CHAPTER 2: Markov Chains (part 3)

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w 35 = 0 w 45 = 1 3 w 15 + 1 3 w 55 + 1 3 w 75 w 55 = 1 + 1 4 w 25 + 1 4 w 45 + 1 4 w 65 + 1 4 w 85 w 65 = 1 3 w 35 + 1 3 w 55 + 1 3 w 95 w 75 = 0 w 85 = 1 3 w 55 + 1 3 w 75 + 1 3 w 95 w 95 = 0 [You can find the answers by solving these set of equations] 4. If the rat starts in com<strong>part</strong>ment 1, what is the probability that the rat is in 4 before it is shocked? Let z i = P (X T −1 = 4|X 0 = i) then z 1 = p 12 z 2 + p 14 z 4 z 2 = p 21 z 1 + p 25 z 5 + p 23 z 3 z 3 = 0; z 4 = p 41 z 1 + p 45 z 5 + p 47 z 5 = p 52 z 2 + p 54 z 4 + p 56 z 6 + p 58 z 8 z 6 = p 63 z 3 + p 65 z 5 + p 69 z 9 z 7 = 0 z 8 = p 85 z 5 + p 87 0 + p 89 0 z 9 = 0 Example A model of Fecundity Changes in sociological patterns such as increase in age at marriage, more remarriages after widowhood, and increased divorce rates have profound effects on overall population growth rates. Here we attempt to model the life span of a female in a population in order to provide a framework for analyzing the effect of social changes on average fecundity. For a typical woman, we may categorize her in one of the follows states E 0 : Prepuberty E 1 : Single E 2 : Married E 3 : Divorced, 4
E 4 : Widowed E 5 : Died or emigration from the population Suppose the transition probability matrix is P = E 0 E 1 E 2 E 3 E 4 E 5 E 0 0 0.9 0 0 0 0.1 E 1 0 0.5 0.4 0 0 0.1 E 2 0 0 0.6 0.2 0.1 0.1 E 3 0 0 0.4 0.5 0 0.1 E 4 0 0 0.4 0 0.5 0.1 E 5 0 0 0 0 0 1 We are interested in the mean duration spent in state E 2 , Married, since this corresponds to the state of maximum fecundity. Let w i2 be the mean duration in state E 2 given the initial state is E i . From the first step analysis, we have w 22 = 1 + p 21 w 12 + p 22 w 22 + p 23 w 32 + p 24 w 42 + p 22 w 52 . For absorbing state 5, we have w 52 = 0 If i is not a state 2 and absorbing state 5, All together, we have w i2 = p i1 w 12 + p i2 w 22 + p i3 w 32 + p i4 w 42 + p i2 w 52 w 02 = 0w 02 + 0.9w 12 + 0w 22 + 0w 32 + 0.1w 42 + 0.1w 52 w 12 = w 02 + w 12 + w 22 + w 32 + w 42 + 0w 52 w 22 = 1 + w 02 + w 12 + w 22 + w 32 + w 42 + 0w 52 w 32 = w 02 + w 12 + w 22 + w 32 + w 42 + 0w 52 w 42 = w 02 + w 12 + w 22 + w 32 + w 42 + 0w 52 w 52 = 0 The solution is w 02 = 4.5, w 12 = 5, w 22 = 6.25, w 32 = 5, w 42 = 5, w 52 = 0. Each female, on the average, spend w 02 = 4.5 periods in the childbearing state E 2 during her lifetime. 5
Page 1 and 2: CHAPTER 2: Markov Chains (part 3) E
Page 3: we have, for the others i = 1, 2, 4
Page 7 and 8: We have P (x n+1 = s|x n−1 = s, x

CHAPTER 2: Markov Chains (part 3)

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