Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does the following property for martingales hold? Given a continuous martingale $(X_t)_{t\leq T}$ that is almost surely strictly positive at time T, i.e. $\mathbb{P}(X_T >0)=1$, we have $P(X_t > 0 \,\, \text{for all} \,\, t \in [0,T])=1$.

I tried fiddling around with the optional stopping theorem and different stopping times like $\inf\{t\leq T\, |\, X_t = 0\} \wedge T$, but have gotten nowhere yet.

share|cite|improve this question
that's the right idea, you know the field associated with the stopping time, (the pre-$\tau$ field) ? Use that $X_0,X_{\tau},X_T$ is martingale. – mike Jan 22 '13 at 21:51
up vote 5 down vote accepted

As you propose, let $\tau = \inf\{t : X_t = 0\} \wedge T$, and $A = \{X_t > 0\,\forall t \in [0,T]\}$. Since $\tau$ is a bounded stopping time, by optional stopping we have $E[X_0] = E[X_\tau] = E[X_T]$. On the other hand, since $X_T > 0$, we have $X_\tau \le X_T$ almost surely. It follows that $X_\tau = X_T$ almost surely ($X_T - X_\tau$ is a nonnegative random variable with expectation zero). On the event $A^c$ we have $0 = X_\tau < X_T$, so it must be that $P(A) = 1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.