P,(0O 111)2
qo(20) -
P,(0 | 11)2 +P(1 2 11)2
and Eq. 9.32
P,(1 11)2
qo(02) .
SP(0I 11)2+P(1 I 11)2
Then, substituting Eq. 4.2 in Eq. 9.32 yields
20 R(0)2
R(0)2 + R(1)2
and Eq. 9.33
qo(02)= -
R (0)2 + R(1)2
The limit for the nonabsorbing state m = (11) is known to be zero by Proposition 7.5. An
identical result to Eq. 9.33 obtains for the three-operator case because for the one-bit
problem, crossover is nullified and P'(i I n) = Pi(i I n) (see Eq. 4.22).
Additional insight into the behavior of the limiting stationary distribution is obtain-
able by examining the one-bit problem with population size 3. These parameters
(L = 1,M = 3) leave S and N unaltered but change the other state space related sets and
parameters to S' = {(30), (21), (12), (03)}, N' = 4, SA' = {(30), (03)} and
S'- SA' = {(21), (12)}. By retracing the previous development (the M = 2 case) with r
limited as indicated by these state space sets, results analogous to Eq. 9.32 and 9.33 are
obtained. Thus, the M=3 counterpart of Eq. 9.32 is
P,(0 1 21)3 [P(0 1 12)3+ 3P,(0 112)2P1(1 12)] +
S P(0 12)\ [P,(1 1 21)3+ 3P,(0 I 21)P,(1 121)2]
qo(30) =
D
and Eq. 9.34
P,(1 1 21) [P,(1 | 12)3+ 3P,(1 I 12)21(0 1 12)] +
P,(1 I 12)3[P(0 1 21)3 + 3P,(1 21)P,(0 1 21)2]
qo(03) = D