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I have a question about proving the optional sampling theorem in discrete setting. I dont know if what I am doing is mathematical justified. Can someone help me with this?

Defenition: Let $\tau$ be a stopping time, then $\mathcal{F}_{\tau}=\{F\subset \Omega: \forall n \in N \cup \{\infty \}, F\cap(\tau\leq n)\in \mathcal{F}_{n} \}$ is a sigma-algebra.

Defenition: Let $[M_{k}, k \in \mathcal{N}]$ be a martingale w.r.t. a filtration. Let $\tau$ be a stopping time that is never $\infty.$ Then $M_{\tau}$ is a function defined by $\forall \omega \in \Omega: M_{\tau}(\omega)=M_{\tau(\omega)}(\omega)$

Defenition Indicator function: Let $F\in \mathcal{F}$, where $\mathcal{F}$ is a sigma algebra. Then the indicator function is a function from $\Omega$ to $[0,1]$. $1_{F}(\omega)=1$ if $\omega \in F$, else $1_{F}(\omega)=0$

Question 1) Prove that $M_{\tau}(\omega)\in m\mathcal{F}_{\tau}$

Question 2) When in addition is given that $\tau \leq K$, where $K$ is a given positive integer, prove then that $E(M_{\tau})=M_{0}$. Here it might help to write $1_{\Omega}=\sum_{k=0}^{K}1_{\tau=k}$.

My attempts to answer these questions.

(Q1) Since $\tau(\omega) \neq \infty, \forall \omega \in \Omega$, we can write for some $t \in N$

$M_{\tau}=\sum_{i=0}^{t}1_{\tau=i}M_{i} \in \mathcal{F}_{t}$, Since the indicator and the martingale are both random variables, and products of random variables are also random variables.

Then $\forall n \in N: [\sum_{i=0}^{n}1_{\tau=i}M_{i}]^{-1}(B(\Re))\cap(\tau \leq n) \in \mathcal{F}_{n}$. Since both parts of the intersections are in $\mathcal{F}_{n}$ and hence $M_{\tau}(\omega)\in m\mathcal{F}_{\tau}$

(Q2) $E(M_{\tau})=E(M_{\tau}1_{\Omega})=E(M_{\tau}\sum_{k=0}^{K}1_{\tau=k})=\sum_{k=0}^{K}E(M_{k}1_{\tau=k})=\sum_{k=0}^{K}E(M_{k}1_{\tau=k}|\mathcal{F}_{0})=\sum_{k=0}^{K}M_{0}E(1_{\tau=k})=\sum_{k=0}^{K}M_{0}P(1_{\tau=k})=M_{0}$

Can someone assure me whether my attempts are legit? Or give legit proofs?

I have a new attempt to answer question 1. I think this one is more general.

Given is that $M:\Omega \times \mathcal{T} \rightarrow \Re$ such that $M(n,\cdot) \in m \mathcal{F}_{n}, \forall n \in \mathcal{N}$

$\tau: \Omega \rightarrow \mathcal{N}$

$\mathcal{F}_{\tau}=\{ F \subset \Omega: \forall n \in \mathcal{N} \cup\{ \infty \}, F \cap (\tau \leq n) \in \mathcal{F}_{n} \}$

Claim: $M_{\tau} \in m \mathcal{F}_{\tau}$


$M_{\tau}(\omega)=M(\tau(\omega),\omega) \forall \omega \in \Omega$.

Since $M(n,\cdot) \in m \mathcal{F}_{n}, M_{n}^{-1}(\mathcal{B}(\Re))\subseteq \mathcal{F}_{n}$ and $\forall \omega \in \Omega: \tau(\omega) \in \mathcal{N}$ it follows that

$M_{\tau(\omega)}^{-1}(\mathcal{B}(\Re)) \subseteq \mathcal{F}_{\tau(\omega)}$

$\rightarrow M_{\tau(\omega)}^{-1}(B) \in \mathcal{F}_{\tau(\omega)}, \forall B \in \mathcal{B}(\Re)$

$\rightarrow M_{\tau(\omega)}^{-1}(B) \cap (\tau \leq n) \in \mathcal{F}_{n}, \forall B \in \mathcal{B}(\Re)$

$\rightarrow M_{\tau}^{-1}(B) \in \mathcal{F}_{\tau}, \forall B \in \mathcal{B}(\Re)$

$\rightarrow M_{\tau} \in m \mathcal{F}_{\tau}$.

