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Let $(\Omega, \mathfrak A, P)$ be a probability space and $\mathcal F$ a filtration with $B \in \mathcal F_t$ for a $t\le s$.

I would like to know if $$\mathbb E[X|\mathcal F_{t1_B+s1_{B^c}}]=\mathbb E[X|\mathcal F_s]1_{B^c}+\mathbb E[X|\mathcal F_t]1_B$$ holds.

I tried it this way: It is true that $$\mathbb E[X|\mathcal F_{t1_B+s1_{B^c}}]=\mathbb E[X1_B|\mathcal F_{t1_B+s1_{B^c}}]+\mathbb E[X1_{B^c}|\mathcal F_{t1_B+s1_{B^c}}]. \tag 1$$

  1. case: $1_B=0$, so $1_B=0 \Rightarrow 1_{B^c}=1$, then due to $(\rm 1)$, $\mathbb E[X| \mathcal F_{t1_B+s1_{B^c}} ]=\mathbb E[0|\mathcal F_s]+\mathbb E[X|\mathcal F_s]=\mathbb E[X|\mathcal F_s]$.

  2. case: $1_B=1$, so $1_B=1 \Rightarrow 1_{B^c}=0$, then again due to $(1)$ $\mathbb E[X| \mathcal F_{t1_B+s1_{B^c}} ]=\mathbb E[X|\mathcal F_t]+\mathbb E[0|\mathcal F_t]=\mathbb E[X|\mathcal F_t]$.

Putting both cases together we get $$\mathbb E[X|\mathcal F_{t1_B+s1_{B^c}}]=\mathbb E[X|\mathcal F_s]1_{B^c}+\mathbb E[X|\mathcal F_t]1_B,$$

since if $1_B=0$ it gives us $\mathbb E[X|\mathcal F_s]$ and if $1_B=1$ we get $\mathbb E[X|\mathcal F_t]$.

Is this correct?

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    $\begingroup$ How is $\mathcal{F}_{t1_B+s1_{B^c}}$ even defined? $\endgroup$
    – Snoop
    Commented Nov 20, 2022 at 1:13
  • $\begingroup$ @Snoop you are right, I should've mentioned that. To make it clear, let first consider $\mathcal T:=\{\tau:\tau \text{ is stopping time with } \tau \le T\}$, then for $\tau \in \mathcal T$ $\sigma:\tau1_B+T1_{B^c}$ is also a stopping time. Then the definition says $\mathcal F_\sigma := \{A \in \mathcal F: A\cap \{\sigma \le t\} \in \mathcal F_t \forall t \in T\}$. Now choose $\tau =t$ a constant stopping time and we get the filtration mentioned in my question. $\endgroup$
    – scholar
    Commented Nov 20, 2022 at 1:31
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    $\begingroup$ When $B$ is in ${\frak A}\setminus\bigcup\limits_{t\le s}{\cal F}_t$ it is not measurable w.r.t. any ${\cal F}_t\,.$ Then I have the strongest doubts that $\sigma=t1_B+s1_{B^c}$ is a stopping time w.r.t. this filtration. OK: you assume $B\in{\cal F}_t$ for all $t\le s$. To me this is almost as strong as $B\in\{\emptyset,\Omega\}$. Then I think your formula is trivial. $\endgroup$
    – Kurt G.
    Commented Nov 20, 2022 at 10:23
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    $\begingroup$ It depends a lot on what ${\cal F}_0$ is and if the filtration is right continuous in the sense that ${\cal F}_t=\bigcap_{s>t}{\cal F}_s\,.$ If ${\cal F}_0=\{\emptyset,\Omega\}$ (the smallest possible $\sigma$-algebra describing the sure and the impossible events) it is more trivial than you think that your equation holds: The stopping time $\sigma$ equals $t$ or $s$ and your equation collapses on both sides to $\mathbb E[X|{\cal F}_t]$ resp. to $\mathbb E[X|{\cal F}_s]\,.$ $\endgroup$
    – Kurt G.
    Commented Nov 20, 2022 at 17:38
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    $\begingroup$ I maybe understood something wrong but isn't this the same stopping time as mentioned in this post Math.SE? The context of this equation is part of a proof containing snell envelopes (Föllmer, Schied, Chapter 6.5 on page 387). There is a Lemma 6.48 in which it is mentioned that for $B \in \mathcal F_\tau$ $\sigma:=\tau1_B+T1_{B^c}$ is a stopping time. $\endgroup$
    – scholar
    Commented Nov 20, 2022 at 18:20

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