Is the sequence $(X_n)$ of $L^1$ random variables uniformly integrable?

  • MHB
  • Thread starter Euge
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    2017
In summary, uniform integrability of a sequence of $L^1$ random variables means that the sequence is well-behaved with respect to integration. Uniform integrability is a stronger condition than convergence in $L^1$, and the conditions for a sequence to be uniformly integrable include uniform boundedness and uniform absolute continuity. A sequence of $L^1$ random variables cannot be uniformly integrable if it is not uniformly bounded, and some examples of non-uniformly integrable sequences include independent random variables with different distributions, random variables with increasing variances, and random variables that converge in probability but not in $L^1$.
  • #1
Euge
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Here is this week's POTW:

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Let $(X_n)$ be a sequence of $L^1$ random variables on a probability space $(\Omega, \Bbb P)$. Let $f$ be a continuous, nondecreasing function from $[0,\infty)$ onto itself such that

1. $\Bbb E[f(|X_n|)]$ is uniformly bounded

2. $\dfrac{f(x)}{x}\to \infty$ as $x\to \infty$

Show that $(X_n)$ is uniformly integrable.
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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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  • #2
Due to the difficulty some may have had with this problem, I'm extending the deadline one more week.
 
  • #3
No one answered this week's problem. You can read my solution below.
Set $M = \sup_{m\in \Bbb N} E[f(\lvert X_n\rvert)]$. Given $\epsilon > 0$, choose $\delta > 0$ such that for all $x$, $x \ge \delta$ implies $f(x) > \frac{x}{\epsilon}$. For all $n\in \Bbb N$,

$$E[\lvert X_n\rvert I_{\lvert X_n\rvert \ge \delta}] \le E[\epsilon f(\lvert X_n\rvert)I_{\lvert X_n\rvert \ge \delta}] \le \epsilon M$$

Since $\epsilon$ was arbitrary, $(X_n)$ is uniformly integrable.
 

FAQ: Is the sequence $(X_n)$ of $L^1$ random variables uniformly integrable?

1. What does it mean for a sequence of $L^1$ random variables to be uniformly integrable?

Uniform integrability of a sequence of $L^1$ random variables means that the sequence is well-behaved with respect to integration. In other words, the sequence satisfies certain conditions that allow us to interchange the order of taking limits and integration.

How is uniform integrability related to convergence in $L^1$?

Uniform integrability is a stronger condition than convergence in $L^1$. A sequence of $L^1$ random variables that is uniformly integrable must also converge in $L^1$, but the converse is not necessarily true. In other words, uniform integrability implies convergence in $L^1$, but convergence in $L^1$ does not necessarily imply uniform integrability.

What are the conditions for a sequence of $L^1$ random variables to be uniformly integrable?

The conditions for a sequence of $L^1$ random variables to be uniformly integrable are:

  • Uniform boundedness: The sequence must be bounded in the $L^1$ norm, i.e. $\sup_{n} \int |X_n| \, dP < \infty$.
  • Uniform absolute continuity: For any $\epsilon > 0$, there exists a $\delta > 0$ such that for all $A \in \mathcal{F}$ with $P(A) < \delta$, we have $\int_A |X_n| \, dP < \epsilon$ for all $n$.

Can a sequence of $L^1$ random variables be uniformly integrable if it is not uniformly bounded?

No, a sequence of $L^1$ random variables cannot be uniformly integrable if it is not uniformly bounded. Uniform boundedness is one of the necessary conditions for uniform integrability.

What are some examples of sequences of $L^1$ random variables that are not uniformly integrable?

Some examples of sequences of $L^1$ random variables that are not uniformly integrable are:

  • A sequence of independent random variables with different distributions.
  • A sequence of random variables with increasing variances.
  • A sequence of random variables that converges in probability but not in $L^1$.

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