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STA3007_0506_t05_ssol

# STA3007_0506_t05_ssol - STA 3007 Applied Probability...

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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 that the process will be in state j. c Mean fraction of transitions: E [ 1 m m - 1 k =0 [ I { X k +1 = 2 or X k = 0 } + I { X k +1 = 3 or X k = 0 } + I { X k +1 = 2 or X k = 1 } + I { X k +1 = 3 or X k = 1 }| X 0 = i ]] = 1 m m - 1 k =0 Pr { X k +1

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