Consider the random walk martingale $S_n=\sum_{k=1}^n X_k$ where $X_k$ are uniformly bounded, iid with $E(X_1)=0,E(X_1^2)=\sigma^2>0$. Let $a>0$ and set $T=\inf\{n:S_n\geq a\}$. Show that $E(\min_n S_{n\wedge T})=-\infty$.
I was thinking of defining $T(k)=\inf\{n:S_n\leq -k\}$ and using the martingale $S_{n\wedge (T\wedge T(k))}^2-(n\wedge T\wedge T(k))\sigma^2$. We will then get (using MCT and boundedness and $S_{n\wedge (T\wedge T(k))}^2$) $E(S^2_{T\wedge T(k)})=\sigma^2(T\wedge T(k))$. This implies $b^2P(T<T(k))+k^2P(T>T(k))=\sigma^2 E(T\wedge T(k))$. I am not sure how to proceed from here.
How about this?
For any $N < \infty$, by the optional sampling theorem, we have $E(S_{T \wedge T(k) \wedge N}) = 0$. And $E(S_{T \wedge T(k) \wedge N} I_{T < T(k)}) = E(S_{T \wedge N} I_{T < T(k)}) \ge a P(T < T(k) \wedge N) \to a$ as $N, k \to \infty$.
So $E(S_{T \wedge T(k) \wedge N} I_{T > T(k)}) = - E(S_{T \wedge T(k) \wedge N} I_{T < T(k)})$ converges to a negative number as $N,k \to \infty$.
Let $U = \min_n S_{n \wedge T}$. Now $U I_{U < -k} \le S_{T \wedge T(k) \wedge N} I_{T > T(k)}$. If $E(U) > -\infty$, then $E(U I_{U < -k}) \to 0$ as $k \to \infty$, which is a contradiction.