This is motivated by [this](https://math.stackexchange.com/questions/1923064/generalizing-an-inequality-involving-a-convex-function) question on math.SE. 


One way to think about Jensen's inequality is that it says that if we have some probability distribution $\mu$ (continuous or discrete) over a space $X$ and $f:X\to\mathbb{R}$ is convex, changing the distribution to a deterministic distribution $\tilde{\mu}$ concentrated at the mean of $\mu$ cannot increase the expected value of $f$, i.e.
$$\mathbb{E}\_{\mu}[f(x)] \geq \mathbb{E}\_{\tilde{\mu}}[f(x)]$$


It seems plausible that there would be other distributions $\nu$ besides $\tilde{\mu}$ with this property (in particular, they would have to be non-deterministic). What I'm wondering is if there are nice criteria that capture what $\nu$ should look like with respect to $\mu$ and work for any convex $f$?



Mass-redistribution generalization of Jensen's inequality
There is a conditional version of the Jensen inequality that may be what you are looking for. 
$$
f(E(X\mid {\cal F})) \leq E(f(X) \mid {\cal F})
$$
Taking the expectation, this gives
$$
E(f(E(X\mid {\cal F}))) \leq E(f(X)).
$$
So you can replace the expectation $E(X) = \int X \, d\mu$ by any averages $E(X\mid {\cal F})$ along the atoms of a $\sigma$-algebra ${\cal F}$. Assuming that there are conditional probabilities associated to ${\cal F}$, we get $E(X\mid {\cal F})(\omega) = \int X(\omega) \, d\mu(\omega \mid {\cal F})$.


The simplest case is given by a $\sigma$-algebra associated to a countable partition $\{E\_i\}\_{i\in {\bf N}}$ of the underlying space. This gives
$$
\sum\_i {\mu(E\_i)} \ f\Bigl({1\over \mu(E\_i)}\int\_{E\_i} X \, d\mu\Bigr) \leq \int f(X) \, d\mu.
$$


