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In page 45 of the book "Financial Derivatives In Theory and Practice by P.J.Hunt and J.E.Kennedy, it seems to me that the author says the stopped Brownian Motion is not a martingale as follows.


Does the martingale property


hold if $T$ is a stopping time? In general the answer is no, as can be seen by taking M to be Brownian Motion and $T=\inf\{t>0: M(t)\ge1\}$ (Unquote)

I do not understand why the martingale property does not hold in this case and appreciate any explanation on this.

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See… to close as this is better suited to one of the sites in the FAQ. – Steve Huntsman Dec 23 '10 at 7:49
Take the expectation of both sides. – Nate Eldredge Dec 23 '10 at 9:45
Optional stopping of a martingale is OK for BOUNDED stopping times, and for some weaker conditions. But not for general stopping times, as this example illustrates. – Gerald Edgar Dec 23 '10 at 19:07

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