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The proof for a stopped discrete-time martingale is shown as follows.

Let $M=(M_n)_{n\ge0}$ be a discrete-time martinglae w.r.t. the filtration $(\mathcal F_n)_{n\ge0}$, and let $M^T=(M_{n\land T})_{n\ge0}$ be the stopped martingale, where $T$ is a stopping time w.r.t. $(\mathcal F_n)_{n\ge0}$. Since:

  1. We have $$\begin{align} M_{n\land T}&=1_{T\ge n+1}M_n+1_{T\le n}M_T\\ &=1_{T\ge n+1}M_n+ \sum_{k=1}^n 1_{T=k}M_k\\ \end{align}$$ where $1_{T\ge n+1}, M_n, 1_{T=k}$, and $M_k$ are all $\mathcal F_n-measurable$, hence $M_{n\land T}$ is $\mathcal F_n-measurable$;

  2. ${\Bbb E}|M_{n\land T}| \le \underbrace{{\Bbb E}|M_1|+…+{\Bbb E}|M_n|}_{{\Bbb E}|M_i| \text{ is integrable, }\forall i\ge0}<\infty$, i.e., $M_{n\land T}$ is integrable $\forall n \ge 0$;

  3. We have $$\begin{align} {\Bbb E}(M_{n+1 \land T}|{\mathcal F_n}) &= {\Bbb E}(M_{n \land T}+1_{T \ge n+1}(M_{n+1}-M_n)|{\mathcal F_n})\\ &=M_{n \land T}+1_{T \ge n+1}{\Bbb E}(M_{n+1}-M_n)\\ &=M_{n \land T}\\ \end{align}$$

Therefore, the stopped martingale satisfies the definition of a discrete martingale. Proof complete.

However, I am not able to extend these three parts of proof to a continuous version, because I cannot devide the time into separate spots with one next to another as the discrete version did. So I really wonder how to give a proof for a continuous-time martingale.

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  • $\begingroup$ What is preventing you to apply the technique shown in the other answer? $\endgroup$
    – Did
    Commented Jan 8, 2015 at 22:06
  • $\begingroup$ @Did I have edited the question. I think the difficulty lies in dealing with the continuity of time. $\endgroup$
    – Zhanbin Du
    Commented Jan 9, 2015 at 3:47
  • $\begingroup$ With dyadics, page 2 : dam.brown.edu/people/huiwang/classes/Am264/Archive/Brownian.pdf This idea is to discretize the time, build discrete stopping times on these time slots, and conclude using backward martingales. $\endgroup$
    – thomasb
    Commented Aug 25, 2019 at 17:09

2 Answers 2

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We are going to prove the following proposition:

If $M$ is a continuous martingale and $T$ a stopping time, the stopped process $M^{T}$, i.e., $\{M(t \wedge T), t \geq 0\}$ is a martingale with respect to $\left(\mathcal{F}_{t}\right)$.

Here we go.

The process $M^{T}$ is obviously continuous and adapted. Firstly we use a weak form of optional stopping theorem, saying that a martingale has equal expectation at any bounded stopping time. If $S$ is a bounded stopping time, so is $S \wedge T$; hence $$ E\left[M_{S}^{T}\right]=E\left[M_{S \wedge T}\right]=E\left[M_{0}\right]=E\left[M_{0}^{T}\right]. $$ Then we use this conclusion twice to get our desired equality. If $s<t$ and $A \in \mathcal{F}_{s}$ the r.v. $T=t \mathbf{1}_{A^{c}}+s \mathbf{1}_{A}$ is a stopping time and consequently $$ E\left[X_{0}\right]=E\left[X_{T}\right]=E\left[X_{t} \mathbf{1}_{A^{c}}\right]+E\left[X_{s} \mathbf{1}_{A}\right] .$$ On the other hand, $t$ itself is a stopping time, and $$ E\left[X_{0}\right]=E\left[X_{t}\right]=E\left[X_{t} \mathbf{1}_{A^{c}}\right]+E\left[X_{t} \mathbf{1}_{A}\right]. $$

