Find the almost sure limit of $X_n/n$, where each random variable $X_n$ has a Poisson distribution with parameter $n$

Question

$X_{n}$ 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 ?

I think we can write $X(n) \sim X(1)+X(1)+\cdots+X(1)$ where the sum is taken on independent identical distribution then use the law of large number. But I am not sure that is it correct or not. Can anyone give me some hints? Thank you in advance!

No you cannot write that, first because no kind of relationship between the random variables $X(\lambda)$ for diferent values of $\lambda$ is postulated, second because $X(1)+X(1)+\cdots+X(1)$ is not Poisson. All this means you need to use more robust tools... such as the first Borel-Cantelli lemma. Any idea to pursue? — Did, Jun 19 '16 at 18:04
@Did I believe the question actually being asked is:
"Let $X_n\sim\mathsf{Pois}(n)$, $n=1,2,\ldots$. Does $\frac{X_n}n$ converge almost surely? If so, what is the limit?"

But I am still puzzled as to what $X(\lambda)$ could actually mean. — Math1000, Jun 19 '16 at 18:08
@Math1000 Yes this is the question. (And I think we can overcome the oddity of the notation $X(\lambda)$ vs $X_\lambda$, if this is the point of your comment.) — Did, Jun 19 '16 at 18:09
Yes, because $X$ is a function, and the notation $X(\cdot)$ suggests the evaluation of $X$ at a point. — Math1000, Jun 19 '16 at 18:14
@ThomasEdison First check the exact statement of the exercise. Second explain how you came to the idea that the LLN was even relevant. (The LLN might apply, but in a very specific situation, which is not when $(X_n)$ is independent.) — Did, Jun 19 '16 at 18:41
@ThomasEdison "It seemed in the comments to OP's question that nobody was listening to me." Funny, this is exactly the impression I got about the way you "listen" to my comments... — Did, Jun 19 '16 at 18:45
@Did Okay, I will try to listen to you. What is wrong with this statement? 'A Poisson random variable with parameter $n\in \mathbb{N}$ being equal to an independent sum of $n$ identical Poisson random variable with parameter $1$?' — Sungjin Kim, Jun 19 '16 at 18:51
@i707107 Nothing. And this is not the statement you would need to make your idea work. — Did, Jun 19 '16 at 18:52
@i707107 Nope, it does not. Let me note that you suddenly stopped to write down the details of the argument you have in mind. Which is a shame because if you did, you would see that no LLN applies here (except in a very specific situation which is most probably not the one the OP or you have in mind). — Did, Jun 19 '16 at 18:56
@Did I would like to see your answer too. Then you would use Chernoff bound for Poisson distribution, right? — Sungjin Kim, Jun 19 '16 at 19:22
@i707107 Indeed, any large deviations bound (say, Chernoff) leads directly to the conclusion. — Did, Jun 19 '16 at 19:27
@ThomasEdison Please see Did's comments and my comments on the answer by Math1000. — Sungjin Kim, Jun 20 '16 at 14:54
@i707107 I saw your comments. Thank you so much for honesty ! — Thomas Edison, Jun 20 '16 at 15:32

Math1000 · Accepted Answer · 2016-06-23T16:26:59.620

3

If $X_n\sim\mathsf{Pois}(n)$ for $n=1,2,\ldots$, then $n^{-1}X_n\stackrel{\mathrm{a.s.}}\longrightarrow 1$. It suffices to show that for all $\varepsilon>0$, $$\mathbb P\left(\liminf_{n\to\infty}\left|n^{-1}X_n-1\right|<\varepsilon\right)=1. $$

