Uniformly Integrable

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Let $\{X_t\}$, $t\in T$ be a fmaily of random variables where $T$ is an arbitrary index set.  

[Definition] $\{X_t\}$ is said to be uniformly integrable iff $$\underset{A\rightarrow\infty}{\lim}\int_{|X_t|>A}|X_t|\,d\mathscr{P}=0$$ uniformly in $t\in T$.


[Theorem] The family $\{X_t\}$ is uniformly integrable if and only if the following two conditions are satisfied:
(1) $\mathscr{E}|X_t|$ is bounded in $t\in T$.
(2) For every $\varepsilon>0$, there exists $\delta(\varepsilon)>0$ such that for any $E\in\mathscr{F}$, $$\mathscr{P}(E)<\delta(\varepsilon)\implies\int_E|X_t|d\mathscr{P}<\varepsilon\mbox{ for every }t\in T.$$

$\bullet$ Proof.

$(\Rightarrow)$ Suppose $\{X_t\}$ is uniformly integrable.
For (1), choose $\varepsilon=1$, then there exists $A(\varepsilon)>0$ and $A(\varepsilon)+1\leq M<\infty$ such that $$\int_{|x_t|>A(\varepsilon)}|x_t|\,d\mathscr{P}<1.$$ Thus, $$\mathscr{E}|X_t|=\int_{|x_t|>A(\varepsilon)}|x_t|\,d\mathscr{P}+\int_{|x_t|\leq A(\varepsilon)}|x_t|\,d\mathscr{P}\leq 1+A(\varepsilon)\leq M\;\forall\,t\in T.$$
For (2), given $\varepsilon>0$, and for an event $E\in\mathscr{F}$ such that $\mathscr{P}(E)<\delta$ for some $\delta$.  Next we would find what is this $\delta$.$$\begin{array}{rl}\int_E|x_t|\,d\mathscr{P}&=\int_{E\cap\{|x_t|>A\}}|x_t|\,d\mathscr{P}+\int_{E\cap\{|x_t|\leq A\}}|x_t|\,d\mathscr{P}\\ &\leq\int_{|x_t|>A}|x_t|\,d\mathscr{P}+A\mathscr{P}(E).\end{array}$$Choose $\delta=\frac{\varepsilon}{2A}$ and choose $A$ such that $\int_{|x_t|>A}|x_t|\,d\mathscr{P}<\frac{\varepsilon}{2}$ due to the assumption of uniformly integrable. Then we have $$\int_E|x_t|\,d\mathscr{P}\leq\frac{\varepsilon}{2}+A\frac{\varepsilon}{2A}=\varepsilon\;\forall\,t\in T.$$

$(\Leftarrow)$ Suppose (1) and (2) hold.  Let $E_t=\{|X_t|>A\}$.  Then by Chebyshev;s inequality, $$\mathscr{P}(E_t)\leq\frac{\mathscr{E}|X_t|}{A}\leq\frac{M}{A}\mbox{ due to (1).}$$Choose $A>\frac{M}{\delta}$, then $\mathscr{P}(E_t)\leq\delta$, we have $$\int_{E_t}|x_t|\,d\mathscr{P}<\varepsilon\mbox{ due to (2).}$$Since $\varepsilon>0$ is arbitrary, $\{X_t\}$ is uniformly integrable.

$\Box$