Let $F_n$ be any filtration and $X$ any random variable with $E|X|<\infty$. Let $X_n=E(X|F_n)$. Give Specific example when $X_\infty=X$ and when $X_\infty \not =X$.

$X_n$ can easily be shown to be a martingale. for both examples I though if I can construct a filtration such that $F_\infty=F$ for the first and $F_\infty \not =F$ for the second then i would get the desired result. Any hints on how to proceed.


1 Answer 1


Since any filtration is acceptable, why not make life easy and put all $\cal F_n$ equal to the same $\sigma$-algebra, and even a simple one such as generated by a single set of measure strictly between 0 and 1? Then for your first example choose $X$ to be a variable that is constant on $A$ and constant on its complement. For the second example let $X$ take two different values on $A$ (each on a subset of nonzero measure).

  • $\begingroup$ thank you. how can we approach the solution if $F_n$ is an increasing filtration. $\endgroup$ Commented Dec 7, 2015 at 21:51
  • 1
    $\begingroup$ Take $\Omega=[0,1]$ and let $\cal F_n$ divide the lower half of the interval in $2^n$ equal parts whereas the upper half remains a solid block (no smaller $\cal F_n$-measurable subsets than the upper half interval itself). For the first example let $X=1$ constant. For the second example let $X=1$ on $[0,\frac34]$ and $X=2$ on $[\frac34,1].$ $\endgroup$ Commented Dec 7, 2015 at 22:09

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