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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 2n ) p 2( np ) 2 = np (1p ) . 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 (1p ). Alternatively, one can compute E ( X 2 )E X = E ( X ( X1)) 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 i1 ( i1)! = . Similarly one can show that E ( X 2 )E X = E X ( X1) = X i =0 i ( i1) e i i ! = 2 e X i =2 i2 ( i2)! = 2 , 12...
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 Spring '09
 wen
 Bernoulli

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