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Let $\{X_n\}$ be a sequence of iid random variables such that $E|X_1| = \mu < \infty$. I would like to show that $\{\bar{X}_n\}$ is uniformly integrable.

Thoughts

By the strong law of large numbers, $\bar{X}_n \rightarrow \mu$ almost surely, so it converges in probability. By the continuous mapping theorem, $|\bar{X}_n| \rightarrow |\mu|$ almost surely. If I could show that $E |X_n| \rightarrow |\mu|$, that would imply $\{\bar{X}_n\}$ is UI. It would also suffice to show $\bar{X}_n \rightarrow \mu$ in $L_1$.

I also tried to use the definition of uniform integrability to show that $$ \lim_{t \rightarrow \infty} \sup_n E(|\bar{X}_n| \, 1[\bar{X}_n > t]) = 0.$$

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Hint: Show this in two steps:

  1. If $(X_n)_{n\geq 1}$ is a sequence of i.i.d. random variables such that ${\rm E}[|X_1|]<\infty$, then $\{X_n\mid n\geq 1\}$ is uniformly integrable.

  2. If $\{X_n\mid n\geq 1\}$ is uniformly integrable, then also $\{\bar{X}_n\mid n\geq 1\}$ is uniformly integrable, where $\bar{X}_n=\frac1n \sum_{i=1}^n X_i$.

The first should be pretty straightforward to show since the sequence is i.i.d. For the second result, I recommend that you use the following equivalent definition of uniformly integrability:

A sequence $(Y_n)$ is uniformly integrable if and only if

a) $\sup_{n\ge 1} {\rm E}[|Y_n|]<\infty$,

b) $\forall \varepsilon>0\;\exists\delta>0: P(A)\le \delta\implies \sup_{n\ge 1}{\rm E}[|Y_n|;A]\le\varepsilon$.

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