Let $X_1,X_2, \ldots$ r.v i.i.d. with geometric distribution and parameter $p = 1 – e^{-1}$, i.e. $P(X_1 = k) = p(1 -p)^k;$ $k = 0, 1, 2, \ldots$. Prove that
$$P\left[\limsup_{n \to \infty}\frac{X_n}{\log(n)}=1\right]=1.$$
I'm stuck with this exercise. I think I have to use Borel Cantelli Lemma. Since r.v. are independent I need to prove that
$$\sum_{n=1}^{\infty}P\left[\frac{X_n}{\log(n)}=1\right]=\infty,$$
then
\begin{align*}
\sum_{n=1}^{\infty}P\left[\frac{X_n}{\log(n)}=1 \right]&=\sum_{n=1}^{\infty}P\left[X_n=\log(n)\right]\\
&=\sum_{n=1}^{\infty} p(1 -p)^{\log(n)}\\
&=p\sum_{n=1}^{\infty} (1 -1+e^{-1})^{\log(n)}\\
&=p\sum_{n=1}^{\infty} (e^{-1})^{\log(n)}\\
&=p\sum_{n=1}^{\infty} \left(\frac{1}{e}\right)^{\log(n)}\\
\end{align*}
I know $\log(n)$ is not always an integer so my procedure is definitely wrong but I do not know how to begin… any help?
Best Answer
In order to show $P(\limsup_n Y_n=1)=1$ for a sequence of random variables $Y_1,Y_2,\dots,$ you always need to show two things:
For all $\newcommand\e{\varepsilon}\e>0$, $P(Y_n>1+\e\;\text{ i.o.})=0$.
For all $\e>0$, $P(Y_n>1-\e\;\text{ i.o.})=1$.
This is because for a deterministic sequence $a_1,a_2,\dots$, we have $\limsup_n a_n=L$ if and only if infinitely many $a_n$ are larger than $L-\e$ for all $\e>0$, while only finitely many $n$ are larger than $L+\e$ for any $\e>0$. These are just undergraduate analysis facts, so I omit proving them.
The Borel-Cantelli lemmas suggest we should look at the convergence of $\sum_n P(X_n/\log n> 1\pm \e)$ to determine points $(1)$ and $(2)$. Using the fact that $P(X_1>n)=(1-p)^n$ for any integer $n$ (which I will apply to some non-integer $n$, and not worry about the details), we have $$ \sum_{n=2}^\infty P\left(\frac{X_n}{\log n}>1\pm \e\right)=\sum_{n=2}^\infty P(X_n>(1\pm \e)\log n)=\sum_{n=2}^\infty (1-p)^{(1\pm \e)\log n}=\sum_{n=2}^\infty n^{(1\pm\epsilon)\log(1-p)} $$ In the last step, I used the property that $a^{\log b}=b^{\log a}$. After substituting $p=1-e^{-1}$, you can determine that this series is indeed convergent for $+\e$ and divergent for $-\e$, which allows you to conclude $(1)$ and $(2)$.