We have a poisson process that has an arrival rate of $\lambda$. If the arrival counter doesn't count certain arrivals at random (with a probability $p$), then how do I go about showing that the arrival process is still Poisson and finding the new arrival rate?

Intuitively it makes sense because the arrivals are independent and random, but I'm not sure how to mathematically approach the problem.

  • $\begingroup$ I answered this same question a few hours ago: math.stackexchange.com/questions/1665672/… $\qquad$ $\endgroup$ – Michael Hardy Feb 21 '16 at 21:42
  • $\begingroup$ What is $Poisson(\lambda)$ process? It's a count in process that has a count in $dx$ with probability $\sim\lambda dx$ independent of other intervals. After thinning, the count in $dx$ happens with probability $\sim p\lambda dx$ - still independent, still a Poisson process. $\endgroup$ – A.S. Feb 21 '16 at 21:45

\begin{align} & \sum_{x=y}^\infty \frac{\lambda^x e^{\lambda }}{x!} \cdot \binom x y p^y (1-p)^{x-y} \\[12pt] = {} & \frac{e^{-\lambda} p^y}{y!} \sum_{x=y}^\infty \frac{\lambda^x(1-p)^{x-y}}{(x-y)!} \\[12pt] = {} & \frac{e^{-\lambda} p^y}{y!} \sum_{w=0}^\infty \lambda^{w+y} \frac{(1-p)^w}{w!} \\[12pt] = {} & \frac{e^{-\lambda} (\lambda p)^y}{y!} \sum_{w=0}^\infty \lambda^w \frac{(1-p)^w}{w!} \\[12pt] = {} & \frac{e^{-\lambda} (\lambda p)^y}{y!} \cdot e^{\lambda(1-p)} \\[12pt] = {} & \frac{e^{-\lambda p} (\lambda p)^y}{y!}. \end{align} So it's a Poisson distribution with expected value $\lambda p$.


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