$\{X_n\}$ is uniformly integrable if $\lim_{M \rightarrow \infty} (\sup_n \mathbb{E}(|X_n| \chi_{|X_n| > M}) = 0$

I would like to know if

$\{X_i\}$ uniformly integrable $\implies \sup_n \mathbb{E}(|X_n|^p) < \infty$, for p > 1.

I know that for p = 1 that is not true! But is it for p > 1?

  • $\begingroup$ That is not the usual definition of uniform integrability. $\endgroup$ – zhw. May 3 '16 at 22:52

You meant $\{X_n\}$ are uniformly integrable if

$$ \lim_{M\to\infty} \sup_n E[ |X_n|\chi_{\{|X_n|>M\}}]=0$$

(e.g. https://en.wikipedia.org/wiki/Uniform_integrability)

Now for your question. Answer is NO.

Why ? In general the random variables are not in $L^p$. Easy example to remember $X_n = X$, where $X$ is a random variable in $L^1$ but not in $L^p$ for any $p>1$. Then the sequence is trivially UI.

The the reverse implication is true: , Given $p>1$, if $\sup E [|X_n|^p] =C <\infty$, then $\{X_n\}$ are uniformly integrable.


By Holder,

$$ E [|X_n| \chi_{\{|X_n| >M\}}]\le (E [|X_n|^p])^{1/p} P(|X_n|>M)^{1/q}=(*)$$

But, $$P(|X_n|>M) \le E [|X_n|^p]/M^p.$$


$$(*) \le C M^{-p/q},$$

and the result follows.

  • $\begingroup$ Yes! I'm sorry! It was a misprint! Thanks for your answer! $\endgroup$ – Luísa Borsato May 3 '16 at 23:58

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