Expectation of positive supermartingale.

Question

Suppose that $\{(X_n, B_n), n \ge 0\}$ is a positive supermartingale and v is a stopping time. Show

$E(X_0) \ge E(X_v1[v<\infty])$. This is exercise 10.51 in textbook a probability path.

Answer 1

Let $\tau$ be a stopping time. By the optional sampling theorem, $(X_{n \wedge \tau})_{n \in \mathbb{N}}$ is a supermartingale and so

$$ \mathbb{E}(X_{n \wedge \tau}) \leq \mathbb{E}(X_0). \tag{1}$$

Since

$$\lim_{n \to \infty} X_{n \wedge \tau} = X_{\tau} \quad \text{on $\{\tau<\infty\}$},$$

and $X_{n \wedge \tau} \geq 0$, it follows from Fatou's lemma that

$$\mathbb{E}(X_{\tau} 1_{\{\tau<\infty\}}) \leq \liminf_{n \to \infty} \mathbb{E}(X_{n \wedge \tau}1_{\{\tau<\infty\}}) \leq \liminf_{n \to \infty} \mathbb{E}(X_{n \wedge \tau}) \stackrel{(1)}{\leq} \mathbb{E}(X_0).$$