Modified Rathbun Analysis
Start with Rathbun’s 4 basic statistical properties.
Bayes Theorem:
P
(
A
|
B
) =
P
(
B
|
A
)
P
(
A
)
P
(
B
)
[1]
Probability for intersecting sets:
P
(
A
&
B
) =
P
(
A
|
B
)
P
(
B
)
[2]
Probability when
A
and
B
are independent :
P
(
A
&
B
) =
P
(
A
)
P
(
B
)
[3]
Decomposing
A
into its intersection with
B
and (
notB
) =
B
−
1
P
(
A
) =
P
(
A
&
B
) +
P
(
A
&
B
−
1
)
[4]
Let
H
be a hypothesis, and
E
i
i
= 1
,
2
, . . . , n
a set of n pieces of evidence that relate to
H
.
Follow Rathbun’s steps within his Eq (1).
Using [1]
P
(
H
|
E
1
&
E
2
. . .
&
E
n
) =
P
(
E
1
&
E
2
. . .
&
E
n
|
H
)
P
(
H
)
P
(
E
1
&
E
2
. . .
&
E
n
)
[5]
Using [4] in the denominator
P
(
H
|
E
1
&
E
2
. . .
&
E
n
) =
P
(
E
1
&
E
2
. . .
&
E
n
|
H
)
P
(
H
)
P
(
E
1
&
E
2
. . .
&
E
n
&
H
) +
P
(
E
1
&
E
2
. . .
&
E
n
&
H
−
1
)
[6]
Using [2] for each of the 2 terms in the denominator
P
(
H
|
E
1
&
E
2
. . .
&
E
n
) =
P
(
E
1
&
E
2
. . .
&
E
n
|
H
)
P
(
H
)
P
(
E
1
&
E
2
. . .
&
E
n
|
H
)
P
(
H
) +
P
(
E
1
&
E
2
. . .
&
E
n
|
H
−
1
)
P
(
H
−
1
)
[7]
Using [3] to distribute the
n
independent pieces of evidence in both numerator and denominator
P
(
H
|
E
1
&
E
2
. . .
&
E
n
) =
R
R
+
S
[8]
R
= [
P
(
E
1
|
H
)
P
(
E
2
|
H
)
. . . P
(
E
n
|
H
)]
P
(
H
)
[8
a
]
S
= [
P
(
E
1
|
H
−
1
)
P
(
E
2
|
H
−
1
)
. . . P
(
E
n
|
H
−
1
)]
P
(
H
−
1
)
[8
b
]
1
Rewriting [8] using Π product notation
P
(
H
|
E
1
&
E
2
. . .
&
E
n
) =
P
(
H
)
Q
n
i
=1
P
(
E
i
|
H
)
P
(
H
)
Q
n
i
=1
P
(
E
i
|
H
) +
P
(
H
−
1
)
Q
n
i
=1
P
(
E
i
|
H
−
1
)
[9]
Equation [9] above is the final identity in Rathbun’s Eq (1). Now depart from the original analysis by
interpretting
p
i
=
P
(
E
i
|
H
)
[10]
as the estimated probability that evidence
E
i
supports that the hypothesis
H
is true and
1
−
p
i
=
P
(
E
i
|
H
−
1
)
[11]
as the estimated probability that evidence
E
i
supports that the hypothesis
H
is false.
P
(
H
) is the prior probability of
H
being true before any of the evidence
E
i
is considered.
P
(
H
−
1
) =
1
−
P
(
H
) is the prior probability of
H
being false before any of the evidence
E
i
is considered.
Substituting [10] and [11] into [9]
P
(
H
|
E
1
&
E
2
. . .
&
E
n
) =
P
(
H
)
Q
n
i
=1
p
i
P
(
H
)
Q
n
i
=1
p
i
+
P
(
H
−
1
)
Q
n
i
=1
(1
−
p
i
)
[12]
If there is no prior knowledge of
H
, then
P
(
H
) =
P
(
H
−
1
) = 0
.
5, and these factors then cancel from the
numerator and denominator of [12] to give
P
(
H
|
E
1
&
E
2
. . .
&
E
n
) =
Q
n
i
=1
p
i
Q
n
i
=1
p
i
+
Q
n
i
=1
(1
−
p
i
)
[13]
2