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Considering a Poisson process with intensity 2, how does one compute the probability that the total number of events in the time interval (0, 2] is exactly twice the number of events during the first half of this interval?

Jonas
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    Then the number of events in $(0,1]$ and the number of events in $(1,2]$ must be the same. – Sungjin Kim Oct 24 '16 at 22:27
  • Hm. Guess I should get something like $\sum_{k=0}^\infty P(N(1)-N(0) = k)\cdot P(N(2)-N(1) = k) = e^{-2\frac{2^k}{k!}\cdot e^{-2}\frac{2^k}{k!}$ then? – Jonas Oct 24 '16 at 23:50

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