# Confidence intervals vs law of the iterated logarithm

Let $X_i$ be iid random variables with mean $\mu$ and variance $1$. Let $S_n=\frac 1n \sum_{k=1}^n X_k$ denote the sample mean. By CLT it is well known that $(S_n - \frac{1.96}{\sqrt n}, S_n + \frac{1.96}{\sqrt n})$ is an asymptotic $95\%$ confidence interval for $\mu$.

By the law of the iterated logarithm, $\frac{\sqrt n}{\sqrt {\log \log n} }(S_n-\mu)$ has $\limsup$ equal to $\sqrt 2$ almost surely. Thus for almost all $w$, we have $\frac{\sqrt n}{\sqrt {\log \log n} }(S_n-\mu) \approx \sqrt 2$ infinitely often, thus $S_n \approx \mu +\frac{\sqrt {2 \log \log n}}{\sqrt n }$ infinitely often.

For large $n$, $\sqrt {2 \log \log n}$ is much bigger than $1.96$, which seems to contradict the confidence interval. Can someone shed light on this paradox (if there is any) ?

• What do you mean by $$\frac{\sqrt{n}}{\sqrt{\log \log n}}(S_n-\mu)\approx \sqrt{2}$$? – Daniel Camarena Perez Nov 28 '18 at 7:21

Let $$X_1,X_2,\dots$$ iid with $$X_1\sim \mathcal{N}(0,1)$$ then $$W_n=\frac{\sqrt{n}}{\sqrt{2\log \log n}}S_n \sim \mathcal{N}\left(0,\frac{1}{2\log \log n}\right).$$ Note $$W_n \to 0$$ in law, but $$W_n\nrightarrow$$ in the sense a.s. convergence. Indeed, by the law of iterated logarithm $$\liminf W_n =-1,\quad \limsup W_n =1$$ with probability one.