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Unformatted text preview: Y i Y j with i 6 = j . Hence Var X = E X 2-( E X ) 2 = np + ( n 2-n ) p 2-( np ) 2 = np (1-p ) . Later we will see that the variance of the sum of independent r.v.s is the sum of the variances, so we could quickly get Var X = np (1-p ). Alternatively, one can compute E ( X 2 )-E X = E ( X ( X-1)) using binomial coecients and derive the variance of X from that. Poisson . X is Poisson with parameter if P ( X = i ) = e- i i ! . Note i =0 i /i ! = e , so the probabilities add up to one. To compute expectations, E X = X i =0 ie- i i ! = e- X i =1 i-1 ( i-1)! = . Similarly one can show that E ( X 2 )-E X = E X ( X-1) = X i =0 i ( i-1) e- i i ! = 2 e- X i =2 i-2 ( i-2)! = 2 , 12...
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- Spring '09