Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 211 / 4 / d

If $X_{T-1}=0$ almost surely, the definition of $T$ gives $\mathbb{P}(X_T>0)>0$. Part c then implies $\mathbb{P}(\xi >0)>0$.

If $\mathbb{P}(X_{T-1}<0)>0$, the second term in the identity of part c is positive, so $\mathbb{P}(\xi >0)>0$.

The assumptions of this case imply $X_{T-1}\ge 0$ almost surely and $\mathbb{P}(X_{T-1}>0)>0$. This would satisfy the defining condition for $T$ one period earlier, contradicting the minimality of $T$. Thus the case is impossible.