final06sol - STAT455/855 Fall 2006 Applied Stochastic...

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STAT455/855 Fall 2006 Applied Stochastic Processes Final Exam, Brief Solutions 1. (15 marks) (a) (7 marks) For 3 j n , starting at the j th best point we condition on the rank R of the point we jump to next. By assumption, we have P ( R = k ) = 1 / ( j - 1) for k = 1 ,...,j - 1. Thus, we have r j = j - 1 X k =1 (1 + r k ) 1 j - 1 = 1 + 1 j - 1 j - 1 X k =1 r k . To solve this first multiply through by j - 1, giving ( j - 1) r j = j - 1 + j - 1 X k =1 r k . Next, write the corresponding equation for r j +1 and subtract from it the equation for r j , giving jr j +1 - ( j - 1) r j = j + j X k =1 r k - ( j - 1) - j - 1 X k =1 r k = 1 + r j . Thus, jr j +1 - jr j = 1, or r j +1 = r j + 1 /j = r j - 1 + 1 j - 1 + 1 j = ... = r 1 + j X k =1 1 k = j X k =1 1 k , since r 1 = 0. Thus, r j = j - 1 k =1 k - 1 , for j = 2 ,...,n . (b) (8 marks) Note that p 1 = 1 and p 2 = 0. Let Y indicate the first pair to combine: for k = 1 ,...,n - 1, Y = k means the pair ( m k ,m k +1 ) was the first to combine. We condition on Y . If Y = 1 then molecule m 1 will not be isolated, if Y = 2 then m 1 will remain isolated for sure, and if Y = 3 then m 1 will not be isolated. For k = 4 ,...,n - 1, if Y = k then we can assume we just started with molecules m 1 ,...,m k - 1 . Since P ( Y = k ) = 1 / ( n - 1) for all k , we have p n = 1 n - 1 ± 0 + 1 + 0 + n - 1 X k =4 p k - 1 ² = 1 n - 1 n - 2 X k =1 p k .
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STAT 455/855 -- Final Exam Solutions, 2006 Page 2 of 7 Therefore, ( n - 1) p n = n - 2 k =1 p k . Also, np n +1 = n - 1 k =1 p k . Subtracting, we have np n +1 - ( n - 1) p n = p n - 1 n ( p n +1 - p n ) = - ( p n - p n - 1 ) p n +1 - p n = - 1 n ( p n - p n - 1 ) . So p n +1 - p n = - 1 n ( p n - p n - 1 ) = ( - 1) 2 1 n ( n - 1) ( p n - 1 - p n - 2 ) . . . = ( - 1) n - 2 1 n ( n - 1) × ... × 3 ( p 3 - p 2 ) = 2( - 1) n - 2 n ! p 3 , since p 2 = 0. But, direct calculation yields p 3 = 1 / 2 and so we have p n +1 - p n = ( - 1) n - 2 /n ! = ( - 1) n /n ! for n 3. Thus, p n +1 = p n + ( - 1) n n ! = p n - 1 + ( - 1) n - 1 ( n - 1)! + ( - 1) n n ! . . . = p 3 + n X r =3 ( - 1) r r ! = n X r =2 ( - 1) r r ! = n X r =0 ( - 1) r r ! . Thus, p n = n - 1 r =0 ( - 1) r /r ! as desired. One can check that the formula works for n = 1 and n = 2 as well.
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STAT 455/855 -- Final Exam Solutions, 2006 Page 3 of 7 2. (15 marks) (a) (2 marks) From the diagram, we get the transition probability matrix as P = 0 0 . 5 0 . 5 0 0 0 . 1 0 0 . 5 0 0 . 4 0 . 1 0 . 1 0
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final06sol - STAT455/855 Fall 2006 Applied Stochastic...

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