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For a homogeneous Poisson process N on [0,$\infty$] show that for 0

$$P(N(s)=k \ | \ N(t))= {{N(t)}\choose{k}} (\frac{s}{t})^k (1-\frac{s}{t})^{N(t)-k} $$ for $k\leq N(t)$

N(t) is Poisson distributed as in the Cramér-lundberg model.

Is first recognised the result as the probability function for a binomial distribution. So what i want to find is probably P(X=k). where X $ \sim$ $Bin(\frac{s}{t},N(t))$. My problem is, I can't seem to find a result that gives the binomial. I can't find the link. I tried to write out the first, but all i get is something that is as well poisson distributed.

Viktor Jeppesen
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1 Answers1

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It's a trivial exercise in conditional probability: since $N(s)$ is Poisson with the mean $\lambda s$, $N(t)-N(s)$ is Poisson with the mean $\lambda(t-s)$, and since they're independent, $P(N(s)=k\mid N(t)=n)=\dfrac{P(N(s)=k, N(t)=n)}{P(N(t)=n)}=\dfrac{P(N(s)=k, N(t)-N(s)=n-k)}{P(N(t)=n)}$, and by independence, $\dfrac{P(N(s)=k)P(N(t)-N(s)=n-k)}{P(N(t)=n)}$. Now substitute Poisson probabilities into the formula and simplify.

Andrei Zorine
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  • Hi, was learning about Poisson Processes and found this question. Was hoping you could explain why $\dfrac{P(N(s)=k, N(t)=n)}{P(N(t)=n)}=\dfrac{P(N(s)=k, N(t)-N(s)=n-k)}{P(N(t)=n)}$. Thanks! – Wei Xiong Yeo Sep 20 '19 at 08:54
  • OK, watch: the in the numerator we have $$P(N(s)=k,N(t)=n)=P(N(s)=k, N(t)-N(s)=n-N(s))=P(N(s)=n,N(t)-N(s)=n-k).$$ Just, equivalent events with equivalent equation systems. – Andrei Zorine Sep 26 '19 at 06:27