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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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