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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 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] = 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 } = . 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 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 { x n +1  x n } = 2 x n and in general we obtain E { x n +...
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 Fall '09
 Jacobs

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