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Here’s an easy one, I hope:

Suppose $\tau$ is a stopping time and $(M_t)$ is a martingale which together satisfy the hypotheses of the optional stopping theorem so that $\mathbb{E}[M_\tau]= \mathbb{E}[M_0]$.

Will it also be the case that $\mathbb{E}[M_\sigma]= \mathbb{E}[M_0]$ for an arbitrary random time $\sigma \leq \tau$, for example $\sigma= \tau-1$?

Thank you very much for your help.

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2 Answers 2

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No. The problem is that $\tau-1$ is not a "stopping time" for the martingale.

Example. Martingale $(X_n), n=0,1,2$ is the standard random walk on $\mathbb Z$. Our sample space is $[0,1)$ with Lebesgue measure. \begin{align} X_0(\omega) &= 0,\qquad \omega \in [0,1). \\ X_1(\omega) &= \begin{cases} 1,\quad & \omega \in [0,1/2), \\ -1,\quad & \omega \in [1/2,1). \end{cases} \\ X_2(\omega) &= \begin{cases} 2,\quad & \omega \in [0,1/4),\\ 0,\quad & \omega \in [1/4,1/2),\\ 0,\quad & \omega \in [1/2,3/4),\\ -2,\quad & \omega \in [3/4,1). \end{cases} \end{align}

The filtration $\mathcal F_0, \mathcal F_1, \mathcal F_2$ for this martingale:
$\mathcal F_0$ is the sigma-algebra generated by $X_0$, so it is the trivial sigma-algebra with atom $[0,1)$.
$\mathcal F_1$ is the sigma-algebra generated by $X_0, X_1$, so it is the sigma-algebra with atoms $[0,1/2), [1/2,1)$.
$\mathcal F_2$ is the sigma-algebra generated by $X_0, X_1, X_2$, so it is the sigma-algebra with atoms $[0,1/4),[1/4,1/2),[1/2,3/4),[3/4,1)$.

Define stopping time $\tau = 2 \wedge \min\{n : X_n < 0\}.$ Then \begin{align} \tau(\omega) &= \begin{cases} 2,\quad & \omega \in [0,1/4),\\ 2,\quad & \omega \in [1/4,1/2),\\ 1,\quad & \omega \in [1/2,3/4),\\ 1,\quad & \omega \in [3/4,1). \end{cases} \\ \sigma(\omega) &= \begin{cases} 1,\quad & \omega \in [0,1/4),\\ 1,\quad & \omega \in [1/4,1/2),\\ 0,\quad & \omega \in [1/2,3/4),\\ 0,\quad & \omega \in [3/4,1). \end{cases} \end{align} Now $\tau$ is a stopping time, so of course $\mathbb E[X_\tau] = 0$. But compute $\mathbb E[X_\sigma] = 1/2 \ne 0$, which shows that $\sigma$ is not a stopping time. Computations: \begin{align}X_\tau(\omega) &= \begin{cases} 2,\quad & \omega \in [0,1/4),\\ 0,\quad & \omega \in [1/4,1/2),\\ -1,\quad & \omega \in [1/2,3/4),\\ -1,\quad & \omega \in [3/4,1). \end{cases} \\ X_\sigma(\omega) &= \begin{cases} 1,\quad & \omega \in [0,1/4),\\ 1,\quad & \omega \in [1/4,1/2),\\ 0,\quad & \omega \in [1/2,3/4),\\ 0,\quad & \omega \in [3/4,1). \end{cases} \end{align} Calculation that shows $\sigma$ is not a stopping time: $$ \{\sigma = 0\} = [1/2,1) \notin \mathcal F_0. $$

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    $\begingroup$ For future reference: An "easy one" like this should be posted at math.stackexchange.com, not here. $\endgroup$ Mar 28, 2020 at 16:45
  • $\begingroup$ My mistake on SE vs here. Thank you for putting that nicely, and for the clear solution. $\endgroup$
    – John
    Mar 28, 2020 at 17:44
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Even if $\sigma$ is a stopping time with $\sigma\le\tau$ the answer is No.

Example: Take your martingale to be a standard 1-dimensional Brownian motion $M$ with $M_0=0$, and then define $\sigma:=\inf\{t: M_t=1\}$ and $\tau:=\inf\{t>\sigma: M_t=0\}$. These are both finite (a.s.) stopping times. And clearly $0<\sigma<\tau$ and $M_\sigma=1$ but $M_\tau=0$ (a.s.), so $\Bbb E[M_\tau]=0=\Bbb E[M_0]$ but $\Bbb E[M_\sigma]=1\not=\Bbb E[M_0]$.

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  • $\begingroup$ This is not quite an example because $M$ and $\tau$ don't satisfy the hypotheses of the optional stopping theorem, even though they do happen to satisfy its conclusion. $\endgroup$ Apr 14, 2020 at 15:47
  • $\begingroup$ Granted. What are the optimal hypotheses for Optional Stopping? $\endgroup$ Apr 17, 2020 at 17:17
  • $\begingroup$ The weakest hypothesis I know is to have $\{ M_{t \wedge \tau} : t \ge 0\}$ uniformly integrable. $\endgroup$ Apr 17, 2020 at 17:59
  • $\begingroup$ I guess the original question is not quite well posed, as "the hypotheses of the optional stopping theorem" is ambiguous. $\endgroup$ Apr 17, 2020 at 19:37

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