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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.

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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).)


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

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    $\begingroup$ I think iid Cauchy random variables are a counterexample to your second "it is well-known" bullet. Are you sure you don't mean to have a limsup there? $\endgroup$ – Chris Janjigian Jun 30 '14 at 21:54
  • $\begingroup$ @ChrisJanjigian, You're right. It should be limsup instead. Thankfully, that does not affect the logic itself. $\endgroup$ – Sangchul Lee Jun 30 '14 at 22:32
  • $\begingroup$ To clarify, exercise 2.2.4 in Durrett provides a counter example for convergence in distribution. As for convergence a.s, the case where there is an expectation is dealt with by the strong law. Additionally Theorem 2.3.7 in Durrett apparently shows that if $\mathbb{E}\left|X_{i}\right|=\infty$ then $\frac{S_{n}}{n}$ can't have a finite limit a.s (this result is simpler than the one in 2.5.9) which proves the claim for the case of convergence a.e. $\endgroup$ – LlamaMan Jun 30 '14 at 22:36
  • $\begingroup$ @NewGuy, I agree that Theorem 2.3.7 is much simpler. I will modify my argument. Thank you. $\endgroup$ – Sangchul Lee Jun 30 '14 at 22:42
  • $\begingroup$ @sos440 could you perhaps add to your reply a proof o the convergence in probability of the suggested sequence given in Durrett 2.2.4? I would very much appreciate it. $\endgroup$ – LlamaMan Jun 30 '14 at 22:49

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