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Does there exist a random variable $X$ with $\mathbb{E}X = \infty$ and some constants $a_n \to \infty$ such that if $X_1, X_2, \ldots$ are iid $\sim X$, then $$\lim_{n \to \infty} \frac{X_1 + X_2 + \cdots + X_n}{a_n} \to Z$$ for some non-trivial random variable $Z$, i.e. $Z \in (0,\infty)$? (Choose your favorite mode of convergence: distribution, probability, almost sure.)

I suspect this is impossible. Can we prove this? (If I've missed something simple, feel free to suggest an additional assumption that rules out a trivial example or a variant that makes this easier.)

Clearly $Z$ would have to have $\mathbb{E}Z = \infty$, and $a_n$ would satisfy $a_n \gg n$.

One thought is something along the lines of $P(X = n!) = 2^{-n}$ for $n \geq 1$, i.e. where the sum of the $X_i$ is basically equal to the maximum of the $X_i$. Then if the maximum scales 'smoothly' enough, a limit would exist.

A possibly interesting generalization is: what if the $a_n$ are replaced by a different iid (independent of the $X$'s) sequence $Y_n$? Could

$$\lim_{n \to \infty} \frac{X_1 + X_2 + \cdots + X_n}{Y_n}$$

actually exist?

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  • $\begingroup$ Why does one "clearly" need $a_n \gg n$? $\endgroup$ – Clement C. Nov 26 '18 at 20:38
  • $\begingroup$ It's a pretty well-known variant of the SLLN that if $E[X] = \infty$ then $(X_1 + \dots + X_n)/n$ diverges to $\pm \infty$ a.s. So by multiplying and dividing by $a_n$ we see that in order to have any hope of getting the conclusion, we must have $n/a_n \to 0$. $\endgroup$ – Nate Eldredge Nov 26 '18 at 21:00
  • $\begingroup$ @NateEldredge Thanks for the clarification. $\endgroup$ – Clement C. Nov 26 '18 at 23:23
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In Durrett's Probability: Theory and Examples, Example 2.2.7 describes the "St. Petersburg Lottery": Let $T \sim \mathrm{Geo}(1/2)$ and let $X = 2^T$. It is easy to check that $E[X] = \infty$. If $X_1, X_2, \dots$ are iid with the same distribution as $X$, and $S_n = X_1 + \dots + X_n$, he shows in Example 2.2.7 that $S_n / (n \log_2 n) \to 1$ in probability.

On the other hand, Durrett's Theorem 2.5.9 (due to Feller) states that whenever $E|X| = \infty$, there cannot exist any sequence $a_n$ such that $S_n / a_n$ converges a.s. to a finite nonzero limit. In fact, $\limsup S_n/a_n$ is either identically zero or identically $\infty$. So you cannot strengthen "in probability" to "almost surely".

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  • $\begingroup$ This is what I was looking for! So any $X$ with $EX = \infty$ is too poorly behaved to hope for a.s. convergence. $\endgroup$ – J Richey Nov 26 '18 at 22:28
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You can take $a_n = n^2$ and $X_1,\dots, X_n,\dots$ to be i.i.d. Lévy r.v.'s.

Then $$ \mathbb{E}[X_i] = \infty $$ but $$ \frac{1}{n^2} \sum_{k=1}^n X_k \stackrel{d}{=} X_1 $$ (Note that the Lévy distribution is 1/2-stable, which explains the latter property.)

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