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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