Approximate normal distribution

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 $$U_{a,n} =\left \lbrace X_n \ge a\times \log(n)\right \rbrace.$$

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.

Should we do this question by thinking about the distribution of $X/\sqrt{\log(n)}$? Or just plug in $ \{X_n \ge a\times \log(n)\}$ into the formula above?

I did it but cannot proceed anymore.

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

1

Hint: for a fixed $a$, the events $(U_{a,n})_{n\geqslant 1}$ are independent. The equivalent given in the opening post allows us to determine the convergence of the series $\sum\limits_{n=1}^\infty\mathbb P(U_{a,n})$. Then we use the $\fbox{B___-__}$ lemma.

For part b), see here.

edited Apr 13 '17 at 12:21

Community

1

answered Dec 15 '14 at 14:01

Davide Giraudo

181,608

what about the part b)? – annimal Dec 15 '14 at 18:50
Does't the answer to part a) help? – Davide Giraudo Dec 15 '14 at 20:23
I try to use a) but there is a sqrt in b) and I dont know where the $\sqrt{2}$ comes from. can u give me more hint? thanks a lot – annimal Dec 15 '14 at 20:28
I included a link. – Davide Giraudo Dec 15 '14 at 20:35
the link said he use the fact that $$\left(\frac{1}{x}-\frac{1}{x^3}\right)e^{-\frac{x^2}{2}}\leq\mathbb P(X_n>x)\leq \frac{1}{x}e^{-\frac{x^2}{2}},$$ and the Borel Cantelli lemmas to prove what I want.but I still cannot see how it is done, can you give me more details? thanks – annimal Dec 15 '14 at 21:29
If there is something specific which is not clear to you (either in this post or in the linked one), please tell me. – Davide Giraudo Dec 16 '14 at 09:32

Approximate normal distribution

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