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Need to find $c$ such that $\text{P}(\limsup\limits_{n\to \infty} X_n/\sqrt{\log n}=c)= 1$, where $X_n$ are a sequence of independent random variables such that $X_k\sim\mathcal{N}(0,1) $ I need to use the fact that

$\sqrt{2\pi}\xi\exp\left({\xi^2\over 2}\right)\text{P}(X\geq \xi)\to 1$ as $\xi \to \infty$

This implies that $\exists \alpha,\beta>0$ such that $\forall k,n\geq1$

$$\frac{\alpha}{\sqrt{2\pi}\gamma n^{\gamma^2} \sqrt{2 \log n}} \leq \text{P}(X\geq\gamma\sqrt{2 \log n})\leq \frac{\beta}{\sqrt{2\pi}\gamma n^{\gamma^2}\sqrt{2\log n}}$$

Just by setting $a=\gamma\sqrt{2\log n}$ in the previous relation:

$$\text{P}(X\geq\gamma \sqrt{2\log n})\to \frac{1}{\sqrt{2\pi}\gamma n^{\gamma ^2}\sqrt{2\log n}}$$

Is this true in general for limits? $$\lim_{n\to\infty}A_n=A\Rightarrow\exists\alpha,\beta>0:\forall n\geq 1, \alpha A \leq A_n\leq\beta A$$

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  • $\begingroup$ The $X_n$ are all $\mathcal{N}(0,1)$, I guess if a sequence converges, it is bounded from above and below in that way? Was your question a hint? $\endgroup$
    – shilov
    Apr 28, 2013 at 20:53
  • $\begingroup$ No, I wanted to know for example whether these random variables are independent or not. $\endgroup$ Apr 28, 2013 at 20:55
  • $\begingroup$ They are independent identically distributed random variables - I didn't put that in the question in error. I thought that the probability question was context for the query about limits $\endgroup$
    – shilov
    Apr 28, 2013 at 20:58
  • $\begingroup$ The question is absurd: $P(\limsup\limits_{n\to\infty}X_n/\sqrt{\log n}=c)$ is a number, it does not depend on $n$, hence it cannot converge to $1$. $\endgroup$
    – Did
    Apr 28, 2013 at 21:14
  • $\begingroup$ Sorry I got a bit mixed up. I have changed the title and question, it makes sense that it does not depend on $n$ and does not converge to $1$ but is equal to $1$ $\endgroup$
    – shilov
    Apr 28, 2013 at 21:17

1 Answer 1

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Hint: One needs to find $c$ such that, for every $a\lt c\lt b$, $$ \mathbb P\left(\limsup\limits_{n\to\infty}X_n/\sqrt{\log n}\leqslant a\right)=0, \qquad \mathbb P\left(\limsup\limits_{n\to\infty}X_n/\sqrt{\log n}\leqslant b\right)=1. $$ Both limits stem from Borel-Cantelli lemma and from the estimations you recalled, stated more weakly as $$ \mathbb P(X\geqslant\gamma\sqrt{2\log n})=n^{-\gamma^2+\varepsilon_n(\gamma)},\qquad\lim\limits_{n\to\infty}\varepsilon_n(\gamma)=0. $$ In particular, if $\gamma\gt1$, there exists some $\delta\gt1$ and a finite $N$ such that, for every $n\geqslant N$, $$ \mathbb P(X\geqslant\gamma\sqrt{2\log n})\leqslant n^{-\delta}, $$ hence the series $\sum\limits_n\mathbb P(X\geqslant\gamma\sqrt{2\log n})$ converges, while, if $\gamma\lt1$, there exists some $\delta\lt1$ and a finite $N$ such that, for every $n\geqslant N$, $$ \mathbb P(X\geqslant\gamma\sqrt{2\log n})\geqslant n^{-\delta}, $$ hence the series $\sum\limits_n\mathbb P(X\geqslant\gamma\sqrt{2\log n})$ diverges. Note finally that in each case, $\sum\limits_n\mathbb P(X\geqslant\gamma\sqrt{2\log n})$ is also $\sum\limits_n\mathbb P(X_n\geqslant\gamma\sqrt{2\log n})$. Can you carry on from here?

Edit:

Is this true in general for limits? $$\lim_{n\to\infty}A_n=A\Rightarrow\exists\alpha,\beta>0:\forall n\geq 1, \alpha A \leq A_n\leq\beta A$$

This is logically unrelated to the rest of the post but a short answer is as follows. If $A=0$, no hope. If $A\gt0$ and $A_n\leqslant0$ for some $n$, no hope either. If $A\gt0$ and $A_n\gt0$ for every $n$, then $A_n/A\to1$ and $A_n/A\gt0$ for every $n$ hence indeed $(A_n/A)_{n\geqslant1}$ is bounded below by a positive constant and above by a finite constant, this the statement you suggest holds. Likewise if $A\lt0$ and $A_n\lt0$ for every $n\geqslant1$.

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  • $\begingroup$ No clue about what is meant by "Is this true in general for limits?" in the last paragraph. $\endgroup$
    – Did
    Apr 28, 2013 at 21:56
  • $\begingroup$ I was asking if, say, a sequence converges to some limit that we can say something about how the sequence itself is bounded based on the limit. It was the best guess I could think of $\endgroup$
    – shilov
    Apr 28, 2013 at 23:01
  • $\begingroup$ See Edit. $ $ $ $ $\endgroup$
    – Did
    Apr 29, 2013 at 7:54

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