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
(k − j )!
j =0 k
j k −j
λ1 λ2 −(λ1 +λ2 )
j !(k − j )!
j =0 k
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
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.
- Spring '08
- The Land