Papoulis ch. 15

# Papoulis ch. 15 - 215.1 The chain represented by P = 1 2 1...

This preview shows pages 1–4. Sign up to view the full content.

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 15-11 and 15-23. From Example 15-23, we get u i +1 = p i,i +1 u i = 1 i + 1 u i = u o ( i + 1)! so that from (15-206) ∞ 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 (15-288)-(15-289), 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 (15-307) 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 (15-287)-(15-289)) 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 +...
View Full Document

{[ snackBarMessage ]}

### Page1 / 13

Papoulis ch. 15 - 215.1 The chain represented by P = 1 2 1...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online