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Klenke defines the Poisson process as a family of non-negative integer valued random variables with independent increments, whose increments are distributed according to the Poisson distribution (Definition 5.33). He proceeds to prove that the Poisson process can be characterized by five properties (Theorem 5.34, the five properties P1-P5). In order to prove the direction "If a process satisfies P1-P5, it is a Poisson process" (proof), he appeals to the Poisson approximation theorem. This theorem assumes a collection of numbers $\left(p_{n,k}\right)\subseteq\left[0,1\right]$ that satisfy two conditions: $$\left(i\right)\space\space \lim_{n\rightarrow\infty}\sum_{k=1}^\infty p_{n,k}\in\left(0,\infty\right)$$ $$\left(ii\right)\lim_{n\rightarrow\infty}\sum_{k=1}^\infty p_{n,k}^2=0$$ Accordingly, in the proof of Theorem 5.34 Klenke defines a collection of numbers $\left(p_{n,k}\right)=\left(p_n\right)$ and shows that they satisfy condition $\left(i\right)$ (this takes place five lines from the end, where it is shown that $\alpha t=\lim_{n\rightarrow\infty}p_n2^n=\lim \sum_{k=1}^{2^n} p_n$). However, condition $\left(ii\right)$ isn't verified, which brings me to wonder: why does condition $\left(ii\right)$ hold?

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We have, as $p_{n,k}=\frac{\lambda}n$ when $k\leq n$ and $0$ otherwise, $$\sum_{k=1}^{+\infty}p_{n,k}^2=\sum_{k=1}^n\frac{\lambda^2}{n^2}=\frac{\lambda^2}n,$$ which gives the wanted result, unless I am missing something.

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Thanks, but what's $\lambda$? Are you referring to Theorem 5.34's proof or to the comment preceding the Poisson approximation theorem? I'm interested in understanding the former. –  Evan Aad Sep 9 '12 at 21:43
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How is $p_{n,k}$ defined in the former? –  Davide Giraudo Sep 9 '12 at 21:58
    
It is defined as $p_{n,k}:=\mathrm{P}\left(N_{\left(0,t/2^n\right]}\geq1\right)$ if $k\in\left\{1,2,\dots,2^n\right\}$ and $p_{n,k}:=0$ otherwise. –  Evan Aad Sep 10 '12 at 3:08
    
OK, i think i get it now. We define $\lambda_n:=2^np_n$, so that $p_n=\lambda_n/2^n$. It is shown in the proof that $\lim_{n\rightarrow\infty}\lambda_n=\alpha t\in\mathbb{R}$. So $$\lim_{n\rightarrow\infty}\sum_{k=1}^{2^n} p_{n,k}^2=\lim_{n\rightarrow\infty}\frac{\lambda_n^2}{2^n}=0$$ Thanks, Davide! –  Evan Aad Sep 10 '12 at 5:30

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