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Strong Law of Large Numbers demonstrates if $X_1,\,X_2,\,\ldots$ are i.i.d. random variables with $\mathbb{E}|X|<\infty$, $S_n:=\sum_{k=1}^n X_k$, then $$\frac{S_n}{n}\to \mathbb{E} X\quad\text{a.s.}$$ My question is can you give an example such that $\frac{S_n}{n}\not\to \mathbb{E} X$ in $L^1$?

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No you cannot find such an example since $$ \frac{S_n}{n}\to E[X]\quad \text{in }L^1. $$ It follows from the fact that it converges almost surely and that $$ \left\{\frac{S_n}{n}\;\bigg|\; n\geq 1\right\} $$ is uniformly integrable.

Hint on how to show the uniform integrability property:

  1. Using the iid assumption, show that $\{X_n\mid n\geq 1\}$ is uniform integrable.

  2. Now show that $\left\{\frac{S_n}{n}\mid n\geq 1\right\}$ is uniformly integrable using for example the equivalent formulation of uniform integrability $(*)$.

$(*)$ A sequence $(X_n)_{n\geq 1}$ is uniformly integrable if and only if $$ \sup_{n\geq 1}\int |X_n|\,\mathrm dP<\infty $$ and for all $\varepsilon>0$ there exists a $\delta>0$ such that if $P(A)\leq \delta$ then $$ \sup_{n\geq 1}\int_A |X_n|\,\mathrm dP\leq \varepsilon. $$

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Perhaps you could add some hint how to prove the uniform integrability. – saz Feb 26 '13 at 8:46
@Stephan Hensan Thanks indeed. I'm not sure if the second condition in your hint is easily applied. But perhaps a better way is to realize that $\frac{S_n}{n}$ is a backward martingale. Another way is to check directly by definition that $\frac{S_n}{n}$ is $U.I.$ Thank you a second time. – Coiacy Feb 28 '13 at 2:55

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