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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? >$$P(N_1=2, N_4=6) = P(N_1=2, N_4-N_1=4) = P(N_1=2) \cdot P(N_3=4)$$ How does the above equation make sense in case of a Poisson process? Why is $(N_3=4)=(N_4-N_1=4)$? 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 step-1, AND 6 items arriving at step-4. 2. $P(N_1=2, N_4-N_1=4)$ means that "the probability of count of 2 items arriving at step-1 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 step-1 AND 4 items at step-3". I understand (2) as (1) is represented as a "difference" term in (2). But, how is (2) and (3) equivalent? Tags case, equation, make, poisson, process, sense, stochastic process Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post girlbadatmath Algebra 2 August 9th, 2014 10:53 AM king.oslo Applied Math 4 September 19th, 2013 11:56 PM perfect_world Elementary Math 3 August 13th, 2013 10:22 AM eliudthorn Algebra 8 January 21st, 2012 01:07 PM

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