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April 12th, 2019, 03:41 PM  #1 
Newbie Joined: Apr 2019 From: Poland Posts: 2 Thanks: 0  How does the following equation make sense in case of a Poisson process?
>$$P(N_1=2, N_4=6) = P(N_1=2, N_4N_1=4) = P(N_1=2) \cdot P(N_3=4)$$ How does the above equation make sense in case of a Poisson process? Why is $(N_3=4)=(N_4N_1=4)$? Can anyone explain? Last edited by user366312; April 12th, 2019 at 03:59 PM. 
April 12th, 2019, 10:28 PM  #2 
Newbie Joined: Apr 2019 From: Poland Posts: 2 Thanks: 0  How does the following equation make sense in case of a Poisson process?
What I understand is: 1. $P(N_1=2, N_4=6)$ means "the probability of count of 2 items arriving at step1, AND 6 items arriving at step4. 2. $P(N_1=2, N_4N_1=4)$ means that "the probability of count of 2 items arriving at step1 AND the 4 items as the difference of steps 4 and 1. 3. which is same as $P(N_1=2, N_3=4)$ i.e. "the probability of count of 2 items at step1 AND 4 items at step3". I understand (2) as (1) is represented as a "difference" term in (2). But, how is (2) and (3) equivalent? 

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case, equation, make, poisson, process, sense, stochastic process 
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