# Is stopped brownian motion not a martingale ?

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.

(Quote)

Does the martingale property

$$M(t)=E[M(T)|F(t)]$$

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 mathoverflow.net/questions/27042/… ...voting 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