Prove/Disprove $E[1_A | \mathscr{F_t}] = 0 \ \text{or} \ 1 \ \text{a.s.} \ \Rightarrow E[1_A | \mathscr{F_{s}}] = E[1_A | \mathscr{F_t}] \ \text{a.s.}$

Given a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F}_n\}_{n \in \mathbb{N}}, \mathbb{P})$, let $A \in \mathscr{F}$.

Suppose $$\exists t \in \mathbb{N} \ \text{s.t.} \ E[1_A | \mathscr{F_t}] = 1 \ \text{a.s.}$$ Does it follow that $$E[1_A | \mathscr{F_{s}}] = E[1_A | \mathscr{F_t}] \ \text{a.s.} \ \forall s > t \ ?$$ What about $\forall s < t$?

What if instead $$\exists t \in \mathbb{N} \ \text{s.t.} \ E[1_A | \mathscr{F_t}] = 0 \ \text{a.s.} \ ?$$ Or what if $$E[1_A | \mathscr{F_t}] = p \ \text{a.s.} \ \text{for some} \ p \in (0,1) \ ?$$

What I tried:

If $\Bbb E[1_A|\mathscr F_t]=1$, then $\Bbb E[1_A]=1$, which is the same as $1_A=1$ (almost surely). In this case $\Bbb E[1_A|\mathscr F_s]=1$ (almost surely) for each $s$.

Likewise, if $\Bbb E[1_A|\mathscr F_t]=0$, then $\Bbb E[1_A]=0$, which is the same as $1_A=0$ (almost surely). In this case $\Bbb E[1_A|\mathscr F_s]=0$ (almost surely) for each $s$.

If $\Bbb E[1_A|\mathscr F_t]=p$, for a constant $p\in(0,1)$, then we have

$\Bbb E[1_A|\mathscr F_s]=E[E[1_A|\mathscr F_t]|\mathscr F_s] = E[p|\mathscr F_s] = p$. This may fail if $s>t$.

Alternatively for $= p$ case:

Let $F$ be a bounded $\mathscr F_t$-measurable random variable.

$$\Bbb E[1_A\cdot F]=\Bbb E[E[1_A\cdot F|\mathscr F_t]]=\Bbb E[F\cdot E[1_A|\mathscr F_t]]$$

$$=\Bbb E[p\cdot F]=p\Bbb E[F]=\Bbb E[1_A]\cdot\Bbb E[F]$$

meaning that $1_A$ and $F$ are independent. In other words, $\sigma(A)$ and $\mathscr F_t$ are independent. So $\sigma(A)$ and $\mathscr F_s$ are also independent if $s<t$ and hence $E[1_A|\mathscr F_s] = E[1_A] = p$ . This may fail if $s>t$.

I guess the idea is that a constant is both independent of $\mathscr F_s$ and $\mathscr F_s$-measurable.


1 Answer 1


Your argument seems to be valid, but you start off by assuming that $E[1_A | \mathscr{F_t}] = 1$. However, the question states that $E[1_A | \mathscr{F_t}] \in \{0, 1\}$, which I would take to mean that the random variable $E[1_A | \mathscr{F_t}]$ takes values in the set $\{0, 1\}$ i.e. $E[1_A | \mathscr{F_t}]=1_B$ where $B\in\mathscr{F_t}$. The defining property of this conditional expectation is that $\int_F 1_B d\mathbb{P}=\int_F 1_A d\mathbb{P}$ for all $F\in\mathscr{F_t}$. In particular, taking $F=B$ leads to $P(B)=P(A\cap B)$, from which we can conclude that $B\subset A$ (except possibly on a set of probability zero). However, we also know (as in the argument you have written) that $E[E[1_A | \mathscr{F_t}]] = E[1_B]$ i.e. $P(A)=P(B)$, so the only possible conclusion is that $A=B$ (except possibly for a set of probability zero).

For $s\gt t$, $\mathscr{F_t}\subset\mathscr{F_s}$, so the tower law for conditional expectations implies that $E[1_A | \mathscr{F_t}]=E[E[1_A | \mathscr{F_t}] | \mathscr{F_s}]$. But $E[1_A | \mathscr{F_t}]=1_A$, so $E[1_A | \mathscr{F_s}]=1_A$. So all the conditional expectations for $s>t$ are equal (to $1_A$). For $s<t$, if $A\in\mathscr{F_s}$ then we will still have $E[1_A | \mathscr{F_s}]=1_A$. On the other hand, if we go back to a time where $A$ is not in $\mathscr{F_s}$, then I don't think anything can be said about $E[1_A | \mathscr{F_s}]$ in general. For a concrete example, see this paper, Figure 1. Taking $A=\{\omega_2\}\in\mathscr{F_2}\setminus\mathscr{F_1}$, for example, gives the sequence of conditional expectations $E[1_A | \mathscr{F_0}]=\frac{1}{8} 1_\Omega$, $E[1_A | \mathscr{F_1}]=\frac{1}{2}1_{\{\omega_1,\omega_2\}}$, $E[1_A | \mathscr{F_2}]=1_{\{\omega_2\}}$, $E[1_A | \mathscr{F_3}]=1_{\{\omega_2\}}$.

  • $\begingroup$ Thanks S. Catterall. How do you know 1 $P(B) = P(A \cup B) \to B \subseteq A$? 2 $E[1_A | \mathscr{F}_t] = 1_A$ ? Also going to edit question. Sorry for any inconvenience. I'm going to use some of your insight for edit $\endgroup$
    – BCLC
    Nov 18, 2015 at 0:37
  • 1
    $\begingroup$ Let me try to summarise in natural language; a filtration corresponds to an increasingly finer subdivision of the outcome space, and the conditional expectation of event $A$ w.r.t. successive elements of the filtration ("as more information becomes available") becomes more peaked around the event (at the initial state of information $\mathscr{F}_0$ it is just the uniform distribution). The stopping time is the stochastic level set surface of the process (in the paper, the outcome variable is binary, and value $0$ was chosen). $\endgroup$
    – ocramz
    Nov 18, 2015 at 1:39
  • 1
    $\begingroup$ In this picture, if we are measuring event $A$, but the sample process ends up in a configuration $\omega_i$ that doesn't belong to $A$, $A$ becomes effectively "unknowable" (measure $0$). Is this description correct? Moreover, how the conditional expectations at consecutive times behave reminds me of the iterative Bayes' process, is there a connection between these concepts? @S. Catterall $\endgroup$
    – ocramz
    Nov 18, 2015 at 1:52
  • 1
    $\begingroup$ In answer to the questions in your first comment: if $P(B) = P(A \cap B)$ then, because $B$ is a disjoint union of $A\cap B$ and $A^c\cap B$ we must have $P(A^c\cap B)=0$, which means that $B\subset A$ a.s. Now, in the same way, we can use $P(A)=P(B)$ to conclude that $A=B\in \mathscr{F_t}$ a.s. $\endgroup$ Nov 18, 2015 at 22:30
  • 1
    $\begingroup$ @BCLC I've checked your edits and it looks much better now, great. I assume that when you write $\Bbb E[1_A\cdot F]$ you actually mean $\Bbb E[1_A\cdot 1_F]$ where $F\in \mathscr{F_t}$. $\endgroup$ Nov 29, 2015 at 12:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.