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s
0
˄
˄
˄
˄
1/2
1/
2
1/2
1/2
s
2
s
1
a
˄
s
3
ˉ
s
4
(a)
(b)
Fig. 4.3
Testing the process
Q
1
1
2
h
(
ˉ
)
.
In fact this equation has a unique solution in [0, 1], namely
h
(
ˉ
). Therefore,
A
1
2
· V
h
(
s
0
)
h
(
s
0
)
V
=
+
h
(
a.ˉ
,
Q
1
)
={
h
(
ˉ
)
}
.
Example 4.3
Consider the process
Q
2
depicted in Fig.
4.4
a and the application of
the test
T
=
a.ˉ
to it; this is outlined in Fig.
4.4
b. In the pLTS of
T
|
Act
Q
2
, for
h
(
R
k
)
1
each
k
≥
1 there is a resolution
R
k
such that
V
=
(1
−
2
k
); intuitively it goes
around the loop (
k
−
1) times before, at last, taking the right-hand
˄
action. Thus
h
(
T
,
Q
2
) contains (1
1
A
1. But it also contains
h
(
ˉ
), because
of the resolution that takes the left-hand
˄
move every time. Therefore,
−
2
k
)
h
(
ˉ
) for every
k
≥
h
(
T
,
Q
2
)
A
includes the set
1
h (ˉ)
,
1
h
(
ˉ
),
...
,
1
h
(
ˉ
),
... h
(
ˉ
)
.
1
2
1
2
2
1
2
k
−
−
−
h
(
T
,
Q
2
) is actually the convex closure of this
From later results it will follow that
A
set, namely [
2
h
(
ˉ
),
h
(
ˉ
)].
4.5
Extremal Reward Testing
In the previous section, our approach to testing consisted of two parts:
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