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$N_1(t)$ and $N_2(t)$ are two independent Poisson processes with intensities $λ_1$ and $λ_2$ respectively. $v_1$ and $v_2$ are two dependent positive random variables, and $p=P(v_1<c)=P(v_2<c)$, where $c$ is a constant.

When some random event of $N_1(t)$ happened and $v1<c$, it means an event of point process $N_3(t)$ happened. When some random event happened in $N_2(t)$ and $v_2<c$, it means an event of point process $N_4(t)$ happened.

Then $N_3(t)$ and $N_4(t)$ are two dependent Poisson processes with intensities $p*λ_1$ and $p*λ_2$ respectively. My questions are:

  1. Is the superposition of $N_3(t)$ and $N_4(t)$ a Poisson process approximately?
  2. Let $N_5(t)$ be a Poisson process with intensity $p(λ_1+λ_2)$, is there $P\{N_3(t)+N_4(t)\le m\} \le P\{N_5(t)\le m\}$? (Here $m$ is a integer.)

All your answers are helpful. I really appreciate them and thanks for gung's editing.

Specially, $v_1$ and $v_2$ be denoted as two dependent log-normal varialbles with correlation $\rho$. In other words, we can describe $v_1$ and $v_2$ as a multivariate log-normal variable. In this time, how to answer those two questions?

I agree that ($v_1$, $v_2$) can be replaced by the dependent Bernoulli pair($b_1$, $b_2$). And now $N_{3}(t)$ and $N_{4}(t)$ are two dependent poisson process. Is their superposition a poisson process approximately?

I really appreciate all your reply.

I have read sveral literature and consulted a professor. $N_{3}(t)$+$N_{4}(t)$ is a Cox process according to reply of the professor. And it converges to a poisson process.(Serfozo 1984, Chen & Xia 2011).

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  • $\begingroup$ Welcome to CV! Thanks for clarifying your problem - I had some concerns but your question is much improved now. $\endgroup$ – soakley May 29 '13 at 16:05
  • $\begingroup$ Not sure to understand. Does that mean that $N_3(t) \equiv 0$ when $v_1\geq c$ ? If so, then $N_3$ is not a Poisson process. $\endgroup$ – Stéphane Laurent May 29 '13 at 16:33
  • $\begingroup$ Can't we just replace $v_1, v_2, c$ with a single Bernoulli random variable $B$ with probability $p$? $\endgroup$ – Neil G May 29 '13 at 17:27
  • $\begingroup$ I don't think so. He wants $v_1$ and $v_2$ to be dependent random variables. Maybe we could get some more detail on that. $\endgroup$ – soakley May 29 '13 at 18:12
  • $\begingroup$ @soakley: what should it matter since you almost surely don't "look at" both variables at the same time. $\endgroup$ – Neil G May 29 '13 at 19:30
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Too long for a comment.

But I do not think v1 and v2 can be replaced with a Bernoulli random variable. Specially, v1 and v2 be denoted as two dependent log-normal varialbles with correlation ρ. In other words, we can describe v1 and v2 as a multivariate log-normal variable. In this time, how to answer those two questions?

I think you're mistaken; they can be.

The only way that $v_1$ and $v_2$ affect anything is through the condition that they're less than $c$. Let $b_1$ and $b_2$ take the value 1 when their corresponding $v_i<c$. You now have that $b_1$ and $b_2$ are dependent Bernoullis. What information relevant to the question other than the information in the pair $(b_1,b_2)$ is there in $(v_1,v_2)$? If there is no additional information of relevance to the question, then $(b_1,b_2)$ clearly can replace $(v_1,v_2)$.

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  • $\begingroup$ I agree that ($v_1$, $v_2$) can be replaced by the dependent Bernoulli pair($b_1$, $b_2$). And now $N_{3}(t)$ and $N_{4}(t)$ are two dependent poisson process. Is their superposition a poisson process approximately? $\endgroup$ – bruce May 30 '13 at 1:57
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    $\begingroup$ I think you should edit to pose this back in your original question. $\endgroup$ – Glen_b May 30 '13 at 2:03
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    $\begingroup$ @bruce: +1, I also think that $b_1, b_2$ can be replaced with a single Bernoulli since the only time that the dependence between them matters is when $N_1$ and $N_2$ have simultaneous events, but this happens almost never. $\endgroup$ – Neil G May 30 '13 at 7:55
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Your question is not well stated

As I claim in my comments, according to your description of the problem, $N_3$ and $N_4$ are not Poisson processes. Indeed (according to your description), $$N_3(t) = \begin{cases} N_1(t) & \text{if } V_1 < c \\ 0 & \text{if } V_1 \geq c \end{cases}.$$ ($V_1$ does not depend on $t$ in your assumptions, there's only one random variable $V_1$.)

