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?


1 Answer 1


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 $(X_{n \wedge \tau})$.

  • $\begingroup$ 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

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