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I see to use the law of the iterated logarithm that tells us:

$\limsup_{n\to\infty}\left|\frac{S_n}{\sqrt {n\log\log n}}\right|=\sigma\sqrt 2$ with probability $1$

where: $S_n:=X_1+\dots+X_n$ , and $ \sigma^2=\mathbb EX_1^2$.

Hence if $\sqrt{n\log\log n}\cdot c_n\to 0$, one has $ c_nS_n\to 0$ with probability $1$.

This is not quite a necessary and sufficient condition, because we can construct decreasing sequences $(c_n) $ that most of the the time are of smaller order than $1/\sqrt{n\log\log n}$, but occasionally are as big as $1/\sqrt{n\log\log n}$ for which $ c_nS_n\to 0 $ still goes to $0$ almost surely, but if we want $c_n$ to be some kind of regularly decaying sequence, then this is exactly the right condition.

I looked at this book, Theorem 2.5.7, p. 71 :http://www.math.duke.edu/~rtd/PTE/pte.html

I want someone to see my answer for my question if it's true, I can improve my question.

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