Assume a Poisson process with parameter 2. How do I compute $$P\{ N(1)=2, N(3) = 4 | N(5) = 6\}?$$ I know from conditional uniformity that we should have $$P\{N(1) = 2, N(2) = 2 | N(5)=6\}$$ so that 2+2 should be equal to 6. But here is the case that the condition is not satisfied. I need help. Could there be a typo in the problem or there is a way around this problem?
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what does $N(t)$ stand for? number of events by time $t$? – Daniel S. Aug 03 '20 at 02:44
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Yes $N(t)$ denotes the number of events at t – holala Aug 03 '20 at 02:48
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Using the definition of conditional probability to write your quantity as a ratio, the numerator is $$ \eqalign{P[N(1)=2, & N(3)-N(1)=2, N(5)-N(3)=2] \cr &=P[N(1)=2]P[N(3)-N(1)=2]P[N(5)-N(3)=2]\cr } $$ The three probabilities on the right can be evaluated directly, resulting in a product of $128e^{-10}$. And the denominator is just ${10^6\over 6!}e^{-10}$.
John Dawkins
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If the number of events by time $5$ is 6, you have 6 events uniformly distributed over the interval $[0,5]$. Now,
$$P(N(1)=2, N(3)=4|N(5)=6) = P(N(3)=4|N(1)=2,N(5)=6) P(N(1)=2|N(5)=6)$$
Then, from memoryless property,
$$ P(N(3)=4|N(1)=2,N(5)=6) P(N(1)=2|N(5)=6)=P(N(2)=2|N(4)=4) P(N(1)=2|N(5)=6) $$
The first term equals the probability that the first 2 events from 4 uniform events in the interval [0,4] occur before 2. I believe that by symmetry this equals 0.5
The second term equals the probability that the first 2 events from 6 uniform events in the interval [0,5] occur before 1. I believe you can compute this from conditional probability poisson process
Daniel S.
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From https://math.stackexchange.com/questions/733487/conditional-probability-poisson-process?rq=1 Observed that the author provided a general case at the bottom where $S = X_1 + X_2 + \cdots + X_n$. This is exactly what I meant earlier. This condition has to be meet. – holala Aug 03 '20 at 03:18