I'm reading Meyer's seminal work Probability and Potentials (1966), in which he states the following "borrowed" theorem from Dubins "Rises and Upcrossings of Nonnegative Martingales" (1961).

LEMMA. (Generalized Jensen's inequality) Let $(\Omega, \mathscr{F}, \mathbf{P})$ be a probability space, $X$ an integrable random variable on this space, and $\mathscr{G}$ a sub-$\sigma$-field of $\mathscr{F}$. We shall denote by $Y$ a version of $\mathbf{E}[X\mid \mathscr{G}]$.

Let $f$ be a measurable mapping of $\Omega \times \mathbf{R}$ (with the natural product $\sigma$-field) into $\mathbf{R}$, such that

(a) The mapping $\omega \mapsto f(\omega,t)$ is $\mathscr{G}$-measurable for each $t\in\mathbf{R}$.

(b) The mapping $t\mapsto f(\omega,t)$ is convex for each $\omega \in \Omega$.

Suppose that the random variable $\omega \mapsto f(\omega,X(\omega))$ is integrable. We then have the inequality $$\mathbf{E}[f(\cdot,X(\cdot))\mid \mathscr{G}] \geq f(\cdot,Y(\cdot))\ \text{a.s.}$$

He indicates that one could prove the theorem for simple $X$ and then pass to a limit "which is a little more delicate than usual" and then omits the proof. As a side note, I have searched Dubins' paper and find nothing even remotely similar in it.

I don't see how assuming $X$ is simple would help, but here's what I've tried so far.

Proof. By the convexity assumption, for all $x,x_0,\omega$ we have $$ f(\omega,x) \geq \partial_2 f(\omega,x_0)(x-x_0) + f(\omega,x_0) $$ where $$ \partial_2 f(\omega,x_0) = \lim_{n\to\infty} \frac{f(\omega,x_0+1/n)-f(\omega,x_0)}{1/n} $$ is the right-handed derivative in $f$'s second variable. Note that $\partial_2 f(\cdot,x_0)$ is a limit of $\mathscr{G}$-measurable functions and hence is as well, for each $x_0$. Now plug in $x=X(\omega)$, so we have $$ f(\omega,X(\omega)) \geq \partial_2 f(\omega,x_0)(X(\omega)-x_0) + f(\omega,x_0) $$ for all $x_0$. At this point I would like to condition both sides relative to $\mathscr{G}$ to get $$ \mathbf{E}[f(\cdot,X(\cdot))\mid \mathscr{G}](\omega) \geq \partial_2 f(\omega,x_0)(Y(\omega)-x_0) + f(\omega,x_0) $$ for almost all $\omega$ in $\Omega_{x_0}$ with $\mathbf{P}(\Omega_{x_0})=1$, and hence for all $\omega \in \Omega^*$ where $\Omega^* = \cap_{q \in \mathbf{Q}} \Omega_{x_0}$ has full measure. Then for all $\omega \in \Omega^*$ I can take a sequence $q_n \searrow Y(\omega)$ to obtain $$ \mathbf{E}[f(\cdot,X(\cdot))\mid \mathscr{G}](\omega) \geq f(\omega,Y(\omega)) $$ where here we make use of the local boundedness of $x_0 \mapsto \partial_2 f(\omega,x_0)$ given to us by convexity, and the continuity of $x_0 \mapsto f(\omega,x_0)$ also given to us by convexity.

I said like up in bold because I do not know that $\omega \mapsto \partial_2 f(\omega, x_0)X(\omega), \omega\mapsto \partial_2 f(\omega,x_0)$, and $\omega \mapsto f(\omega, x_0)$ are integrable, so I cannot just pull out the $\mathscr{G}$-measurable parts. Is there some way I can resolve this?

EDIT: It may be useful to note that $f$ is actually $\mathscr{G}\otimes \mathscr{B}(\mathbf{R})$-measurable: define $f_n(\omega,x) := f(\omega, d_n(x))$ where $d_n(x) = \inf\{ \frac{k}{2^n} : \frac{k}{2^n} \geq x, k \in \mathbf{Z}\}$. Then $\{ f_n \in B \} = \cup_{\frac{k}{2^n}, k \in \mathbb{Z}}( \{\omega:f(\omega,\frac{k}{2^n}) \in B\} \times [\frac{k}{2^n},\frac{k+1}{2^n}))$. The $f_n$ are demonstrably $\mathscr{G} \otimes \mathscr{B}(\mathbf{R})$ measurable and $f_n \to f$ by continuity of $f$.

