Analyse combinatoire, tome 1 by Comtet L. PDF

By Comtet L.

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By j -/> k, it means that it is not possible to find a n such that Pjk(0, n) > 0. A closed set consists of a single state is called an absorbing class and the element the absorbing state. 2, the state space itself is a closed set. If the state space contains a proper closed subset, then the chain is said to be reducible; otherwise not reducible or irreducible. 5 are irreducible homogeneous Markov chains.

6] W. Y. Tan and C. W. Chen, Stochastic models of carcinogenesis, Some new insight, Math Comput. Modeling 28 (1998) 49-71. [7] CDC, 1993 Revised Classification System for HIV Infection and Expanded Surveillance Case Definition for AIDS Among Adolescents and Adults, MMWR 4 1 (1992), No. RR17. [8] W. Y. Tan, Stochastic Modeling of AIDS Epidemiology and HIV Pathogenesis, World Scientific, Singapore (2000). [9] W. Y. Tan and H. Wu, Stochastic modeling of the dynamics of CD4 T cell infection by HIV and some Monte Carlo studies, Math.

Let {X(t), £ = 1 , 2 , . . , } denote the above ten genotypes at generation t. Then X(t) is a finite Markov chain with discrete time and with state space given by S = {AB/AB, Ab/Ab, aB/aB, ab/ab, AB/Ab, AB/aB, Ab/ab, aB/ab, AB/ab, aB/Ab}. Letting p be the recombination value between the two loci (0 < p < 5), then under self-fertilization with selection, the one step transition matrix is given by: R I Q Examples from Genetics and AIDS 35 where R={R1, R2, CN-i R 3, t^j S^J R A) r^j xi +x2 xi +2/2 Xi 0 0 ci Cl +X2 0 0 C2 C2 xi + y2 0 y\ + 2/2 0 C3 C3 (X1 + X2) C5 0 2/i + ^2 c4 2/1 +2/2 c4 — (xi + 2/2) c5 —(2/1 + 32) —(2/1 + 2/2) C5 2 ,2 Lc5 (11+12) 2 ( l + :ci) — (zi+2/2) cs ^2 —(2/1 + 3:2) cs —(2/1+2/2) cs 0 0 0 0 2 ( 1 + aa) 0 0 0 0 0 0 2(1 + 2/1) c4 0 0 Cl 0 2p<7 C2 0 0 0 0 2(1+2/2) C3 1 + si 2pg C5 l + a:2 C5 C5 l + o:i 0 2pq 1 + 32 C5 2M 1 + 2/2 C5 2p<7 1 + 2/2 C5 1 + 2/1 4g 2 C5 c5 l + 2/i 4p 2 C5 C5 2pg 2pq and where ci = 4a; 1 + X2 + 2/2 + 2 , c 2 = 4aj2 + a;i + 2/1 + 2 , C3 = 42/2 + xi + 2/1 + 2 , c 4 = 42/i + x2 + 2/2 + 2 , c 5 = 3 i + x2 + 2/1 + 2/2 + 4 .

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Analyse combinatoire, tome 1 by Comtet L.


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