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Let $\{ X_i \}_{i=1}^n$ be sequence of i.i.d. mean zero ($\mathbb{E} X_i = 0$) random variables with bounded second moment ($\mathbb{E} X_i^2 < \infty$) . Now what would be the limiting behavior of

$$S^a_n := \frac{\sum_{i=1}^n X_i}{n^a},$$

as $n \to \infty$ where $a \in [0,1]$ or even $\mathbb{R}$.

When $a = 1$, we have the Strong Law of Large Numbers which say that $S^1_n$ almost surely converges to $0$.

When $a = 1/2$, we have the Central Limit Theorem which say that $S^{1/2}_n$ converges to a normal distribution ($\mathcal{N} (0, \mathbb{E}X_i^2 )$) in distribution.

But what happens for other choices of $a$ and how can one prove it converges or diverges in those scenarios?

This is Terence Tao's comment on it (which I didn't fully understand) - https://terrytao.wordpress.com/2008/06/18/the-strong-law-of-large-numbers/

Assuming the random variables have bounded second moment, the limit will be almost surely zero for a > 1/2, converge (in a distributional sense) to a normal distribution for a=1/2, and diverge to infinity almost surely for a < 1/2, all thanks to the (strong) law of large numbers. For heavy-tailed random variables with infinite second moment, the situation is going to be more complicated, but can be worked out for any specific distribution by a variety of tools (e.g. Fourier analysis)"

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Using Slutsky's lemma, it is immediate that if $a\gt 1/2$, then $S_n^a=S_n^{1/2}n^{1/2 -a}\to 0$ in probability. More can be said thank s to the law of the iterated logarithms. Indeed, since $\limsup_n S_n^{1/2} /\sqrt{2\log\log n}$ and $\liminf_n S_n^{1/2} /\sqrt{2\log\log n}$ are almost surely finite, we derive that $S_n^a\to 0$ almost everywhere. This can also be derived from Marcinkiewicz law of large numbers.

When $a\lt 1/2$, we cannot even expect a convergence in distribution. If $\left(Y_n\right)_{ n\geqslant 1}$ is a non-negative sequence of random variables which converges in distribution to a non-degenerated random variable $Y$, then for any sequence $\left(R_n\right)_{n\geqslant 1}$ going to infinity, the sequence $\left(R_n Y_n\right)_{n\geqslant 1}$ cannot be convergent in distribution.

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  • $\begingroup$ Thank you for the answer, @Davide Giraudo. I looked at Marcinkiewicz law of large numbers, it gives exactly what we want for $a \in (1/2, \infty)$. Just to make it clear I understand by Slutsky's lemma we can say that $S^a_n \overset{d}{\to} 0$ and convergence in probability follows from the fact that $X_n \overset{d}{\to} c \implies X_n \overset{p}{\to} c$, where $c$ is a constant. $\endgroup$
    – John Doe
    Dec 5, 2016 at 0:05
  • $\begingroup$ But I have two follow up questions. (1) By law of iterated logarithms I understand that $S_n^{1/2}/\sqrt{2 \log \log n}$ finite a.s, but how does it imply (along with the "in probability" convergence proved by you) that $S^a_n = 0$ a.s. for $a > 1/2$? $\endgroup$
    – John Doe
    Dec 5, 2016 at 0:06
  • $\begingroup$ (2) I didn't understand why $(R_n Y_n)_{n \geq 1}$ does not converge and why does it imply that $S_n^a$ doesn't converge for any $a \in (1, 1/2)$, could you give me some hints please. $\endgroup$
    – John Doe
    Dec 5, 2016 at 0:06

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