Trying to find a closed form summation related to Poisson distribution

Question

If I have a equipment which can detect the arrival of particles with probability $p$, and the number of particles generated in a second is governed by a Poisson distribution.$$P_r[N=n]=\frac{\lambda^n e^{-\lambda}}{n!}$$ I can calculate the probability distribution of number of detected particles.$$P_d[N=n]=\sum_{k=0}^n \frac{\lambda^n e^{-\lambda}}{(n+k)!}p^n(1-p)^k$$ Does this series $$\sum_{k=0}^n \frac{\lambda^n e^{-\lambda}}{(n+k)!}p^n(1-p)^k $$ have closed form formula?

Answer 1

There are three errors in your sum. The sum should extend to infinity; the exponent of $\lambda$ should be $n+k$ instead of $n$; and you're missing the binomial coefficient $\binom{n+k}n$ that counts the number of ways of selecting $n$ detected particles from $n+k$ arriving particles. If you correct these errors, the result is

\begin{eqnarray*} P_d[N=n] &=& \sum_{k=0}^\infty\frac{\lambda^{n+k}\mathrm e^{-\lambda}}{(n+k)!}\binom{n+k}np^n(1-p)^k \\ &=& \frac{\lambda^n\mathrm e^{-\lambda}p^n}{n!}\sum_{k=0}^\infty\frac{\lambda^k(1-p)^k}{k!} \\ &=& \frac{\lambda^n\mathrm e^{-\lambda}p^n}{n!}\mathrm e^{\lambda(1-p)} \\ &=& \frac{(\lambda p)^n\mathrm e^{-\lambda p}}{n!}\;. \end{eqnarray*}

This is a Poisson distribution with parameter $\lambda p$. This result could be obtained more directly by noting that a Poisson process with rate $\lambda$ from which a fraction $p$ of events are selected is again a Poisson process, with rate $\lambda p$.