# How to prove that $\mathbb{E}[ \varphi(X, Y) | \mathcal{G}] = \psi(Y)$ where $\psi(y) := \mathbb{E}[ \varphi(X, y)]$?

$$\newcommand{\diff}{\mathrm d}$$

I'm trying to prove Proposition 12.4. (given without proof) in this note.

Let $$\mathcal{G}$$ be a sub-$$\sigma$$-field of $$\mathcal{F}$$ and $$X, Y$$ two random variables such that $$X$$ is independent of $$\mathcal{G}$$ and $$Y$$ is $$\mathcal{G}$$-measurable. Let $$\varphi: \mathbb{R}^2 \rightarrow \mathbb{R}$$ be Borel-measurable such that $$\mathbb{E}[|\varphi(X, Y)|] < \infty$$. Then $$\mathbb{E}[ \varphi(X, Y) | \mathcal{G}] = \psi(Y) \quad \text { a.s.} \quad \text{where} \quad \psi(y) := \mathbb{E}[ \varphi(X, y)].$$

In below attempt, I'm stuck at showing $$\int_\mathbb R \left[ \int_\mathbb R \varphi (x,y) \diff \color{blue}{\mu'}(x) \right ] \mathrm d \nu'(y) = \int_\mathbb R \left[ \int_\mathbb R \varphi (x,y) \diff \color{blue}{\mu}(x) \right ] \mathrm d \nu'(y).$$

Could you elaborate on how to finish the proof?

Proof Let $$\mu, \nu$$ be the distributions under $$\mathbb P$$ of $$X, Y$$ respectively. We have $$X, Y$$ are independent and thus the distribution $$\lambda$$ of $$(X, Y)$$ is the product measure of $$\mu$$ and $$\nu$$, i.e., $$\lambda = \mu \otimes \nu$$. By Fubini's theorem, $$\varphi(X, y)$$ is integrable for $$\nu$$-a.e. $$y\in Y$$ and the map $$y \mapsto \mathbb{E}[ \varphi(X, y)]$$ is Borel. Clearly, $$\psi(Y)$$ is $$\mathcal G$$-measurable.

Fix $$A \in \mathcal G$$. Let's prove that $$\int_A \varphi (X, Y) \diff \mathbb P = \int_A \psi(Y) \diff \mathbb P.$$

Let $$\mathcal G'$$ be the sub-$$\sigma$$-algebra of $$\mathcal G$$ induced by $$A$$ and $$\mathbb P'$$ the restriction of $$\mathbb P$$ to $$\mathcal G'$$. We now consider integration w.r.t. $$(A, \mathcal G', \mathbb P')$$. Clearly, $$\int_A \varphi(X, Y) \diff \mathbb P = \int_A \varphi(X, Y) \diff \mathbb P' \quad \text{and} \quad \int_A \psi(Y) \diff \mathbb P = \int_A \psi(Y) \diff \mathbb P'.$$

Notice that $$X,Y$$ are still independent under $$(A, \mathcal G', \mathbb P')$$. Let $$\mu', \nu'$$ be the distributions under $$\mathbb P'$$ of $$X, Y$$ respectively. Then the distribution of $$(X, Y)$$ under $$\mathbb P'$$ is $$\lambda' :=\mu' \otimes \nu'$$. By change of variables formula and Fubini's theorem, $$\int_A \varphi (X, Y) \diff \mathbb P' = \int_{\mathbb R^2} \varphi (x,y) \diff \lambda(x, y) = \int_\mathbb R \int_\mathbb R \varphi (x,y) \diff \mu'(x) \diff \nu'(y).$$

By change of variables formula, $$\int_A \psi(Y) \diff \mathbb P' = \int_\mathbb R \psi(y) \diff \mathbb \nu'(y) = \int_\mathbb R \left[ \int_\Omega \varphi (X,y) \diff \mathbb P \right ] \mathrm d \nu'(y) = \int_\mathbb R \left[ \int_\mathbb R \varphi (x,y) \diff \mu(x) \right ] \mathrm d \nu'(y).$$

You are over complicating things. It is much simpler to use a monotone class argument, but if you insist on using this approach here's how you can do it.

