Close property of conditional expectation under convexity

Suppose we have a random vector {Z\in \mathbb{R}^n} and we know that {Z} only takes value in a convex set {C}, i.e.,

\displaystyle \mathbb{P}(Z\in C) =1.

Previously, we showed in this post that as long as {C} is convex, we will have

\displaystyle \mathbb{E}(Z) \in C,

if {\mathbb{E}(Z)} exists. It is then natural to ask how about conditional expectation. Is it true for any reasonable sigma-algebra {\mathcal{G}} that

\displaystyle \mathbb{P}(\mathbb{E}(Z\mid \mathcal{G}) \in C )=1?

The answer is affirmative. Let us first recall our previous theorem.

Theorem 1 For any borel measurable convex set {C\subset {\mathbb R}^n}, and for any probability measure {\mu} on {(\mathbb{R}^n,\mathcal{B})} with

\displaystyle \int_C \mu(dx) =1, \quad \int_{\mathbb{R}^n} \|x\|\mu(dx) <\infty,

we have

\displaystyle \int_{\mathbb{R}^n} x\mu(dx) \in C.

By utilizing the above theorem and regular conditional distribution, we can prove our previous claim.

Theorem 2 Suppose {Z} is a random vector from a probability space {(\Omega, \mathcal{F},\mathbb{P})} to {(\mathbb{R}^n,\mathcal{B})} where {\mathcal{B}} is the usual Borel sigma algebra on {\mathbb{R}^n}. If {C} is Borel measurable and

\displaystyle \mathbb{P}(Z\in C) =1, \mathbb{E}\|Z\| <\infty ,

then we have

\displaystyle \mathbb{P}( \mathbb{E}(Z\mid \mathcal{G})\in C) =1

for any sigma algebra {\mathcal{G} \subset \mathcal{F}}.

Proof: Since {Z} takes value in {\mathbb{R}^n}, we know there is a family of conditional distribution {\{\mu(\omega,\cdot)\}_{\omega \in \Omega}} such that for almost all {\omega \in \Omega}, we have for any {A\in \mathcal{B}}

\displaystyle \mathbb{E}(1_{\{Z\in A\}}\mid \mathcal{G})(\omega) = \int_{A} \mu(\omega, dz)

and for any real valued borel function {f} with domain {\mathbb{R}^n} and {\mathbb{E}(|f(Z)|)}<+\infty, we have

\displaystyle \mathbb{E}(f(Z)\mid \mathcal{G}) (\omega)= \int_{\mathbb{R}^n} f(z) \mu(\omega, dz).

The above two actually comes from the existence of regular conditional distribution which is an important result in measure-theoretic probability theory.

Now take {A=C} and {f(Z) = Z}, we have for almost all {\omega\in \Omega},

\displaystyle \mathbb{E}(1_{\{Z\in C\}}\mid \mathcal{G})(\omega) = \int_{C} \mu(\omega, dz)

and

\displaystyle \mathbb{E}(Z\mid \mathcal{G}) (\omega)= \int_{\mathbb{R}^n} z \mu(\omega, dz).

But since {\mathop{\mathbb P}(Z\in C)=1}, we know that for almost all {\omega \in \Omega},

\displaystyle 1= \mathbb{E}(1_{\{Z\in C\}}\mid \mathcal{G})(\omega) = \int_{C} \mu(\omega, dz).

Thus for almost all {\omega} we have a probability distribution {\mu(\omega, dz)} on {{\mathbb R}^n} with mean equals to {\mathop{\mathbb E} (Z\mid \mathcal{G} )(\omega)} and {\int_C \mu (\omega ,dz) =1}. This is exactly the situation of Theorem 1. Thus by applying Theorem 1 to the probability measure {\mu(\omega, dz)}, we find that

\displaystyle \mathbb{E}(Z\mid \mathcal{G})(\omega) = \int _{\mathbb{R}^n} z\mu (\omega, dz) \in C,

for almost all {\omega \in \Omega}. \Box

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s