Given any integrable random variable $X : (\Omega, \mathcal F, \mathbb P) \to \mathbb R$ (where $(\Omega, \mathcal F, \mathbb P)$ is not necessarily a standard probability space) and a sub-sigma-algebra $\mathcal D \subset \mathcal F$, is it always possible to build another random variable $X'$ living on a standard probability space $(\Omega', \mathcal F', \mathbb P')$ and a sub-sigma-algebra $\mathcal D' \subset \mathcal F'$ such that $(X, E(X | \mathcal D))$ is equal in distribution to $(X', E(X'| \mathcal D'))$ (i.e., same joint distribution)?

Also, is the same true for any countable collection of random variables $(X_i)_{i \in \mathbb N}$? That is, given $(X_i)_{i \in \mathbb N}$ and $\mathcal D \subset \mathcal F$, is it always possible to build $(X'_i)_{i \in \mathbb N}$ and $\mathcal D'$ living on a standard probability space such that the probability distribution of $((X_i, E(X_i | \mathcal D)))_{i \in \mathbb N}$ and that of $((X'_i, E(X'_i | \mathcal D')))_{i \in \mathbb N}$ is the same probability measure on $(\mathbb R^2)^{\mathbb N}$?

Some cases

If $\mathcal D$ is equal to $\sigma(Y)$ (up to null sets) for some random variable $Y$, then we can build $Y'$ and $(X'_i)_i$ living on a standard probability space such that $(Y, X_1, X_2, \cdots)$ is equal in distribution to $(Y', X'_1, X'_2, \cdots)$ (for example, we can build them on $\mathbb R \times \mathbb R^{\mathbb N}$ with the obvious push forward measure on it.) Then $((X_i, E(X_i | \mathcal D)))_{i \in \mathbb N}$ is equal in distribution to $((X'_i, E(X'_i | \mathcal D')))_{i \in \mathbb N}$ (where $\mathcal D' := \sigma(Y')$) because $E(X_i | \mathcal D) = g_i(Y)$ while $E(X'_i | \mathcal D') = g'_i(Y') = g_i(Y')$.

I guess in the end it boils down to the issue of whether there is a pathological $\mathcal D$ on which conditional expectations behave irreparably bad.

At first, I thought the perforated interval would give me a counter example, but the sigma algebra associated with the perforated interval, which is pathological in some sense, is generated by a random variable so it's not irreparably pathological.


If the answer is no, the counter example would be an interesting example.


2 Answers 2


The conditional expectation $E(X | \mathcal D)$ is always almost surely equal to the conditional expectation with respect to a random variable, namely the random variable $E(X | \mathcal D)$.

  • $\begingroup$ It helped me to rephrase that a little bit in order to get what you mean: $E(X | \mathcal D)=E(X | Y)$ a.s. for $Y=E(X | \mathcal D)$. Is that right? $\endgroup$ Sep 17, 2014 at 9:49
  • $\begingroup$ @ElmarZander Yes, that is right. $\endgroup$ Sep 17, 2014 at 10:35

To build on the other answer and address the case of countably infinite collection of random variables:

Let $\mathcal D' = \bigvee_{i \in \mathbb N} \sigma(E(X_i | \mathcal D))$, that is, the sigma algebra generated by all $E(X_i | \mathcal D)$.

Then $E(X_i | \mathcal D) = E(X_i | \mathcal D')$ (because $E(X_i | \mathcal D)$ is $\mathcal D'$-measurable).

On the other hand, we can construct one random variable $Y$ such that $\mathcal D' = \sigma(Y)$ (because any sigma algebra generated by a countable family of (real-valued) random variables is generated by just one real-valued random variable, which can be shown, for example, by realizing that a real number is (measurably) described by countably many binary bits and vice versa).

So the answer is yes, even for countably infinite collection of random variables.


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