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Let $(X_1,X_2,...)$ be i.i.d random variables with mean $0$ and variance $1$. By the Law of the Iterated Logarithm, for all $\epsilon >0$,

\begin{equation} P\left[ \frac{1}{t}\sum_{i=1}^{t}X_i \geq (1+\epsilon)\sqrt{\frac{2 \log \log t}{t}}\text{ i.o.}\right] =0 \end{equation}

Let \begin{equation} Y :=\left\{ \frac{1}{t}\sum_{i=1}^{t}X_i<(1+\epsilon)\sqrt{\frac{2 \log \log t}{t}}\text{ for all }t \geq 3 , \;\epsilon >0\right\} \end{equation} be the set of sample paths such that $\frac{1}{t}\sum_{i=1}^{t-1}X_i$ never exceeds $(1+\epsilon)\sqrt{\frac{2 \log \log t}{t}}$ for all $\epsilon > 0$.

I am trying to show that $P(Y) >0$.

Here is an equivalent formulation which may be useful. Let $I_t = I\{A_t\}$ denote the indicator rv for the event $A_t$, where $A_t$ is the event corresponding to $$\frac{1}{t}\sum_{i=1}^{t}X_i \geq (1+\epsilon)\sqrt{\frac{2 \log \log t}{t}} $$ and let $N = \sum_{t=1}^{\infty} I_t$ denote the total number of the events to occur. Then this is equivalent to showing $P(N=0)>0$.

Matt
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  • Of course, this is not true, at least you should write $t\ge 3$. But otherwise, why not use the Borel-Cantelli lemma? – zhoraster Jan 22 '16 at 05:17
  • Indeed I meant for $t \geq 3$. Can you give me some guidance on how the Borel-Cantelli lemma applies to this case? Thank you. – Matt Jan 22 '16 at 05:23
  • Why do you write the definition of $Y$ this way? $Y$ equals $$\left{ \frac{1}{t} \sum_{i=1}^t X_i \leq \sqrt{\frac{2 \log \log t}{t}} , , \text{for all $t \geq 3$} \right}$$ or am I missing something obvious? – saz Jan 24 '16 at 20:15
  • Yep, you can think of it that way too. I wrote it using the formal definition for the Law of the Iterated Logarithm from a probability textbook. – Matt Jan 24 '16 at 20:48

1 Answers1

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Let

\begin{equation} Y(t_1,t_2) :=\left\{ \frac{1}{t}\sum_{i=1}^{t}X_i<(1+\epsilon)\sqrt{\frac{2 \log \log t}{t}}\text{ for all }t\in[t_1,t_2) , \;\epsilon >0\right\} \end{equation}

Then $$P(Y(t_1,\infty))=P(Y(t_1,t_2))P(Y(t_2,\infty)|Y(t_1,t_2))\ge P(Y(t_1,t_2))P(Y(t_2,\infty))$$ and $P(Y(t_1,t_2))>0$ while $P(Y(t_2,\infty))\to 1$ by LIL.

A.S.
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  • I am guessing that you are taking the limit of $P(Y(t_2,\infty))$ with respect to $t_2 \rightarrow \infty$. But then don't you still need to prove that $\lim_{t_2 \rightarrow \infty} P(Y(t_1,t_2))\neq 0$, which essentially brings us back to the initial question? – Matt Feb 01 '16 at 17:31
  • @sisyphus No. You just need existence of $t_2<\infty$ s.t. $P(Y(t_2,\infty))>0$ which is guaranteed by LIL. I'm not taking the limit, but using the limits to justify existence of such a finite $t_2$. – A.S. Feb 01 '16 at 17:35