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The state transitions between the two states Γ 1 and Γ 2 are analyzed in the
following.
Γ 1
Γ 1 . The probability for a success is p , which is relatively low, see
lemma 7.1. A success results in a step size increase σ t +1 = γσ t . A successful
mutation may lie arbitrarily close to the constraint boundary ( d/σ
0) and
therefore decrease the success probability p rapidly. But a success may also
lead to an increase of the distance to the constraint boundary, at most by
σ tan β . The constrained case is not left, if the step size increase is higher
than the distance increase to guarantee d/σ < 1:
d + σ tan β
σγ
d
σ +tan β<γ
< 1
(7.15)
As d/σ < 1itmusthold γ> 1 + (tan β ) to fulfill the above condition for the
proof of step size reduction. So, the probability for staying in state Γ 1 is
P ( Γ 1
Γ 1 |
Γ 1 )= p,
σ t +1 = γσ t .
(7.16)
Γ 2
Γ 1 : The probability for a success if the last step was a failure is again p ,
the step size is increased and the constrained case is not left for γ> 1+tan β .
P ( Γ 2
Γ 1 |
Γ 2 )= p,
σ t +1 = γσ t .
(7.17)
Γ 1
p . It results in a step decrease
σ t +1 = γ 1 σ t . If the step decrease leads to d/σ t +1 > 1, the constraint bound-
ary is left. In this case step size decrease and increase occur with the same
probability and the expected change of σ becomes E ( γ )= γ
Γ 2 : The probability for a failure is 1
γ 1 =1.But
the constraint boundary will be reached again within the following steps.
Hence, we summarize
·
σ t +1 = γ 1 σ t .
P ( Γ 1
Γ 2 |
Γ 1 )=1
p,
(7.18)
Γ 2
Γ 2 : Similar to transition Γ 1
Γ 2 the probability to stay in the state
Γ 2 is the probability 1
p for a failure.
σ t +1 = γ 1 σ t .
P ( Γ 2
Γ 2 |
Γ 2 )=1
p,
(7.19)
This yields the following state transition probability matrix T for states Γ 1 and
Γ 2 :
T = p
(1
p )
(7.20)
p
(1
p )
It is worth to mention that in the case of a success with an overwhelming prob-
ability of p =(1
β/ (2 π ))
1for β
0, the distance d to the constraint
 
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