# Thinning a Poisson Process

Suppose that events are produced according to a Poisson process with an average of lambda events per minute. Each event has a probability $p$ of being Type A event, independent of other events.

Let the random variable $Y$ represent the number of Type A events that occur in a one-minute period. Prove that $Y$ has a Poisson distribution with mean $\lambda p$.

I read over it and I feel like I'm missing something because I still see it as having the mean as lambda not $\lambda p$.

• What did you do to see this? A good approach with Poisson variables is to work with the generating function. – Siméon Nov 25 '13 at 18:53
• This answer by Zoli will be useful: math.stackexchange.com/questions/2254770/… – L.V.Rao Apr 27 '17 at 15:29

Let $X \sim \mathcal{P}(\lambda)$ denote the number of events during a give one-minute period. Each of these events being of Type A with probability $p$ (independently with respect to the others), if $Z_1,Z_2,Z_3,\dots$ is a sequence of independent Bernoulli variables with $P(Z_i = 1)=p$, then the following identity holds in distribution: $$Y \overset{(d)}{=} \sum_{i=1}^X Z_i = \sum_{i=1}^\infty Z_i\,1_{i \leq X}.$$
By independence of $Z_i$ and $X$ for every $i$, one has $$E(Y) = \sum_{i=1}^\infty E(Z_i)E(1_{i \leq X}) = p \sum_{i=1}^\infty E(1_{i \leq X}) = pE(X) = p\lambda.$$
To prove that $Y$ is Poisson with parameter $p\lambda$, we use the generating function: $$E[s^Y] = \sum_{n=0}^\infty E[s^Y \mid X = n] \frac{\lambda^n}{n!}e^{-\lambda}$$ where $E[s^Y \mid X = n] = E[s^{Z_1}]^n = (1+p(s-1))^n$. So finally, one has $$E[s^Y] = e^{\lambda(1+p(s-1))}e^{-\lambda} = e^{p\lambda(s-1)}.$$