Suppose that the number of events that occur in a specified time is a Poisson random variable with parameter $\lambda$. If each event is counted with probability $p$, independently of every other event, show that the number of events that are counted is a Poisson random variable with parameter $\lambda p$. Also, give an intuitive argument as to why this should be so. As an application of the preceding result, suppose that the number of distinct uranium deposits in a given area is a Poisson random variable with parameter $\lambda =10$. If, in a fixed period of time, each deposit is discovered independently with probability $\frac{1}{50}$, find the probability that

(a) exactly ,

(b) at least $1$, and

(c) at most $1$deposit is discovered during that time.

Define $X~Pois\left(\lambda \right)$. Define $Y$as the random variable that counts the events that are accepted. We are required to show that $Y~Pois\left(p\lambda \right)$. Take any $k\in {\mathrm{\mathbb{N}}}_0$and using LOTP we have that

$P(Y=k)=\sum _{n=k}^{\infty }P(Y=k\mid X=n)P(X=n)$

Observe that if we are given $X=n,Y$has binomial distribution since every of $n$events are independently accepted with the probability $p$.

$P(Y=k)=\sum _{n=k}^{\infty }\left(\begin{array}{l}n\\ k\end{array}\right)p^k(1-p{)}^{n-k}·\frac{{\lambda }^n}{n!}{e}^{-\lambda }$

$=\sum _{n=k}^{\infty }\frac{n!}{k!(n-k)!}{p}^k(1-p{)}^{n-k}·\frac{{\lambda }^n}{n!}{e}^{-\lambda }$

$=\frac{p^k}{k!}{e}^{-\lambda }\sum _{n=k}^{\infty }\frac{1}{(n-k)!}(1-p{)}^{n-k}·{\lambda }^n$

$=\frac{p^k}{k!}{e}^{-\lambda }\sum _{n=0}^{\infty }\frac{1}{n!}(1-p{)}^n·{\lambda }^{n+k}$

$=\frac{p^k}{k!}{e}^{-\lambda }{\lambda }^k\sum _{n=0}^{\infty }\frac{1}{n!}(1-p{)}^n·{\lambda }^n$

$=\frac{p^k}{k!}{e}^{-\lambda }{\lambda }^k\sum _{n=0}^{\infty }\frac{\left(\lambda \right(1-p){)}^n}{n!}$

$=\frac{p^k}{k!}{e}^{-\lambda }{\lambda }^k·e^{\lambda (1-p)}$

$=\frac{(\lambda p{)}^k}{k!}{e}^{-\lambda p}$

Since it holds for every $k$, we have proved that $Y~Pois\left(\lambda p\right)$. Observe that it is consistent with our intuition: the number of accepted events also behaves like Poisson process since the original distribution is Poisson and the average rate is $p\lambda$. For the last question, we are given that $\lambda p=10·\frac{1}{50}=0.2$.

(a) $P(Y=1)=0.2·e^{-0.2}$

(b) $P(Y\ge 1)=1-P(Y=0)=1-e^{-0.2}$

(c)$P(Y\le 1)=P(Y=0)+P(Y=1)=e^{-0.2}+0.2·e^{-0.2}$