Question #171355

Suppose (Ω, 𝒜, 𝑃) is a probability space and 𝐵 is an event with 𝑃(𝐵) > 0. Prove that (Ω, 𝒜, 𝑃(⋅ |𝐵)), where 𝑃(⋅ |𝐵) is the conditional probability given 𝐵, is a probability space.

Please provide step by step solutions and logical explanations :)


1
Expert's answer
2021-03-16T07:30:11-0400

A probability space is defined if the following components are defined and the following relations/properties are satisfied:

1) A sample space is to be defined as a set. The sample space is Ω\Omega.

2) An event space is to be defined as a sigma-algebra of subsets of the sample space. The event space is 𝒜. This is a sigma-algebra of subsets of Ω\Omega.

3) A probability function is to be defined as a non-negative sigma-additive function P' which has the event space as a domain and which is normalized by the condition P'(Ω)=1. The candidate to this function is 𝑃(⋅ |𝐵), the necessary properties of which are to be verified:

3.1) The domain of P'=𝑃(⋅ |𝐵) must be the event space 𝒜. - Valid.

3.2) The function P' is non-negative. - Valid, because of P(A)=P(AB)/P(B)P'(A)=P(AB)/P(B), P(AB)0P(AB)\geq 0 and P(B)>0P(B)>0.

3.3) The function P' is normalized. - Valid, because of P(Ω)=P(ΩB)/P(B)=1P'(\Omega)=P(\Omega B)/P(B)=1.

3.4) The function P' is additive. Indeed, if A1,A2A_1, A_2 are two events such that A1A2=A_1\cap A_2=\emptyset, then A1BA2B=A_1B\cap A_2B=\emptyset and P(A1A2)=P((A1A2)B)/P(B)=P((A1B)(A2B))/P(B)=(P(A1B)+P(A2B))/P(B)=P(A1B)/P(B)+P(A2B)/P(B)=P(A1)+P(A2)P'(A_1\cup A_2)=P((A_1\cup A_2)\cap B)/P(B)=P((A_1\cap B)\cup (A_2\cap B))/P(B)=(P(A_1\cap B)+P(A_2\cap B))/P(B)=P(A_1\cap B)/P(B)+P(A_2\cap B)/P(B)=P'(A_1)+P'(A_2)

(the 3rd equality is provided by additivity of the probability function P).

3.5) The function P' is sigma-additive. Indeed, if A1,A2,A3,A_1, A_2, A_3, \dots is a countable sequence od non-intersecting sets, then AiBAjB=(AiAj)B=A_iB\cap A_jB=(A_i\cap A_j)\cap B=\emptyset and

P(i=1Ai)=P((i=1Ai)B)/P(B)=P(i=1AiB)/P(B)=(i=1P(AiB))/P(B)=i=1P(AiB)/P(B)=i=1P(Ai)P'(\bigcup\limits_{i=1}^\infty A_i)=P((\bigcup\limits_{i=1}^\infty A_i)\cap B)/P(B)=P(\bigcup\limits_{i=1}^\infty A_i\cap B)/P(B)=(\sum\limits_{i=1}^\infty P(A_i\cap B))/P(B)=\sum\limits_{i=1}^\infty P(A_i\cap B)/P(B)=\sum\limits_{i=1}^\infty P'(A_i)

(the 3rd equality is provided by sigma-additivity of the probability function P).

As all conditions are satisfied, therefore, (Ω, 𝒜, 𝑃(⋅ |𝐵)) is indeed a probability space.


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