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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.