LLN when $E X = \infty$

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?

• Why does one "clearly" need $a_n \gg n$? – Clement C. Nov 26 '18 at 20:38
• 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 $n/a_n \to 0$. – Nate Eldredge Nov 26 '18 at 21:00
• @NateEldredge Thanks for the clarification. – Clement C. Nov 26 '18 at 23:23

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".
• This is what I was looking for! So any $X$ with $EX = \infty$ is too poorly behaved to hope for a.s. convergence. – J Richey Nov 26 '18 at 22:28
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.)