Conditional expectation of a random vector taking values in convex sets

Question

on a probability space $(\Omega, \mathcal{F},\mathbb{P})$ i have a random vector $X\in L^1_{\mathbb{P}}(\mathbb{R}^d)$ (integrable with values in $\mathbb{R}^n$), such that $\mathcal{P}-a.s.$ $$X\in C$$ where $C$ is a convex set of $\mathbb{R}^n$. I want to know if it implies that $\mathcal{P}-a.s.$ $$E[X| \mathcal{F}_0] \in C$$ for any sigma algebra $\mathcal{F}_0\subset \mathcal{F}$. It seems natural and intuitive but i am not able to write the clean proof explicitely in the general case. Do we need to put some further assumptions of $C$ ? Do you have any idea of how we can wite the proof explicitely ? Thank you.

Answer 1

If the convex set $C$ is closed, then $C$ is an intersection of half spaces $H_u^c=\{x\mid \langle x,u\rangle\leqslant c\}$ for some family $\mathcal H$ of pairs $(u,c)$ in $\mathbb R^n\times\mathbb R$. For every such pair $(u,c)$ in $\mathcal H$, $\langle X,u\rangle\leqslant c$ almost surely and $\langle E(X\mid \mathcal F_0),u\rangle=E(\langle X,u\rangle\mid\mathcal F_0)$ almost surely hence $\langle E(X\mid \mathcal F_0),u\rangle\leqslant c$ almost surely. If $\mathcal H$ is at most countable, this implies that $\langle E(X\mid \mathcal F_0),u\rangle\leqslant c$ almost surely for every $(u,c)$ in $\mathcal H$, that is, $P(E(X\mid \mathcal F_0)\in C)=1$.

The space $\mathbb R^n$ is separable hence, if $C$ is a closed convex set, then $C$ is constructible, that is, $C$ is the intersection of countably many halfspaces.

All this proves that if $X$ is in a convex set $C$ almost surely then $E(X\mid\mathcal F_0)$ is in $\bar C$ almost surely.

Answer 2

More can be said when $C$ is an open interval, say $C=(0,1)$. Let $A$ be the event on which $E(X|{\mathcal F}_0) = 0$. Then $$ 0=E(1_AE(X|{\mathcal F}_0))=E(1_AX), $$ where $1_A$ is the indicator of the event $A$. Consequently, $$ P(A\cap\{X>1/n\})\le nE(1_AX) =0. $$ for $n=1,2,3,\ldots$. It follows that $P(A\cap\{X>0\}=0$. Because $P(X>0)=1$, we must have $P(A)=0$. That is, $P(E(X|{\mathcal F}_0)=0)=0$. Likewise, $P(E(X|{\mathcal F}_0)=1)=0$. So $P(E(X|{\mathcal F}_0)\in C)=1$ in case $C$ is an open interval.

For open convex $C\subset R^n$ I need to use regular conditional distributions. Let $(\Omega,{\mathcal F},P)$ be the probability space on which $X$ is defined. There is a kernel $\mu(\omega,B)$ ($\omega\in\Omega$, $B$ in the Borel subsets of $R^n$) that is an ${\mathcal F}$-measurable function of $\omega$ for each fixed $B$ and a probability measure as a function of $B$ for each fixed $\omega\in\Omega$, such that $$ E(1_A f(X)) =\int_A\left[\int_{R^n}\,\mu(\omega,dx)f(x)\right]\,P(d\omega), $$ for all $A\in{\mathcal F}_0$ and Borel $f:R^n\to R$ such that $f(X)$ is integrable. In other words, for $f$ as above, $\omega\mapsto \int_{R^n}\mu(\omega,dx)f(x)$ is a version of the random variable $E(f(X)|{\mathcal F}_0)$.

For example, the choice $f=1_C$ yields $$ P(A) = \int_A \mu(\omega,C)\,P(d\omega), $$ and thereby the knowledge that $\mu(\omega,C)=1$ for $P$-a.e. $\omega\in\Omega$. I shall write $G$ for $\{\omega\in\Omega:\mu(\omega,C)=1\}$.

Let's now focus on $E(X|{\mathcal F}_0)$. I claim the following: If $\mu$ is a probability measure concentrated on the open convex set $C$, then the barycenter $x^*:=\int_C x\,\mu(dx)$ is an element of $C$. (The integrability of $x$ with respect to $\mu$ is assumed.) By Didier's argument provided above, $x^*$ is necessarily an element of $\overline C$. Arguing by contradiction, suppose that $x^*$ is in the boundary of $C$; that is $x^*\in\overline C\setminus C$. The convex set $C$ admits a support hyperplane at $x^*$, embodied by an affine function $f:R^n\to R$ of the form $f(x)=a+b\cdot x$ ($a\in R, b\in r^n$) such that $f(x^*)=0$ but $f(x)<0$ for all $x\in C$ (because $C$ is open). We have $$ \int_C f(x)\,\mu(dx) =a+b\cdot\left[\int_C x\mu)dx)\right]=a+b\cdot x^*=f(x^*)=0. $$ On the other hand, $\int_C f(x)\,\mu(dx)<0$ because $f(x)<0$ for all $x\in C$, and we have our contradiction. Finally, if $\omega\in G$ (defined at the end of the last paragraph) then the preceding discussion applies to $\mu(\omega,\cdot)$, and so $E(X|{\mathcal F}_0)(\omega) =\int_C x\,\mu(\omega,dx)\in C$, for $\omega\in G$, hence a.s.

It may be that a refinement of this argument yields the desired result for general convex $C$.