Exercise 4.8.3 Durret states
Let $S_n= \xi_1 + \ldots \xi_n$, $(\xi_i)_i$ independent with $0$ mean and $Var(\xi_i)=\sigma ^2$. Then $S_n^2 – n \sigma^2$ is a martingale. Let consider the stopping time $T= \inf \{ n: |S_n|> a \}$.
Show $E[T]> \frac{a^2}{\sigma^2}$.
I need to know if my argument is okay, especially the application of the optional stopping theorem!
Of course the filtration is the canonical one $F_n = \sigma(\xi_i, i \leq n) $ and is easy to show that $S_n^2 – n \sigma^2$ is a $(F_n)_n$-adapted martingale.
Morever, $T$ is a stopping time since $\{T= n \}$ can be written as \begin{align}
( \cap_{i=1}^{N-1} \{ |S_i|<a \}) \cup \{ |S_n|>a \}
\end{align}
and by construction this belongs to $F_n$
To show the claim, I need to use the optional stopping theorem applied to the given martingale. The standard argument is that $T$ is not bounded, so I consider the stopped martingale $ S_{n ∧ T} – (n ∧ T) \sigma^2$.
For every $n$, $n∧T$ is bounded and $(n∧T) \rightarrow_n T$ Here I should show that $P(T< +\infty)=1$ a.s, but don't know how
Hence, by OST applied to stopping times $T ∧ n$ and $0$:
\begin{align}
E[S_T – T] = E[\lim_n S_{T ∧ n} – (T ∧ n) \sigma^2] = \lim_n E[S_{T ∧ n} – (T ∧ n) \sigma ^2] = E[S_0]=0
\end{align}
where I used DCT.
Hence $E[T] \sigma^2= E[S_T^2]$, but when I stop at $T$ I have by definition: $|S_T|>a$ and hence follows $E[T] \sigma^2 = E[S_T^2]>E[a^2]=a^2$ and hence
\begin{align}
E[T]> \frac{a^2}{\sigma^2}
\end{align}
EDIT
To apply DCT, I note that $S_{ n \wedge T}^2 \leq (a+1)^2$ and hence, by OST, $E[T \wedge n]=E[S_{ n \wedge T}^2] \leq (a+1)^2$ and hence I can take the limit in $E[S_{ n \wedge T}^2 – T \wedge n]=0$ and obtain the thesis
Best Answer
I just want to summarise the discussion in the comments (as some of them have been deleted), give an argument that allows you to interchange taking limits and expectations, and make a remark about your application of the DCT.
First, you don't need to show that $P(T<\infty)=1$. You are able to apply the optional sampling theorem since $T\wedge n \le n$, which guarantees that $T\wedge n$ is a bounded stopping time.
We will now justify the step $$ E\left[\lim_{n\to \infty} S_{T\wedge n}^2 -(T\wedge n)\sigma^2\right] = \lim_{n\to \infty} E\left[ S_{T\wedge n}^2 -(T\wedge n)\sigma^2\right], \tag{*}\label{*} $$ in two separate steps using the dominated and monotone convergence theorems respectively.
As you argue in the comments, we observe first that $$\vert S_{T\wedge n} \vert \le \vert S_{(T\wedge n) -1} \vert + \vert S_{T\wedge n} - S_{(T\wedge n) -1}\vert \le a + \vert \xi_{T\wedge n} \vert.$$ Thus, $$ \vert S_{T\wedge n} \vert^2 = S_{T\wedge n}^2 \le \left(a + \left\vert \xi_{T\wedge n} \right\vert\right)^2, $$ where we know that the right-hand side is integrable. This means that $$ E\left[\lim_{n\to \infty} S_{T\wedge n}^2 \right] = \lim_{n\to \infty} E\left[ S_{T\wedge n}^2 \right]. $$
Now, observe that $T\wedge n$ is non-negative, and non-decreasing in $n$. Thus, we can use the monotone convergence theorem to conclude that $$ E\left[\lim_{n\to \infty}(T\wedge n)\sigma^2\right] = \lim_{n\to \infty} E\left[(T\wedge n)\sigma^2\right]. $$
Putting these two steps together gives us \eqref{*}.
Note that the argument you make in your edit is not correct, because it does not allow you to apply the DCT. The DCT requires that you find an integrable random variable $Z$ such that $T\wedge n \le Z$. What you've done is instead found an upper bound for $E[T\wedge n]$, which is not sufficient to apply the DCT.
The rest of your post looks fine.