{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol5 - STA3007 Applied probability(08-09 Chapter 4 Solution...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
STA3007 Applied probability (08-09) Solution to Assignment 5 Chapter 4 Problem 1.2 Let X n be the number of balls in urn A , the state space is { 0 , 1 , 2 , 3 , 4 , 5 } . The transition matrix is as follows P = 0 1 0 0 0 0 1 5 0 4 5 0 0 0 0 2 5 0 3 5 0 0 0 0 3 5 0 2 5 0 0 0 0 4 5 0 1 5 0 0 0 0 1 0 . By solving the following linear system equations P T π = π, 5 i =0 π i = 1 , we have π 0 = π 5 = 1 32 , π 1 = π 4 = 5 32 , π 2 = π 3 = 10 32 . Therefore, in the long run, the fraction of time that the urn A is empty is π 0 = 1 / 32 = 0 . 0313. Problem 1.3 The stationary distribution π satisfies the following system equations π 0 = α 1 π 0 + π 1 π 1 = α 2 π 0 + π 2 π 2 = α 3 π 0 + π 3 π 3 = α 4 π 0 + π 4 π 4 = α 5 π 0 + π 5 π 5 = α 6 π 0 5 i =0 π i = 1 , Representing π i ( i = 0 , 1 , 2 , 3 , 4 , 5) by α and π 0 from the first six equations, and substi- tuting them into the last equation, we get the limiting probability of being in state 0 1
Background image of page 1

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

View Full Document Right Arrow Icon
is π 0 = 1 6 k =1 k . Problem 1.4 Let Z n = ( X n , X n +1 ), then it is easily to show that { Z n } n 0 is a Markov chain. The question is to find the fraction of the transitions which are from a prescribed state k to a prescribed state m (note that they are adjacent). It is the limiting distribution of
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}