# Conditional Expectation of Composite Function

Preliminaries

Let $$(\Omega, \mathcal{G}, \mathbb{P})$$ be a complete probability space.

Let $$D$$ be a complete, separable, metrizable topological space with Borel $$\sigma$$-algebra $$\mathcal{B}(D)$$ (such as $$D = \mathbb{R}^q$$ with $$\sigma$$-algebra $$\mathcal{B}(D) = \mathcal{B}(\mathbb{R}^d)$$).

Let $$\mathbb{R}$$ be equipped with its canonical Borel $$\sigma$$-algebra $$\mathcal{B}(\mathbb{R})$$.

Let $$g: \Omega \times D \rightarrow \mathbb{R}$$ be a bounded $$(\mathcal{G} \otimes \mathcal{B}(D) ) / \mathcal{B}(\mathbb{R})$$-measurable function.

Let $$\Pi: \Omega \rightarrow D$$ be a $$\mathcal{G}/\mathcal{B}(D)$$-measurable random variable.

Let $$H : \Omega \rightarrow \mathbb{R}$$ be a $$\mathcal{G}/\mathcal{B}(\mathbb{R})$$-measurable random variable, defined by $$H(\omega) := g(\omega, \Pi(\omega)).$$ Note, that, since $$g$$ is bounded, we have $$H \in \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P})$$.

Let $$j: D \rightarrow \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P})$$ be defined by $$j(\pi)(\omega) := g(\omega, \pi)$$

For all $$\pi \in D$$, let $$j(\pi)$$ be independent of $$\Pi$$.

Question

I am interested in the conditional expectation $$\mathbb{E}[H \mid \Pi] :\Omega \rightarrow \mathbb{R}$$ of $$H$$ with respect to $$\Pi$$. More specifically, I suspect that (a $$\mathbb{P}$$-unique version of) this condititional expectation is given by

$$\mathbb{E}[H \mid \Pi] (\omega) = \mathbb{E}[j(\Pi(\omega))], \quad (\dagger)$$ whereby $$\mathbb{E}[j(\Pi(\omega))]$$ can of course also be written as $$\mathbb{E}[j(\Pi(\omega))] = \int_{\Omega} j(\Pi(\omega))(\tilde{\omega}) d\mathbb{P}(\tilde{\omega}) .$$

How can I prove, that $$(\dagger)$$ is the case? I have tried, tracking the definition of conditional expectation and using Fubini, but with little success so far.

$$\def\om{\omega}$$ $$\def\Om{\Omega}$$ $$\def\bR{\mathbb{R}}$$ $$\def\si{\sigma}$$ $$\def\cB{\mathcal{B}}$$ $$\def\cF{\mathcal{F}}$$

My original answer (below) contains an error, since $$\Phi$$ is not necessarily measurable. In fact, that original proof sketch does not use the fact that $$g$$ is a measurable stochastic process, only that it is a stochastic process. Right now, I cannot see a way to fix this without adding additional assumptions on $$g$$. In fact, I do not believe it is true without additional assumptions.

Let $$\Om=[0,1]$$ with $$\cF$$ the Lebesgue $$\si$$-algebra and $$P$$ Lebesgue measure. Let $$D=[0,1]$$. Let $$G(\om,t)=1_{\{\om=t\}}$$ and $$\Pi(\om)=\om$$. For fixed $$t\in D$$, we have $$G(t)=0$$ a.s., so the random variable $$G(t)$$ is independent of everything, and $$h(t):=E[G(t)]=0$$ for all $$t$$. On the other hand, $$G(\Pi)=1$$ a.s. So $$G(\Pi)$$ is independent of everything, which gives $$E[G(\Pi)\mid\Pi]=E[G(\Pi)]=1.$$

First, let me point out a small confusion in notation. Under normal usage, $$E[j(\Pi)] = \int j(\Pi(\omega))(\omega)\,dP(\omega),$$ without any tildes, which is of course not what you want. One way of carefully notating what you intend is to say that $$E[H\mid\Pi]=h(\Pi)$$, where $$h(\pi)=E[j(\pi)]$$.

This is indeed the correct answer. Heuristically, $$g$$ and $$\Pi$$ are independent, so in the conditional expectation, you can treat $$\Pi$$ like a constant and just use the ordinary expectation. For a rigorous formulation of this, you can do the following.

First, we may regard $$g$$ as a function from $$\Omega$$ to $$\mathbb{R}^D$$, the set of functions from $$D$$ to $$\mathbb{R}$$, with $$g(\omega)(\pi)=g(\pi,\omega)$$. With this identification, it follows that $$g$$ is $$\mathcal{G}/\mathcal{B}(\mathbb{R})^D$$-measurable. Here $$\mathcal{B}(\mathbb{R})^D=\bigotimes_{\pi\in D}\mathcal{B}(\mathbb{R})$$ is the product $$\sigma$$-algebra.

Next, show that since $$j(\pi)$$ and $$\Pi$$ are independent for all $$\pi\in D$$, it follows that $$g$$ and $$\Pi$$ are independent. (The $$\pi$$-$$\lambda$$ theorem should do the trick here.)

Now define $$\Phi:\mathbb{R}^D\times D\to\mathbb{R}$$ by $$\Phi(f,\pi)=f(\pi)$$, so that $$H=\Phi(g,\Pi)$$, and verify that $$\Phi$$ is $$(\mathcal{B}(\mathbb{R})^D \otimes \mathcal{B}(D))/\mathcal{B}(\mathbb{R})$$-measurable.

Finally, use the following.

Theorem. Let $$(\Omega,\mathcal{F},P)$$ be a probability space and $$(S,\mathcal{S})$$ a measurable space. Let $$X$$ be an $$S$$-valued random variable, $$\mathcal{G}\subset\mathcal{F}$$ a $$\sigma$$-algebra, and suppose $$X$$ and $$\mathcal{G}$$ are independent. Let $$(T,\mathcal{T})$$ be a measurable space and $$Y$$ a $$T$$-valued random variable. Let $$f:S\times T\to\mathbb{R}$$ be $$(\mathcal{S}\otimes\mathcal{T},\mathcal{B}(\mathbb{R}))$$-measurable with $$E|f(X,Y)|<\infty$$. If $$Y$$ is $$\mathcal{G}/\mathcal{T}$$-measurable, then $$E[f(X,Y) \mid \mathcal{G}] = \int_S f(x,Y)\,\mu(dx) \quad\text{a.s.},$$ where $$\mu$$ is the distribution of $$X$$.

This theorem is a special case of Theorem 6.66 in these notes: http://math.swansonsite.com/19s6245notes.pdf.

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