Conditional Expectation Relative to “Random Time” - Consistency of the Substitution Rule

I am thinking of the following situation:

On a probability space $\left( \Omega, \mathscr{F}, \cal{P} \right)$ with arbitrary structure, suppose we are given a random function (as it is called in the Stochastic Programming literature) $g: \Omega \times \mathbb{R}^N \rightarrow \mathbb{R}$, which is additionally jointly measurable relative to the product $\mathscr{F} \times \mathscr{B}\left( \mathbb{R}^N \right)$.

Consider a sub $\sigma$-field $\mathscr{Y} \subset \mathscr{F}$. We are also given another random element $X: \Omega \rightarrow \mathbb{R}^N$, measurable relative to $\mathscr{Y}$.

Then, we know that $g\left(\cdot, X \right)$ is an $\mathscr{F}$-measurable random variable.

Now, suppose additionally that $\mathbb{E} \left[ g\left( \cdot, X \right) \right]$ exists.

Under this general setting, can I assert that the conditional expectation $\mathbb{E} \left[ g\left( \cdot, X \right) | \mathscr{Y} \right]$ obeys the substitution rule, that is, is it true that

$$\mathbb{E} \left[ g\left( \cdot, X \right) | \mathscr{Y} \right] \left( \omega \right) = \mathbb{E} \left[ g\left( \cdot, X \left( \omega \right) \right) | \mathscr{Y} \right] \left( \omega \right) \quad\quad?$$

What I mean with this equation is that, if I want to evaluate this conditional expectation, I can do that first by fixing $X$, taking it as a parameter through the evaluation, and let it vary again in $\omega$ after the result has been determined.

For simplicity, we may also assume that $\mathscr{Y}$ is the $\sigma$-field generated by another random element $Y:\Omega \rightarrow \mathbb{R}^M$.

My guess is that, because $\left( \Omega, \mathscr{F}, \cal{P} \right)$ is "non-standard", further conditions on the structure of $g$ should be imposed, in order to assert that the substitution rule holds in this case.

Thanks!

EDIT: So, the answer is provided in the discussion below. The result holds generally; no conditions need to be imposed. The result may be proved by invoking Dynkin’s Multiplicative System Theorem and, in particular, Corollary 8.3 in these notes. See also Exercise 14.7 also in these notes, for a very related result, which makes use of the aforementioned Corollary.

Many Thanks to Nate Eldredge for pointing this out!

• I think you could prove it with the multiplicative system lemma, since it's true when $g$ is of the form $g(\omega, x) = g_1(\omega) g_2(x)$. – Nate Eldredge Oct 19 '16 at 16:52
• I was not aware of this result. Thanks for pointing out, I will take a look to see how it can be applied to this problem. – underpi Oct 19 '16 at 17:08
• However, this seems to prove the result only for bounded measurable functions, except if I am missing something... – underpi Oct 19 '16 at 21:38
• Once you have bounded functions, you can get nonnegative functions by truncation and conditional monotone convergence, then get general integrable functions by taking positive and negative parts. The "standard mantra". – Nate Eldredge Oct 19 '16 at 21:43
• Yes, that is what I meant by "integrable", i.e. integrable with respect to the measure $\mathcal{P}$. – Nate Eldredge Oct 19 '16 at 22:15

I have taken my time to work out a proof based on Dynkin's Multiplicative System Theorem, in the fashion on Exercise 14.7 in these notes. From the discussion that follows, I think that, unfortunately, the substitution rule cannot be proved for the general case, where the base probability space $\left( \Omega,\mathscr{F},\cal{P} \right)$ is arbitrary.

Here is why:

In order to use the Dynkin's Multiplicative System Theorem, we have to show that the class

$$\mathbb{F}\triangleq \left\{ g\in \left[ \mathscr{F}\times\mathscr{B}\left( \mathbb{R}^N\right) \right]_b \left| \mathbb{E} \left[ g\left( \cdot, X \right) | \mathscr{Y} \right] \left( \omega \right) = \mathbb{E} \left[ g\left( \cdot, X \left( \omega \right) \right) | \mathscr{Y} \right] \left( \omega \right), \quad \cal{P}-a.e. \right. \right\},$$

where $\left[ \mathscr{F}\times\mathscr{B}\left( \mathbb{R}^N\right) \right]_b$ denotes the set of all bounded, jointly $\mathscr{F}\times\mathscr{B}\left( \mathbb{R}^N\right)$-measurable functions, is closed under bounded convergence, which means that, if $\left\{ g_n \in \mathbb{F}\right\}_{n\in\mathbb{N}}$ is a sequence of functions, such that

$$\sup_{\left( \omega, x, n \right) \in \Omega \times \mathbb{R}^N \times \mathbb{N}} \left|g_n\left( \omega, x \right)\right|=M<\infty,$$

then

$$g_n \left( \omega,x \right) \underset{n\rightarrow\infty}{\longrightarrow} g\left( \omega,x \right),\quad\forall\omega,x\in\Omega\times\mathbb{R}^N \quad\Rightarrow \quad g\in\mathbb{F}.$$

