where ap = Pr {xgp_1 = | up | | Bp_1 }. For ap, we have

$$\begin{array}{c}{{a_{p}=\frac{P r\left\{B_{p-1},x_{g_{p-1}}=\left|u_{p}\right|\right\}}{P r\left\{B_{p-1}\right\}}}}\\ {{=\frac{1}{P r\left\{B_{p-1}\right\}}(P r\left\{B_{p-1},x_{g_{p-1}}=\left|u_{p}\right|,x_{g_{p-2}}=\left|u_{p-1}\right|\right\}}}\\ {{\quad+P r\left\{B_{p-1},x_{g_{p-1}}=\left|u_{p}\right|,x_{g_{p-2}}\neq\left|u_{p-1}\right|\right\})}}\\ {{\quad1}}\end{array}$$
+ Pr {Bp-1; xgp-1 = {Up|, xgp-2 F {Up-1|}) (4.8) (Pr{xgp-1 = |Up|| Bp-1;xgp-2 = |Up-1|} · Pr{Bp-1;xgp-2 = |Up-1|} Pr{Bp-1} + Pr {xgp_1 = {up|| Bp-1,xgp_2 ≠ {up_1|} · Pr {Bp-1,xgp_2 ≠ {up_1|}) { f Pr {Bp-1, xgp-2 = {up-1|} + f Pr {Bp-1, xgp-2 ≠ {up-1|} Pr{Bp-1}
The last equation in the above formula is because that given Bp-1 and xgp-2 = 1 Up-1 |
(or xg,,,,, + | Up-1 |), we have two (three, respectively) choices in selecting the local fitness function gp-1. Consider the two terms Pr{Bp-1,xgp-2 = |Up-1|} and Pr{Bp-1,xgp-2 F
| up-1|} in (4.8), we have

$$Pr\left\{B_{p-1},x_{g_{p-2}}=\left|u_{p-1}\right|\right\}\tag{4.9}$$ $$=Pr\left\{g_{p-2}\neq g_{p-1}\mid B_{p-2},x_{g_{p-2}}=\left|u_{p-1}\right|\right\}\cdot Pr\left\{B_{p-2},x_{g_{p-2}}=\left|u_{p-1}\right|\right\}$$ $$=\frac{2}{3}Pr\left\{x_{g_{p-2}}=\left|u_{p-1}\right|\mid B_{p-2}\right\}\cdot Pr\left\{B_{p-2}\right\}$$ $$=\frac{2}{3}a_{p-1}\cdot Pr\left\{B_{p-2}\right\}$$

and

$$Pr\left\{B_{p-1},x_{y_{p-2}}\neq\left|u_{p-1}\right|\right\}\tag{4.10}$$ $$=Pr\left\{g_{p-2}\neq g_{p-1}\mid B_{p-2},x_{y_{p-2}}\neq\left|u_{p-1}\right|\right\}\cdot Pr\left\{B_{p-2},x_{y_{p-2}}\neq\left|u_{p-1}\right|\right\}$$ $$=Pr\left\{x_{y_{p-2}}\neq\left|u_{p-1}\right|\right|\mid B_{p-2}\right\}\cdot Pr\left\{B_{p-2}\right\}$$ $$=\left(1-a_{p-1}\right)\cdot Pr\left\{B_{p-2}\right\}$$

By plugging (4.9) and (4.10) into (4.8), we get

$$a_{\,p}=\frac{P r\left\{B_{p-2}\right\}}{P r\left\{B_{p-1}\right\}}\left(\frac{1}{3}a_{\,p-1}+\frac{1}{3}(1-a_{\,p-1})\right)=\frac{1}{3b_{p-1}}.$$  (4, 7) gives us. 
This, together with (4.7), gives us

$$b_{p}=1-{\frac{1}{9b_{p-1}}}.$$
$$(4\,,11\,)$$

It is not difficult to show that the sequence { b } is decreasing and lower bounded by 0. Letting lim bp = b and taking the limit on both sides, we get

$$b=1-{\frac{1}{9\,b}},$$
$$(4\,,12)$$
