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In-Depth Information
7.7 Probability, Possibility and Necessity
In the case where
F
is a Boolean sub-algebra of
P (
X
)
, also probabilities p
: F ₒ
[
0
,
1
]
can be taken into account. Once a dual-pair
(
N
, ˀ)
is given, it appears the
problem of which probabilities are consistent with
(
N
, ˀ)
.
A C
A C
The property N
(
A
) +
N
(
)
1, equivalent to N
(
A
)
1
N
(
) = ˀ(
A
)
,
showsthatforall A
∈ F
it is
N
(
A
) ˀ(
A
),
provided the pair
(
N
, ˀ)
is dual. A probability p is said consistent with the dual pair
(
N
, ˀ)
if
N
(
A
)
p
(
A
) ˀ(
A
)
,
for all
A
∈ F .
In this hypothesis, if N
(
A
)>
0, it is also p
(
A
)>
0, and
ˀ(
A
)>
0, and if
ˀ(
A
) =
0itis N
(
A
) =
p
(
A
) =
0. That is,
If something is just a little bit necessary, it is probable and possible.
If something is not possible at all, it is neither necessary nor probable.
Analogously, if p
(
A
)>
0itis
ˀ(
A
)>
0 although it could be N
(
A
) =
0.
That is,
If something is just a little bit probable, it is possible, but not necessarily necessary.
A C
Notice that from N
(
A
)
p
(
A
) ˀ(
A
)
, or equivalently from 1
ˀ(
)
A C
A C
p
(
A
) ˀ(
A
)
, follows 1
ˀ(
A
)
p
(
) ˀ(
)
.
Example 7.7.1 Let is X
={
1
,
2
,
3
}
, and
μ =
0
.
7
|
1
+
1
|
2
+
0
.
5
|
3. Then, with
, and N ( A )
ˀ μ (
) =
(μ(
), μ A (
))
=
ˀ μ (
)
A
Sup
i
min
i
i
1
A
, we get:
μ
X
ˀ μ (
) = μ(
) =
.
ˀ μ (
) = μ(
) =
ˀ μ (
) = μ(
) =
.
1.
1
1
0
7,
2
2
1,
3
3
0
5.
ˀ μ ( {
1
,
2
} ) =
max
(μ(
1
), μ(
2
)) =
1,
ˀ μ ( {
1
,
3
} ) =
max
(
0
.
7
,
0
.
5
) =
0
.
7,
ˀ μ ( {
2
,
3
} ) =
1,
ˀ μ {
X
}=
1.
2.
N
μ (
1
) =
1
ˀ μ ( {
2
,
3
} ) =
1
1
=
0, N
μ (
2
) =
1
ˀ μ ( {
1
,
3
} ) =
1
0
.
7
=
0
.
3,
N
μ (
3
) =
1
ˀ μ ( {
1
,
2
} ) =
0
N
μ ( {
1
,
2
} ) =
1
ˀ μ (
3
) =
0
.
5, N
μ ( {
1
,
3
} ) =
1
ˀ μ (
2
) =
0, N
μ ( {
2
,
3
} ) =
1
ˀ μ (
1
) =
0
.
3, N
μ (
X
) =
1.
Hence, the consistent probabilities are given by triplets p
(
1
)
, p
(
2
)
, p
(
3
)
in
[
0
,
1
]
such that p
(
1
) +
p
(
2
) +
p
(
3
) =
1, and verifying:
0
p
(
1
)
0
.
7, 0
.
3
p
(
2
)
1, 0
p
(
3
)
0
.
5
.
 
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