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I need the following result in my bachelor's thesis, and want to better understand the proof of it.

$X_{n}$ independent and $X_n \sim \mathcal{P}(n) $ meaning that $X_{n}$ has Poisson distributions with parameter $n$. What is the $\lim\limits_{n\to \infty} \frac{X_{n}}{n}$ almost surely ?

This is from this question: Find the almost sure limit of $X_n/n$, where each random variable $X_n$ has a Poisson distribution with parameter $n$

The answers were helpful, still there are two things I don't understand:

  1. (the more important one) Math1000, in his/her answer uses the Chernoff bound, but I don't know how they use it/which version, to get to the estimated result. If I use the definition of the bound from wikipedia, I get a different result.

  2. How do we get from the lim inf to the limit, i.e. why do we know that the limit exists?

Help would be much appreciated

PS: would have commented on that thread, but I don't have enough reputation for that yet.

blubby
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  • If you scroll down to the section on the multiplicative form of the Chernoff bound here, you'll find the exact form used in the answer (note that the mean of a Poiss$(n)$ RV is $n$.)
  • The only place that the $\lim\inf$ is used is in the first line of the answer, and that is a standard form of almost sure convergence. Read the section on a.s. convergence here for confirmation, and give it a good think if you want a proof, it's essentially definition chasing.
    • – stochasticboy321 Oct 03 '17 at 23:03
  • Hey thanks for the response. I have never seen that definiton for almost sure convergence before. But it answers my question. Still, about the bound, I believe the multiplicative version only makes sense for Bernoulli variables. At least all the resources I found use that version with Bernoulli variables. So I doubt we could use it for the Poisson distribution. Still, I'll try to calculate the Chernoff bound by hand and will post later whether I got a convergence result. – blubby Oct 04 '17 at 10:52
  • My Chernoff bound of $\mathbb: \forall t>0 P(X_n\leqslant n(1-\varepsilon))$ is $\leqslant exp(tn(1-\epsilon))exp(n(e^{-t}-1))=(\frac{1}{{exp(e^{-t}-1)+t(1-\epsilon)})^{-n}$, we have to show that the denominator is bigger than 1 for all epsilon. It is possible for all $\epsilon$ except for $\epsilon>0$. – blubby Oct 04 '17 at 16:04
  • On the other hand, does the multiplicative Chernoff bound work for Poisson because of the Poisson approximation? – blubby Oct 04 '17 at 16:06
  • You're quite right, my apologies for being too dismissive on first read. In any case, I think the bounds still hold, although not due to the material in the link I provided earlier. I haven't the time for an answer just now, but I stumbled across this note by Cannone which may be helpful. In particular, theorem 1 in the note has precisely the inequalities used by Math1000. – stochasticboy321 Oct 04 '17 at 22:44
  • Yes, I calculated it and it is exactly what is written in the other thread. One just uses $n(1-\epsilon)=n-n\epsilon$ and sets $n=\lambda$ and $n\epsilon =x$. Thanks for your help! – blubby Oct 05 '17 at 13:19