We have $$\left\{\left|n^{-1}X_n-1\right|<\varepsilon\right\}^c = \left\{X_n\leqslant n(1-\varepsilon) \right\}\cup\left\{X_n\geqslant n(1+\varepsilon) \right\}, $$ so the Chernoff bounds yield $$\mathbb P(X_n\leqslant n(1-\varepsilon)) \leqslant \frac{e^{-n}(ne)^{n(1-\varepsilon)}}{(n(1-\varepsilon))^{n(1-\varepsilon)}}= \left(e^{\varepsilon}(1-\varepsilon)^{1-\varepsilon} \right)^{-n} $$ and $$\mathbb P(X_n\geqslant n(1+\varepsilon)) \leqslant \frac{e^{-n}(ne)^{n(1+\varepsilon)}}{(n(1+\varepsilon))^{n(1+\varepsilon)}}= \left(e^{-\varepsilon}(1+\varepsilon)^{1+\varepsilon} \right)^{-n} . $$

Now

$$e^\varepsilon(1-\varepsilon)^{1-\varepsilon} = e^{\varepsilon + (1-\varepsilon)\log(1-\varepsilon)}=e^{\frac{\varepsilon^2}2+O(\varepsilon^3)}>1 $$ and $$e^{-\varepsilon}(1+\varepsilon)^{1+\varepsilon} = e^{-\varepsilon+(1+\varepsilon)\log(1+\varepsilon)}=e^{\frac{\varepsilon^2}2+O(\varepsilon^3)}>1, $$ so $$\sum_{n=1}^\infty \mathbb P(X_n\leqslant n(1-\varepsilon))\leqslant\sum_{n=1}^\infty \left(e^{\varepsilon}(1-\varepsilon)^{1-\varepsilon} \right)^{-n}<\infty $$ and $$\sum_{n=1}^\infty \mathbb P(X_n\geqslant n(1+\varepsilon))\leqslant\sum_{n=1}^\infty \left(e^{-\varepsilon}(1+\varepsilon)^{1+\varepsilon} \right)^{-n}<\infty.$$ It follows from the first Borel-Cantelli lemma that $$\mathbb P\left(\limsup_{n\to\infty} \{X_n\leqslant n(1-\varepsilon)\}\right)=\mathbb P\left(\limsup_{n\to\infty} \{X_n\geqslant n(1+\varepsilon)\}\right)=0,$$ and so we conclude.

