Stopping Theorem - Counterexample

Question

I am looking at the stopping theorem in my script:

Let $(X_n)_{n\in\mathbb{N}}$ be a martingale and let $T$ be a stopping time. Then $(X_{\min(n,T)})_{n\in\mathbb{N}}$ is also a martingale. In particular if $T$ is bounded, then $X_T\in L^1$ and we have that $\mathbb{E}(X_T) = \mathbb{E}(X_0)$.

Then after this there is an example that aims to point out that the hypothesis that $T$ is bounded is crucial. They consider the random walk $X_n = Y_1+\cdots+Y_n$ starting at zero with $Y$'s having the Rademacher distribution (which is a martingale). Now if $T = \inf\{n\geq0:X_n=1\}<\infty$ then we would have $1=\mathbb{E}(X_T)\neq\mathbb{E}(X_0)=0$. But $T$ is not bounded and hence there is no contradiction with the theorem.

My question: How does one see that $T=\inf\{n\geq0:X_n=1\}$ is not bounded?

Answer 1

For any $n \in \mathbb{N}$ we have

$$\mathbb{P}(X_k \leq 0 \, \, \text{for all $k \leq n$}) \geq \mathbb{P}(Y_1 = \ldots = Y_k = -1) = \frac{1}{2^n} ;$$

since

$$\{X_k \leq 0 \, \, \text{for all $k \leq n$}\} \subseteq \{T>n\}$$

this implies

$$\mathbb{P}(T>n) \geq \mathbb{P}(X_k \leq 0 \, \, \text{for all $k \leq n$}) \geq \frac{1}{2^n}>0.$$

As $n \in \mathbb{N}$ is arbitrary, this proves that $T$ is unbounded.

Answer 2

Proof by contradiction.

Suppose $T$ is bounded. Then there exists some finite $k$ such that $P(T \leq k)=1$.

But consider that $P(Y_1=Y_2=\dots=Y_k=Y_{k+1}=-1)=\frac{1}{2^{k+1}}$.

Therefore we see $P(T \leq k)=1-P(T>k)< 1-\frac{1}{2^{k+1}}<1 \Rightarrow P(T\leq k)\neq1\Rightarrow T$ is not bounded.

Answer 3

In my very first MSE answer, I show that $\mathbb{P}(T\geq 2t)={2t\choose t}\left({1\over 2}\right)^{2t}.$ Since this probability is positive for every positive $t$, the random variable $T$ is not bounded. There is at least a tiny chance that $T$ takes very large values.