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Let $X_{1},X_{2},...$ be independent random variables such that $X_{k}$ is Po(k)-distributed for k=1,2... Show that:

$$Z_{n}=\frac{1}{n}\sum_{k=1}^{n}\left(X_{k}-\frac{n^{2}}{2}\right)$$

converges to a N($\frac{1}{2}$,$\frac{1}{2}$) distribution as $n \to > \infty $

I know that $X_{k} \overset{\underset{\mathrm{d}}{}}{=} \sum_{j=1}^{k}Y_{j,k}$ for every $k\geq1$, are independent Po(1)-distributed random variables.

My two immediate thoughts to solving this problem involve the central limit theorem OR characteristic function convergence. However, I'm not sure if the central limit theorem can be used since the RV's are not i.i.d. All examples that I've seen in the past have involved converging to a normal distribution with 0 mean.

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    $\begingroup$ Shouldn't it be $(X_k - k^2 / 2)$? (i.e. not $n^2$) $\endgroup$ – Felipe Gerard Dec 7 '15 at 2:42
  • $\begingroup$ @FelipeGerard no the problem is written that way $\endgroup$ – user1988 Dec 7 '15 at 2:47
  • $\begingroup$ Got it. You should delete your answer, though, as the comment is the right way to "chat" here. $\endgroup$ – Felipe Gerard Dec 7 '15 at 2:51
  • $\begingroup$ The problem is wrongly written. It should be $$Z_{n}=\frac{1}{n}\sum_{k=1}^{n}(X_{k}-\frac{n}{2})$$ $\endgroup$ – Alecos Papadopoulos Dec 7 '15 at 3:41
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You are quite close to solving the problem:

Using the representation $X_k=\sum_{j=1}^k Y_{kj}$ where $Y_{ij}\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$, you have $$\sum_{k=1}^n X_k= \sum_{k=1}^n\sum_{j=1}^k Y_{kj} = \sum_{u=1}^{n(n+1)/2} \xi_{u}$$where $\xi_u\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$ $(u=1,\ldots,n(n+1)/2)$. Therefore, if you normalise the above sum, you get $$\eqalign{\dfrac{\sum_{k=1}^n X_k-\mathbb{E}[\sum_{k=1}^n X_k]}{\text{var}(\sum_{k=1}^n X_k)^{1/2}} &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\mathbb{E}[\sum_{u=1}^{n(n+1)/2} \xi_{u}]}{\text{var}(\sum_{u=1}^{n(n+1)/2} \xi_{u})^{1/2}}\\ &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\frac{n(n+1)}{2} }{(n(n+1)/2)^{1/2}}\\ &=\sqrt{2}\,\dfrac{\sum_{k=1}^n X_k-\frac{n^2}{2}-\frac{n}{2}}{n(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{\frac{1}{n}\sum_{k=1}^n \left[X_k-\frac{n^2}{2}\right]-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{Z_n-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ }$$ which should help you conclude, along with a CLT on the above.

The theoretical references for a Central Limit Theorem for independent but not i.i.d. random variables are Liapounov's and Lindeberg's versions of the CLT. The former requires moments of order $2+\epsilon$ with $\epsilon>0$ and the latter for vanishing tail second moments. Both conditions apply for Poisson variates.

That the result holds (and hence that there is no mistake in the formulation) can be checked by a quick R experiment, as illustrated by the following that compares an histogram of 10³ $Z_n$'s with the $\text{N}(1/2,1/2)$ density:

enter image description here

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  • $\begingroup$ Could you please provide a reference/link where one could study the case of sum of Poissons whose parameter increases with their number (like we have here)? I would like to study the proof that the CLT holds in such a case. $\endgroup$ – Alecos Papadopoulos Dec 7 '15 at 12:52
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    $\begingroup$ @alecos-papadopoulos I think you should examine CLT using Lyapunov or Lindeberg form. $\endgroup$ – Alexey Zaytsev Dec 7 '15 at 13:18
  • $\begingroup$ @xian You are right. Just to point this thing out to Alecos. $\endgroup$ – Alexey Zaytsev Dec 7 '15 at 13:22
  • $\begingroup$ @Alexey Indeed, I just got confused and used $k$ as the standard deviation and not as the variance, in proving that the Lindeberg condition holds. $\endgroup$ – Alecos Papadopoulos Dec 7 '15 at 14:06

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