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I have read the statement of Law of iterated logarithm is like for some random variable $Y_i$ $$\lim\sup_{n\rightarrow\infty}\frac{\sum_{i=1}^nY_i}{\sqrt{2\log\log n}}=1\ a.s.$$

But I have also found another version using big O notation like $$\sum_{i=1}^nY_i=O((\log\log n)^{1/2})$$

Can someone tell me the relation between these two equations? It's for me not easy to understand from Definitions, since there is no limsup in Definition for big O

Thanks a lot for any hints

toki
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  • If I am not mistaken the first statement implies the second, but not the other way around, since the second statement would also follow from a hypothetical law stating that the left hand side of the first equality equals 2 rather than 1. – Vincent Nov 01 '23 at 17:04

1 Answers1

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The Landau notation $f(n) = O(g(n))$ as $n\to \infty$ means that $\limsup_{n\to\infty} f(n)/g(n) < \infty.$ If we define $f(n) = \sum_{i \le n} Y_i$, and $g(n) = \sqrt{\log\log n},$ then the first statement is precisely giving a bound on $\limsup f(n)/g(n),$ and so proves the second statement (up to measure $0$). Of course, the second statement is as such weaker, since it's only claiming the existence of a constant $C$ such that $f(n) \le C g(n)$ for large $n,$ while the first statement is saying that $C = \sqrt{2}$ works.

stochasticboy321
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  • @stochasicboy321 sorry I still have a question. I noticed that some version of LIL write $|\sum_{i=1}^nY_i|$, but some are simply $\sum_{i=1}^nY_i$. But the definition of big O almost surely I found is $\lim\sup\frac{|\sum_{i=1}^nY_i|}{a_n}<\infty$. Do you know how can they match with each other? – toki Nov 02 '23 at 15:32
  • Usage varies somewhat: typically, if a quantity can be both positive or negative, then it's common to take absolute value, but this can vary from author to author, depending on what they're trying to do. That said, the value of the big-O notation is mostly in that it captures an upper bound on the scaling behaviour and simplifies a bunch of algebraic manipulations, and I'd recommend keeping that in mind. In any case, if only a bound on $\limsup \sum Y_i$ is shown, we can still say that $(\sum Y_i)+ = O(\cdot)$, where $(z)+ = \max(z,0),$ and this might be enough for the purposes of the study. – stochasticboy321 Nov 02 '23 at 17:16
  • Also note that if you say "the LIL", then it's assumed that you mean the result that for iid random variables $Y_i$ with zero mean and unit variances, $\limsup |\sum_{i \le n} Y_i|/\sqrt{2n\log\log n} = 1$. More generally, any result of this type (some control on a stochastic process with an iterated logarithm showing up) is called 'a' LIL. In other words, LILs (much like CLTs) are sort of generic labels for types of results, but with one exemplar result that's often treated as the standard one, and is usually meant when no other details are given. – stochasticboy321 Nov 02 '23 at 17:19