# What does it mean "a conditional expectation given a stochastic process"?

Let $$(\Omega,\mathscr{F},P)$$ be a probability space.

Let $$X,Y$$ be random variables on $$\Omega$$.

Then, we say $$Z\sim X|Y$$ iff (i) $$\int_{Y^{-1}(A)} X dP = \int_{Y^{-1}(A)} Z dP$$ and (ii) $$Z$$ is $$\sigma(Y)$$-measurable.

Now, let $$S:\mathscr{\mathbb{R}} \times \Omega\rightarrow \mathbb{R}$$ be a stochastic process.

What does it mean by $$X|S$$?

There are numerous papers saying like "... because $$X|S \sim S$$, $$P(X\in A|S)= S(A)...$$.

I think this is NOT actually a conditional expectation, but it is just a way to denote De Finneti theorem. Isn't it?

Note that $$S$$ can be seen as a measurable map $$\Omega\rightarrow \prod_{A\in \mathscr{B}_{\mathbb{R}}} \mathbb{R}$$. If the definition $$X|S$$ is consistent with the standard conditional expectation definition, $$X|S$$ is a random variable taking values in $$\mathbb{R}$$, same as $$X$$. However, since $$X|S\sim S$$, $$X|S$$ must take values in $$\prod_{A\in \mathscr{B}_{\mathbb{R}}} \mathbb{R}$$. Do you see inconsistency here?

This makes me confusing, so I am curious what's the definition of $$X|S$$.

What does $$X|S$$ mean?

** EDIT **

Here's the usage of this in "Theory of statistics - Mark J. Schervish" As you can see here, the author says "$$X_n$$'s are independent and identically distributed as $$P$$ conditional on $$\mathbb{P}=P$$."

This means that $$X|\mathbb{P}\sim \mathbb{P}$$, which I do not get how to formally define it.

And Last EDIT The author says that it is a fact that $$P(X\in A|\mathbb{P}=P)=P(A)$$. So there must be another definition the author is referring to..

– whuber
Oct 14, 2019 at 13:49
• @whuber I know the general sigma algebra definition of conditional expectation. However, as I wrote in my post, saying "$X|S\sim S$" is inconsistent with that standard definition. Oct 14, 2019 at 13:51
• Can you give us an accessible reference to investigate the context of such usage?
– whuber
Oct 14, 2019 at 13:55
• It's conditional on a given probability measure P. For instance, suppose B is [0,1], and your P is $\mathcal N(0, 1)$, then $Pr(X_1\in [0,1])=\frac 1 {2\pi}\int_0^1 e^{-x^2/2}dx$ Oct 14, 2019 at 14:22
• Pollard's A User's Guide to Measure Theoretic Probability has a good coverage of disintegrations and regular conditional distributions (these are more flexible than Kolmogorov-style conditional expectations, but require some topological conditions for their existence) Jun 6, 2020 at 6:36

Let $$Prob(\mathbb{R})$$ be the collection of Borel probability measures on $$\mathbb{R}$$ and let $$\mathfrak{M}$$ be the $$\sigma$$-algebra generated by the set of sets $$\{\mu\in Prob(\mathbb{R}):\mu(A) < t\}$$
Define $$\bar{S}(w)(A):=S(w,A)$$. Then $$\bar{S}:(\Omega,\mathscr{F}) \rightarrow (Prob(\mathbb{R}),\mathfrak{M})$$ is a measurable function.
Let $$\mu_{X|\bar{S}}$$ be a regular conditional distribution of $$X$$ given $$\bar{S}$$. (I am not sure if this regular version exists in this case. It must exist if $$\mathfrak{M}$$ is a sub $$\sigma$$-algebra of the Borel $$\sigma$$-algebra of the topology of weak convergence on $$Prob(\mathbb{R})$$. I think this is true by Portmanteau theorem and the property that every Borel measure on $$\mathbb{R}$$ is inner regular, but I have to check the details...)
Now, we write $$X|S\sim S$$ to mean $$\mu_{X|\bar{S}}(\cdot,\lambda) = \lambda$$ for almost every $$\lambda\in Prob(\mathbb{R})$$ with respect to $$\bar{S}_*P$$ (the push-forward measure).