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Suppose that $X_1,\dots, X_n$ are positive iid continuous random variables satisfying $E(X_i)=\infty$, and denote the sample average by $\bar{X}_n=n^{-1}\sum_{i=1}^nX_i$, then it is straightforward to show that $$\limsup_{n\rightarrow\infty}\ \bar{X}_n=\infty \quad\text{almost surely.}$$ This is somewhat of a converse Strong Law of Large Numbers (SLLN), although it only describes the limit superior. I am interested in what can be said of the limit inferior, which I think is required to show \begin{align} \lim_{n\to\infty}P\left[\bar{X}_n> 1+o(1)\right] = 1. \end{align} If I can show that $\liminf_{n\to\infty}\bar{X}_n$ diverges in some probabilistic sense, then I believe that Fatou's Lemma completes the rest.

  1. Did I overlook something?
  2. How do I show that $\liminf_{n\to\infty}\bar{X}_n$ diverges?
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Under your assumptions it will hold that in fact $\lim_{n\to \infty} \bar{X}_n=\infty$ . Use truncation to get integrability and apply the LLL.

In case, that X can be infinite with positive probability you are done, since if one of the variable is infinite the arithmetic mean will be infinite.

Assume that $X$ is finite a.s.. Then, $X_i 1_{[0,N]} \to X_i$ for $N\to\infty$ monotonically, so $EX_i 1_{[0,N]} \to E X_i$. Finally,

$$\frac{1}{n} \sum_i X_i \ge \frac{1}{n} \sum_i X_i 1_{[0,N]} \to_n E [X_i 1_{[0,N]}] \to_N E X_1 = \infty.$$

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