 21:48
  • $\begingroup$ E[S] is the mean of the observed data. E[X|S] is the conditional mean of X given the observed S. $\endgroup$
    – Niki Kotsenko
    Feb 15, 2019 at 21:07
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    $\begingroup$ I'm struggling to see why anything like this should hold in such generality. As an extreme case, suppose prior "$X^\prime$" puts all probability of the mean on $0$ and $S=0$ is observed. Your claim implies that no matter what the prior "$X$" might be, the posterior expectation must be zero. That's obviously untrue. $\endgroup$
    – whuber ♦
    Feb 16, 2019 at 21:55

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