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Let $x_1,x_2 \dots$ be adapted to the filtration $\mathcal{F}_1, \mathcal{F}_2, \dots$. Let $\tau$ be a stopping time that is also adapted to the filtration. Say that $\mathbb{E}[x_i \mid \mathcal{F}_{i-1}] = 0$. Is it true that $$\mathbb{E}[x_{\tau}] = 0?$$ One idea I had was to write $x_{n} = x_0 + (x_1 - x_0) + \dots + (x_{n}- x_{n-1}) = x_0 + \sum\limits_{i=1}^{n}z_i$ where $z_i = x_{i} - x_{i-1}$. I thought perhaps the partial sums $S_n = \sum\limits_{i=1}^{n}z_i$ could be a martingale, and I could use the optional sampling theorem, but it doesn't seem to be the case.

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2 Answers 2

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Now I see this couldn't be the case: Suppose each $x_i$ is random variable that is $-1$ with probability $1/2$ and $+1$ with probability $1/2$. Then define $\tau = \inf\{ n \geq 1 \mid x_n =1\}$. Let $\mathcal{F}_n = \sigma(x_1,\dots, x_n)$. Then the assumptions hold but $\mathbb{E}[x_{\tau}] = 1$.

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    $\begingroup$ If the $x_i$'s are iid and have distribution Ind(p), then $\mathbb E[x_i| \mathcal F_{i-1}] = \mathbb E[x_i] =1-2p \ne 0$ if $p\neq 1/2$. $\endgroup$
    – Mick
    Commented Feb 29, 2020 at 20:23
  • $\begingroup$ True, thank you. Do you believe the counter example be valid if $p=1/2$? $\endgroup$
    – ttb
    Commented Feb 29, 2020 at 20:28
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    $\begingroup$ Yes, then the assumptions hold, and $x_\tau = 1$ with probability $1$. $\endgroup$
    – Mick
    Commented Feb 29, 2020 at 20:32
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Suppose $\tau$ is finite with probability one. Then $$ \mathbb E x_{\tau} = \mathbb E [ \mathbb E[x_\tau | \tau]] = \sum_n \mathbb E [x_\tau | \tau = n]P(\tau = n) $$ If $\tau$ is not finite, i.e. $P(\tau <\infty) <1$, then let $N$ be a positive integer and let $\tau_N := \min(\tau,N)$. Then $\tau_N$ is finite with probability 1, due to Fatou's lemma $$ \mathbb E[x_\tau I(\tau<\infty)] = \mathbb E[\liminf_N x_{\tau_N} ]\le \liminf_N \mathbb E[x_{\tau_N}]. $$

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  • $\begingroup$ $\tau$ finite a.s. is not enough to get the desired conclusion. You should require $\tau \leq M$ a.s. for some constant $M$ (which happens rarely) or some less stringent conditions as the ones described in the corresponding wikipedia article en.wikipedia.org/wiki/Optional_stopping_theorem $\endgroup$
    – Olivier
    Commented Mar 10, 2020 at 20:31

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