Suppose we have an integrable random variable $X:\Omega\to\mathbb{R}$. Denote partition of $\mathbb{R}$ by $\Pi^n = \{\Pi^n (k) \subset \mathbb{R} : k = 1,\dots, n\}$ and the corresponding generated sigma-algebra $\sigma^{n} = \sigma(\{X \in \Pi^n(k)\}, k=1,\dots,n)$. Assume $\Pi^{n+1}$ is a refinement of $\Pi^n$ i.e. every component in $\Pi^{n+1}$ is a subset of a component in $\Pi^n$.

Define random variable $X^n:= E[X\mid\sigma^n]$. For a fixed sample path $w \in \Omega$, is the set $\{ X^n (w)\}_{n\in\mathbb{N}}$ bounded?

My thoughts:

I recently asked a similar question here which was answered. However, the counterexample provided involved independently and identically copies of the random variable. So I was curious if the set is now bounded if we condition the same variable on finer and finer partitions.

  • $\begingroup$ Are you considering $$X^n=E(X\mid\sigma^n)\ ?$$ $\endgroup$
    – Did
    Commented Nov 25, 2017 at 10:57
  • $\begingroup$ Yes. Is my notation not good? $\endgroup$ Commented Nov 25, 2017 at 17:53
  • 1
    $\begingroup$ Indeed, if $G$ is a sigma-algebra, then $E(X:G)$ does not exist but $E(X\mid G)$ does. $\endgroup$
    – Did
    Commented Dec 9, 2017 at 21:32
  • $\begingroup$ Thanks. I was under the mistaken assumption that $E[X:\mathcal{G}]$ represents the usual conditional expectation. So what does it really mean? $\endgroup$ Commented Dec 11, 2017 at 5:29
  • $\begingroup$ Nothing (see previous comment). $\endgroup$
    – Did
    Commented Dec 11, 2017 at 7:08

1 Answer 1


I'm going to answer my own question. I hope this is right and helps someone in the future.

I'm going to assume that $\sigma(X) = \sigma(\cup_{n\in\mathbb{N}} \sigma^n)$. Then we know that $X^n$ converges to $X$ with probability one. Therefore, for all paths $w$ from the set of non-zero probability paths, the sequence $(X^n(w))_{n\in\mathbb{N}}$ converges and is therefore bounded.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .