STAT 333 a2sol

STAT 333 a2sol - Stat 333 - Assignment 2 Solutions - Fall...

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Unformatted text preview: Stat 333 - Assignment 2 Solutions - Fall 2011 1. (a) We have an even number of S s in n trials iff there is a failure in the 1st trial, and we have an even number of successes in remaining n- 1 trials, or there is a success in the 1st trial, and we have an odd number of successes in remaining n- 1 trials. Therefore, e n = P ( even number of Ss in n-1 trials | 1st trial is a failure ) P ( 1st trial is a failure ) + P ( odd number of Ss in n-1 trials| 1st trial is a success ) P ( 1st trial is a success ) = P ( even number of Ss in n-1 trials ) P ( 1st trial is a failure ) + P ( odd number of Ss in n-1 trials ) P ( 1st trial is a success ) = e n- 1 (1- p ) + (1- e n- 1 ) p = e n- 1 q + (1- e n- 1 ) p, n 1 where the second expression follows from the fact that we have independent trials and q = 1- p . (b) Note that e = 1 since we assume zero is even. Also, before we proceed, we rewrite the above expression as e n = ( q- p ) e n- 1 + p . We have G ( z ) = X n =0 e n z n = 1 + X n =1 e n z n = 1 + X n =1 [( q- p ) e n- 1 + p ] z n = 1 + ( q- p ) X n =1 e n- 1 z n + p X n =1 z n = 1 + ( q- p ) z X k =0 e k z k + p X n =1 z n where k = n- 1 = 1 + ( q- p ) zG ( z ) + p z 1- z Therefore, G ( z ) = 1 + p z 1- z 1- ( q- p ) z = 1- z (1- p ) (1- z )[1- ( q- p ) z ] (c) When p = q = 0 . 5 , G ( z ) = 1- . 5 z 1- z = 1 + . 5 z 1- z = 1 + 0 . 5 X n =1 z n which implies that e n = 1 , n = 0 . 5 , n 1 1 2. Since d = 0 , D ( z ) = X n =0 d n z n = X n =1 d n z n = X n =1 P ( occurs on trial n ) z n Now, P ( occurs on trial n ) = n X k =1 P ( occurs on trial n | occurs first on trial...
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STAT 333 a2sol - Stat 333 - Assignment 2 Solutions - Fall...

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