I want to prove the conditionnal Jensen's inequality. Let $(\Omega, \mathcal H, \mathbb P)$ be a probability space, $\mathcal G \subset \mathcal H$ a sub sigma algebra, $\varphi : \mathbb R \longrightarrow \mathbb R$ convex and $X \in L^1(\mathcal H)$. Then the Jensen inequality is $$ \mathbb E [\varphi(X) | \mathcal G] \geq \varphi(\mathbb E [X | \mathcal G]). $$

For the proof the classical argument is that by basic properties of convex functions, forall $x \in \mathbb R$, $$ \tag{1}\varphi(x) = \sup_{(a,b) \in E_\varphi} ax+b = \sup_{(a,b) \in E_\varphi \cap \mathbb Q^2} ax+b $$ where $$ E_\varphi = \{ (a,b) \in \mathbb R^2 : \forall x \in \mathbb R ~~~\varphi(x) \geq ax+b \}. $$ The second equality of $(1)$ is false when $\varphi$ is affine with irrational growth ratio and I don't understand why we would need $(a,b)$ to vary in a countable set. This is how I proceed from the first equality of $(1)$ :

Fix $(a,b) \in E_\varphi$, forall $x \in \mathbb R$ we have $\varphi(x) \geq ax+b$ so $\varphi(X) \geq aX+b$. Taking the conditionnal expectation, which is non negative and linear, yields $$ \mathbb E [\varphi(X) | \mathcal G] \geq a \mathbb E [X | \mathcal G] + b. $$ Now taking the supremum of the left hand side on $(a,b)$ and using the first equality of $(1)$ yields the result. Is my proof correct?

  • $\begingroup$ Your proof makes sense to me. If you don't understand the second equality of $(1)$ I think you should remove it. Especially since you never use it. $\endgroup$
    – jakobdt
    Commented Nov 18, 2021 at 13:13
  • $\begingroup$ @jakobdt thank you for your answer. The second equality of $(1)$ is false in the case $\varphi$ is a line with irrational growth ratio. This is because in this case $E_\varphi$ is empty. In fact my proof is false as I detailed in my answer. $\endgroup$ Commented Nov 18, 2021 at 17:51
  • $\begingroup$ That’s an interesting point. What you mean is that $E_\varphi\cap\mathbb{Q}^2$ is empty, right? Also, I finally understood why you need the countability. $\endgroup$
    – jakobdt
    Commented Nov 18, 2021 at 20:19
  • $\begingroup$ @jakobdt if $\varphi (x) = ax+b$ then $E_\varphi = \{a\} \times (-\infty,b]$ so asking $a$ irrationnal yields $E_\varphi \cap \mathbb Q^2$ empty. $\endgroup$ Commented Nov 18, 2021 at 20:34
  • 1
    $\begingroup$ Perfect, then we agree. $\endgroup$
    – jakobdt
    Commented Nov 18, 2021 at 21:08

1 Answer 1


In fact my proof is not correct and we should ask for a countable dense subset of $E_\varphi$. Indeed, fix a representant of $X$, it is true that for any $(a,b) \in E_\varphi$ we have everywhere on $\Omega$ $$ \varphi(X) \geq aX+b. $$ But when taking the conditionnal expectation only get an inequality almost surely because the conditionnal expectation is only defined almost surely. Thus we have in fact $$ \forall (a,b) \in E_\varphi,~~~\mathbb P -a.s., ~~~\mathbb E[\varphi(X) | \mathcal G] \geq a \mathbb E [X | \mathcal G] +b. $$ To swap the quantificators and since the application $(a,b) \mapsto ax+b$ is continuous we only need a countable dense subset of $E_\varphi$ which is possible because $E_\varphi \subset \mathbb R^2$ which is metric separable.


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