Is stopped brownian motion not a martingale ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T21:29:31Z http://mathoverflow.net/feeds/question/50228 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50228/is-stopped-brownian-motion-not-a-martingale Is stopped brownian motion not a martingale ? unknown (google) 2010-12-23T07:05:41Z 2010-12-23T11:20:21Z <p>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.</p> <p>(Quote)</p> <p>Does the martingale property </p> <p>\$\$M(t)=E[M(T)|F(t)]\$\$</p> <p>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)</p> <p>I do not understand why the martingale property does not hold in this case and appreciate any explanation on this. </p>