STA3007_0506_t05_ssol

STA3007_0506_t05_ssol - STA 3007 Applied Probability...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
STA 3007 Applied Probability 2005 Tutorial 5 Suggested Solution 1. The Long Run Behavior of Markov Chains i. (Problem 1.11) a π 2 = (0 . 2) π 1 + (0 . 4) π 2 + (0 . 4) π 3 π 3 = 0 . 1636 π 0 = (1 - π 1 - π 2 - π 3 ) π 0 = 0 . 3087 b Assume the process last for 0,1,. ..,m-1 steps, Fraction of time in state 2 or 3: 1 m m - 1 k =0 I { X k = 2 or X k = 3 } I { X k = 2 or X k = 3 } = 1 if X k = 2 or X k = 3 , 0 otherwise The mean fraction of time spend in state 2 or 3: E [ 1 m m - 1 k =0 I { X k = 2 or X k = 3 }| X 0 = i ] = E [ 1 m m - 1 k =0 I { X k = 2 }| X 0 = i ] + E [ 1 m m - 1 k =0 I { X k = 3 }| X 0 = i ] = 1 m m - 1 k =0 Pr { X k = 2 | X 0 = i } + 1 m m - 1 k =0 Pr { X k = 3 | X 0 = i } = 1 m m - 1 k =0 P ( k ) i 2 + 1 m m - 1 k =0 P ( k ) i 3 lim m →∞ E [ 1 m m - 1 k =0 I { X k = 2 or X k = 3 }| X 0 = i ] = lim m →∞ 1 m m - 1 k =0 P ( k ) i 2 + lim m →∞ 1 m m - 1 k =0 P ( k ) i 3 = π 2 + π 3 = 0 . 3773 The limiting probability π j is interpreted as the long run proportion of time
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/21/2011 for the course STA 3007 taught by Professor Kb during the Spring '11 term at CUHK.

Page1 / 2

STA3007_0506_t05_ssol - STA 3007 Applied Probability...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online