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Let $X_t=\int_0^t W_s ds$ where $W_s$ is Brownian motion, so $E[W_s]=0$.

Then $E[X_t]=\int_0^t E[W_s] ds=\int_0^t 0 ds=0$.

So $E[X_t|{\cal F}_s]=0\neq X_s$, almost everywhere. So by previous sentence, $X_t$ is not a martingale.

Question: Is the above a correctly stated and argued proof that $X_t$ is not a martingale?

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    $\begingroup$ Please do a search. This has been discussed many times. $\endgroup$ – Gordon Oct 24 '17 at 20:29
  • $\begingroup$ See, for example, here. $\endgroup$ – Gordon Oct 24 '17 at 20:36
  • $\begingroup$ @Gordon, you did not address my specific argument, which is not the same as the arguments used in your answer in the linked question. It would be more helpful if you pointed out the holes in my argument. My question is really about what constitutes a good proof and whether this particular argument is a good proof or not. It is not really about whether the question has been answered in other ways. I already know that the answer is No and I have seen other proofs. $\endgroup$ – Lars Ericson Oct 24 '17 at 21:04
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    $\begingroup$ Your answer is not correct: zero expectation does not imply zero conditional expectation. $\endgroup$ – Gordon Oct 25 '17 at 0:00
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    $\begingroup$ With respect to your comment: as @Gordon has said, $0$ expectation does not imply $0$ conditional expectation, take for example the case of Brownian Motion, $E[W_t]=0$ but clearly $E[W_t|\mathcal{F}_s]=W_s$. $\endgroup$ – Daneel Olivaw Oct 25 '17 at 0:40

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