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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?
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 $$\sum_{n=1}^\infty\frac{e^{-\frac{\epsilon^2}{2}\ln^2n}}{\epsilon \ln n} < \infty.$$
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 $$ \frac{X_{n}}{n} \to_n 0 \ a.s. $$
