# 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. Commented Feb 6, 2016 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
Commented Feb 6, 2016 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 $$E[\varphi(X)|\mathcal F]\ge \varphi(Y)$$, 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 $$E[X|\mathcal F](\omega)$$.
• It seems to me that to formalize Method 1 one also needs to use the existence of a regular conditional distribution. Furthermore, to conclude the argument in Method 2 one should notice that $\mathbb{E}\phi(X)=\mathbb{E}\mathbb{E}(\phi(X)|\mathscr{F})=\mathbb{E}\phi(\mathbb{E}(X|\mathscr{F}))$ for all measurable positive $\phi$, to conclude that $X=\mathbb{E}(X|\mathscr{F})$ and hence is $\mathscr{F}$-measurable. Commented Feb 12, 2023 at 21:48
• In Method 1, as noted, we have $\{X\not=Y\}\subset \{E[\varphi(X)\mid\mathcal F]>\varphi(Y)\}$, a.s. Therefore, because $\{E[\varphi(X)\mid\mathcal F]>\varphi(Y)\}\in\mathcal F$, $$P[X\not= Y\mid\mathcal F]\}\le 1_{\{E[\varphi(X)\mid\mathcal F]>\varphi(Y)\}},$$ almost surely. This proves the final assertion in Method 1, with no use of regular conditional distributions. In Method 2, as you note, if $E[\varphi(X)\mid\mathcal F]=\varphi(E[X\mid\mathcal F])$ then $X=E[X\mid \mathcal F]$ a.s., so $X$ is $\mathcal F$-measurable, assuming completeness. Commented Feb 13, 2023 at 23:28
• Thanks for answering. I now think my previous comment was wrong and that in both cases to conclude we need to argue as follows: from $\mathbb{E}(1_{X=\mathbb{E}(X|\mathscr{F})}|\mathscr{F})=1$ we get $\mathbb{E}(1_{X=\mathbb{E}(X|\mathscr{F})})=1$, hence $1_{X=\mathbb{E}(X|\mathscr{F})}=1$ a.e., hence $X=\mathbb{E}(X|\mathscr{F})$ a.e. Commented Jun 5, 2023 at 23:59