MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
show/hide this revision's text 2 Texified

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)]

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

hold if T $T$ 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)>=1} $T=\inf\{t>0: M(t)\ge1\}$ (Unquote)

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

show/hide this revision's text 1

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)>=1} (Unquote)

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