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Suppose that we have a probability space $(\Omega, \mathcal{F}, P)$. Let $X,Y$ be real-valued random variables defined on this space, and let $\mathcal{H} \subset \mathcal{F}$ be a sub-sigma-algebra.

Suppose $X$ is $\mathcal{H}$-measurable (i.e, $X^{-1}(B) \in \mathcal{H}$ for all Borel $B \subset \mathbb{R}$).

Also suppose that $Y$ is independent of $\mathcal{H}$ (which implies that $X$ and $Y$ are independent).

Then is it true that for any Borel-measurable function $g: \mathbb{R}^2 \to \mathbb{R}$, we have that $\mathbb{E}(g(X,Y)|\mathcal{H})=\mathbb{E}(g(X,Y)|X)$?

Observations: It seems to be true for functions of the form $g(x,y)=f(x)h(y)$, because $\mathbb{E}(f(X)h(Y)|\mathcal{H}) = \mathbb{E}(h(Y)) \cdot f(X)=\mathbb{E}(f(X)h(Y)|X)$ by independence of $Y$ to $X$ and $\mathcal{H}$. But I can't seem to prove it for the general case. Maybe we can approximate arbitrary $g$ by functions of this form from below, and then use MCT?

Is it possible to show that any measurable $g: \mathbb{R}^2 \to \mathbb{R}$ can be written as the (upward) limit of linear combinations of functions of the form $\chi_{A \times B}$, for Borel $A,B \subset \mathbb{R}$? Because then we can just apply MCT and use the preceding comment, and we're done. (The $\chi_E$ denotes characteristic function of $E$.)

shalop
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  • Actually, we don't have to assume that $X,Y$ are real valued. They can take values in any measurable space $(E, \mathcal{M})$, and then $g$ would map $E^2 \to \mathbb{R}$ and the result would still hold... – shalop May 14 '15 at 18:19

2 Answers2

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Hints

  1. Show that the claim holds for $g(x,y) = 1_B(x) 1_C(y)$ where $B,C \in \mathcal{B}(\mathbb{R})$ are Borel sets.
  2. Show that $$\mathcal{D} := \{D \in \mathcal{B}(\mathbb{R}^2); \text{claim holds for} \, g(x,y) = 1_D(x,y)\}$$ is a Dynkin system. Conclude that $\mathcal{D} = \mathcal{B}(\mathbb{R}^2)$.
  3. Use Beppo Levi to extend the statement from simple functions (aka elementary functions) to non-negative measurable functions $g$.
  4. Conclude.
saz
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  • Oh, I posted the same exact solution as you were posting yours. I think that I just get confused when dealing with Dynkin systems and pi-systems, etc. Just so much to remember. But I'm glad I got it right! – shalop Mar 18 '15 at 07:41
  • @Shalop Yes, your answer is fine. – saz Mar 18 '15 at 16:06
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I think I figured it out. Can someone please confirm that the proof is correct? Is this the special case of a more general result??

It suffices to prove the claim when $g=\chi_E$, for Borel sets $E \subset \mathbb{R}^2$. The reason is that any $g$ can be written as the monotone limit of linear combinations of such functions, and then we can apply MCT.

First consider the case when $E=A \times B$ for Borel sets $A,B \subset \mathbb{R}$. Then it is true. Indeed, it is easy enough to check directly that if $g=\chi_{A \times B}$, then $g(X,Y) = \chi_{X \in A} \cdot \chi_{Y \in B}$, and so it follows that $\mathbb{E}(g(X,Y)|\mathcal{H})=P(Y \in B) \cdot \chi_{X \in A} = \mathbb{E}(g(X,Y)|X)$.

Let $\mathcal{I}= \{ A \times B : A,B \subset \mathbb{R} \text{ are Borel} \}$. Then $\mathcal{I}$ is a $\pi$-system that generates the Borel sets in $\mathbb{R}^2$. Therefore the Dynkin-system generated by $\mathcal{I}$ is the same as the sigma-algebra generated by $\mathcal{I}$, i.e, the Borel sets in $\mathbb{R}^2$.

Now it is easy enough to check that $\{ F \subset \mathbb{R}^2 : F \text{ is Borel and } \mathbb{E}(\chi_F (X,Y) | \mathcal{H}) = \mathbb{E} (\chi_F(X,Y)|X) \}$ is itself a Dynkin-system containing $\mathcal{I}$, from which the result follows for $g=\chi_F$, and then we extend as stated above...

shalop
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