# Conditional expectation and product measure space

Let $$X$$ and $$Y$$ be $$\mathbb{R}^n$$- and $$\mathbb{R^m}$$-valued random variables on the probability space $$(\Omega, \mathcal{F}, P)$$. Further assume that $$X$$ is $$\mathcal{F}_1$$-measurable and $$Y$$ is $$\mathcal{F}_2$$-measurable, where $$\mathcal{F}_1$$ and $$\mathcal{F}_2$$ are independent sub-$$\sigma$$-algebras of $$\mathcal{F}$$.

Now, for $$A \in \mathcal{B}(\mathbb{R}^n)$$ and $$B \in \mathcal{B}(\mathbb{R}^m)$$ consider the maps \begin{align} f(x, y) &= 1_{A \times B} (x, y), \\ f( x, Y) &= 1_{A \times B} (x, Y), \\ g(x) &= E[f(x, Y )] = E[1_{A \times B} (x, Y)], \\ g(X) &= E[f(x, Y)] \big|_{x = X} = E[1_{A \times B} (x, Y)] \big|_{x = X}. \end{align} Suppose I know that $$E[1_{A \times B}(X, Y) | \mathcal{F}_1] = E[1_{A \times B} (x, Y)] \big|_{x = X}.$$ Then by the definition of the conditional expectation this means that for any $$F \in \mathcal{F}_1$$ $$\tag{1} \int_{F} 1_{A \times B}(X, Y) dP = \int_F E[1_{A \times B} (x, Y)] \big|_{x = X} dP.$$ I want to show that we also have $$\tag{2} \int_{F} h(X, Y) dP = \int_F E[h (x, Y)] \big|_{x = X} dP$$ for all $$\mathcal{B}( \mathbb{R^n} ) \otimes \mathcal{B}( \mathbb{R^m} )$$-measurable positive step functions $$h$$. The latter of course will follow by linearity if we can show that $$\tag{3} \int_{F} 1_D (X, Y) dP = \int_F E[1_D (x, Y)] \big|_{x = X} dP$$ for any $$D \in \mathcal{B}( \mathbb{R^n} ) \otimes \mathcal{B}( \mathbb{R^m} )$$.

An argument I have come across says:

Both sides of $$(1)$$ can be extended from $$\mathcal{B}( \mathbb{R^n} ) \times \mathcal{B}( \mathbb{R^m} )$$ to define measures on $$\mathcal{B}( \mathbb{R^n} ) \otimes \mathcal{B}( \mathbb{R^m} )$$. By linearity $$(1)$$ becomes $$(2)$$.

How exactly should one understand this argument? For instance, if we take $$F = \Omega$$, the LHS of $$(1)$$ is $$P ( (X, Y ) \in A \times B )$$. The independence of $$X$$ and $$Y$$ then gives $$P ( (X, Y ) \in A \times B ) = P ( X \in A) P ( X \in B),$$ and for the distribution functions we have $$P_{( X, Y)} ( A \times B ) = P_X ( A ) \times P_Y ( B )$$, where $$P_{(X, Y)}$$ is a probability measure on $$\mathcal{B}( \mathbb{R^n} ) \otimes \mathcal{B}( \mathbb{R^m} )$$. Does this have any connection with the quoted argument? What about the RHS of $$(1)$$. And how does one obtain $$(2)$$ or $$3$$? A detailed demonstration would be very much appreciated.

1 to 3 is a standard $$\pi$$-$$\lambda$$ lemma argument. The collection $$\mathcal{P}$$ of sets of the form $$A \times B$$ is certainly closed under intersection, i.e. is a $$\pi$$-system. Now show that the collection $$\mathcal{L}$$ of all sets $$D$$ satisfying (3) is a $$\lambda$$-system. (The monotone convergence theorem will be useful.) You conclude that $$\mathcal{L}$$ contains $$\sigma(\mathcal{P})$$, which by definition equals $$\mathcal{B}( \mathbb{R^n} ) \otimes \mathcal{B}( \mathbb{R^m} )$$, and so you have shown that (3) holds for all $$D \in \mathcal{B}( \mathbb{R^n} ) \otimes \mathcal{B}( \mathbb{R^m} )$$.

It can also be done with the monotone class theorem.

• Thanks for the answer. I tried following these steps and posted an answer below. Could you please take a look to see if something could be improved? Nov 10, 2019 at 1:15
• I have also posted a follow-up question, in case you would be interested: math.stackexchange.com/questions/3429193/… Nov 10, 2019 at 2:53

Following the steps described in Nate Eldrege's answer:

Let $$\mathcal{L}$$ denote the family of sets $$D \in \mathcal{B}( \mathbb{R^n} ) \otimes \mathcal{B}( \mathbb{R^m})$$ for which we have

$$\int_{F} 1_D (X, Y) dP = \int_F E[1_D (x, Y)] \big|_{x = X} dP, \quad \text{for all } F \in \mathcal{F}_1.$$

According to $$(1)$$, the family $$\mathcal{P} := \{ A \times B : A \in \mathcal{B}( \mathbb{R^n} ), \, B \in\mathcal{B}( \mathbb{R^m} ) \} \subset \mathcal{L}$$. Moreover, $$\mathcal{P}$$ is intersaection-stable ($$\pi$$-system) and $$\sigma(\mathcal{P}) = \mathcal{B}( \mathbb{R^n} ) \otimes \mathcal{B}( \mathbb{R^m})$$.

