December 2003

The Bayesian Group

Definition: Bayesian Addition

The Bayesian Group is the set = x | 0 < x <1 with the operator defined as:

              A*B
A B = ----------------
        A*B + (1-A)(1-B)

From the definition we should notice the identity element .

½  A = A     A  

Theorem (proof left to reader):

A  (B  C) = (A  B)  C

Definition: Bayesian Subtraction

The definition of  is meant to be the inverse operator of , and is defined as

            A*(1-B)
A  B = ----------------
        A*(1-B) + (1-A)B

Theorems: (proof left to reader)

          ½ = A  A
      A  B = A  (1 - B)
(A  B)  C = A  (B  C)
 A  B = C <=> A = C  B

The proofs for the third and fourth theorems are easily done by using the second.

Definition: Bayesian Multiplication

We define a multiplication operator, where n  , and P :

             Pⁿ
n  P = ------------
        Pⁿ + (1 - P)ⁿ

This does not make the Bayesian group into a ring because n is not in .

Theorem:

(n  P)  (m  P) = (n + m)  P

Proof: (be sure your browser is displaying wide enough)

                               (n  P)(m  P)
(n  P)  (m  P) = ----------------------------------------
                     (n  P)(m  P) + (1-(n  P))(1-(m  P))

               /     Pⁿ      \ /      P^m     \
               | ----------- | | ----------- |
               \ Pⁿ + (1-P)ⁿ / \ P^m + (1-P)^m /
= ------------------------------------------------------------------
  /      Pⁿ     \ /      P^m     \    /    (1-P)ⁿ   \  /   (1-P)^m   \
  | ----------- | | ----------- | + | ------------ | | ----------- |
  \ Pⁿ + (1-P)ⁿ / \ P^m + (1-P)^m /    \ Pⁿ + (1-P)ⁿ /  \ P^m + (1-P)^m /

          Pⁿ P^m
= ---------------------
  Pⁿ P^{m +} (1-P)ⁿ (1-P)^m

       P^n+m
= ---------------
  P^{n+m +} (1-P)^n+m

= (n + m)  P

Intuitive Multiplication Theorem:

The definition for was originally derived from the desire to have n P mean "add P to itself n times". This theorem proves that this intuitive meaning still holds.

n  P = P  P  ...  P       n  Naturals
           (n times)

Proof:

Base Case:

             P¹
1  P = ------------ = P
         P¹ + (1-P)¹

Assume true for n-1:


n  P = (n - 1 + 1)  P = (1  P)  ((n-1)  P)

      = P  (P  P  ...  P)
               (n-1 times)

      = P  P  ...  P
           (n times)

More Theorems:

      0  P = ½
n  (P  Q) = (n  P)  (n  Q)
n  (m  P) = (n * m)  P
(n + m)  P = (n  P)  (m  P)
(n - m)  P = (n  P)  (m  P)

December 2003, first draft