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.
\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 $\lambda p$.