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Unformatted text preview: X has a Poisson distribution with parameter λ if P ( X = i ) = e-λ λ i i ! . Note ∑ ∞ i =0 λ i /i ! = e λ , so the probabilities add up to one. To compute expectations, E X = ∞ X i =0 ie-λ λ i i ! = e-λ λ ∞ X i =1 λ i-1 ( i-1)! = λ. Similarly one can show that E ( X 2 )-E X = E X ( X-1) = ∞ X i =0 i ( i-1) e-λ λ i i ! = λ 2 e-λ ∞ X i =2 λ i-2 ( i-2)! = λ 2 , 16...
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This note was uploaded on 12/29/2011 for the course MATH 316 taught by Professor Ansan during the Spring '10 term at SUNY Stony Brook.
- Spring '10