Large Deviations and the Law of the Iterated Logarithm*

Prove (6.2) by using (9.9) and the fact that cumulants add in the presence of independence.

In the Bernoulli case, (9.21) gives

where $p<a<1$ and $x_{n}=n(a-p)$. Theorem 9.4 gives

where $x_{n}=a_{n} \sqrt{n p q}$. Resolve the apparent discrepancy. Use (9.25) to compare the two expressions in case $x_{n} / n$ is small. See Problem 27.17.

Relabel the binomial parameter $p$ as $\theta =f(p)$, where $f$ is increasing and continuously differentiable. Show by (9.27) that the distinguishability of $\theta$ from $\theta +\Delta \theta$, as measured by $K$, is $(\Delta \theta )^{2} / 8 p(1-p)\left(f^{\prime }(p)\right)^{2}+O(\Delta \theta )^{3}$. The leading coefficient is independent of $\theta$ if $f(p)=\arcsin \sqrt{p}$.

From (9.35) and the same result for $\left\{ -X_{n}\right\}$, together with the uniform boundedness of the $X_{n}$, deduce that with probability 1 the set of limit points of the sequence $\left\{ S_{n}(2 n \log \log n)^{-1 / 2}\right\}$ is the closed interval from -1 to +1.

$\uparrow$ Suppose $X_{n}$ takes the values $\pm 1$ with probability $\frac{1}{2}$ each, and show that $P\left[S_{n}=0\right.$ i.o. $]=1$. (This gives still another proof of the persistence of symmetric random waik on the line (Example 8.6).) Show more generally that, if the $X_{n}$ are bounded by $M$, then $P\left[\left|S_{n}\right| \leq M\right.$ i.o. $]=1$.

Weakened versions of (9.36) are quite easy to prove. By a fourthmoment argument (see (6.2)), show that $P\left[S_{n}>n^{3 / 4}(\log n)^{(1+\epsilon ) / 4}\right.$ i.o. $]=0$. Use (9.29) to give a simple proof that $P\left[S_{n}>(3 n \log n)^{1 / 2}\right.$ i.o. $]=0$.

Show that (9.35) is true if $S_{n}$ is replaced by $\left|S_{n}\right|$ or $\max_{k \leq n} S_{k}$ or $\max_{k \leq n}\left|S_{k}\right|$.