So, suppose that

$$g_n \left( \omega,x \right) \underset{n\rightarrow\infty}{\longrightarrow} g\left( \omega,x \right),\quad\forall\omega,x\in\Omega\times\mathbb{R}^N.$$

Since the convergence is pointwise, for every random element $X:\Omega \rightarrow\mathbb{R}^N$, it must be true that

$$g_n \left( \omega,X\left(\omega\right) \right) \underset{n\rightarrow\infty}{\longrightarrow} g\left( \omega,X\left(\omega\right) \right),\quad\forall\omega\in\Omega.$$

Now, since $g_n\in\mathbb{F}$, for all $n\in\mathbb{N}$, the desired property holds, that is,

$$\mathbb{E} \left[ g_n\left( \cdot, X \right) | \mathscr{Y} \right] \left( \omega \right) = \mathbb{E} \left[ g_n\left( \cdot, X \left( \omega \right) \right) | \mathscr{Y} \right] \left( \omega \right)\equiv \mathbb{E} \left[ g_n\left( \cdot, x \right) | \mathscr{Y} \right] \left( \omega \right)|_{x=X\left( \omega \right)},\quad\forall\omega\in\Omega_{\Pi_1},$$

where ${\cal{P}} \left( \Omega_{\Pi_1} \right)=1$, with $\Omega_{\Pi_1}$ being independent of $n$. Note that such an $\Omega_{\Pi_1}$ exists, since $\mathbb{N}$ is countable.

We would like to show that the desired property holds also for the limit $g$. On the one hand, since everything is bounded and $\cal{P}$ is finite, invoking the Dominated Convergence Theorem for conditional expectations, we get

$$\mathbb{E} \left[ g_n\left( \cdot, X \right) | \mathscr{Y} \right] \left( \omega \right) \underset{n\rightarrow\infty}{\longrightarrow} \mathbb{E} \left[ g\left( \cdot, X \right) | \mathscr{Y} \right] \left( \omega \right),\quad\forall\omega\in\Omega_{\Pi_2},$$

where ${\cal{P}} \left( \Omega_{\Pi_1} \right)=1$. This in turn implies that

$$\mathbb{E} \left[ g_n\left( \cdot, x \right) | \mathscr{Y} \right] \left( \omega \right)|_{x=X\left( \omega \right)} \underset{n\rightarrow\infty}{\longrightarrow} \mathbb{E} \left[ g\left( \cdot, X \right) | \mathscr{Y} \right] \left( \omega \right),\quad\forall\omega\in\Omega_{\Pi_1}\cap\Omega_{\Pi_2}\triangleq\Omega_{\Pi_3},$$

where, of course, ${\cal{P}} \left( \Omega_{\Pi_3} \right)=1$.

And this is the point where the problems start.

Again invoking the Dominated Convergence Theorem for conditional expectations, it is true that, for each fixed $x\in\mathbb{R}^N$, there exists an event $\Omega_x$ with ${\cal{P}} \left( \Omega_x \right)=1$, such that

$$\mathbb{E} \left[ g_n\left( \cdot, x \right) | \mathscr{Y} \right] \left( \omega \right) \underset{n\rightarrow\infty}{\longrightarrow} \mathbb{E} \left[ g\left( \cdot, x \right) | \mathscr{Y} \right] \left( \omega \right),\quad\forall\omega\in\Omega_x,$$

But $\mathbb{R}^N$ is uncountable! Therefore, in general, we may not be able to find a global, independent of the particular $x$ event of full measure, such that the above holds. As a result, the above cannot be guaranteed to hold if $x$ is replaced by $X\left(\omega\right)$.

Consequently, under this general setting, the class $\mathbb{F}$ cannot be guaranteed to be closed under bounded convergence. Maybe further assumptions are needed.

I really hope I am wrong...

But my conclusion is reinforced by Jason Swanson's Remark 3.11, in these other notes.

Now, we now that the result is true if a regular conditional distribution of $\omega\triangleq Z\left(\omega\right) | \mathscr{Y}$ exists.

But existence of such a regular conditional distribution cannot be guaranteed in general, when the structure of $\left( \Omega,\mathscr{F},\cal{P} \right)$ is arbitrary.