final07sol - STAT455/855 Fall 2007 Applied Stochastic...

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STAT455/855 Fall 2007 Applied Stochastic Processes Final Exam, Solutions 1. (15 marks) (a) (11 marks) Let q = P ( X i > m ). This is the same for all X i , and is equal to q = X j = m +1 p (1 - p ) j - 1 = (1 - p ) m p X j =0 (1 - p ) j = (1 - p ) m p 1 - (1 - p ) = (1 - p ) m . (i) (2 marks) The event { N k = k } occurs if and only if the first k observations are all greater than m . The probability of this is P ( N k = k ) = q k = (1 - p ) mk . (ii) (4 marks) N 1 has a Geometric( q ) distribution and so E [ N 1 ] = 1 q = 1 (1 - p ) m . Conditioning on N 1 , and then on the next observation after N 1 , we have E [ N 2 ] = X j =1 E [ N 2 ± ± N 1 = j ] P ( N 1 = j ) = X j =1 [( j + 1) q + ( j + E [ N 2 ])(1 - q )] P ( N 1 = j ) = X j =1 jP ( N 1 = j ) + ( q + (1 - q ) E [ N 2 ]) X j =1 P ( N 1 = j ) = E [ N 1 ] + q + (1 - q ) E [ N 2 ] = 1 q + q + (1 - q ) E [ N 2 ] or E [ N 2 ] = 1 + 1 q 2 = 1 + 1 (1 - p ) 2 m .
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Page 2 of 7 (iii) (5 marks) Conditioning on N k - 1 and the next observation after N k - 1 , we have E [ N k ] = X j =1 E [ N k ± ± N k - 1 = j ] P ( N k - 1 = j ) = X j =1 [( j + 1) q + ( j + E [ N k ])(1 - q )] P ( N k - 1 = j ) = E [ N k - 1 ] + q + (1 - q ) E [ N k ] or E [ N k ] = 1 + 1 q E [ N k - 1 ] . Recursing, we obtain E [ N k ] = 1 + 1 q E [ N k - 1 ] = 1 + 1 q + 1 q 2 E [ N k - 2 ] . . . = 1 + 1 q + 1 q 2 + . .. + 1 q k - 2 + 1 q k - 1 E [ N 1 ] = 1 + 1 q + 1 q 2 + . .. + 1 q k - 2 + 1 q k = 1 - 1 /q k - 1 1 - 1 /q + 1 q k = q k - 1 - 1 q k - 1 - q k - 2 + 1 q k = (1 - p ) m ( k - 1) - 1 (1 - p ) m ( k - 1) - (1 - p ) m ( k - 2) + 1 (1 - p ) mk . (b) (4 marks) Let N i , i = 1 , .. . , 6, denote the number of biased dice that show i , and let X denote the outcome of the fair die. Conditioning on X , we have E [ N ] = 6 X i =1 E [ N ± ± X = i ] 1 6 = 1 6 6 X i =1 E [ N i ] = 1 6 E [ N 1
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final07sol - STAT455/855 Fall 2007 Applied Stochastic...

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