ddcon4r

2022-07-01

I have two questions about the uniform integrability.

The definition I am using is that a class of random variables $\chi $ is uniformly integrable if given an $\u03f5>0$, there exists a $k$ such that for any $x$ in $\chi $ we have $\mathbb{E}[|x|\mathbb{I}\{x\ge k\}]<\u03f5$.

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 $({x}_{n}{)}_{n\ge 1}$ (with varying density function w.r.t n). For the two functions $f,g$, I know that $f(x)\le g(x)$ for all $x$. Does the uniform integrability of the sequence $(g({x}_{n}){)}_{n\ge 1}$ imply the uniform integrability of $(f({x}_{n}){)}_{n\ge 1}$?

The definition I am using is that a class of random variables $\chi $ is uniformly integrable if given an $\u03f5>0$, there exists a $k$ such that for any $x$ in $\chi $ we have $\mathbb{E}[|x|\mathbb{I}\{x\ge k\}]<\u03f5$.

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 $({x}_{n}{)}_{n\ge 1}$ (with varying density function w.r.t n). For the two functions $f,g$, I know that $f(x)\le g(x)$ for all $x$. Does the uniform integrability of the sequence $(g({x}_{n}){)}_{n\ge 1}$ imply the uniform integrability of $(f({x}_{n}){)}_{n\ge 1}$?

Ordettyreomqu

Beginner2022-07-02Added 22 answers

A family of random variables $({X}_{\alpha}{)}_{\alpha \in A}$ is u.i. if

$\underset{\alpha \in A}{sup}\mathsf{E}|{X}_{\alpha}|1\{|{X}_{\alpha}|>M\}\to 0$

as $M\to \mathrm{\infty}$. 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 $\alpha \in A$,

$\mathsf{E}|{X}_{\alpha}|1\{|{X}_{\alpha}|>M\}\le \mathsf{E}|{Y}_{\alpha}|1\{|{Y}_{\alpha}|>M\}.$

$\underset{\alpha \in A}{sup}\mathsf{E}|{X}_{\alpha}|1\{|{X}_{\alpha}|>M\}\to 0$

as $M\to \mathrm{\infty}$. 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 $\alpha \in A$,

$\mathsf{E}|{X}_{\alpha}|1\{|{X}_{\alpha}|>M\}\le \mathsf{E}|{Y}_{\alpha}|1\{|{Y}_{\alpha}|>M\}.$

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