Question

Suppose that Xi, i = 1, 2, 3 are independent Poisson random variables with respective means λi, i = 1, 2, 3. Let X = X₁ + X₂ and Y = X₂ + X₃. The random vector X, Y is said to have a bivariate Poisson distribution. Find its joint probability mass function. That is, find P{X = n, Y = m}.

Short answer

$P(X=n,Y=m)=e^{-({\lambda }_1+{\lambda }_2+{\lambda }_3)}\sum _{k-0}^{min(n,m)}\frac{{{\lambda }_1}^{n-k}}{(n-k)!}.\frac{{{\lambda }^k}_2}{k!}.\frac{{{\lambda }_3}^{m-k}}{(m-k)!}$

Step-by-step solution

Content Introduction

Since X and Y are sums of independent Poisson variables, we have that$X~Pois({\lambda }_1+{\lambda }_2),Y~Pois({\lambda }_2+{\lambda }_3)$

Content Explanation

Use conditional probability to obtain that

$$\begin{gathered} P(X=n,Y=m)=\sum _{k=0}^{\infty }P(X=n,Y=m,X_2=k)P(X_2=k) \\ =\sum _{k=0}^{\infty }P(X_1+X_2=n,X_2+X_3=m/X_2=k)P(X_2=k) \\ =\sum _{k=0}^{n}P(X_1=n-k)P(X_3=m-k)P(X_2=k) \\ =\sum _{k=0}^n\frac{{{\lambda }_1}^{n-k}}{(n-k)!}{e}^{-\lambda }.\frac{{{\lambda }_3}^{m-k}}{(m-k)!}{e}^{-{\lambda }_3}.\frac{{{\lambda }_2}^k}{k!}{e}^{-{\lambda }_2} \\ =e^{-({\lambda }_1+{\lambda }_2+{\lambda }_3)}\sum _{k=0}^n\frac{{{\lambda }_1}^{n-k}}{(n-k)!}.\frac{{{\lambda }_3}^{m-k}}{(m-k)!}.\frac{{{\lambda }_2}^k}{k!} \end{gathered}$$