Conditional expectation: commuting integration and supremum

Question

Let $X$ and $A$ be compact Polish spaces endowed with Borel $\sigma$-algebras. Let $\mathcal{A} = X\times \mathcal{B}(A)$ be the $\sigma$-algebra consisting of cylinders whose projections on $A$ are Borel sets. Let $u: X\times A\rightarrow\mathbb{R}$ be a continuous function. Let $\pi\in \Delta(X\times A)$ be a probability measure.

Let $\mathcal{F}\equiv \{f:A\rightarrow A|\;f \text{ is } \mathcal{A} \text{ measurable}\}$ denote the collection of all measurable functions from $A$ to $A$.

Question: Is it true that $$\sup_{f\in \mathcal{F}}\int_{X\times A} E \big[u\big(x,f(a)\big) \big| \mathcal{A}\big] d\pi(x,a) = \int_{X\times A}\; \sup_{f\in \mathcal{F}}E \big[u\big(x,f(a)\big) \big| \mathcal{A}\big] d\pi(x,a)?$$

An economic interpretation: As Michael Greinecker pointed out in the comment, if we interpret $u$ as the a payoff function, $x\in X$ as an unknown state, $a\in A$ as an action, and $\pi$ as a system of stochastic action recommendations, then the claim I am trying to establish can be interpreted as saying that choosing an optimal contingent plan ex ante leads to the same expected utility as maximizing for each recommended action at the interim stage.

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My thoughts so far:

My first instinct is to invoke the Measurable Selection Theorem, which would be similar to the arguments in Theorem 14.60 of Rockafella and Wets' "Variational Analysis". However, I do not know how to work with the conditional expectations in the expression above, which is itself a random variable that is only unique almost surely.

Specifically, to use the Measurable Selection Theorem as Rockafellar and Wets did, I would need to somehow establish that $$ \Big\{k: E \big[u\big(x,k\big) \big| a\big] \ge c \Big\} $$ is a closed set for each $a\in A$ and $c\in \mathbb{R}$, but I'm not sure why that would be true, especially since conditional expectation is only pinned down for almost all $a\in A$.

Any pointers would be greatly appreciated!

Answer 1

Note: This answer used to be a counterexample that missed the mark.

The way to get around he definitional issues with conditional expectations is to work with regular conditional probabilities in product form, which guarantee that all conditional expectations fit together well. In particular, there exists a measurable function (or a transition probability, essentially the same thing) $\kappa:A\to X$ such that for $\pi_A$ the $A$-marginal of $\pi$, we have for every Borel set $E\subseteq X\times A$ that $$\pi(E)=\int\int 1_E(x,a)~\mathrm d\kappa_a(x)~\mathrm d\pi_A(a).$$ The function $\kappa$ is unique up to $\pi_A$-null sets.

That way, one can show that $$\max_{f\in \mathcal{F}}\int_{X\times A} u\big(x,f(a)\big)~\mathrm d\pi(x,a) =\max_{f\in \mathcal{F}}\int_A \int_X u\big(x,f(a)\big)~\mathrm d\kappa_a(x) ~\mathrm d\pi_A(a)$$ $$= \int_A \max_{f\in \mathcal{F}} \int_X u\big(x,f(a)\big)~\mathrm d\kappa_a(x) ~\mathrm d\pi_A(a).$$ The left side is trivially no larger than the right side. For the other direction, you show that the correspondence that associates to each $a$ the argmax of $\int_X u\big(x,\cdot\big)~\mathrm d\kappa_a(x)$ is measurable with nonempty compact values. So you can use the Kuratowski-Ryll-Nardzewski measurable selection theorem to turn a solution for the problem on the right to a solution of the problem on the left, which must, therefore, give the same value.

The argument does not require $X$ to be compact, any Polish space will do, and $u$ need not be continuous in $X$, any bounded (or integrably bounded) Carathéodory function will do.

Answer 2

ADDENDUM. Vokram told me that I answered another question, not his question. Therefore, I give another counterexample (not so different from the previous one) disproving the equality.

Choose a function $u$ depending only on the second variable, namely $u(x,a) = v(a)$ for all $(x,a) \in X \times A$, so the conditional expectations with regard to $\mathcal{A}$ have no effect on the random variables $U_f : (a,x) \mapsto u(x,f(a)) = v(f(a))$ and $\sum_f U_f$.

We are led to compare $$\sup_{f\in \mathcal{F}} \int_{X\times A} v(f(a)) \; d\pi(x,a) \text{ and } \int_{X\times A}\; \sup_{f\in \mathcal{F}} v(f(a)) \; d\pi(x,a),$$ Calling $\nu$ the second marginal of $\pi$, we compare $$\sup_{f\in \mathcal{F}} \int_{A} v(f(a)) \; d\nu(a) \text{ and } \int_{A} \sup_{f\in \mathcal{F}} v(f(a)) \; d\nu(a).$$

If $A=\mathbb{U}$, unit circle in $\mathbb{C}$, $\nu$ is the Haar measure on $\mathbb{U}$, $\mathcal{F}$ be the family of all rotations on $\mathbb{U}$, and $v : z \mapsto \Re(z)$, then the left-hand side is $0$, whereas the right-and side is $1$ because for all $a \in \mathbb{U}$ $$\sup_{f \in \mathcal F} \Re(f(a)) = |f(a)| = 1.$$

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The notations are not good. For example, when you write $$\int_{X\times A}\; \sup_{f\in \mathcal{F}}E \big[u\big(x,f(a)\big) \big| \mathcal{A}\big] d\pi(x,a),$$ $u(x,f(a))$ in the integral behaves like a constant so $E \big[u\big(x,f(a)\big) \big| \mathcal{A}\big] = u(x,f(a))$.

You should write instead $$\int_{X\times A}\; \sup_{f\in \mathcal{F}} E \big[U_f \big| \mathcal{S}\big] d\pi,$$ where $U_f$ the random variable defined bu $U_f(x,a) = u(x,f(a))$ and $\mathcal{S} = \{\emptyset,X\} \times \mathcal{A}$.

Anyway, the equality cannot be always true. For example, consider $X=A=\mathbb{U}$, unit circle in $\mathbb{C}$. Endow $X \times A = \mathbb{U}^2$ with its Borel $\sigma$-field and $\eta \otimes \eta$, where $\eta$ is the Haar measure on $\mathbb{U}$. Then taking conditional expectation with regard to $\mathcal{S}$ is just taking averages over the fist component.

Let $\mathcal{F}$ be the family of all rotations on $\mathbb{U}$, and $u : (x,y) \mapsto \Re(xy)$. Then for each $f \in \mathcal{F}$, the average over $x$ of $U(x,f(a)) = \Re(xf(a))$ is $0$. Yet, since $$\sup_{f \in \mathcal F} u(x,f(a)) = \sup_{y \in \mathbb U} \Re(xy) = |x| = 1,$$ the average over $x$ of $\sup_{f \in \mathcal F} u(x,f(a))$ is $1$.