If $X_{i}$ are I.I.D and $n^{-1}\sum_{i=1}^{n}X_{i}$ converges a.s/in-distribution to a constant $a$ is it true that $a=\mathbb{E}[X_{i}] $?

Question

The question itself is in the title. It is immediate by the strong law of large numbers that if $X_{i}$ had a finite first moment then we would have a.e convergence (and thus in probability and in distribution) such that $a=\mathbb{E}[X_{i}]$. So essentially the question comes down to whether the sample mean can converge a.s or in distribution to a constant for a sequence of variables with an infinite first moment or without first moment?

To clarify, in case one of the statements is true I'd appreciate a proof or a reference to one and otherwise a counterexample would be great.

Answer 1

Let $X_{i}$ be i.i.d. and $Y_{n} = \sum_{i=1}^{n} X_{i}$. It is well-known that

• If $\Bbb{E} X$ exists in $[-\infty, \infty]$, then $S_{n}/n \to \Bbb{E} X$ a.s.
• If $\Bbb{E}|X| = \infty$, then in probability 1, $S_{n}/n$ does not converge in $(-\infty, \infty)$.

(The first case is SLLN when $\Bbb{E}|X| < \infty$, and the case $\Bbb{E}X = \pm\infty$ are dealt with in Theorem 2.4.5 of Durrett. The second case is dealt with in Theorem 2.3.7 of Durrett as well.[1]) Therefore $S_{n}/n \to a$ a.s. guarantees $\Bbb{E}X = a$.

On the other hand, neither $S_{n}/n \to a$ in probability nor in distribution guarantees the existence of $\Bbb{E}X$. Indeed, there exists $X_{i}$ with $\Bbb{E}|X_{i}| = \infty$ but still $S_{n}/n \to 0$ in probability and hence also in distribution (see Exercise 2.2.4 in Durrett).

But in this pursuit, it is proven that

Theorem. $S_{n}/n \to a$ in probability if and only if $\phi'(0) = ia$, where $\phi(t) = \Bbb{E}\exp(itX)$.

(According to Durrett, this result is due to E. J. G. Pitman (1956).)

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[1] Thank you @NewGuy for pointing out this simpler argument.