Imagine a 1-dimensional Bayesian update, with a prior $P(\Theta = \theta)$ and a likelihood function for evidence $D$ written as $\mathcal{L}_D: \theta \mapsto f(\theta)$.
The posterior PDF $P(\Theta=\theta \mid D) \propto P(\Theta=\theta) \times f(\theta)$.
Suppose we now consider evidence $D'$ with a likelihood function shifted by $x$, i.e. $\mathcal{L}_{D'}: \theta \mapsto f(\theta-x) $.
For concreteness say $x>0$, so we're shifting the likelihood function to the right. My question is, what's a sufficient condition to imply that the expected value of the posterior is greater under $D'$ than under $D$? i.e. what's a sufficient condition for $\mathbb{E}[\Theta \mid D'] > \mathbb{E}[\Theta \mid D]$? Presumably it would be some condition on the shape of the prior PDF and likelihood function.
It's clear this isn't always the case. For example:
Note that this counter-example is not just driven by the restricted domain of the uniform prior. You could just put arbitrarily thin tails on the ends of the uniform distribution I pictured (making it defined on $\mathbb{R}$ or anything else you want), without affecting the posterior expected value significantly.
I realise this question is very general, I would be interested in any sufficient conditions we can find, even if they are not the weakest achievable sufficient condition. However, a class of very strong sufficient conditions is not useful to me: conjugate priors. In the case of conjugate priors there is no need for the simulation I am doing (as the entire question is then answerable analytically).
One specific narrowing that would be particularly useful to me is: what if we add the restriction that the likelihood function is the normal likelihood function (the one that arises if the evidence is normally distributed)? This cannot be combined with the restriction that the prior should be normal, because then we are just back to conjugate priors.
The context is: I have a simulation where, in each iteration, we only care about whether the posterior expected value is greater than some bar, we needn't care what value it takes exactly. The answer to this question may help me make the simulation radically more efficient computationally.
If I am making an incorrect assumption or my notation is wrong, a correction is welcome.
Best Answer
Here's a very restrictive sufficient condition: both the prior and likelihood are normal.
If the likelihood is normal with known variance $\sigma^2$, and the prior is normal, then the posterior is normal. Say the likelihood is normal centred on $z$ with variance $\sigma^2$, and the prior has mean $\mu_0$ and variance $\sigma_0^2$, then the posterior has mean $\Big(\frac{1}{\sigma_0^2}+\frac{1}{\sigma^2}\Big)^{-1}\Big(\frac{\mu_0}{\sigma_0^2}+\frac{z}{\sigma^2}\Big)$, which is strictly increasing in $z$.
If this is too restrictive, you could consider other conjugate priors: https://en.wikipedia.org/wiki/Conjugate_prior.