edited Jun 23 '16 at 16:26

answered Jun 19 '16 at 18:35

Math1000

36,983

This proves that $Y_n/n\to1$ almost surely, for some sequence $(Y_n)$ such that, for each $n$, $Y_n$ and $X_n$ have the same distribution. Nothing can be deduced from this about the almost sure convergence of $X_n/n$... – Did Jun 19 '16 at 18:43
@Did it is better to write your solution too. – Thomas Edison Jun 19 '16 at 18:46
@Did Ah, I see now. Allow me to revise the answer. – Math1000 Jun 19 '16 at 18:54
@Did Then can we not say that if $X_n \sim B(n,p)$, then $X_n/n \rightarrow p$ a.s? because $X_n$ and $Y_n= B_1+\cdots +B_n$ are the same distribution where $B_i$'s are iid Bernoulli random variables with parameter $p$, so it just says that $Y_n/n \rightarrow p$ a.s, but not for $X_n/n$? – Sungjin Kim Jun 19 '16 at 19:20
@i707107 It happens that $U_n/n\to p$ almost surely, for every random sequence $(U_n)$ such that each $U_n$ is binomial $(n,p)$, but this is not for the reason you say. – Did Jun 19 '16 at 19:26
@Did I am still not sure why my argument cannot apply. It seems that it works for binomial distribution as you say, but why not for Poisson? – Sungjin Kim Jun 19 '16 at 19:32
@i707107 The result holds for Poisson and binomial. But the argument to prove it that you explain, based on the LLN, fails for Poisson and binomial. – Did Jun 19 '16 at 19:32
@Did Can you please explain why it fails? It seems that you don't use LLN to prove that if $X_n\sim B(n,p)$ then $X_n/n\rightarrow p$ a.s. I see that large deviation bound can prove these results, but still don't see why specifically LLN does not apply. – Sungjin Kim Jun 19 '16 at 19:40
@i707107 Again: please explain specifically the sequence you want to apply the LLN to (which random variables, which independences), otherwise I fail to see what we are discussing. – Did Jun 19 '16 at 20:01
@Did I see your point now. You are concerned about how random variables are made of. So, just by assuming $X_n\sim \textrm{Poisson}(n)$ would not imply that $X_n = Y_1+\cdots + Y_n$ with iid random variables $Y_i \sim \textrm{Poisson}(1)$. Sorry for my misunderstanding. – Sungjin Kim Jun 19 '16 at 20:14
My argument would apply if it is assumed that $X_t$ is a Poisson process. That's what I had understood OP's question in the first place, but looking back OP's problem again, it is not assumed to be Poisson process. It is my misunderstanding, and possibly mislead OP. I will delete all of my comments based on this misunderstanding. – Sungjin Kim Jun 19 '16 at 20:14
@Did Would there be any example that $X_n\sim Y_n$ with $Y_n\rightarrow a \ \textrm{a.s.}$ and $X_n\not\rightarrow a \ \textrm{a.s.}$? – Sungjin Kim Jun 19 '16 at 20:54
@i707107 Consider $X_n=\mathbf 1_{nU<1}$, where $U$ is uniform on $(0,1)$, and $(Y_n)$ independent with $P(Y_n=1)=\frac1n=1-P(X_n=0)$. Then $P(X_n\to0)=1$ but $\sum\frac1n$ diverges hence $Y_n=1$ for infinitely many $n$, almost surely, by Borel-Cantelli second lemma, in particular, $P(Y_n\to0)=0$. – Did Jun 19 '16 at 21:01
@i707107 Indeed, your argument show that if $(X_t)$ is a Poisson process with intensity $1$ then $X_n/n\to1$ almost surely, by the LLN. – Did Jun 19 '16 at 21:02
@Did Thank you for the clarification, and I am very sorry for all of cynically sounding comments that I had previously. Great to know that I still have much to learn. – Sungjin Kim Jun 19 '16 at 21:09
@Did I think that $Y_n$ with indicator on $(h_n, h_n+\frac1n) \ \mathrm{mod} \ 1$ would also work, where $h_n= 1 + 1/2 + \cdots + 1/n$. – Sungjin Kim Jun 19 '16 at 21:23
1

@Math1000 The new proof is complete, provided one shows that $$e^{\varepsilon}(1-\varepsilon)^{1-\varepsilon}>1,\qquad e^{-\varepsilon}(1+\varepsilon)^{1+\varepsilon}>1.$$ – Did Jun 20 '16 at 05:40
1

Correction: the new proof is incomplete, since the step "It follows that $\mathbb P\left(\left|k^{-1}X_k-1\right|<\varepsilon \right)=1$" does not hold. The preceding construction only shows that $\mathbb P\left(\left|k^{-1}X_k-1\right|<\varepsilon \right)>1-\varepsilon$, which does not allow to conclude. Please modify this step and the computations which prepare it, using Borel-Cantelli lemma (as mentioned in the very first comment to the OP...). – Did Jun 20 '16 at 09:20
@Did For the first point, $e^{\varepsilon}(1-\varepsilon)^{1-\varepsilon}=e^{\varepsilon^2/2+O(\varepsilon^3)},\qquad e^{-\varepsilon}(1+\varepsilon)^{1+\varepsilon}=e^{\varepsilon^2/2+O(\varepsilon^3)}.$ – Sungjin Kim Jun 20 '16 at 14:16
1

@Did For the second point, by the above, we have $\sum_n P(X_n\leqslant n(1-\varepsilon)) <\infty$ and $\sum_n P(X_n\geqslant n(1+\varepsilon)) <\infty$. Then by first Borel-Cantelli lemma, we have the result. – Sungjin Kim Jun 20 '16 at 14:19
@i707107 I know, thanks. Please address your computations to the OP... :-) – Did Jun 20 '16 at 14:41
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@Did I knew that you would know, just wanted to make sure what I have is correct. I think the OP will get a notification by my comments anyways. – Sungjin Kim Jun 20 '16 at 14:49

Find the almost sure limit of $X_n/n$, where each random variable $X_n$ has a Poisson distribution with parameter $n$

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