STAT 333 a3sol

# STAT 333 a3sol - Stat 333 - Assignment 3 Solutions - Fall...

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Unformatted text preview: Stat 333 - Assignment 3 Solutions - Fall 2011 1. (a) From the transition diagram, we can see that the states , 1 ,...,N- 1 ,N belong to the same (equivalence) class, and are aperiodic ( P 00 > ) and positive recurrent (finite state space). 1 2 3 ...... N- 1 N (b) For this problem, it is better to list the transition probabilities. Let q = 1- p . P ij = q ( i N ) , j = i- 1 q ( 1- i N ) + p ( i N ) , j = i p ( 1- i N ) , j = i + 1 , otherwise (c) When N = 2 , the one-step transition matrix is given by P = q p q/ 2 1 / 2 p/ 2 q p Solving ~P = ~ such that 2 j =0 j = 1 , we obtain = q 2 , 1 = 2 pq and 2 = p 2 . (d) From part c), when N = 2 , the limiting distribution is binomial with n = 2 and prob- ability of success, p . In general, for any N Z + , the limiting distribution is binomial with n = N and probability of success, p . Hence, j = ( N j ) p j q N- j , j = 0 ,...,N . (e) The equilibrium equations are = q + q N 1 = = q p N 1 (1) N = p N N- 1 + p N = N = p q N N- 1 (2) j = p 1- j- 1 N j- 1 + q 1- j N + p j N j + q j + 1 N j +1 , j = 1 ,...,N- 1 (3) 1 Equations (1) and (2) are trivial to show. Lets consider the right hand side of (3): p 1- j- 1 N j- 1 + q 1- j N + p j N j + q j + 1 N j +1 = p 1- j- 1 N N j- 1 p j- 1 q N- j +1 + q 1- j N + p j N N j p j q N- j + q j + 1 N N j + 1 p j +1 q N- j- 1 = 1- j- 1 N q j N- j + 1 N j p j q N- j + q...
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## This note was uploaded on 01/21/2012 for the course STAT 333 taught by Professor Chisholm during the Fall '08 term at Waterloo.

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STAT 333 a3sol - Stat 333 - Assignment 3 Solutions - Fall...

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