Spin-off from my previous question: Prove/Disprove $E[1_A | \mathscr{F_t}] = 0 ~\text{or} ~ 1 \ \Rightarrow E[1_A | \mathscr{F_{s}}] = E[1_A | \mathscr{F_t}]$

Apparently the conclusion holds true almost surely. Is it really only almost surely? What is an example of a filtered probability space and event $A$ s.t.

$\exists t \in \mathbb N$ s.t. $P(P(A|\mathscr F_t) = 0) = 1$ or $P(P(A|\mathscr F_t) = 1) = 1$


$\exists \omega \in \Omega$ s.t. $P(A|\mathscr F_t)(\omega) \ne P(A|\mathscr F_s)(\omega)$ for some $s > t$?

  • 2
    $\begingroup$ Conditional expectations are only unique up to null-sets, so, of course, we can only hope that the equality holds almost surely. $\endgroup$ Jan 25, 2016 at 8:39

1 Answer 1


I'm not sure if this is what you need, but I will try to give a constructive example where your conditions hold true.

Consider $U_t$~ $DiscreteUniform\{0,1\}$ for $ k \geq t \geq 0, t \neq k$ and $U_t = 2$ for $t > k$. Define $Z_t = 1_{\{U_t = 2\}}$. Define filtration $\mathbb{F_t} = \sigma\{U_m:t \leq m\leq 0\}$.

Clearly, $P(P(Z_t| \mathbb{F_{t-1}}) = 2) = 0$ for $t<k$. Choose $s = k+1$ - now $P(P(Z_s| \mathbb{F_{s-1}}) = 2) = 1 \neq 0$. Hence, it suffices to choose $\omega = \{2\}$ and $A_t = \{U_t = \omega\}$ to see that $P(A_t|F_{t-1})(\omega) \neq P(A_s|F_{t-1})(\omega)$.

This answers your question if and only if $A$ can be time-dependent, which it usually should be (since one of your tags is 'filter').

  • $\begingroup$ Filter can refer to filtration? Or not? $\endgroup$
    – BCLC
    Jan 24, 2016 at 19:58
  • $\begingroup$ @BCLC Yep, but also to something else (e.g., Kalman Filter). Does the answer help you? $\endgroup$
    – Jeremias K
    Jan 24, 2016 at 22:17
  • $\begingroup$ Lol forgot all about this. Well...should it be $Z_t = \{U_t = 2\}$? Not sure what it means to say $P(Z_t | F) = 2$ if $Z_t$ is a random variable $\endgroup$
    – BCLC
    Feb 5, 2016 at 16:02

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