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Unformatted text preview: Stat 333 : Midterm 2 Solutions  Fall 2011 1. [12 points] A blue jay sits outside Sheldons window and chirps every minute. Sheldon eventually realizes that the length of the chirps are either 1, 2, or 5 seconds; denote these chirps by A, B and C respectively. Additionally, it appears that each type of chirp is equally likely to occur and that chirps occurs independently of each other. (a) [3 points] Using the renewal theorem, on average, how many minutes must Sheldon wait until he hears a sequence of chirps given by AABAACAABAA? Solution: E ( T AABAACAABAA ) = E ( T A ) + E ( T g AA ) + E ( T ^ AABAA ) + E ( T ^ AABAACAABAA ) = 3 + 3 2 + 3 5 + 3 11 = 177 , 402 . (b) Raj notices that Sheldon is actually wrong and points out to Sheldon that the sequence of chirps CA occurs with probability 1 4 . Let denote this sequence. i. [4 points] Show that the pgf of the waiting time (in minutes) until first occurs is F ( s ) = 1 4 s 2 ( 1 1 2 s ) 2 Solution: We have r = 1 , r 1 = 0 , and r n = 0 . 25 , n 2 . Therefore, R ( s ) = X n =0 r n s n = 1 + X n =2 . 25 s n = 1 + . 25 s 2 1 s ,  s  < 1 and using the renewal theorem, F ( s ) = 1 1 R ( s ) = . 25 s 2 1 s + 0 . 25 s 2 = 1 4 s 2 ( 1 1 2 s ) 2 ii. [2 points] Hence, find f = P ( T < ) . Is recurrent or transient? Solution: is recurrent since f = P ( T < ) = F (1) = 1 iii. [3 points] Find an expression for P ( T = k ) , k 2 . You can use the fact that 1 ( 1 1 2 s ) 2 = X k =0 ( k + 1) 1 2 k s k ,  s  < 2 Solution: Using the hint, we have F ( s ) = 1 4 s 2 X k =0 ( k + 1) 1 2 k s k = X k =0 ( k + 1) 1 2 k +2 s k +2 = X k =2 ( k 1) 1 2 k s k and so P ( T = k ) = ( k 1) 1 2 k , k 2 . 1 2. [16 points] Federer and Nadal are currently playing a series of games for the Wimbledon mens tennis championship. The first player to have a 2 game lead over the other will win the championship. A score of x y implies that Federer has won x games, whereas Nadal has won y games. For example, the following table provides some possible events thatgames....
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 Fall '08
 Chisholm
 Probability

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