Lim sup of independent identically distributed Poisson random variables

Question

I want to prove that if $X_n$ is sequence of collectively independent Poisson random variables and $\mathbb{E}(X_n) = 1$ than $$\mathbb{P}\left(\lim{\sup{}_{n\to\infty}\frac{X_n\ln{\ln{n}}}{\ln{n}}} = 1\right)=1$$I don't know where to start. I would be very grateful for help!

Answer 1

In general, to show that $P(\limsup_n Y_n=M)=1$ for random variables $Y_n$, you need to show two things:

• For all $\epsilon>0$, $P(Y_n>M+\epsilon \text{ infinitely often})=0.$

• For all $\epsilon>0$, $P(Y_n>M-\epsilon \text{ infinitely often})=1.$

To prove both of these, we will use the Borel-Cantelli lemmas, which involve looking the convergence of $\sum_{n\ge 1}P(X_n \ln\ln n/\ln n>1\pm \epsilon)$. Therefore, you will need a good estimate on $P(X_n\ge k)$, and apply this to $k=(\ln n/\ln \ln n)(1\pm \epsilon)$.

Assume $k$ is an integer for now. On the one hand, $P(X\ge k)\ge P(X=k)=\frac{e^{-1}}{k!}$. On the other hand, you can upper bound $P(X\ge k$ by a geometric series: $$ P(X\ge k) =\frac{e^{-1}}{k!}\sum_{i\ge k}{1 \over (k+1)(k+2)\cdots (i-i)\cdot i} \le \frac{e^{-1}}{k!}\sum_{i\ge k}\frac1{k^{i-k}} =\frac{e^{-1}}{k!}\cdot \frac{1}{1-1/k} $$ This shows that $\lim_{k\to\infty}\frac{P(X\ge k)}{P(X=k)}=1$, so you can use the simple estimate $P(X\ge k)\approx e^{-1}/k!$ to determine the summability of $P(X_n\ge \ln n/(\ln \ln n)(1\pm \epsilon))$.

Therefore, you just need to show that the sum $$ \sum_{n\ge 1}P\left(X_1\ge (1\pm \epsilon)\frac{\ln n}{\ln \ln n}\right) $$ is finite when the sign is $+$, and infinite when the sign is $-$. This is equivalent to showing the same about the sums $$ e^{-1}\sum_{n\ge 1}\frac1{\Big((1\pm \epsilon)\ln n/\ln \ln n\Big)!} $$ To prove this sum is infinite/finite, use Stirlings approximation, which says $\lim_{k\to\infty}\frac{k!}{k^ke^{-k}\sqrt{2\pi k}}=1$, so you can replace $k!$ with $k^ke^{-k}\sqrt{2\pi k}$ without affecting convergence. The math will get a bit messy, but everything works out nicely in the end. There is also the small issue that the argument to the factorial is not an integer, so you really need to round down, but this does not affect things in the long run.