# Equality in Conditional Jensen's Inequality

Conditonal Jensen's Inequality says that for a convex function $\varphi$, a random variable $X$, and a sub-sigma-field $\mathcal{F}$, $E[\varphi(X)\mid \mathcal{F}] \geq \varphi(E[X\mid \mathcal{F}])$. In ordinary Jensen's Inequality, $E[\varphi(X)]\geq \varphi(E[X])$, and we have equality if and only if $X$ is degenerate (i.e., almost surely a constant) or $\varphi$ is linear. I'm wondering if an analogous result holds for the conditional version. Is it the case that $E[\varphi(X)\mid\mathcal{F}]=\varphi(E[X\mid\mathcal{F}])$ if and only if $X \in \mathcal{F}$ or $\varphi$ is linear? (Certainly the "if" is true, but I'm wondering about the "only if.")

• The conditions you state for equality in ordinary Jensen are not correct. $\varphi$ need not be linear everywhere, just on the essential range of $X$. That will include the degenerate case. – GEdgar Feb 6 '16 at 15:22
• Okay, but the question still stands. If $\varphi$ is strictly convex on the essential range of $X$, must it be that $X \in \mathcal{F}$? – kccu Feb 6 '16 at 15:43

Method 1: Abbreviate $Y:=E[X|\mathcal F]$. Let $g(x)$ denote the right-hand derivative of $\varphi$ at $x$. Because $\varphi$ is strictly convex, we have $\varphi(x)>g(m)(x-m)+\varphi(m)$ for all $x\not=m$. Thus, $$\varphi(X)\ge g(Y)(X-Y)+\varphi(Y)$$ with strict inequality off $\{X=Y\}$ (almost surely). Taking conditional expectations in the inequality above we obtain $E[X|\mathcal F]\ge \varphi(Y)$, and $$\{E[\varphi(X)|\mathcal F]=\varphi(Y)\}\subset\{P[X\not=Y|\mathcal F]=0\}$$ almost surely.
Method 2: Let $\mu(\omega,dx)$ be a regular conditional distribution of $X$ given $\mathcal F$. (Such exists because $X$ is real valued.) That is, for each Borel set $B\subset\Bbb R$, $\omega\mapsto \mu(\omega,B)$ is $\mathcal F$-measurable, for each $\omega\in\Omega$, $B\mapsto \mu(\omega,B)$ is a probability measure on $\Bbb R$, and $\int_{\Bbb R} f(x)\,\mu(\omega,dx)$ is a version of $E[f(X)|\mathcal F](\omega)$ for suitably integrable $f$. Now apply Jensen's inequality (for the strictly convex function $\varphi$) to the probability measure $\mu(\omega,\cdot)$ for each fixed $\omega$. The conclusion is that for $P$-a.e. $\omega\in\Omega$, the equality of $E[\varphi(X)|\mathcal F](\omega)$ and $\varphi(E[X|\mathcal F](\omega))$ forces $\mu(\omega,\cdot)$ to be a unit point mass at $E[X|\mathcal F](\omega)$.