# On the requirements of the Jensen inequality for conditional expectation

I'm often reading this version of the Jensen inequality for conditional expectation:

Let $$(\Omega,\mathcal A, P)$$ a probability space and $$X$$ a integrable random variable. Then for any convex function $$\phi:\mathbb R\to\mathbb R$$ holds $$\phi\big(E(X|\mathcal A)\big)\leq E\big(\phi(X)|\mathcal A\big).$$

I'm wondering if there isn't the requirement "$$\phi(X)$$ is integrable" missing? Otherwise I can't imagine why the RHS should even exist.

For the ordinary case (not conditioned) I understand why we don't need $$\phi(X)$$ to be integrable, as the RHS would be $$\infty$$ and the inequality would hold trivially. But here we might have something on the RHS that doesn't even exist.

• Even if $\varphi$ is not integrable, it still holds.
– tiko
Commented Feb 24, 2021 at 16:59

Because $$\phi$$ is convex, it is bounded below by a function of the form $$x\mapsto mx+b$$, so the integrability of $$X$$ implies that of $$\phi(X)^-$$ (the negative part of $$\phi(X)$$). Even though $$\phi(X)^+$$ may not be integrable, the conditional expectation $$E[\phi(X)\mid\mathcal A]$$, as a random variable with values in $$(-\infty,+\infty]$$, exists in a generalized well-defined sense (for example, as the increasing limit $$\lim_nE[\phi(X)\wedge n\mid\mathcal A]$$). As such Jensen's inequality holds, even on the event $$\{E[\phi(X)\mid\mathcal A]=+\infty\}$$ because $$\phi(E[X\mid\mathcal A])$$ is a.s. finite.