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This is motivated by this 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 $f$?

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    $\begingroup$ If we write $\mu$ as a weighted average of some other distributions $\nu_i$, then the distribution formed by the weighted average of $\tilde{\nu}_i$ has this property. One could probably state this as Jensen's inequality on $X$ where $X$ is some random variable in a joint distribution with some other random variable $Y$ which we are conditioning on. I guess $\mathbb E\left(f\left(X\right)\right)\geq\mathbb E\left(f\left(\mathbb E\left(X|Y\right)\right)\right)$. $\endgroup$
    – Oscar Cunningham
    Commented Apr 14, 2017 at 8:19
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    $\begingroup$ Should probably be on Math.SE rather than here... I also wouldn't really call this "convex analysis"... $\endgroup$
    – user541686
    Commented Apr 14, 2017 at 8:55

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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 $$ \sum_i {\mu(E_i)} \ f\Bigl({1\over \mu(E_i)}\int_{E_i} X \, d\mu\Bigr) \leq \int f(X) \, d\mu. $$

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Yes, this is called 'majorization' or 'second order stochastic dominance' of measures (first term is used in analysis, second in probability). The idea is very simple: we partition the measure $\mu$ on several summands $\mu=\sum \mu_i$ and replace each $\mu_i$ to its mean value.

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    $\begingroup$ Is there a reference or an easy proof that these are the only such measures? $\endgroup$
    – Oscar Cunningham
    Commented Apr 14, 2017 at 8:32
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    $\begingroup$ Not quite: we may consider continuous sums, that is, integrals. Maybe, the book of Marshall and Olkin contains the proof. $\endgroup$
    – Fedor Petrov
    Commented Apr 14, 2017 at 8:49

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