Image Processing Reference
In-Depth Information
From equation [A.12], we get the additivity of the probabilities of disjoint events.
The second step of the argument focuses on determining conditional probabilities.
To do this, we will consider three bets:
1) a bet on
E
1
E
2
(
E
1
and
E
2
), with the stake
S
1
, and the wager
p
1
S
1
;
2) a bet on
E
2
, with the stake
S
2
, and the wager
p
2
S
2
;
3) a bet on
E
1
|
E
2
, with the stake
S
, and the wager
pS
, the gain for this bet being:
- (1
−
p
)
S
if
E
1
|
E
2
is true,
E
2
is false,
-0if
E
2
does not occur (the game is considered void in this case, and the wager
is paid back).
- (
−
p
)
S
if
E
1
|
Three outcomes are possible:
1) if
E
1
and
E
2
occur, then the gain is:
G
1
=
1
p
1
S
1
+
1
p
2
S
2
+(1
−
−
−
p
)
S
;
[A.14]
2) if
E
2
occurs but not
E
1
, then the gain is:
p
1
S
1
+
1
p
2
S
2
−
G
2
=
−
−
pS
;
[A.15]
3) if
E
2
does not occur, then the gain is:
−
p
1
S
1
−
G
3
=
p
2
S
2
.
[A.16]
Consider equations A.14, A.15 and A.16 as a system of 3 equations with 3
unknowns, as before. The determinant is equal to:
p
1
−
pp
2
[A.17]
and, again for reasons of consistency, has to be equal to 0. We then get the relation:
E
2
=
p
E
1
E
2
p
E
1
|
p
E
2
[A.18]
from which we infer Bayes' theorem.
Note that in this case, the expected gain is equal to:
p
1
G
1
+
p
2
−
p
1
G
2
+
1
p
3
G
3
=
p
1
−
p
2
p
S
=0
.
−
[A.19]
Therefore, the expected gain is zero for any
S, S
1
,S
2
.
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