Equality in Jensen's inequality for non convex functions

Jensen's inequality, $$\mathbb{E}[\phi(X)]\geq\phi(\mathbb{E}[X])$$, holds for any convex function $$\phi$$ and random variable $$X$$.

Given that $$\mathbb{E}[\phi(X)]=\phi(\mathbb{E}[X])$$ for some non-degenerate random variable $$X$$, what can we say about $$\phi$$? any why? (assuming that it is not necessarily convex). What about the case in which $$\phi$$ is differentiable? I wonder if there is an interesting universal property on all function $$\phi$$ that lead to such equality.

• Do you mean "holds for all measurable (and integrable) $f$", not for "some"? Otherwise, $\phi(0)=0$ is a sufficient condition (it will hold for $f=0$). Commented Jan 12, 2021 at 22:41
• Yes, sorry. I edited my post. I am looking for a non-trivial property. Commented Jan 13, 2021 at 7:21
• Is this what you are looking for? math.stackexchange.com/q/3007917 – Oops, the question changed again... Commented Jan 13, 2021 at 15:01
• Yeah sorry, iterating for conciseness. Commented Jan 13, 2021 at 15:04
• Are you assuming that equality holds for all $X$ (or for all functions $f$ in your previous version) or just that it holds for some $X$? Commented Jan 13, 2021 at 15:06

I don't think there is much that can be said without more restrictions. If $$X$$ is some random variable such that $$P(X=E[X]) = 0$$, then for any function $$\phi$$ you can define $$\tilde{\phi}(t) = \begin{cases} \phi(t) & t \ne E[X] \\ E[\phi(X)] & t = E[X] \end{cases}$$ to get $$\tilde{\phi}(E[X]) := E[\phi(X)] = E[\tilde{\phi}(X)]$$.
• What if $\phi$ is differentiable? Commented Jan 13, 2021 at 15:46
• @GuyOhayon If $P(E[X] - \epsilon < X < E[X] + \epsilon) = 0$ for some $\epsilon > 0$ (this might happen for discrete $X$, for example), then I think you can similarly deform $\phi$ smoothly on the interval $[E[X] - \epsilon, E[X] + \epsilon]$ so that $\tilde{\phi}(E[X])=E[\phi(X)]$. Commented Jan 13, 2021 at 15:52