Question about probability measure

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Homework Help Overview

The discussion revolves around the properties of probability measures, specifically examining functions derived from a given probability measure M. Participants are tasked with demonstrating certain properties of these functions in relation to the axioms of probability measures.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of modifying a probability measure by scaling and squaring, questioning whether the resulting functions satisfy the axioms of probability measures. There is a request for a counterexample to illustrate a specific case.

Discussion Status

The discussion is ongoing, with some participants attempting to provide examples and counterexamples. There is a request for further assistance in finding a counterexample related to item (b) of the original problem, indicating a collaborative effort to clarify understanding.

Contextual Notes

Participants note the need for examples to support their reasoning, and there is an emphasis on the requirement to demonstrate specific properties of the functions derived from the probability measure M.

Alexsandro
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An probability measure on same space [itex]\Omega[/itex] is a function of subsets of [itex]\Omega[/itex] satisfying three axioms:

(i) For every set [itex]A \subset \Omega[/itex], the value of the function is a non-negative number: P(A) [itex]\geqslant[/itex] 0.

(ii) For any two disjoint sets A and B, the value of the function for their union A + B is equal to the sum of its value for A and its value for B:

P(A + B) = P(A) + P(B) provided A.B = [itex]{\O}[/itex].

(iii) The value of the function for [itex]\Omega[/itex] (as a subset) is equal to 1:

P([itex]\Omega[/itex]) = 1.


Now, reply these questions:

If M is a probability measure, show:

(a) that the function M/2 satisfies Axiom(i) and (ii) but not (iii).

(b) the function [itex]M^2[/itex] satisfies (i) and (iii) but not necessary (ii); give a counterexample to (ii).
 
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Omega={A,B}
M(gurnisht)=0
M({A})=1/2
M({B})=1/2
M({A,B})=1
then
M^2({A,B})!=M^2({A})+M^2({B})
(Either it's that simple or i don't understand the problem)
 

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