slides_4_ranvecs

# X m e j 1 m exp tx j j 1 m e exp tx j j 1 m j t 102

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X m  E j 1 m exp tX j j 1 m E exp tX j  j 1 m j t 102

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EXAMPLE : Suppose X 1 , X 2 , ..., X n are independent Poisson random variables and define the sum Y X 1 X 2 ... X n . It follows from previous properties of expected value and variance that E Y n and Var Y n . Because the variance equals the mean, it is tempting to conclude Y has the Y Poisson n distributions. But we need to show more than that the first two moments of Y match with the Poisson n distribution. 103
To show Y Poisson n we use the MGF approach with i t exp exp t 1  for each i : Y t i 1 n exp exp t 1  exp i 1 n exp t 1 exp n exp t 1  and the final expression is the MGF of the Poisson n distribution. We can extend this result to allow different means: if X i has the Poisson i distribution then the sum has the Poisson 1 2 ... n distribution. 104

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EXAMPLE : Suppose X 1 , X 2 , ..., X n are independent Exponential random variables. Then Y X 1 X 2 ... X n has the Gamma n , distribution, where the gamma distribution is parameterized such that E Y n (which obviously must be true because E X i for each i ). 105
Proof : For the Gamma , distribution, the MGF is 1 t , t 1/ (with 1 giving the exponential case). With Y defined as the sum we have Y t i 1 n 1 t 1 1 t n , whichistheMGFofthe Gamma n , distribution. 106

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An extension that is often useful is to define Y a X b , where again the elements of X are independent with MGFs j t . Then Y t E exp t a 1 X 1 ... a m X m b  E j 1 m exp ta j X j exp tb exp tb j 1 m E exp ta j X j  exp bt j 1 m j a j t 107
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X m E j 1 m exp tX j j 1 m E exp tX j j 1 m j t 102 EXAMPLE...

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