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

Expert verified

P(X=n,Y=m)=e-(λ1+λ2+λ3)k-0min(n,m)λ1n-k(n-k)!.λk2k!.λ3m-k(m-k)!

Step by step solution

01

Content Introduction

Since X and Y are sums of independent Poisson variables, we have thatX~Pois(λ1+λ2),Y~Pois(λ2+λ3)

02

Content Explanation

Use conditional probability to obtain that

P(X=n,Y=m)=k=0P(X=n,Y=m,X2=k)P(X2=k)=k=0P(X1+X2=n,X2+X3=m/X2=k)P(X2=k)=k=0nP(X1=n-k)P(X3=m-k)P(X2=k)=k=0nλ1n-k(n-k)!e-λ.λ3m-k(m-k)!e-λ3.λ2kk!e-λ2=e-(λ1+λ2+λ3)k=0nλ1n-k(n-k)!.λ3m-k(m-k)!.λ2kk!

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