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$S,T$ are stopping times and $M$ is a (right) continuous martingale. My lecturer set this as an exercise and I am given a solution(essentially split $M_T = M_T \mathbf{1}_{S≤T} + M_T \mathbf{1}_{S>T}$ and use this property) but there is a step I cannot justify, namely

$$ \mathbb{E}[ M_T \mathbf{1}_{T>S} \mid \mathcal{F}_T ] \overset{?}{=} \mathbb{E}[ M_T \mathbf{1}_{T>S} \mid \mathcal{F}_{T\color{red}{\wedge S}} ] $$

More generally, given an $L^1(Ω)$ random variable $X$, what conditions the on $\sigma$-algebras $\mathcal{F},\mathcal{G}$ guarantee the following equality? $$ \mathbb{E}[X \mid \mathcal{F}] = \mathbb{E}[X \mid \mathcal{G}] $$

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Calvin Khor
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1 Answers1

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Should the relationship in question be $$ \mathbb{E}[ M_T \mathbf{1}_{T>S} \mid \color{red}{\mathcal{F}_S} ] \overset{?}{=} \mathbb{E}[ M_T \mathbf{1}_{T>S} \mid \mathcal{F}_{T\color{red}{\wedge S}} ]\,, $$ in line with the primary problem of determining the general form of $\mathbb E[M_{T}\mid\mathcal F_{S}]$? If this is the case, notice that, on the event $[T\geqslant S]$, $M_T=M_{T\vee S}$ and $M_S=M_{T\wedge S}$. Also, $\mathcal F_{T\wedge S}\subseteq\mathcal F_T$ and $\mathcal F_S\subseteq \mathcal F_{T\vee S}$. Consequently,

$$\mathbb E[M_T{\bf 1}_{T\geqslant S}\mid\mathcal F_S]=\mathbb E[M_{T\vee S}{\bf 1}_{T\geqslant S}\mid\mathcal F_S]=\mathbb E[M_{T\vee S}\mid\mathcal F_S]{\bf 1}_{T\geqslant S}=M_S{\bf 1}_{T\geqslant S}=M_{T\wedge S}{\bf 1}_{T\geqslant S}=\mathbb E[M_{T}\mid\mathcal F_{T\wedge S}]{\bf 1}_{T\geqslant S}=\mathbb E[M_{T}{\bf 1}_{T\geqslant S}\mid\mathcal F_{T\wedge S}]$$

Remark: The relationships above hold because of the interplay between the martingale property and the random variables being measurable with respect to multiple sigma-algebras.

  1. By the martingale property (for martingale $X$ and $t,r>s$), $$ \mathbb{E}[X_t \mid \mathcal{F}_s]=X_s=\mathbb{E}[X_r \mid \mathcal{F}_s] $$

  2. Also, for random variable $X$, measurable with respect to both $\mathcal G$ and $\mathcal F$, we have $$\mathbb{E}[X \mid \mathcal{F}] =X= \mathbb{E}[X \mid \mathcal{G}]$$

Michael Hardy
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ki3i
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    "for $[T\geqslant S]$, $\mathcal F_S\subseteq\mathcal F_T$" What does that mean when, like here, one does not assume that $T\geqslant S$ almost surely? – Did Apr 15 '15 at 11:20
  • Thanks for the answer and for spotting my mistake. Although I feel its an `obvious' step I still can't follow the second equality $\mathbb{E}[M_T|\mathcal{F}S]\mathbf{1}{T\geq S} = M_S\mathbf{1}_{T\geq S}$. Messing around with conditional expectations is getting me nowhere because I don't see a step where I can apply the martingale property.

    When you say for $[T \geq S]$, $\mathcal{F}_S\subseteq \mathcal{F}_T$, do you mean $\mathcal{F}_S ∩ [T\geq S] \subseteq \mathcal{F}_T ∩ [T\geq S]$?

     – Calvin Khor Apr 15 '15 at 11:24
  • @Did, You are right to be confused, thank you. I mean $\mathcal F_S\cap[T\geqslant S]\subseteq\mathcal F_T\cap[T\geqslant S]$. I will amend the post. – ki3i Apr 15 '15 at 11:39
  • @Did, please see edits – ki3i Apr 15 '15 at 12:05
  • @ki3i I have a question about your first remark. The martingale property holds under deterministic times by definition. A stopped martingale is still a martingale but you cannot use that result here since that is what the OP is trying to prove in the first place. I am a bit confused as well by your answer. – Calculon Apr 15 '15 at 12:16
  • @Calculon, The martingale property holds once the times are ordered. The solution, as presented now, involves only ordered times. – ki3i Apr 15 '15 at 12:19
  • re:your edits thank you very much I understand fully now. – Calvin Khor Apr 15 '15 at 12:20
  • @CalvinKhor, You are Welcome. – ki3i Apr 15 '15 at 12:22
  • @ki3i Ok but then the proof of the main statement would be as simple as this: $E[M_T\mid \mathcal{F}_S] = M_S$ if $T\geq S$ due to the martingale property and $E[M_T\mid \mathcal{F}_S] = M_T$ if $S >T$ due to $\mathcal{F}_T \subseteq \mathcal{F}_S$ and $M_T$ being $\mathcal{F}_T$-measurable. – Calculon Apr 15 '15 at 12:24
  • @Calculon its not true that either of the events $[S>T]$ or $[S≤T]$ holds almost surely so reasoning by cases (at least like that) isn't justified (If I understand correctly). – Calvin Khor Apr 15 '15 at 12:27
  • @ki3i I didn't say what I did was correct. I just said what you did in your answer implies what I did. – Calculon Apr 15 '15 at 12:28
  • Indeed -- and this was precisely the mistake (now corrected) in a first version of this answer. – Did Apr 15 '15 at 12:29
  • @Did thank you. I will read the revised answer. I may have missed the changes while typing my comment. – Calculon Apr 15 '15 at 12:30
  • @Calculon, What is it that you are confused about? Have you read the updated answer? – ki3i Apr 15 '15 at 12:34
  • @ki3i it is fine now, really. I have read the updated answer. It looks good. – Calculon Apr 15 '15 at 12:36
  • @Calculon, Thank you. Feel free to upvote ;) – ki3i Apr 15 '15 at 12:39
  • @Did, I forgot to say thank you for acknowledging the correction I made. – ki3i Apr 15 '15 at 14:44