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I have a problem in understanding the derivation of the Poisson Process in the second edition of Klenke "Probability Theory". At the end of page 124 (screenshot), he explains the meaning of P5 that is $\limsup_{\epsilon\to 0}P[N_\epsilon\geq 2]=0$, by defining $$\lambda=\limsup_{\epsilon\to 0}P[N_\epsilon\geq 2].$$

And writes: For any $n\in\mathbb{N}$ and $\epsilon>0$, we have

$$ P[N_{2^{-n}}\geq 2]\geq\lfloor\frac{2^{-n}}{\epsilon}\rfloor P[N_{\epsilon}\geq2]-\lfloor\frac{2^{-n}}{\epsilon}\rfloor^{2}P[N_{\epsilon}\geq2]^{2}. $$ Hence, $$2^n P[N_{2^{-n}}\geq 2]\geq \lambda-2^{-n}\lambda^2\xrightarrow{n\to\infty} \lambda.$$

I cannot fully derive none of the above inequalities.

For the first one this is what I did:

Let $R_{n,\epsilon}\triangleq\lfloor\frac{2^{-n}}{\epsilon}\rfloor$ and $\underline{R}_{n,\epsilon}\triangleq2^{-n}\mod\epsilon$, so that $2^{-n}=R_{n,\epsilon}\epsilon+\underline{R}_{n,\epsilon}$, with $0\leq\underline{R}_{n,\epsilon}<\epsilon.$ We use the first three proprieties P1, P2, and P3 to derive the above relation as follows,

$$\begin{align*} P[N_{2^{-n}}\geq2]= & P\left[N_{\bigcup_{r=0}^{R_{n,\epsilon}-1}(r\epsilon,(r+1)\epsilon]\cup(R_{n,\epsilon}\epsilon,2^{-n}]}\geq2\right]=P\left[\sum_{r=0}^{R_{n,\epsilon}-1}N_{(r\epsilon,(r+1)\epsilon]}+N_{(R_{n,\epsilon}\epsilon,2^{-n}]}\geq2\right]\\ \geq & P\left[\bigcup_{r=0}^{R_{n,\epsilon}-1}\left\{ N_{(r\epsilon,(r+1)\epsilon]}\geq2\right\} \right]=R_{n,\epsilon}P\left[N_{(r\epsilon,(r+1)\epsilon]}\geq2\right]=\lfloor\frac{2^{-n}}{\epsilon}\rfloor P\left[N_{\epsilon}\geq2\right] \end{align*},$$

Which only gives the first term of the inequality but not the second squared term. Given the second line we do not really need it as it goes to 0 by taking $n$ to infinity, but I am afraid I might have done a mistake in the above result.

I appreciate if someone could verify it and also explains how the second relation after the "hence" comes about (I can see that but not exactly).

RozaTh
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1 Answers1

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For notational convenience, write $A_r := \left\{ N_{(r\epsilon,(r+1)\epsilon]}\geq2\right\} $. The argument for the first inequality goes like this: $$ P(N_{2^{-n}}\ge2)\stackrel{(1)}\ge P(\bigcup_r A_r)\stackrel{(2)}\ge \sum_r P(A_r) - \sum_{r<s}P(A_r\cap A_s) $$ In the above, (1) follows because the event $\bigcup_rA_r$ implies the event $\{N_{2^{-n}}\ge2\}$. Inequality (2) is a Bonferroni inequality, namely the one obtained by truncating inclusion-exclusion at two terms. Next, replace $\sum_r P(A_r)$ with $R_{n,\epsilon}P(A_0)$ as you've done. Simplify $\sum_{r<s}P(A_r\cap A_s)$ to $\sum_{r<s}P(A_r)P(A_s)=\sum_{r<s}P(A_0)^2$ by properties P3 and P2; and clearly $\sum_{r<s}P(A_0)^2$ is bounded above by $R_{N,\epsilon}^2$ times $P(A_0)^2$, which completes the proof.

