1. ## Probability Function

Let Y1 and Y2 be independent Poisson random variables with means x1 and x2 respectively. Find the

a) probability function of Y1 + Y2

b) conditional probability function of Y1, given that Y1 + Y2 = m

2. Originally Posted by jzellt
Let Y1 and Y2 be independent Poisson random variables with means x1 and x2 respectively. Find the

a) probability function of Y1 + Y2

b) conditional probability function of Y1, given that Y1 + Y2 = m

Here is (a). (b) is left for you to do.

$\Pr(Y_1+Y_2 =n) = \sum_{k=0}^n \Pr(Y_1=k, Y_2 = n-k)$

(since the events are disjoint, that is, mutually exclusive)

$= \sum_{k=0}^n \Pr(Y_1 = k) \Pr(Y_2 = n-k)$

(using independence)

$= \sum_{k=0}^n \frac{e^{-\lambda}\lambda^k}{k!} \frac{e^{-\mu} \mu^{n-k}}{(n-k)!}$

(using the pmf of a Poisson random variable)

$= e^{-(\lambda+\mu)} \sum_{k=0}^n \frac{\lambda^k \mu^{(n-k)}}{k!(n-k)!}$

$= \frac{e^{-(\lambda+\mu)}}{n!} \sum_{k=0}^n {n \choose k} \lambda^k \mu^{(n-k)}$

$= \frac{e^{-(\lambda+\mu)}}{n!} (\lambda+\mu)^n$

(using binomial theorem)

which is recognised as the pmf of a Poisson random variable with parameter $\lambda+\mu$.