Let X be a square integrable martingale with square variation process $\langle X\rangle$. Let $\tau$ be a finite stopping time. Show the following:
If $E(\langle X\rangle_{\tau})=\infty$, then both equalities $E((X_\tau-X_0)^2)=E(\langle X\rangle_\tau)$ and $E(X_\tau)=E(X_0)$ may fail.
The following answer was given by saz in another thread(here is the link Ex 10.2.1 in Klenke's Probability Theory textbook ):
Consider a simple random walk $(X_n)_{n \in \mathbb{N}}$ started at $X_0=0$ and the stopping time $$\tau := \inf\{n \in \mathbb{N}; X_n=1\}.$$
However I think this answers contradicts the fact $E(X_\tau)=\infty$ since $\tau := \inf\{n \in \mathbb{N}; X_n=1\}$ can be finite. For example $\tau$ could be $1$ if $X_1=1$.
Questions:
Am I misunderstanding this example? How should I solve this question?
Thanks in advance!
The first hitting time $$\tau:= \inf\{n \in \mathbb{N}; X_n=1\}$$ of a simple random walk $(X_n)_{n \in \mathbb{N}}$ (started at $X_0=0$) is almost surely finite, i.e. $\mathbb{P}(\tau<\infty)=1$. This implies that $$\mathbb{E}(X_{\tau}^2)=1 \quad \text{and} \quad \mathbb{E}(X_{\tau})=1.$$ On the other hand, it can be shown that $\mathbb{E}(\tau)=\infty$. (Note that this does not contradict $\tau<\infty$ almost surely. Finite random variables may have an infinite expectation.) Since the compensator $\langle X \rangle_n$ of $(X_n)_{n \in \mathbb{N}}$ equals $\langle X \rangle_n=n$, this means that $$\mathbb{E}(\langle X \rangle_{\tau}) = \mathbb{E}(\tau)=\infty.$$ Hence $$\mathbb{E}((X_{\tau}-X_0)^2) = 1 \neq \infty=\mathbb{E}(\langle X \rangle_{\tau})$$ and $$\mathbb{E}(X_{\tau})=1 \neq 0 = \mathbb{E}(X_0).$$