# Strictly positive martingales

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.

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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

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 $P(A) = 1$.