  • $\begingroup$ One of the ways to prove Jenson's inequality in the classical case is by proving it for affine functions $f(x)=ax+b$ and then using the fact that any convex function can be expressed as the envelope of affine functions. A similar argument should work here. $\endgroup$ – Alex R. Aug 13 '15 at 1:21
  • $\begingroup$ @Alex I tried that but was not successfull. The classical proof works because you can choose a countable set of $a_n,b_n$ such that $f(x) = \sup_n a_n x + b_n$. I don't see how that could work in this case because the $a_n,b_n$ you would need to choose have to depend on $\omega$. If you could elaborate on how you mean to overcome this issue I would be interested in that alternative proof. $\endgroup$ – nullUser Aug 13 '15 at 1:32
  • $\begingroup$ To pull out a $G$ measurable function $f$ out of the conditional expectation, it should suffice to know that $f$ is quasi-integrable. You know this, because the $f$ in question is bounded above by $f(\omega, X(\omega))$. $\endgroup$ – PhoemueX Aug 13 '15 at 3:30
  • $\begingroup$ @PhoemueX what is the definition of quasi-integrable, and do you have a link or could you post a proof of statement? $\endgroup$ – nullUser Aug 13 '15 at 3:33

I think I've got it, thanks to @PhoemueX for the idea.

The key is that all I really need is that $$ \mathbf{E}[f(\cdot,X(\cdot))\mid \mathscr{G}](\omega) \geq \partial_2 f(\omega,x_0)(Y(\omega)-x_0) + f(\omega,x_0) $$ a.s., I don't need to necessarily get there by taking the conditional expectation relative to $\mathscr{G}$ of the previous $$ f(\omega,X(\omega)) \geq \partial_2 f(\omega,x_0)(X(\omega)-x_0) + f(\omega,x_0). $$ Instead, take any $G \in \mathscr{G}$ and compute $$ \int f(\omega,X(\omega))1_G d\mathbf{P}(\omega) \geq \int(\partial_2 f(\omega,x_0)(X(\omega)-x_0) + f(\omega,x_0))1_Gd\mathbf{P}(\omega) $$ and since the integral on the left is finite, the one on the right can be split into the finite integral of its positive part, and the possibly infinite integral of its negative part. Thus the previous integral equals $$ = \lim_n\int(\partial_2 f(\omega,x_0)(X(\omega)-x_0) + f(\omega,x_0))1_G 1_{|\partial_2 f(\omega,x_0)|, |f(\omega,x_0)| \leq n}d\mathbf{P}(\omega) $$ which, since now $\partial_2f, f$ are bounded, we can then apply the condition expectation property for $X$ to the previous line $$ = \lim_n\int(\partial_2 f(\omega,x_0)(Y(\omega)-x_0) + f(\omega,x_0))1_G 1_{|\partial_2 f(\omega,x_0)|, |f(\omega,x_0)| \leq n}d\mathbf{P}(\omega) $$ and for the same reason as before (that reason being quasi-integrability, i.e. $\int h d\mu $ exists but may be infinite), the previous equals $$ = \int(\partial_2 f(\omega,x_0)(Y(\omega)-x_0) + f(\omega,x_0))1_G d\mathbf{P}(\omega). $$ Thus, we have shown that for all $G \in \mathscr{G}$ $$ \int f(\omega,X(\omega))1_G d\mathbf{P}(\omega) \geq \int(\partial_2 f(\omega,x_0)(Y(\omega)-x_0) + f(\omega,x_0))1_G d\mathbf{P}(\omega). $$ Since the integrand on the right is itself $\mathscr{G}$-measurable, this implies the desired inequality $$ \mathbf{E}[f(\cdot,X(\cdot))\mid \mathscr{G}](\omega) \geq \partial_2 f(\omega,x_0)(Y(\omega)-x_0) + f(\omega,x_0), $$ even though the right side might not be integrable. (If not, compare integrals over the $\mathscr{G}$-measurable set where the right side strictly exceeds the left to arrive at a contradiction.) The rest of the proof goes through.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.