Thus there is a positive probability that the trajectory of $N_3$ is identically equal to $0$, whereas this almost never occur for a Poisson process.

A possible rephrasement of your question

Maybe you need to rephrase your problem ? We could imagine the following variant. Two sequences $(X_n)$ and $(Y_n)$ of independent Bernoulli random variables are given. For each $n$, we color the $n$-th event of $N_1$ in blue if $X_n=1$, in red if $X_n=0$. Then $N_3$ is the Poisson process defined by the blue events. We do the similar construction for $N_4$ with $(Y_n)$. Moreover you assume a dependence between $X_n$ and $Y_n$. @NeilG's construction does not match your problem because this dependence appears nowhere in his answer; you clearly start with a construction based on a dependent "coloring".

I have not tried to assess the distribution of $N_3+N_4$ with the construction above.

Showing the dependency between $N_3$ and $N_4$

Clearly, @NeilG's answer cannot be correct because it does not take into account the dependence assumption (actually his strangely upvoted answer has nothing to do with your question). So let's have a first look at my rephrasement of your question and let's show that $N_3$ and $N_4$ are dependent. Denote by $T_1$, $T_2$, $\ldots$ the jumping times of $N_1$ and by $T'_1$, $T'_2$, $\ldots$ the jumping times of $N_2$. With my notations above, the first jumping time of $N_3$ is $T_S$ where $S=\min\{n\geq 1 \mid X_n=1\}$ and similarly the first jumping time of $N_3$ is $T'_{S'}$ where $S'=\min\{n\geq 1 \mid Y_n=1\}$. Let's have a look at the joint distribution of $T_S$ and $T'_{S'}$: $$\Pr(T_S \in \mathrm{d}x, T'_{S'} \in \mathrm{d}x') = \sum_{k,k'} \Pr(T_k \in \mathrm{d}x, T'_{k'} \in \mathrm{d}x') \Pr(S=k,S'=k').$$ One has $\Pr(T_k \in \mathrm{d}x, T'_{k'} \in \mathrm{d}x')=\Pr(T_k \in \mathrm{d}x) \Pr(T'_{k'} \in \mathrm{d}x')$ but $\Pr(S=k,S'=k') \neq \Pr(S=k)\Pr(S'=k')$ in general by your dependence assumption. I don't continue the calculation, but we see that the equality $\Pr(T_S \in \mathrm{d}x, T'_{S'} \in \mathrm{d}x') = \Pr(T_S \in \mathrm{d}x) \Pr(T'_{S'} \in \mathrm{d}x')$ is not expected, which implies that $N_3$ and $N_4$ are not independent.

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    $\begingroup$ Stéphane, $N_{3}(t)$ is a classical thinning Poisson process in this problem. $N_{3}(t)$ is thinned by the following procedure: $N_{1}(t)$ is retained with probability $p$, here $p=P{v_{1}<c}$. $\endgroup$ – bruce May 31 '13 at 1:53
  • $\begingroup$ @bruce Don't you agree that according to the statement you have given one has $N_3(t) \equiv 0$ when $V_1 \geq c$ ? $\endgroup$ – Stéphane Laurent May 31 '13 at 5:43
  • $\begingroup$ @bruce To construct a thinning Poisson process one need to classify each event as "red" or "blue", each independently of the other. This is not possible with your assumptions: you have only one classifier (the random variable $V_1$). The second part of my answer is a construction of thinning Poisson processes with correlated classifiers for $N_1$ and $N_2$, that seems to be close to what you have in mind (and NeilG use independent classifiers). $\endgroup$ – Stéphane Laurent May 31 '13 at 5:52
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    $\begingroup$ @bruce If there's only one $V_1$, that makes no sense to say that there is a realization of $V_1$ for each event. Do you understand ? So I really think the second part of my answer clarifies your question. $\endgroup$ – Stéphane Laurent May 31 '13 at 7:58
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    $\begingroup$ @ Stéphane You statement describe my problem more clearly. Thank you very much. The question seems difficult for me now, and I will read more literatures on Poisson process approximation and Stein's method. $\endgroup$ – bruce May 31 '13 at 8:14
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I am assuming that we can just replace $v_1, v_2, c$ with a Bernoulli random variable $B$ with probability $p$, which we use to accept events since $v_1$ and $v_2$ are almost surely not sampled at the same time.

