Approximate normal distribution(this is different from what I asked earlier $\log(n)$ is replaced by $\sqrt{\log{n}}$)

Question

Let $ X \sim N (0, 1)$. For $x$ large enough, the tail of the distribution of $X$ may be approximated as $$P(X > x) \sim e^{-x^2/2}/(x\sqrt{2\pi})$$ Consider a sequence of independent r.v. all having a standard normal distribution. For every $a > 0$, $n = 1, 2, \ldots$, define the event $$U_{a,n} =\left \lbrace X_n \ge a\times \sqrt{\log(n)}\right \rbrace.$$

a) Calculate $P(\limsup \, U_{a,n})$, which may depend on $a$.
b) Prove $\limsup \left(X_n/\sqrt{\log(n)} \right) = \sqrt{2} $ a.e.

for a) should i still consider B-C lemma to do it?
and anyone can give me some hint for b)?

Answer 1

a) Yes, B-C does it. Now, \begin{align*} \sum_{n\ge 1}\mathbf{P}(U_{a,n})&\leq\sum_{n\geq 1}\frac{\exp(-{a^2\log n}/2)}{a\sqrt{2\pi\log n}}\\ &=\sum_{n\geq 1}\frac{1}{a n^{a^2/2}\sqrt{2\pi\log n}}\\ &<\infty, \quad\text{if}\quad a> \sqrt{2}. \end{align*} Hence, if $a> \sqrt{2}$, \begin{equation*} \mathbf{P}(\limsup_{n\rightarrow\infty} U_{a,n})=\mathbf{P}(U_{a,n},\text{i.o.})=0. \end{equation*} Similarly, you can show (either using the usual Gaussian lower bound $\frac{x}{x^2+1}e^{\dots}\dots$ or being careful with the approximation--the latter's easier) that \begin{equation*} \sum_{n\ge 1}\mathbf{P}(U_{a,n})=\infty, \quad\text{if}\quad a\le \sqrt{2}. \end{equation*} Since the $U_{a,n}$s are independent, for $a\le\sqrt{2}$, \begin{equation*} \mathbf{P}(\limsup_{n\rightarrow\infty} U_{a,n})=\mathbf{P}(U_{a,n},\text{i.o.})=1. \end{equation*}

b) From the first part of (a): Let $a> \sqrt{2}$. Then there is some $N$ such that $n\ge N$ implies: \begin{equation*} \frac{X_{n}}{\sqrt{\log n}}\le a,\quad\text{a.s.}, \end{equation*} since the $\limsup$ is $0$. Now let $n\rightarrow\infty.$ It follows that (since it holds for all $a>\sqrt{2}$) \begin{equation*} \limsup_{n\rightarrow\infty}\frac{X_{n}}{\sqrt{\log n}}\le \sqrt{2},\quad\text{a.s.}. \end{equation*} Showing $\ge$ follows from using the definition of $\limsup$ and the second part of (a).

Summary: Show that the series in step (a) converges for certain values of a and diverges otherwise. Then, use B-C to find the $\limsup$. Third, follow the definition of $\limsup$.