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There is a sequence of random variables $X_1,X_2,...$ such that for each $i$, $X_i\sim N(0,1)$.

Does $ \frac{X_n}{n} \rightarrow 0 $ almost surely?

Does $ \frac{X_n}{\ln n} \rightarrow 0 $ almost surely?

Use Borel Cantelli lemma on solution.

I would glad if you can give me a solution on that question because i have no idea what to do here... Borel Cantelli lemma is (very!) not intuitive to me...

Also it will be helpful if you can tell me what "almost surely convergence" means?

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  • $\begingroup$ I consider that but i thought this law is work only on the summation x1+x2+.. also the hint of borel cantelli seem important... can you show how can i use the strong law in the first one? $\endgroup$
    – aviadch
    Feb 8, 2014 at 14:47
  • $\begingroup$ Sorry, I actually misread the question -- my bad. Ignore my comment... $\endgroup$
    – Clement C.
    Feb 8, 2014 at 14:48

2 Answers 2

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I'm going to give it a shot -- Borel-Cantelli is a tad hazy in my mind, though, so you definitely should doublecheck.

Fix any $\epsilon > 0$, and define the event $$E_n\stackrel{\rm{}def}{=} \mathbb{1}_{\left\{\frac{\lvert X_n\rvert}{n}\geq \epsilon\right\}}$$ so that $$ \mathbb{P} E_n = \mathbb{P}\left\{ \lvert X_n\rvert \geq n\epsilon \right\} = \operatorname{erfc}\left(\frac{\epsilon n}{ \sqrt{2} }\right) \operatorname*{\sim}_{n\to\infty} \sqrt{\frac{2}{\pi}} \frac{e^{-\frac{\epsilon^2n^2}{2}}}{\epsilon n}$$ and thus $\sum_{n=1}^\infty \mathbb{P} E_n < \infty$. By Borel-Cantelli, $$ \mathbb{P}\left(\limsup_{n\to\infty} E_n\right) = 0 $$ that is, $$ \forall \epsilon > 0, \forall_{\rm{}a.s.} \omega\in\Omega,\ \exists N\geq 0,\forall n\geq N,\ \frac{\lvert X_n\rvert}{n}< \epsilon$$ or, equivalently, $\displaystyle\lim_{n\to\infty}\frac{\lvert X_n\rvert}{n}\to0$ a.s.

This also seems to work for $\frac{X_n}{\ln n}$, as $$\sum_{n=1}^\infty\frac{e^{-\frac{\epsilon^2}{2}\ln^2n}}{\epsilon \ln n} < \infty.$$

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  • $\begingroup$ what did you calculated for $\mathbb P\{|X_n|\ge n\epsilon\}$ and why do we need $|X_n|$ there as one can directly apply Markov's Inequality to conclude $\mathbb P\{X_n\ge n\epsilon\}\le \frac{\mathbb E(X_n)}{n\epsilon}=0$ each $X_i$ follows $N(0,1)$ $\endgroup$
    – bunny
    Oct 17, 2017 at 18:39
  • $\begingroup$ @bunny Markov's inequality is for nonnegative random variables. $\endgroup$
    – Clement C.
    Oct 18, 2017 at 1:00
  • $\begingroup$ but in the proof we do not use any non negative conditions $\endgroup$
    – bunny
    Oct 18, 2017 at 4:48
  • $\begingroup$ @bunny I don't understand what you mean. In the proof of Markov's inequality? $\endgroup$
    – Clement C.
    Oct 18, 2017 at 4:49
  • $\begingroup$ i was reading from Amir Dembo notes and there in the statement itself he has no restriction on $X_n$ to be non negative $\endgroup$
    – bunny
    Oct 18, 2017 at 4:51
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For $Z_n \sim N(0,1)$ actually it's even easier to show that $$ \sum_{n=1}^{\infty}P(|Z_n|\geq n \varepsilon) = 2\sum_{n=1}^{\infty}P(Z_n \geq n \varepsilon) $$ if you consider MGF for $Z_n, \varphi_{Z_{n}}(s) = e^{\frac{s^2}{2}}, s>0$. Since $\varphi$ is positive, you can use Markov inequality directly and it's easy to show that $$ P(Z_n> r) \leq e^{\frac{s^2}{2} - sr} $$ By taking the derivative of the upper bound, you can show that it achieve its minumum at $s=r$, in this case $r= n \varepsilon$, so $$ P(Z_n > n \varepsilon ) \leq e^{-\frac{(n \varepsilon)^2}{2}} $$ Therefore, $$ \sum_{n=1}^{\infty}P(|Z_n|\geq n \varepsilon) \leq\sum_{n=1}^{\infty}e^{-\frac{(n \varepsilon)^2}{2}} < \infty $$ because the corresponding integral converges (error function). As a result, $$ P(\limsup |\frac{X_n}{n}|> \varepsilon) = P( |\frac{X_n}{n}|> \varepsilon \ i.o.) = 0 $$ and $$ \frac{X_{n}}{n} \to_n 0 \ a.s. $$

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