Let $\{X_n \}$ be a uniformly integrable martingale w.r.t. the natural filtration $\{ \mathcal{F}_n \}$.

Is $\{ X_\tau : \tau \text{ is a stopping time w.r.t. } \{ \mathcal{F}_n \} \}$ uniformly integrable?

I've tried using this theorem that conditional expectations of integrable random variables forms a uniformly integrable family, but this isn't working because I don't see how $X_\tau$ is actually the same thing as $\mathbb{E}(X_n | \mathcal{F}_\tau)$.

I'm really not sure how to proceed here... can anyone help?

Note that $X_\tau := \sum\limits_{n=0}^\infty X_n \textbf{1} \{ \tau = n\}$.

  • $\begingroup$ See Theorem 9.3.5 in Chung's 'A Course in Probability Theory'. $\endgroup$ Sep 6, 2018 at 8:41
  • $\begingroup$ You have to apply the theorem in Chung with $\tau$ as one stopping time and $\infty$ as the other. Define $X_{\infty}$ as the limit of the martingale (so $X_n=E(X|\mathcal F_n)$). $\endgroup$ Sep 6, 2018 at 9:08

1 Answer 1


Since the martingale $(X_n)_{n \in \mathbb{N}}$ is uniformly integrable, the limit $X_{\infty} = \lim_{n \to \infty} X_n$ exists pointwise and in $L^1$; in particular, $$\forall n \in \mathbb{N}_0: \quad \mathbb{E}(X_{\infty} \mid \mathcal{F}_n) = X_n. \tag{1}$$

Claim: For any stopping time $\tau$ it holds that $\mathbb{E}(X_{\infty} \mid \mathcal{F}_{\tau}) = X_{\tau}$.

Indeed: For $F \in \mathcal{F}_{\tau}$ we have $F \cap \{\tau=n\} \in \mathcal{F}_n$, and so

$$\int_F X_{\tau} \, d\mathbb{P} = \sum_{n \geq 0} \int_{F \cap \{\tau=n\}} \underbrace{X_{\tau}}_{=X_n} \, d\mathbb{P} \stackrel{(1)}{=} \int_{F \cap \{\tau=n\}} X_{\infty} \, d\mathbb{P} = \int_F X_{\infty} \, d\mathbb{P}.$$

Since $X_{\tau}$ is $\mathcal{F}_{\tau}$-measurable, this proves $\mathbb{E}(X_{\infty} \mid \mathcal{F}_{\tau}) = X_{\tau}$.

Applying the result mentioned in your question, we conclude that the family $\{X_{\tau}; \tau$ stopping time$\}$ is uniformly integrable.


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