Hence $M_{\tau}$ is a random variable on $\mathcal{F}_{\tau}$.

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(Q1) What you do works when $\tau$ is bounded. You only know that $\tau$ is almost surely finite, which is more general. (Q2) Apply the answer to your other question. – Did Jun 26 '13 at 14:35
@Did (Q1) Do you maybe know how to prove this then in the general case? (Q2) I think I did use here that $M_{\tau} \in m\digamma_{\tau}$? So what do you exactly mean? – Niels Jun 26 '13 at 15:00
What is mF? If you are referring to the boundedness of $M_\tau$, no this is not necessary. – Did Jun 26 '13 at 15:01
I mean that $M_{\tau}$ is measurable in $\mathcal{F}$ – Niels Jun 26 '13 at 15:42
@Did In question 1 I want to prove that $M_{\tau}$ is a random variable w.r.t. $\mathcal{F}_{\tau}$. This is equivalent to the statement that $M_{\tau} \in m\mathcal{F}_{\tau}$ right? – Niels Jun 26 '13 at 15:49

Note: I just saw your updated answer for question (1), and it is essentially the same as the one I have written here. Maybe the alternate presentation will be useful anyways.

Question 1:

So, an equivalent definition for $\mathcal{F}_{\tau}$ is: $$ \mathcal{F}_{\tau} = \{ F \subset \Omega: F \cap \{\tau = n\} \in \mathcal{F}_n, \; \forall n \in \mathbb{N} \cup \{\infty\}\} $$

To show that $M_{\tau}$ is adapted to $\mathcal{F}_{\tau}$, we need to show that for an arbitrary Borel set $B$, the event $\{M_{\tau} \in B\}$ is $\mathcal{F}_{\tau}$-measurable. By the definition of $\mathcal{F}_{\tau}$, this is equivalent to showing that: $$ \{M_n \in B\} \cap \{\tau = n\} \in \mathcal{F}_n $$ $\{M_n\}$ is a martingale, so $\{M_n \in B\} \in \mathcal{F}_n$. Also, $\tau$ is a stopping time so $\{\tau = n\} \in \mathcal{F}_n$ also. Thus their intersection is also $\in \mathcal{F}_n$.

Question 2:

OK, I'm not sure about the hints you've given, but here is one easy method. I'm also assuming that $\mathcal{F}_0 = \{\emptyset, \Omega\}$ but you can correct me if I'm wrong. If it's wrong then... I'm pretty sure there's some information still missing about the problem.

Denote the process stopped at time $n$ by $M_{\tau \wedge n}$. As Did mentioned in the comments, you have previously shown that $\{M_{\tau \wedge n}\}$ is a martingale. You can verify that $E M_{\tau \wedge n} = M_0$ for all $n$. Then since $\tau \leq K$, setting $n = K$ gives the result.

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First of all, thanx for the alternative for Q1, it's nice. Secondly, I thought I could do that, since we have a filtration, and I apply the tower property and $M$ is a martingale. Then again, maybe I can't do this since we do not take the expectation of $M$ only, but from M times the indicator function. So, that is the part of the proof I doubted myself the most too! Maybe you know how to avoid this, or an other proof? – Niels Jun 27 '13 at 8:47
By the way, my teacher gave the following hint (which we should prove if we use it) $EM_{k}1_{\tau =k} =^{?} EM_{K}1_{\tau =k}$ and $E(M_{K}|\mathcal{F}_{k}=M_{k}$) by defenition of a martingale – Niels Jun 27 '13 at 9:12
Can you be a little more careful in your notation? I think you made a a typo. Your last assertion is not obvious to me. Also, are you sure you're not trying to prove that $E(M_{\tau}) = E(M_0)$? – gogurt Jun 27 '13 at 16:23
Yes, I am sure. We (I) need to prove that $E(M_{\tau})=M_{0}$. The hint that my teacher gave is not obvious to me neither, however he told us that we could use that. – Niels Jun 27 '13 at 20:43
Do we know anything about $\mathcal{F}_0$? I'm guessing that it's defined here to be $\{\emptyset, \Omega\}$, but you never explicitly stated it. That would also imply that $M_0$ is constant, so $E M_0 = M_0$. – gogurt Jun 27 '13 at 21:27

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