Question #288034

If X and Y are independent Poisson variates with mean lambda 1 and lambda 2 respectively, find the probability that x+y=k

1
Expert's answer
2022-01-17T16:03:58-0500

Sums of independent Poisson random variables are Poisson random variables. Let XX and YY be independent Poisson random variables with parameters λ1λ_1 and λ2,λ_2, respectively. Define λ=λ1+λ2λ = λ_1 + λ_2 and Z=X+Y.Z = X + Y. We claim that ZZ is a Poisson random variable with parameter λ.λ.

Then


P(X+Y=k)=P(Z=k)=eλ(λ)kk!P(X+Y=k)=P(Z=k)=\dfrac{e^{-\lambda}(\lambda)^k}{k!}

P(X+Y=k)=eλ1+λ2(λ1+λ2)kk!P(X+Y=k)=\dfrac{e^{-\lambda_1+\lambda_2}(\lambda_1+\lambda_2)^k}{k!}


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