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"

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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.


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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..

  • 1
    $\begingroup$ Please see stats.stackexchange.com/…. $\endgroup$
    – whuber
    Commented Oct 14, 2019 at 13:49
  • $\begingroup$ @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. $\endgroup$
    – Rubertos
    Commented Oct 14, 2019 at 13:51
  • $\begingroup$ Can you give us an accessible reference to investigate the context of such usage? $\endgroup$
    – whuber
    Commented Oct 14, 2019 at 13:55
  • 1
    $\begingroup$ 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 $Pr(X_1\in [0,1])=\frac 1 {2\pi}\int_0^1 e^{-x^2/2}dx$ $\endgroup$
    – Aksakal
    Commented Oct 14, 2019 at 14:22
  • 1
    $\begingroup$ 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) $\endgroup$ Commented Jun 6, 2020 at 6:36

1 Answer 1


Now, I get it.

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).


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