Given that $X_1,\ldots,X_n$ are $n$ independent and identically distributed random variables. We know that they have finite moments up to third order e.g. $EX_i=0$, $EX_i^2<\infty$ and $EX_i^3<\infty$. The question is to find accurate probability bound for the quantity $\frac{n^2\sum_{i=1}^n X_i}{\sum_{i=1}^n X_i^3}$?

I have two results for this quantity. First is that since $\sum_{i=1}^n X_i = O_p(n^{1/2})$ based on central limit theorem (CLT) and by law large number we have $\sum_{i=1}^n X_i^3=O_{p}(n)$, then I have $$\frac{n^2\sum_{i=1}^n X_i}{\sum_{i=1}^n X_i^3} = \frac{n^2O_p(n^{1/2})}{O_p(n)}=O_p(n^{3/2}).$$ Second is that since $\sum_{i=1}^n X_i=O_p(n^{1/2})$ based on central limit theorem (CLT) and by CLT we have $\sum_{i=1}^n X_i^3=O_p(n^{1/2})$, thus

$$\frac{n^2\sum_{i=1}^n X_i}{\sum_{i=1}^n X_i^3} = \frac{n^2 O_p(n^{1/2})}{O_p(n^{1/2})} = O_p(n^2).$$

which one is correct probability bound? What's more, could I find exact order or more accurate order of the quantity?

  • $\begingroup$ The central limit theorem does not apply in the way you state unless $E[X_i]=0$. Is the mean zero? $\endgroup$ – Michael Feb 26 '16 at 4:52
  • $\begingroup$ Thx, Michael, I could add this assumption. $\endgroup$ – lzstat Feb 26 '16 at 4:54
  • $\begingroup$ It makes the problem harder if $E[X_i]=0$. What about $E[X_i^3]$? Is that zero? $\endgroup$ – Michael Feb 26 '16 at 4:55
  • $\begingroup$ we only know $EX_{i}^3$ is finite. $\endgroup$ – lzstat Feb 26 '16 at 4:57
  • $\begingroup$ Your CLT application to $X_i^3$ does not work unless $E[X_i^3]=0$. $\endgroup$ – Michael Feb 26 '16 at 5:04

Let $c = E[X_i^3]$ and assume $c\neq 0$. Then for large $n$ we have $$\frac{n^2\sum_{i=1}^nX_i}{\sum_{i=1}^nX_i^3} = \frac{n^{1.5}\frac{1}{\sqrt{n}}\sum_{i=1}^nX_i}{\frac{1}{n}\sum_{i=1}^n X_i^3} \approx \frac{n^{1.5}G_n}{c}$$ where $G_n$ is Gaussian with mean $0$ and variance $Var(X_1)$.

  • $\begingroup$ Note that, in this problem, we apply the LLN when the mean is nonzero (denominator), and the CLT when the mean is zero (numerator). $\endgroup$ – Michael Feb 26 '16 at 5:13
  • $\begingroup$ Yea, I understand. It makes sense. thank you $\endgroup$ – lzstat Feb 26 '16 at 5:16
  • $\begingroup$ You need to specify that $\approx$ is in probability only. $\endgroup$ – A.S. Feb 26 '16 at 6:07

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