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:

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

  • 1
    $\begingroup$ $X$ 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 $Y$. $\endgroup$
    – Andrew
    Commented Dec 12, 2023 at 23:07
  • $\begingroup$ But didn‘t I write exactly that? I‘m a bit confused. $\endgroup$
    – Henry T.
    Commented Dec 12, 2023 at 23:33

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Henry T.
    Commented Dec 13, 2023 at 0:30

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