As for the second inequality, you multiply the first inequality through by $2^n$ (for fixed $n$), then apply $\limsup$, which preserves the inequality. Then bound the RHS of the result using the inequality $$\limsup (a_\epsilon-b_\epsilon)\ge \limsup a_\epsilon - \limsup b_\epsilon,\tag3$$ (which follows from $\limsup (x_\epsilon + y_\epsilon)\le \limsup x_\epsilon + \limsup y_\epsilon$). Finally use P5 to argue that the limsups on the RHS of (3) are equal to the stated values.

For example, to see why $$\limsup_\epsilon 2^n\epsilon\lfloor 2^{-n}/\epsilon\rfloor \frac1\epsilon P(N_\epsilon\ge2)=\lambda,$$ you can use the inequality $x-1<\lfloor x\rfloor\le x$ to argue that $2^n\epsilon\lfloor 2^{-n}/\epsilon\rfloor\to1$ as $\epsilon\to0$. (Remember $n$ is still fixed.)

grand_chat
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  • Thanks for the answer. I just have one question regarding the second inequality. When we multiply the first inequality by $2^n$, what happens to the floor function? It appears to me that we cannot just simplify $2^n\lfloor 2^{-n}/\epsilon\rfloor$ to $1/\epsilon$ and taking away the floor function increases the RHS of the first inequality. Would you please explain that part? – RozaTh Aug 31 '18 at 06:12
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    @Soufa Right, you cannot just take away the floor function. See my edit for a careful argument. – grand_chat Aug 31 '18 at 13:31
  • Thanks!, another question if I may. The book uses $\lim_{n\to\infty} 2^n P[N_{2^{-n}}\geq 2]=\lambda$, in the next page. However, the second inequality is only from below. I guess we need we need an argument like $$\liminf_{n\to\infty} \lambda-2^{-n}\lambda^2\leq \liminf_{n\to\infty} 2^nP[N_{2^{-n}}\geq 2] \leq \limsup_{n\to\infty} 2^nP[N_{2^{-n}}\geq 2]\leq \lambda.$$ I guess we have $\limsup_{n\to\infty} 2^nP[N_{2^{-n}}\geq 2]= \lambda$, since if the limsup is true for $\epsilon\to 0$ it is true for any sequence, which here happens to be $2^{-n}$. Am I right? – RozaTh Aug 31 '18 at 21:05
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    @Soufa Your argument that $\lim_{n\to\infty}=\lambda$ is correct! As to your second point, it is clear that the limsup along a sequence is less than or equal to the limsup as $\epsilon\to0$, but it can be strictly less (think of an example!) . To prevent the strictly less possibility, it is enough to show that $\limsup_{n\to\infty}\ge\limsup_{\epsilon\to0}$, as the author has done. There would be no need for this entire exercise if the limsup along a sequence always equals the limsup as $\epsilon\to0$! – grand_chat Sep 01 '18 at 01:07
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    @Soufa ... although I don't know what's happening on the next page. What is the end result? – grand_chat Sep 01 '18 at 01:18
  • Thanks got it! I can see that $\limsup_{n\to\infty} 2^nP[N_{2^{-n}}\geq 2]$ could be strictly less than $\lambda$ (I guess one example would be the typical $\sin(\frac{1}{x})$ when $x\to0$ and we pick a sequence like $\frac{1}{2n\pi}$). But here either using $\liminf_{n\to\infty} \lambda-2^{-n}\lambda^2\leq \liminf_{n\to\infty} 2^nP[N_{2^{-n}}\geq 2] $ or taking $\limsup$ of the original inequality brings us home. ... The book uses $\lim_{n\to\infty} 2^n P[N_{2^{-n}}\geq 2]=\lambda$ to claim that $\lim_{n\to\infty} (1- P[N_{2^{-n}}\geq 2])^{2^n}=e^{-\lambda}$. Thanks again! – RozaTh Sep 01 '18 at 09:17