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

Question

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?

Answer 1

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.