# Conditional Jensen's inequality proof correctness. Queries regarding convex functions.

Let $$(Ω, \mathcal{F}, P)$$ be a probability space and let $$\mathcal{G} ⊂ \mathcal{F}$$ be a sub-$$σ$$-algebra.

Conditional Jensen's inequality:

Let $$φ : R → R$$ be a convex function, $$X$$ and $$φ(X)$$ be integrable random variables. Prove the conditional Jensen inequality $$φ(E[X|\mathcal{G}]) ≤ E[φ(X)|\mathcal{G}].$$

Hint: Convexity of $$φ$$ implies that for $$x$$, $$y$$$$R$$, there exists a measurable function $$c : R → R$$ such that $$φ(x) ≥ φ(y) + c(y)(x − y).$$

I have made an attempt on the this question but I am sure that it is not fully correct. My attempt:

Attempt 1:

Taking conditional expectation both sides and using monotonocity, we get:

$$E[φ(x)|\mathcal{G}] ≥ E[φ(y)|\mathcal{G}] + E[c(y)(x − y)|\mathcal{G}]$$, where $$x,y$$ are integrable random variables.

Now, put $$X$$ instead of $$x$$ and $$E[X|\mathcal{G}]$$ instead of $$y$$. Then, $$E[φ(X)|\mathcal{G}] ≥ E[φ(E[X|\mathcal{G}])|\mathcal{G}] + E[c(E[X|\mathcal{G}])(X − E[X|\mathcal{G}])|\mathcal{G}]$$.

Now, $$E[φ(E[X|\mathcal{G}])|\mathcal{G}]=φ(E[X|\mathcal{G}])$$ (Using the fact the random variable $$φ(.)$$ is $$\mathcal{G}$$ measurable.) Also, $$c(E[X|\mathcal{G}])$$ is a $$\mathcal{G}$$-measurable function by the definition of the function $$c$$ and the fact that $$E[X|\mathcal{G}]$$ is $$\mathcal{G}$$ measurable.

So, $$E[c(E[X|\mathcal{G}])(X − E[X|\mathcal{G}])|\mathcal{G}]=c(E[X|\mathcal{G}])E[0]$$=$$0$$.

Hence, we get that:

$$E[φ(X)|\mathcal{G}] \geq φ(E[X|\mathcal{G}])$$.

Is this solution alright?

Solution 2:

This is a solution which I found in notes:

While it is fairly easy to see the value at any point of a convex function as the supremum of the values at that point of the tangent lines to the convex function, I am unable to see how to obtain a countable set over which the supremum is taken. Also, why do I need a countable set? Also, could you please explain why $$E[sup_{i}(a_{i}X+b_{i})|\mathcal{G}] \geq a_{i}E[X|\mathcal{G}]+b_{i}$$?

Also, is it possible to prove the conditional Jensen's inequality using conditional Markov's inequality?

• Sorry for that. I have removed the Markov question. Commented Aug 3, 2023 at 11:24
• One way is to use disintegration / regular conditional distribution, from which the desired results are an immediate consequence of the unconditional results. Commented Aug 3, 2023 at 11:24
• @AndrewZhang You are suggesting this solution for the conditional Markov's inequality question.?I have noted your solution. I removed that question as the entire question was becoming very long. Commented Aug 3, 2023 at 11:27
• The countable set is to obtain the equality, since that is the characterization of a convex function. The question about why the supremum inequality holds is just using linearity of conditional expectation. Also, I am suggesting the solution using disintegration to prove Jenssens as well, though it can of course be proven using more elementary methods. Commented Aug 3, 2023 at 11:28
• I got the inequality part. Thank you Andrew. Commented Aug 3, 2023 at 11:30

1. Your proof 1 is the correct idea, but I would modify your first two lines to be more clear what inequality you are taking expectations of: Using the hint and replacing $$x$$ with $$X$$ and $$y$$ with $$E[X|G]$$ gives the following (which holds "surely"): $$\phi(X) \geq \phi(E[X|G]) + c(E[X|G])(X-E[X|G])$$ and we can take conditional expectations of both sides (given $$G$$) to get another inequality. However, that new inequality only holds "almost surely."
2. In the second proof, one reason that it is important that the sup is taken over a countable set, say, $$A$$, is that your inequality only holds for each individual $$i \in A$$ "almost surely." Then, we can ask if all of those inequalities simultaneously hold for all $$i \in A$$ almost surely. The answer is "yes" if $$A$$ is a countable set but "not necessarily" if $$A$$ is uncountable. This uses the idea that if $$\{B_i\}_{i=1}^{\infty}$$ is a sequence of events that satisfy $$P[B_i]=1$$ for all $$i \in \{1, 2, 3, ...\}$$, then we can conclude $$P[\cap_{i=1}^{\infty} B_i]=1$$. Another reason that $$A$$ being countable is important is that it means $$\sup_{i\in A}[a_iX+b_i]$$ is a valid (extended) random variable (where "extended" allows the possibility that it takes value $$\infty$$).
3. In the second part of your question of part 2, for all $$j \in A$$ we have $$\sup_{i \in A} [a_iX+b_i]\geq a_jX+b_j$$ so you can take conditional expectations of both sides.