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I am confused with the way uniform integrability is defined in the context of random variables. Keeping with the analysis idea, I had expected this definition:

If $X$ is a random variable, given $\epsilon \ge 0$ and $A \subseteq \Omega$, there must be a $\delta$ such that $\int_{A} d\mu \le \delta$ and $\int_{A} X(\omega) d \omega \le \epsilon$.

Instead this is what I find in most places including Wiki (when $K \subset L^1(\mu)$):

$$\lim_{c \to \infty} \sup_{X \in K} \int_{|X|\ge c} |X| d \mu = 0$$

What is the reason for this different definition? Why does the bound on the value of $X$ even make an appearance?

Did
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Bravo
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2 Answers2

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Three conditions are involved:

(C) For every $\varepsilon\gt0$, there exists a finite $c$ such that, for every $X$ in $\mathcal H$, $\mathrm E(|X|:|X|\geqslant c)\leqslant\varepsilon$.

(C1) There exists a finite $C$ such that, for every $X$ in $\mathcal H$, $\mathrm E(|X|)\leqslant C$.

(C2) For every $\varepsilon\gt0$ there exists $\delta\gt0$ such that, for every measurable $A$ such that $\mathrm P(A)\leqslant\delta$ and every $X$ in $\mathcal H$, $\mathrm E(|X|:A)\leqslant\varepsilon$.

Then (C) is equivalent to (C1) and (C2). To wit:

  • (C) implies (C1) since $\mathrm E(|X|)\leqslant c+\mathrm E(|X|:|X|\geqslant c)$.

  • (C) implies (C2) since $\mathrm E(|X|:A)\leqslant c\mathrm P(A)+\mathrm E(|X|:|X|\geqslant c)$.

  • (C1) and (C2) imply (C) since $\mathrm P(|X|\geqslant c)\leqslant\mathrm E(|X|)/c$.

The intuition might lie in the decomposition $A=(A\cap[|X|\lt c])\cup(A\cap[|X|\geqslant c])$ used to show that (C) implies (C2): for any given $c$, the expectation of $|X|$ on the first part is controlled uniformly over $X$ by something of the order of $\mathrm P(A)$ and the expectation of $|X|$ on the second part is controlled uniformly over $A$ thanks to condition (C).

Community
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Did
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  • Thanks @Didier. A question here: the condition (C) is what is mentioned in Wiki as $\epsilon \to 0$. What is the reason for $c \to \infty$? Instead shouldn't we have the relation for all finite $c$? – Bravo Mar 20 '12 at 09:38
  • Since the functions $c\mapsto\mathrm E(|X|:|X|\geqslant c)$ and $c\mapsto\sup\limits_{X\in\mathcal H}\mathrm E(|X|:|X|\geqslant c)$ are nonincreasing and positive, they converge to zero when $c\to\infty$ if and only if for every positive $\varepsilon$ they are at most $\varepsilon$ for some $c$. – Did Mar 20 '12 at 09:43
  • +1 Nice to know. – Tim Mar 21 '12 at 12:15
  • Thm 4.5.3 in Chung's. – Tim Mar 21 '12 at 12:59
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There is an equivalent definition of uniform integrability which goes: Let $(\Omega,\mathcal{F},P)$ be a probability space. A set $\mathcal{H}$ of random variables is uniformly integrable if and only if $$ \sup_{X\in \mathcal{H}}\int_\Omega |X| dP < \infty $$ and $$ \forall \epsilon >0\, \exists \delta >0\;\forall A\in \mathcal{F}: P(A)\leq \delta \Rightarrow \sup_{X\in\mathcal{H}} \int_A |X| dP\leq \epsilon. $$ Maybe this it what you are looking for.

Stefan Hansen
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  • Thanks Stefan. I understand this definition, but do you have any intuition for the one from Wiki that I have indicated in the question? – Bravo Mar 10 '12 at 03:37
  • I'm not sure what you mean. The intuition for defining uniform integrability as Wikipedia does rather than the way you do? – Stefan Hansen Mar 11 '12 at 00:23
  • Yes, that's what I meant. Especially I have no clue why the integral is taken over $|X|\ge c$. Could you explain? – Bravo Mar 11 '12 at 11:43
  • I'm not sure there is any intuition to it. But if the "analysis" definition is intuitive to you, then just stick with that, since they're equivalent. – Stefan Hansen Mar 12 '12 at 15:25