final06sol

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

This preview shows pages 1–4. Sign up to view the full content.

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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online