If a stochastic process has the same expectation value for all pisitive t, then is it a martingale? I don’t know how to show it whether that is right.
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The Brownian motion $W_t$ is a martingale because for all $u>t$: $$ W_t = E(W_u | \mathcal{F}_t), $$
However, the process $X_t$
$$X_t = 1_{\{a<t<b\}} (2 . 1_{\{W_a < 0\}} -1) $$ has 0 expectation but is not a martingale.
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$\begingroup$ Hmm... Why 21? If you give more information , I'll very thanks for you. $\endgroup$– HobongDec 10, 2018 at 15:33
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$\begingroup$ I just added the dot between 2 and 1, this process takes value +1 or -1 between a and b, but it is not a martingale as it goes back to 0 for $t>b$ $\endgroup$– SebapiDec 10, 2018 at 15:50