# Mass-redistribution generalization of Jensen's inequality

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 $f$?

• 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 $\mathbb E\left(f\left(X\right)\right)\geq\mathbb E\left(f\left(\mathbb E\left(X|Y\right)\right)\right)$. – Oscar Cunningham Apr 14 '17 at 8:19
• Should probably be on Math.SE rather than here... I also wouldn't really call this "convex analysis"... – user541686 Apr 14 '17 at 8:55

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