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

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

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

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    $\begingroup$ Why is $E[1_A|\mathscr{F}_t] = 1$? $\endgroup$ – Olorun Nov 6 '15 at 8:35
  • $\begingroup$ @Olorun ? $E[1_A | \mathscr{F_t}]$ by assumption. $\endgroup$ – BCLC Nov 6 '15 at 8:36
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    $\begingroup$ Where did you state that assumption? $\endgroup$ – Olorun Nov 6 '15 at 8:36
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    $\begingroup$ Actually $E[1_A | \mathscr{F_t}] $ is a random variable. And by assumption $\{0,1\}$ its values. $\endgroup$ – Nikita Evseev Nov 6 '15 at 8:42
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    $\begingroup$ @BCLC I missed the tower property (the wrong comment is removed). $\endgroup$ – Nikita Evseev Nov 6 '15 at 15:25
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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.

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  • $\begingroup$ Same for p?.... $\endgroup$ – BCLC Nov 23 '15 at 23:32
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    $\begingroup$ See my revised answer. $\endgroup$ – John Dawkins Nov 23 '15 at 23:44
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    $\begingroup$ $1_{F_t}$ is a special kind of "bounded $\mathscr F_t$-measurable random variable" (for $F_t\in\mathscr F_t$). $\endgroup$ – John Dawkins Nov 24 '15 at 14:16
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    $\begingroup$ 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 $B$ $\endgroup$ – John Dawkins Nov 24 '15 at 17:58
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    $\begingroup$ Yes, the argument still works. $\endgroup$ – John Dawkins Nov 24 '15 at 19:32

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