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Unformatted text preview: 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 random variables with P ( Y i = 1) = p and P ( Y i = 0) = 1p , so E X = E Y 1 + + E Y n = np . The cumbersome way is as follows. E X = n X k =0 k n k p k (1p ) nk = n X k =1 k n k p k (1p ) nk = n X k =1 k n ! k !( nk )! p k (1p ) nk = np n X k =1 ( n1)! ( k1)!(( n1)( k1))! p k1 (1p ) ( n1)( k1) = np n1 X k =0 ( n1)! k !(( n1)k )! p k (1p ) ( n1)k = np n1 X k =0 n1 k p k (1p ) ( n1)k = np. 15...
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 Spring '10
 ansan
 Binomial

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