Durrett Q4.8.3 on martingales and the Optional Stopping Theorem

Question

I'm working through this question from Durrett's probability textbook.

Let $S_n = \xi_1+\cdots +\xi_n$ where the $\xi_i$ are independent with $E\xi_i = 0$ and var($\xi_i) = \sigma^2$. $S_n^2 - n\sigma^2$ is a martingale. Let $T = \min\{n : |S_n| > a\}$. Use Theorem 4.8.2 to show $ET \geq a^2/\sigma^2$.

(Theorem 4.8.2 refers to the following result: If $X_n$ is a submartingale and $N$ is a stopping time such that $E|X_N| < \infty$ and $X_n 1_{\{N > n\}}$ is uniformly integrable, then $X_{N \wedge n}$ is also uniformly integrable and hence $EX_N \geq EX_0$)

If I define $X_n = S_n^2 - n \sigma^2$, then $X_n$ is a martingale and we have that $$EX_n1_{\{T > n\}} \leq ES_n^2 1_{\{T > n\}} \leq a^2$$ So, $X_n1_{\{T > n\}}$ is uniformly integrable.

I'm having trouble with the other condition needed - that is, $E|X_T| < \infty$.

Since the final result is immediate if $ET = \infty$, we can assume that $ET < \infty$.

$$ E|X_T| \leq ES_T^2 + \sigma^2 ET $$ So, it is enough to show that $ES_T^2 < \infty$.

Now, as if $T < \infty$, $S_{T-1}^2 \leq a^2$ and if $T = \infty$, $S_n^2 \leq a^2$ for all $n$. $$ ES_N^2 \leq a^2 + E\xi_N^2 + a^2$$ This reduces to showing that $E\xi_N^2 < \infty$, but I have not been able to show this.

Answer 1

As noted, we may assume that $\Bbb E[T]<\infty$.

Because the martingale $S_n^2-n\sigma^2$ when stopped at time $T$ is still a martingale (with value $0$ at time $0$) you have $0=\Bbb E[S_{n\wedge T}^2] -\sigma^2\Bbb E[n\wedge T]$. Therefore, $$ \sigma^2\Bbb E[T]\ge \sigma^2\Bbb E[n\wedge T]=\Bbb E[S_{n\wedge T}^2]\ge\Bbb E[S_T^2; n>T]\ge a^2\Bbb P[n>T]. $$ Because $\Bbb E[T]<\infty$, we have $\Bbb P[T<\infty]=1$, so $\lim_n\Bbb P[n>T]=1$.

The request to appeal to Theorem 4.8.2 looks like pedagogical overkill.

Answer 2

You are right that if $\mathbb E[T]=+\infty$ then the inequality is trivial, so we can assume without loss of generatily that $\mathbb E[T]<+\infty$.

I am not sure that your justification of uniform integrability is correct (and you conclude that a sequence of real numbers is uniformly integrable, which is weird). I would say that by definition, $\vert X_n1_{\{T>n\}}\vert\le a^2+T\sigma^2$. As $(X_n1_{\{T>n\}})_{n\in\mathbb N^*}$ is dominated by an integrable random variable, it is uniformly integrable.

Then, it is indeed sufficient to show that $\mathbb E[S_T^2]<+\infty$. To show that, I would use that $(X_n)_{n\in\mathbb N^*}$ and therefore $(X_{n\wedge T})_{n\in\mathbb N^*}$ is a martingale, hence $$ \mathbb E[S_{n\wedge T}^2]=\sigma^2\mathbb E[n\wedge T]\le\sigma^2\mathbb E[T]. $$ Then you have several ways to conclude, e.g. by Doob's inequality ($\mathbb E[S_T^2]\le4\sup_{n\in\mathbb N^*}\mathbb E[S_{n\wedge T}^2]\le\sigma^2\mathbb E[T]$), or by Fatou's lemma ($\mathbb E[S_T^2]=\mathbb E[\liminf_{n\to+\infty}S_{n\wedge T}^2]\le\liminf_{n\to+\infty}\mathbb E[S_{n\wedge T}^2]\le\sigma^2\mathbb E[T]$).