Define on $$\mathcal{F}$$ the new measure $$\mathbb{P}'$$ given by $$\mathbb{P}'(E) = \frac{\mathbb{P}(A \cap E)}{\mathbb{P}(A)}$$, and note that it can be shown $$\frac{\mathbb{E}(Z1_A)}{\mathbb{P}(A) } = \mathbb{E'}(Z) \$$ for all random variables $$Z$$. Then, as $$Y$$ is $$\mathcal{G}$$ measurable, $$X$$ and $$Y$$ are also independent under $$\mathbb{P}'$$.

It follows then :

$$\mathbb{E}(\varphi(X,Y)1_A ) \\ = \mathbb{P}(A) \mathbb{E'}(\varphi(X,Y)) = \mathbb{P}(A) \int_\mathbb R \int_\mathbb R \varphi (x,y) d\mu'(x) d\nu'(y) = \mathbb{P}(A) \int_\mathbb R \underbrace{\left(\int_\mathbb R \varphi (x,y) d\mu'(x)\right)}_{ = \frac{\mathbb{E}(\varphi(X,y)1_A)}{\mathbb{P}(A) }\ = \ \psi(y) } d\nu'(y) = \mathbb{P}(A) \mathbb{E'}(\psi(Y)) = \mathbb{E}(\psi(Y)1_A)$$

• I got your argument. Could you elaborate on a proof using monotone class argument? Commented Jan 15, 2023 at 11:08
• Sure. Let $A \in \mathcal{G}$ and $V$ be the set of measurable maps $\varphi :\mathbb{R}^2 \to \mathbb{R}$ with the property $\mathbb{E}(\varphi(X,Y)1_{A}) = \mathbb{E}(\psi(Y)1_A)$. Then $V$ is a vector space; contains indicators of cylinder sets (independence) and is closed under increasing limits of non-negative functions, and so contains all non-negative measurable $\varphi$. Commented Jan 15, 2023 at 11:17
• This is elegant. Thank you so much for your help! Commented Jan 15, 2023 at 11:19

For any $$A\in \mathcal G$$ we show that $$E[1_A \varphi(X,Y)]=E[1_A \psi(Y)]$$. This will establish that $$E[\varphi(X,Y)|\mathcal G]=\psi(Y)$$ a.s.

Let $$Z:(\Omega,\mathcal F)\to (\Omega,\mathcal G)$$, $$\omega\mapsto \omega$$ and note that $$\mathcal G = \sigma(Z)$$. Since $$Y$$ is $$\mathcal G$$-measurable, by the Doob–Dynkin lemma, there is some measurable $$g:(\Omega,\mathcal G)\to (\mathbb R, \mathcal B(\mathbb R))$$ such that $$Y = g(Z)$$.

Note that $$A=Z^{-1}(A)$$ thus

\begin{align} E[1_A \psi(Y)] &= E[1_{A}(Z) \psi(g(Z))] \\ &= \int_\Omega 1_A(z) \Big(\int_{\mathbb R} \varphi\big(x,g(z)\big) dP_X(x) \Big)dP_Z(z) \tag {1}\\ &= \int_{\mathbb R \times \Omega} 1_A(z) \varphi\big(x,g(z)\big) d(P_{X}\otimes P_{Z})(x,z) \tag {2}\\ &= \int_{\mathbb R \times \Omega} 1_A(z) \varphi\big(x,g(z)\big) dP_{(X,Z)}(x,z) \tag {3}\\ &= E[1_A(Z)\varphi\big(X,g(Z) \big)] \tag{4}\\ &= E[1_A\varphi\big(X,Y \big)] \end{align}

$$(1)$$: Law of the unconscious statistician (or integration w.r.t. pushforward measure) and $$\psi:y\mapsto \int_{\mathbb R} \varphi(x,y) dP_X(x)$$
$$(2)$$: Fubini's theorem
$$(3)$$: $$X$$ is independent of $$\mathcal G$$, hence $$X$$ and $$Z$$ are independent
$$(4)$$: Law of the unconscious statistician (or integration w.r.t. pushforward measure)