A question about conditional expectations and sub-sigma fields

Question

Let $\sigma$ be a probability measure over $X\times Y$ and let $f:X\to\mathbb R$ and $g:Y\to\mathbb R$ respectively be measurable functions of $x$ and $y$. We define $\tilde f : (x,y)\mapsto f(x)\mathbf 1_{Y}(y) $ and $\tilde g : (x,y)\mapsto \mathbf 1_{X}(x)g(y)$ as the "extensions" of $f$ and $g$ to the product space $X\times Y$.

Let $\mathcal F$ and $\mathcal G$ be $\sigma$-algebras over $X \times Y$ such that $\mathcal F \cap \mathcal G = \{\emptyset, X \times Y\}$. Is it true that $$\int E[\tilde f\mid\mathcal F](x,y)\ E[\tilde g\mid\mathcal G](x,y) \sigma(dx,dy) = \int f(x) g(y) \sigma(dx,dy)$$?

Answer 1

Here is how you can get to the desired conclusion (I let you fill in the details) :

• First, prove that $E[\tilde f\mid\mathcal F] $ and $E[\tilde g\mid\mathcal G]$ are independent, which is a clear consequence of your assumptions on $\mathcal F$ and $\mathcal G$, and the definition of conditional expectation.
• Then, use independence to write the integral on the left hand side as a product of two integrals.
• Lastly, apply Fubini's theorem (add integrability assumptions as required) and conclude.

Answer 2

Here is a shot at a counterexample: Consider $ X \times Y = \{0,1\}^2$ and $\mathcal F = \mathcal G = \{ \emptyset,X \times Y\}$. Let $σ = \frac 1 2 (δ_{(0,0)} + δ_{(1,1)}) $. Since $ \mathcal F$ and $\mathcal G$ are trivial, $ \mathbb E[\tilde f | \mathcal F] = \mathbb E[\tilde f] = \frac 1 2( f(0) + f(1)) $ and similarly for $g$. Consequently, $$ \int \mathbb E[\tilde f\mid\mathcal F](x,y)\ \mathbb E[\tilde g\mid\mathcal G](x,y) \sigma(dx,dy) = \mathbb E[\tilde f] \mathbb E[\tilde g] = \frac 1 4 (f(0)+f(1))(g(0)+g(1)). $$ Taking $ f = g$, this simplifies to $ \frac 1 4 (f(0)+ f(1))^2$. On the other hand, $$ \int f(x) g(y) \sigma(dx,dy) = \frac 1 2 (f(0)^2 + f(1)^2). $$ In general, these are not the same (e.g. with $f(0) = -f(1) = 1$).

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This counterexample works, since $σ$ is not a product measure. If it were, and $\mathbb E[\tilde f\mid\mathcal F] $ and $ \mathbb E[\tilde g\mid\mathcal G]$ are independent (I am unsure), the claim holds, as pointed out in the other answer, by factorization of expectation and Fubini's theorem.