1
$\begingroup$

I have two questions about the uniform integrability.

The definition I am using is that a class of random variables $\mathbb{\chi}$ is uniformly integrable if given an $\epsilon >0$, there exists a $k$ such that for any $x$ in $\chi$ we have $\mathbb{E}[|x| \mathbb{I}\{x \geq k\}] < \epsilon$.

First, is that in the definition of uniform integrability, can the density function which we compute the expectation with respect to it, vary with $n$?

Second, suppose I have a sequence of random variables $(x_n)_{n\geq 1}$ (with varying density function w.r.t n). For the two functions $f,g$, I know that $f(x) \leq g(x)$ for all $x$. Does the uniform integrability of the sequence $(g(x_n))_{n \geq 1}$ imply the uniform integrability of $(f(x_n))_{n \geq 1}$?

$\endgroup$
1
  • 1
    $\begingroup$ There are varying definitions of uniform integrability, context and author dependent. It would be good to edit in the definition and context you are using $\endgroup$
    – FShrike
    Dec 27, 2021 at 16:48

1 Answer 1

2
$\begingroup$

A family of random variables $(X_{\alpha})_{\alpha\in A}$ is u.i. if $$ \sup_{\alpha\in A}\mathsf{E}|X_{\alpha}|1\{|X_{\alpha}|>M\}\to 0 $$ as $M\to\infty$. Note that $X_{\alpha}$'s may have different distributions (densities if exist). If $(Y_{\alpha})_{\alpha\in A}$ is a u.i. family of r.v.s. satisfying $|X_{\alpha}|\le |Y_{\alpha}|$ a.s. for all $\alpha\in A$, then $(X_{\alpha})_{\alpha\in A}$ is u.i. as well because for any $\alpha\in A$, $$ \mathsf{E}|X_{\alpha}|1\{|X_{\alpha}|>M\}\le \mathsf{E}|Y_{\alpha}|1\{|Y_{\alpha}|>M\}. $$

$\endgroup$
2
  • $\begingroup$ Thanks! I know intuitively that the second statement should be true, but I don't see why we can write the last inequality you wrote. Because if we consider the the integral representation of the expectation, then we should also know the comparison of probability density of Y and the one of X, to deduce such an inequality. $\endgroup$
    – Rostam22
    Dec 27, 2021 at 17:25
  • 1
    $\begingroup$ @Rostam22 That's because $|X_{\alpha}|1\{|X_{\alpha}|>M\}\le |Y_{\alpha}|1\{|Y_{\alpha}|>M\}$ a.s. without resorting to "the integral representation". $\endgroup$
    – user140541
    Dec 27, 2021 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.