# Bound on expectation of minimum of two stopping times (proving integrability)

Let $$(M_n)_{n\geq0}$$ be a non-negative martingale with filtration $$(\mathcal{F}_n)_{n\geq0}$$. Suppose $$M_0=1$$ and set $$T=\min\{n\geq0:M_n=0\}.$$ Also, for $$R>0$$, consider the stopping time $$T_R=\min\{n\geq0:M_n\geq R\}.$$ In previous parts of the question we have shown that $$M_n=0$$ for all $$n\geq T$$ almost surely, and that $$\mathbb{P}(T_R<\infty)\leq 1/R$$. Now suppose further that, for some constants $$\sigma\in(0,\infty)$$ and $$\tau\in[1,\infty)$$ we have $$\mathbb{E}((M_{n+1}-M_n)^2|\mathcal{F}_n)\geq\sigma^21_{\{T>n\}}\hspace{0.5cm}\text{and}\hspace{0.5cm}M_{T_R}\leq\tau R.$$ Set $$Y_n=M_n^2-\sigma^2n$$ and $$X_n=Y_{n\wedge T\wedge T_R}$$. I have shown that $$X_n$$ is a submartingale, and am now required to show that $$\mathbb{E}(T\wedge T_R)\leq\frac{\tau^2R}{\sigma^2}.$$

My attempt so far: For any $$n\geq0$$ we have $$|X_n|=|Y_{n\wedge T\wedge T_R}|=|M_{n\wedge T\wedge T_r}^2-\sigma^2(n\wedge T\wedge T_R)|\leq\tau^2R^2+\sigma^2(T\wedge T_R).$$ Thus, if I can show that $$(T\wedge T_R)$$ is integrable, we can use the dominated convergence theorem on $$X_n$$, which as $$n\to\infty$$ would converge almost surely to $$X_{T\wedge T_R}$$ (because integrability also implies that $$T\wedge T_R$$ is almost surely finite), and the result would follow by the fact that $$\mathbb{E}(X_n)\geq\mathbb{E}(X_0)=1$$ by the optional stopping theorem.

However, I can't see how to prove integrability of $$(T\wedge T_R)$$, so any advice and hints would be greatly appreciated!

$$X$$ is a submartingale with $$X_0 = 1$$. So you have $$\mathbf E[M^2_{T \wedge T_R \wedge n} - \sigma^2 (T \wedge T_R \wedge n)] \ge 1 >0$$. Rearrange this to get an inequality for $$\sigma^2 \mathbf E[T \wedge T_R \wedge n]$$. You can take $$n \to \infty$$ into the expectations using the Monotone Convergence Theorem on the LHS. (which sidesteps the issue of integrability and then gives it for free) On the right, use the Dominated Convergence Theorem. (either $$T_R < \infty$$, in which case $$M^2_{T \wedge T_R \wedge n} \le \tau^2 R^2$$ or $$T_R = \infty$$, in which case $$M$$ is bounded above by $$R$$)
So we now have $$T \wedge T_R$$ is integrable, so $$T, T_R$$ are simultaneously infinite with probability $$0$$. Now, you should note that we (a.s.) cannot have $$T < T_R$$ for any $$R$$ such that $$T_R$$ is finite, since once $$M$$ hits $$0$$, it a.s. stays there. (from part a) So $$T \ge T_R$$ a.s. whenever $$T_R$$ is finite. Now use the information given and part b to conclude.