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This is probably a stupid question, but am I mislead if I think that as soon as a stochastic process indexed by $t$ (continuous time) is not uniformly integrable (UI) for a certain range of $t$, say on $[0,1]$, it is also not UI on $[0,\infty]$? And if that is wrong, then why so?

Thank you in advance!

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Oh, I made a mistake, very sorry, I wanted to know if it holds for a process, which is NOT UI, changed my question. –  kelu Apr 25 '12 at 8:20

2 Answers 2

up vote 2 down vote accepted

For any collections $C$ and $D$ of random variables, if $D$ is uniformly integrable and $C\subseteq D$, then $C$ is uniformly integrable.

To see this, recall that $D$ is uniformly integrable if and only if $S(x,D)\to0$ when $x\to+\infty$, where $$ S(x,D)=\sup\{\mathrm E(|X| ; |X|\geqslant x)\mid X\in D\}. $$ Now, if $C\subseteq D$, then $S(x,C)\leqslant S(x,D)$ for every $x$, hence the conclusion follows.

Thus, the contraposition holds: if $C$ is not uniformly integrable and $C\subseteq D$, then $D$ is not uniformly integrable.

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I guess the definition of a stochastic process on an interval $I$ you are using is the following:

$\{X_t,t\in I\}$ is uniformly integrable if and only if $\lim_{R\to\infty}\sup_{t\in I}\int_{\{|X_t|\geq R\}}X_t(\omega)\mathrm d\mathcal P(\omega)=0$.

With this definition, we can have that $\{X_t,0\leq t\leq 1\}$ is uniformly integrable but $\{X_t,0\leq t<\infty\}$ is not. We can take a deterministic example, like $$X_t(\omega)=\begin{cases}0&\mbox{ if }0\leq t\leq 1\\\ t&\mbox{ if }1<t<+\infty. \end{cases}$$

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