Let us now show that $$\mathcal{L}$$ is a Dynkin system ($$\lambda$$-system) on $$\mathbb{R}^n \times \mathbb{R}^m$$:

1. $$\mathbb{R}^n \times \mathbb{R}^m \in \mathcal{P} \Rightarrow \mathbb{R}^n \times \mathbb{R}^m \in \mathcal{L}$$.
2. Let $$D_1, D_2 \in \mathcal{L}$$ such that $$D_2 \subset D_1$$. Then $$1_{ \{ D_1 \setminus D_2 \} } = 1_{D_1} - 1_{D_2}$$ and \begin{align*} \int_{F} 1_{ \{ D_2 \setminus D_1 \} } (X, Y) dP &= \int_{F} 1_{D_1} (X, Y) dP - \int_{F} 1_{D_2} (X, Y) dP \\ &= \int_F E[1_{D_1} (X(\omega), Y)] P(d \omega ) - \int_F E[1_{D_2} (X(\omega), Y)] P(d \omega ) \\ &= \int_F \{ E[1_{D_1} (X(\omega), Y)] - E[1_{D_2} (X(\omega), Y)] \}P(d \omega ) \\ &= \int_F \{ E[1_{D_1} (X(\omega), Y) - 1_{D_2} (X(\omega), Y)] \}P(d \omega ) \\ &= \int_F E \left[ 1_{ \{ D_1 \setminus D_2 \} } (X(\omega), Y) \right] P(d \omega ), \end{align*} hence $$( D_1 \setminus D_2 ) \in \mathcal{F}_1$$
3. Let $$D_n \in \mathcal{L}$$, $$n \in \mathbb{N}$$, such that $$D_1 \subset D_2 \subset D_3 \subset \ldots$$ and $$D := \bigcup_{n \in N} D_n$$. Fix some $$x \in \mathbb{R}^n \times \mathbb{R}^m$$. If $$x \notin D$$, then $$x \notin D_n$$ for all $$n \in \mathbb{N}$$. If $$x \in D$$, then there is some $$n_0 \in N$$ such that $$x \in \bigcup_{n=1}^{n_0} D_n = D_{n_0} \subset D$$. In either case, $$1_{D_n} (x) = 1_D (x)$$ for all $$n \geq n_0$$. Therefore, necessarily, $$1_{D_n} (x) \uparrow 1_D (x)$$ for every $$x \in \mathbb{R}^n \times \mathbb{R}^m$$ (pointwise). Similarly, for every $$\omega \in \Omega$$ there is some $$n_0 \in \mathbb{N}$$ $$1_{D_n} ( X ( \omega ), Y ( \omega ) ) = 1_D ( X(\omega), Y(\omega))$$ for all $$n \geq n_0$$. Trivially, $$1_{D_n} ( X ( \omega ), Y ( \omega ) ) \uparrow 1_D ( X(\omega), Y(\omega))$$ for every $$\omega \in \Omega$$ (pointiwse). Using the monotone convergence theorem, we can write \begin{align} \int_{F} 1_D (X, Y) dP &= \int_{F} \lim_{n \rightarrow \infty} 1_{D_n} (X, Y) dP = \lim_{n \rightarrow \infty} \int_{F} 1_{D_n} (X, Y) dP \\ &= \lim_{n \rightarrow \infty} \int_F E[1_{D_n} ( X(\omega), Y)] P(d \omega) = \ldots \end{align}

For $$x = X(\omega)$$, $$\omega \in \Omega$$ fixed, using the continuity of the measure $$P$$ from below, we can write \begin{align} \lim_{n \rightarrow \infty} E[1_{D_n} (x, Y)] &= \lim_{ n \rightarrow \infty} P ( \{ \omega' \in \Omega: ( x, \omega' ) \in D_n \} ) = P \left( \bigcup_{n \in N} \{ \omega' \in \Omega: ( x, \omega' ) \in D_n \} \right) \\ &= P ( \{ \omega' \in \Omega: ( x, \omega' ) \in D \} ) = E[1_D (x, Y)]. \end{align} Thus, $$E[1_{D_n} (x, Y)] \big|_{x = X(\omega)} \uparrow E[1_D (x, Y)] |_{x = X(\omega)} \quad \text{for every } \omega \in \Omega \text{ (pointwise)}.$$ Applying the monotone convergence theorem once again, we get $$\ldots = \int_F \lim_{n \rightarrow \infty} E[1_{D_n} ( X(\omega), Y)] P(d \omega) = \int_F E[1_D ( X(\omega), Y)] P(d \omega).$$ This shows that $$\mathcal{L}$$ is a Dynkin system. By Dynkin's lemma $$\mathcal{L} = \sigma(\mathcal{P}) = \mathcal{B}( \mathbb{R^n} ) \otimes \mathcal{B}( \mathbb{R^m})$$ and thus $$(3)$$ hols for all $$D \in \mathcal{B}( \mathbb{R^n} ) \otimes \mathcal{B}( \mathbb{R^m})$$.

• Looks good to me! Nov 10, 2019 at 9:29