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I was doing some past exams on distributions, and I can not do part e), nor do I understand the mark scheme. The first 4 questions I did as workings below, but can anyone help me try to understand and show part e)? enter image description here

Aurora Borealis
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    Strictly speaking the question is un-answerable since they say it is a "Poisson distribution" (which is for a single random variable that does not specify timing) rather than a "Poisson process" (which is for a process with i.i.d. inter-arrival times). Nevertheless they want you to use the fact of Poisson processes that, conditioned on $k$ arrivals occuring in a given interval $[0,T]$, the arrivals are independent and are uniformly distributed over that interval. – Michael Jan 30 '19 at 08:00
  • Michael, I'm not sure I see your point. The total number of arrivals in a Poisson process during a fixed time interval is a Poisson random variable, no? And the sum of two Poisson variables is another Poisson variable. That's all that's required here. – nkm Jan 30 '19 at 17:18

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We suppose Archie's emails per day $(X_1,X_2)$ are distributed as a Poisson($\lambda$), so his email count over two days $(Y = X_1+X_2)$ is distributed as Poisson($2\lambda$). The probability that Archie received exactly 10 emails on each of the first two days, given that he received a total of 20 emails over the two days, is $$P = \frac{P(X_1=10) P(X_2=10)}{P(Y=20)} = \frac{\left(\frac{\lambda^{10} e^{-\lambda}}{10!}\right)^2}{\frac{(2\lambda)^{20} e^{-2\lambda}}{20!}} = \frac{20!}{(10!)^2 2^{20}}$$ which does not depend on $\lambda$.

nkm
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