The superposition of $N_3$ and $N_4$ are a Poisson process and are equal to $N_5$ by the colouring and superposition theorems.

Imagine a single Poisson process with rate $T = \lambda_1 + \lambda_2$. Colour each event red with probability $\frac{p\lambda_1}{T}$, green with probability $\frac{p\lambda_2}{T}$, and blue otherwise. Then, the red events are $N_3$ and the green events are $N_4$. This construction matches yours, yet in this one it is clear that $N_3$ and $N_4$ are independent (contrary to your statement).

The superposition of independent Poisson processes is a Poisson process.

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  • $\begingroup$ Thinks for your answer and I really appreciate your help. $\endgroup$ – bruce May 30 '13 at 0:55
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    $\begingroup$ @bruce I'm sorry to insist but $N_3$ (as well as $N_4$) is not a Poisson process because there is a positive probability that $N_3(t) \equiv 0$ (this probability is $\Pr(V_1 \geq c)$). Your definition of $N_3$ is not the same as Neil G's construction: with your construction $N_3$ has no event when $V_1 \geq c$. $\endgroup$ – Stéphane Laurent May 30 '13 at 8:31
  • $\begingroup$ @StéphaneLaurent: I don't understand at all what you mean. $N_3$ is the number of events that have happened between time 0 and time $t$. The condition $v_1>c$ at time $t$ doesn't reset $N_3$; it merely discards a potential event at some time $t'$. This is called "thinning" (Lewis, P. A. W., & Shedler, G. S. (2006). Simulation of nonhomogeneous Poisson processes by thinning. Naval Research Logistics Quarterly, 26(3), 403–413.) or "colouring" (Kingman, J. F. C. (1993). Poisson Processes. Oxford University Press, USA.) I am assuming that $v_1$ and $v_2$ depend on $t$. $\endgroup$ – Neil G May 30 '13 at 18:21
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    $\begingroup$ I only consider the statement of the problem given by the OP. $\endgroup$ – Stéphane Laurent May 30 '13 at 18:26
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    $\begingroup$ Dependence means that $V_1$ and $V_2$ are not independent. $\endgroup$ – Stéphane Laurent May 31 '13 at 5:45
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The process $N_3(t) + N_4(t)$ cannot be a Poisson process if $N_3(t)$ and $N_4(t)$ have nonzero correlation. The variance of the sum of two random variables is given by $\sigma^2_{X+Y} = \sigma^2_X + \sigma^2_Y + 2\sigma_{XY}.$ But the variance of the number of events occurring in time $t$ in a Poisson process must be equal to the mean since the number of events is a Poisson random variable and the Poisson distribution has $\sigma^2 = \mu.$

So if $N_3(t)$ and $N_4(t)$ have positive correlation, then the variance of their sum will be higher than that for a Poisson process with the same mean. Correspondingly, if they have negative correlation, the variance will be smaller than that for a Poisson process.

Generally the sum does approach a Poisson process as the two component processes approach independence. An interesting case would be if they are uncorrelated but not independent. In that case we do know the first two moments would match.

The dependence issue was discussed in an earlier question: Dependent thinning Poisson process

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  • $\begingroup$ Why do you think that the $N_3, N_4$ would have positive correlation? $\endgroup$ – Neil G May 31 '13 at 17:24
  • $\begingroup$ Look at Stephane Laurent's "Update" section where she shows the dependence. Also the OP declares they are dependent in the problem statement. I didn't claim any positive correlation. I'm claiming any nonzero correlation implies that $N_3(t) + N_4(t)$ cannot be a Poisson process. $\endgroup$ – soakley May 31 '13 at 22:24
  • $\begingroup$ His proof works with his rephrasing of the question in which $v_1, v_2$ are correlated for events having the same index, but bruce has made it clear that $v_1, v_2$ are correlated for events having the same time. $\endgroup$ – Neil G May 31 '13 at 22:53
  • $\begingroup$ If you read Bruce's comment (6th one under Stephane's answer), I interpret it as index related. You need realizations of $v_1$ or $v_2$ every time an event occurs. $\endgroup$ – soakley May 31 '13 at 23:36
  • $\begingroup$ You're right that it could be interpreted either way (you would also get realizations at each time if $v_1$ and $v_2$ were correlated for the same time). I wish we could get clarification on this. $\endgroup$ – Neil G May 31 '13 at 23:38

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