# Conditional expectation of random variable on product space.

First one definition:

Conditional Expectation For $$X\in L^1(\Omega, \mathcal{F}, \mathbb{P})$$. Let $$\mathcal{A} \subseteq \mathcal{F}$$ be a $$\sigma$$-Algebra. Then we define $$\mathbb{E}(X| \mathcal{A})$$ as the unique random variable in $$L^1(\Omega, \mathcal{A}, \mathbb{P})$$ s.t. for all $$\mathcal{A}$$-measurable, bounded random variables $$Z$$ we have $$\mathbb{E}(ZX)=\mathbb{E}(Z \mathbb{E}(X| \mathcal{A}))$$.

I now have to consider $$\mathcal{A} \subseteq \mathcal{F}$$ two $$\sigma$$-Algebras and two random variables $$X: (\Omega, \mathcal{F}) \rightarrow E$$ and $$Y: (\Omega, \mathcal{F}) \rightarrow F$$. With $$X$$ independent from $$\mathcal{A}$$ and $$Y$$ being $$\mathcal{A}$$ measurable. Then for any measurable $$g: E \times F \rightarrow \mathbb{R}_+$$ I need to show:

$$\mathbb{E}(g(X,Y)|\mathcal{A})=h(Y):=\left[ \omega \mapsto \mathbb{E}(g(X,Y(\omega))) \right]$$

Now first of all I dont understand how $$g(X,Y)$$ is defined, do we mean:

$$g(X,Y): \Omega \rightarrow \mathbb{R}_+, \ \omega \mapsto g(X(\omega), Y(\omega))$$

Or is it:

$$g(X,Y): \Omega \times \Omega \rightarrow \mathbb{R}_+, \ (\omega, \nu) \mapsto g(X(\omega), Y(\nu))$$

My first instinct was the first meaning but then the equality we need to show wouldn’t make sense. So it seems to me that I need to show:

$$\int_{\Omega} g(X(\omega), Y(\omega)) Z(\omega) d \mathbb{P}( \omega)$$ is equal to: $$\int_{\Omega} \int_{\Omega}g(X(\nu), Y(\omega)) d \mathbb{P}(\nu) Z(\omega) d \mathbb{P}( \omega)$$

But I dont see at all how this fits together with my official solution:

I‘d be very grateful if someone could explain this solution to me.

• $X$ and $Y$ are random variables on the probability space $(\Omega,\mathcal F,P)$. The random variable $h(Y)$ is defined as the composition of $y\mapsto E[g(X,y)]$ with $Y$. Commented Dec 12, 2023 at 23:07
• But didn‘t I write exactly that? I‘m a bit confused. Commented Dec 12, 2023 at 23:33

The expression $$g(X,Y)$$ is defined over one copy of $$\Omega$$, so it's the random variable $$(g(X,Y))(\omega)=g(X(\omega),Y(\omega))$$.
However, the expression $$g(X,Y(\omega))$$ is defined differently. Here, the $$\omega$$ is the "fixed" omega in the preceding notation "$$\omega \mapsto$$", and this is defining a new random variable for this fixed $$\omega$$ that's "random" in another $$\omega'$$, so $$g(X,Y(\omega)) = g(X(\omega'),Y(\omega))$$ with the expectation integrating over $$\omega'$$ not the fixed $$\omega$$.
Yes, it's horrifying, but you're being asked to show that the random variable $$\mathbb{E}(g(X,Y)|\mathcal{A})$$ which can be defined as the unique $$\mathcal{A}$$-measurable and $$L_1$$ random variable $$U$$ satisfying: $$\int_\Omega Z(\omega)g(X(\omega),Y(\omega)) d\mathbb{P}(\omega) = \int_\Omega Z(\omega)U(\omega) d\mathbb{P}(\omega)$$ for all $$\mathcal{A}$$-measurable, bounded $$Z$$ is given by the expression: $$U(\omega) = h(Y(\omega)), \forall\omega\in\Omega\qquad (*)$$ where $$h(y)$$ is defined by: $$h(y) := \int_\Omega g(X(\omega),y) d\mathbb{P}(\omega)$$ or equivalently: $$h(Y(\omega)) := \int_\Omega g(X(\omega'),Y(\omega))d\mathbb{P}(\omega')$$ It is unfortunate that the problem tried to "define" $$h(Y)$$ instead of just defining $$h(y)$$ which would have made things much clearer.
Anyway, the official solution shows that: $$\mathbb{E}[g(X,Y)Z] \equiv \int_\Omega Z(\omega)g(X(\omega),Y(\omega)) d\mathbb{P}(\omega)$$ is equal to: $$\mathbb{E}[h(Y)Z] \equiv \int_\Omega Z(\omega)h(Y(\omega))d\mathbb{P}(\omega)$$ In the process, this shows that $$h(Y(\omega))$$ is the $$U(\omega)$$ we're looking for and establishes (*) by uniqueness.
• Thank you very much for your answer, this now makes sense! Could you maybe help me a bit and show me how the integral on the first line of the official solution is derived from your definition of $\mathbb{E}(g(X,Y)Z)$? I dont get why we integrate over „three dimensions“ all of a sudden. Commented Dec 13, 2023 at 0:30