3
$\begingroup$

If $\{X_n\}$ is a uniformly integrable discrete time Martingale, and if $\tau$ is a (possibly $\infty$-valued) stopping time, then $\{X_{\tau \wedge n}\}$ is uniformly integrable (note I don't care that $\{X_{\tau \wedge n}\}$ is a martingale). The only proof I've seen of this uses stuff about test functions. Is it possible to give an elementary proof without this test function definition of uniform integrability?

$\endgroup$

1 Answer 1

4
$\begingroup$

Since $(X_n)$ is UI, we know there exists $X_\infty \in L^1(\mathcal F_\infty)$ such that $(X_n) \rightarrow X_\infty$ a.s. and in $L^1$. Furthermore, the uniform integrability and optional stopping theorem guarantee that $X_{n \wedge \tau} = \mathbb{E}[X_\infty | \mathcal F_{n \wedge \tau}]$. For the definition of uniform integrability, we want to show that $\sup_{n} \mathbb{E}[|X_{n \wedge \tau}| 1_{|X_{n \wedge \tau}| > R}]$ tends to $0$ as $R$ tends to $\infty$. We compute

\begin{align*} \mathbb{E}[|X_{n \wedge \tau}| 1_{|X_{n \wedge \tau}| > R}] &= \mathbb{E}[|\mathbb{E}[X_\infty|\mathcal F_{n \wedge \tau}]| 1_{|X_{n \wedge \tau}| > R}] \\ &\le \mathbb{E}[\mathbb{E}[|X_\infty||\mathcal F_{n \wedge \tau}] 1_{|X_{n \wedge \tau}| > R}] \\ &= \mathbb{E}[\mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}|\mathcal F_{n \wedge \tau}]] \\ &= \mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}] \end{align*}

and by Markov's inequality $P(|X_{n \wedge \tau}| > R) \le \frac 1R \mathbb{E}[|X_{n \wedge \tau}|] \le \frac 1R \mathbb{E}[|X_{\infty}|]$, where the last inequality comes from the same argument as above. Therefore we can choose $R$ such that $P(|X_{n \wedge \tau}| > R)$ is arbitrarily small for all $n$, and therefore $\mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}]$ can be made arbitrarily small as well. Since $$\mathbb{E}[|X_{n \wedge \tau}| 1_{|X_{n \wedge \tau}| > R}] \le \mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}]$$ this proves the uniform integrability of $(X_{n \wedge \tau})$.

$\endgroup$
1
  • $\begingroup$ Thank you! I was actually trying to prove OST with my question but I already have the bounded version and thankfully that bounded version suffices to prove $X_{n \wedge \tau} = \mathbb{E}[X_\infty | \mathcal F_{n \wedge \tau}]$ since we have $$X_{n \wedge \tau} = \mathbb{E}[X_n | \mathcal F_{n \wedge \tau}]= \mathbb{E}[ \mathbb{E}[X_\infty| \mathcal F_{n}]| \mathcal F_{n \wedge \tau}]= \mathbb{E}[X_\infty | \mathcal F_{n \wedge \tau}] $$ $\endgroup$
    – bart
    Dec 10, 2020 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.