Comparing the two equalities yields $X_{s}=E\left[X_{t} | \mathcal{F}_{s}\right].$

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  • $\begingroup$ What is $M_0$? Why you would have $$E[M_{S \wedge T}] = E[M_0]$$? $\endgroup$
    – Fucio
    Commented May 31 at 11:00
  • $\begingroup$ @Fucio because it's a martingale. $\endgroup$ Commented Jun 5 at 21:20
  • $\begingroup$ I don't get is how comparing the two equalities, we get the last equation... $\endgroup$ Commented Jun 5 at 21:21
  • $\begingroup$ @Anoldmaninthesea. Just use the definition of conditional expectation $Y=E[X|\mathcal{F}]$ if $E[Y; A] = E[X; A]$ for every $A\in\mathcal{F}$. $\endgroup$ Commented Jul 1 at 13:18
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With dyadics, page 2 : http://www.dam.brown.edu/people/huiwang/classes/Am264/Archive/Brownian.pdf. The idea is to discretize the time, build discrete stopping times on these time slots, and conclude using backward martingales.

Optional sampling theorem : If $X = (X_t,\mathcal{F}_t)$ is a supermartingale and $T$ is an arbitrary stopping time, then the stopped process $X^T=(X_{T\wedge t},\mathcal{F}_t)$ is also a supermartingale.

Proof : Here we give a sketch of the proof. Assume $X$ is a supermartingale. First, for a fixed $n\ge 1$, let $$ D_n = \left\{\frac{k}{2^n}, k= 0,1,2,\ldots\right\}\subset D_{n+1}\subset \cdots $$ be the set of non-negative dyadic rationals of order no greater than $n$. It follows that $$ X = (X_t,\mathcal{F}_t; t\in D_n) $$ is a super martingale (discrete time).

Second, we construct a stopping time $T_n$ such that $T_n\ge T$ and $T_n$ only take values in $D_n$. Indeed, let $$ T_n(\omega) = \inf\left\{t\in D_n; t\ge T(\omega)\right\}. $$ Then $T_n\ge T_{n+1} \ge \cdots $ and $T_n$ is a stopping time. Fix $0\le s\le t$, we wish to show that $$ \mathbb{E}\left[X_{t\wedge T}|\mathcal{F_s}\right]\le X_s $$ almost surely. Similarly define $$ t_n = \inf\left\{u\in D_n; u\ge t\right\} \ge t_{n+1}\ge\cdots $$ and $$ s_n = \inf\left\{u\in D_n; u\ge s\right\} \ge s_{n+1}\ge\cdots $$ It follows from the discrete time optional sampling theorem that $$ \mathbb{E}\left[X_{t_n\wedge T_n}|\mathcal{F}_{s_m}\right] \le X_{s_m \wedge T_n} $$ for any integers $m\ge n$. Letting $m\to\infty$, we have $s_m \to s$ decreasing and $\mathcal{F}_{s_m}\to\mathcal{F}_s$. By Lévy's Downward theorem and the continuity of process $X$, we have $$ \mathbb{E}\left[X_{t_n\wedge T_n}|\mathcal{F}_{s}\right] \le X_{s_m \wedge T_n} $$ Observe that $(X_{t_n\wedge T_n},\mathcal{F}_{t_n\wedge T_n})$ is a backward martingale, whence it is uniformly integrable. Letting $n\to\infty$, we arrive at $$ \mathbb{E}\left[X_{t\wedge T}|\mathcal{F}_{s}\right] \le X_{s\wedge T}. $$ This completes the proof.

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    $\begingroup$ Why downvote? I think this solution is good... $\endgroup$
    – William
    Commented Feb 6, 2022 at 4:14
  • $\begingroup$ There are down-voting trolls all over the place... This week-end I asked a question about NoSQL databases and got down-voted twice. Thank you for your support though $\endgroup$
    – thomasb
    Commented Feb 7, 2022 at 7:00

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