Why Product of Probabilities (Masses) for Independent Events? A Remark

Authors

Vladik Kreinovich, The University of Texas at El PasoFollow
Scott Ferson

Publication Date

6-2005

Comments

UTEP-CS-04-24b.

Published in International Journal of Pure and Applied Mathematics, 2005, Vol. 22, No. 3, pp. 341-347.

Abstract

For independent events A and B, the probability P(A&B) is equal to the product of the corresponding probabilities: P(A&B)=P(A)*P(B). It is well known that the product f(a,b)=a*b has the following property: once P(A1)+...+P(An)=1 and P(B1)+...+P(Bm)=1, the probabilities P(Ai&Bj)=f(P(Ai),P(Bj)) also add to 1: f(P(A1),P(B1))+...+f(P(An),P(Bm))=1. We prove that the product is the only function that satisfies this property, i.e., that if, vice versa, this property holds for some function f(a,b), then this function f is the product. This result provided an additional explanation of why for independent events, we multiply probabilities (or, in the Dempster-Shafer case, masses).

In this paper, we strengthen this result by showing that it holds for arbitrary (not necessarily continuous) functions f(a,b).