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Let $X_i$ be a sequence of iid rv with finite variance and $\hat \sigma_n = \sqrt{\frac{1}{n}\sum_{i=1}^n(X_i - \bar X)^2}$. By a LLN, $\hat\sigma_n\rightarrow \sigma$ as $n\rightarrow\infty$. But what about $$\frac{\hat\sigma_n}{\log(\log(n))}?$$ Can I conclude that the limit is equal to zero since $\log(\log(n))\rightarrow\infty$ and $\hat\sigma_n\rightarrow\sigma$ (a real nonnegative constant)?

Syd Amerikaner
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  • Are you looking for the law of iterated logarithm? – lcv Mar 20 '20 at 21:01
  • No, I am aware of that law (but in my case the $X_i$ are not 0-1 distributed but follow rather a normal distribution). I can bind an inequalty by the above expression and thus want to prove that it converges zo $0$ (which is hopefully true). – Syd Amerikaner Mar 20 '20 at 21:08

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$$\frac{1}{n}\sum_{k=1}^n(X_k-\overline{X}_n)^2=\frac{1}{n}\sum_{k=1}^n(X_k^2-2X_k\overline{X}_n+\overline{X}_n^2)=\frac{1}{n}\sum_{k=1}^nX_k^2-\overline{X}_n^2$$ which converges a.s, using the strong law of large number to, $\sigma^2=Var(X_1),$ and since $\frac{1}{\ln(\ln(n))}$ converges to $0$ then $\frac{\hat\sigma_n}{\log(\log(n))}$ converges a.s to $0$

mathex
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