elemprob-page12 - Y i Y j with i 6 = j . Hence Var X = E X...

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5. Some discrete distributions. Bernoulli . A r.v. X such that P ( X = 1) = p and P ( X = 0) = 1 - p is said to be a Bernoulli r.v. with parameter p . Note E X = p and E X 2 = p , so Var X = p - p 2 = p (1 - p ). Binomial . A r.v. X has a binomial distribution with parameters n and p if P ( X = k ) = ± n k ² p k (1 - p ) n - k . The number of successes in n trials is a binomial. After some cumbersome calculations one can derive E X = np . An easier way is to realize that if X is binomial, then X = Y 1 + ··· + Y n , where the Y i are independent Bernoulli’s, so E X = E Y 1 + ··· + E Y n = np . We haven’t defined what it means for r.v.’s to be independent, but here we mean that the events ( Y k = 1) are independent. The cumbersome way is as follows. E X = n X k =0 k ± n k ² p k (1 - p ) n - k = n X k =1 k ± n k ² p k (1 - p ) n - k = n X k =1 k n ! k !( n - k )! p k (1 - p ) n - k = np n X k =1 ( n - 1)! ( k - 1)!(( n - 1) - ( k - 1))! p k - 1 (1 - p ) ( n - 1) - ( k - 1) = np n - 1 X k =0 ( n - 1)! k !(( n - 1) - k )! p k (1 - p ) ( n - 1) - k = np n - 1 X k =0 ± n - 1 k ² p k (1 - p ) ( n - 1) - k = np. To get the variance of X , we have E X 2 = n X k =1 E Y 2 k + X i 6 = j E Y i Y j . Now E Y i Y j = 1 · P ( Y i Y j = 1) + 0 · P ( Y i Y j = 0) = P ( Y i = 1 ,Y j = 1) = P ( Y i = 1) P ( Y j = 1) = p 2 using independence. The square of Y 1 + ··· + Y n yields n 2 terms, of which n are of the form Y 2 k . So we have n 2 - n terms of the form
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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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