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Unformatted text preview: 215 Chapter 15 15.1 The chain represented by P = 1 / 2 1 / 2 1 / 2 1 / 2 1 / 2 1 / 2 is irreducible and aperiodic. The second chain is also irreducible and aperiodic. The third chain has two aperiodic closed sets { e 1 ,e 2 } and { e 3 ,e 4 } and a transient state e 5 . 15.2 Note that both the row sums and column sums are unity in this case. Hence P represents a doubly stochastic matrix here, and P n = 1 m + 1 1 1 ··· 1 1 1 1 ··· 1 1 . . . . . . . . . . . . . . . 1 1 ··· 1 1 lim n →∞ P { x n = e k } = 1 m + 1 , k = 0 , 1 , 2 , ··· m. 15.3 This is the “success runs” problem discussed in Example 1511 and 1523. From Example 1523, we get u i +1 = p i,i +1 u i = 1 i + 1 u i = u o ( i + 1)! so that from (15206) ∞ X k =1 u k = u ∞ X k =1 1 k ! = e · u = 1 216 gives u = 1 /e and the steady state probabilities are given by u k = 1 /e k ! , k = 1 , 2 , ··· 15.4 If the zeroth generation has size m , then the overall process may be considered as the sum of m independent and identically distributed branching processes x ( k ) n , k = 1 , 2 , ··· m, each corresponding to unity size at the zeroth generation. Hence if π represents the probability of extinction for any one of these individual processes, then the overall probability of extinction is given by lim n →∞ P [ x n = 0  x = m ] = = P [ { x (1) n = 0  x (1) = 1 } T { x (2) n = 0  x (2) = 1 } T ···{ x ( m ) n = 0  x ( m ) = 1 } ] = Q m k =1 P [ x ( k ) n = 0  x ( k ) = 1] = π m 15.5 From (15288)(15289), P ( z ) = p + p 1 z + p 2 z 2 , since p k = 0 , k ≥ 3 . Also p + p 1 + p 2 = 1, and from (15307) the extinction probability is given by sloving the equation P ( z ) = z. Notice that P ( z ) z = p (1 p 1 ) z + p 2 z 2 = p ( p + p 2 ) z + p 2 z 2 = ( z 1)( p 2 z p ) and hence the two roots of the equation P ( z ) = z are given by z 1 = 1 , z 2 = p p 2 . Thus if p 2 < p , then z 2 > 1 and hence the smallest positive root of P ( z ) = z is 1, and it represents the probability of extinction. It follows 217 that such a tribe which does not produce offspring in abundence is bound to extinct. 15.6 Define the branching process { x n } x n +1 = x n X k =1 y k where y k are i.i.d random variables with common moment generating function P ( z ) so that (see (15287)(15289)) P (1) = E { y k } = μ. Thus E { x n +1  x n } = E { ∑ x n k =1 y k  x n = m } = E { ∑ m k =1 y k  x n = m } = E { ∑ m k =1 y k } = mE { y k } = x n μ Similarly E { x n +2  x n } = E { E { x n +2  x n +1 , x n }} = E { E { x n +2  x n +1 } x n } = E { μ x n +1  x n } = μ 2 x n and in general we obtain E { x n +...
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 Fall '09
 Jacobs
 Trigraph, Markov chain, pij, Doubly stochastic matrix, Stochastic matrix, rj pji

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