Express fr in terms of fxy solution clearly r is a

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Unformatted text preview: small, with the product (m + n)p = mp + np = λ1 + λ2 . Therefore, the distribution of X + Y is well approximated by the Poisson distribution with parameter λ1 + λ2 . The approximations become better and better as p → 0, so we conclude that X + Y has the Poisson distribution with parameter λ1 + λ2 . Let’s derive the solution again, this time using the convolution formula (4.14). The support of X + Y is the set of nonnegative integers, so select an integer k ≥ 0. Then (4.14) becomes k λj e−λ1 λk−j e−λ2 2 1 j! (k − j )! j =0 k j k −j λ1 λ2 −(λ1 +λ2 ) = e j !(k − j )! j =0 k kj λk e−λ = p (1 − p)k−j k! j pX +Y (k ) = j =0 = λk e−λ k! , where λ = λ1 + λ2 , and p = λ1 . In the last step we used the fact that the sum over the binomial λ pmf with parameters k and p is one. Thus, we see again that X + Y has the Poisson distribution with parameter λ1 + λ2 . Example 4.5.3 Suppose X and Y represent the numbers showing for rolls of two fair dice. Thus, each is uniformly distributed ov...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.

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