2

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

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

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

What I tried:

Case 1: $E[1_A | \mathscr{F_t}] = 1$

$$E[1_A | \mathscr{F_t}] = 1$$

$$\to E[E[1_A | \mathscr{F_t}]] = E[1]$$

$$\to E[1_A] = 1$$

and maybe for this reason $$E[1_A | \mathscr{F_{s}}] = E[1_A | \mathscr{F_t}] \ \text{QED for case 1?}$$

If so, I suspect for similar reasons, we can deduce:

Case 2: $E[1_A | \mathscr{F_t}] = 0$

$$E[1_A | \mathscr{F_t}] = 0$$

$$\to E[1_A | \mathscr{F_{s}}] = E[1_A | \mathscr{F_t}] \ \text{QED for case 2?}$$

Community
  • 1
BCLC
  • 14,197

1 Answers1

0

If $\Bbb E[1_A|\mathscr F_t]=1$, then $\Bbb E[1_A]=1$ (almost surely), which is the same as $1_A=1$. 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 $\Bbb E[1_A]=p$, and (for a fixed bounded $\mathscr F_t$-measurable random variable $F$)

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

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

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 $s>t$.

OP edit: Is independence needed to say $\Bbb E[1_A|\mathscr F_s]=p$ if s < t?

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

$\Bbb E[1_A|\mathscr F_s]=E[E[1_A|\mathscr F_t]|\mathscr F_s] = E[p|\mathscr F_s] = p$

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

Community
  • 1
John Dawkins
  • 29,845
  • 1
  • 23
  • 39
  • Same for p?.... – BCLC Nov 23 '15 at 23:32
  • 1
    See my revised answer. – John Dawkins Nov 23 '15 at 23:44
  • Wait. How do you know that 'In this case $\Bbb E[1_A|\mathscr F_s]=1$ (almost surely) for each $s$.' and 'In this case $\Bbb E[1_A|\mathscr F_s]=0$ (almost surely) for each $s$.'? Is that because $1_A = 1$ and $1_A = 0$, respectively? Edited such. Feel free of course to rollback – BCLC Nov 24 '15 at 00:06
  • Um John Dawkins how does zero covariance imply independence in this case? I think instead of $F$, you should use $1_{F_t}$ where $F_t \in \mathscr{F_t}$ ? It seems that that would indeed show that $\sigma(A)$ and $\mathscr F_t$ are independent – BCLC Nov 24 '15 at 00:27
  • 1
    $1_{F_t}$ is a special kind of "bounded $\mathscr F_t$-measurable random variable" (for $F_t\in\mathscr F_t$). – John Dawkins Nov 24 '15 at 14:16
  • If you don't use that how do the last equations prove Independence? – BCLC Nov 24 '15 at 17:50
  • 1
    I am using the fact that two $\sigma$-algebras $\mathscr A$ and $\mathscr B$ are independent if and only if $\Bbb E[AB]=\Bbb E[A]\Bbb E[B]$ for all bounded $\mathscr A$-measurable real-valued functions $A$ and all bounded $\mathscr B$-measurable real-valued functions $B$ – John Dawkins Nov 24 '15 at 17:58
  • wow. That's really interesting. If I use indicator 1_F_t though, the proof still holds? – BCLC Nov 24 '15 at 18:00
  • 1
    Yes, the argument still works. – John Dawkins Nov 24 '15 at 19:32
  • John Dawkins, is independence needed? Edited – BCLC Nov 25 '15 at 13:23