Does this function define a probability function?

Question

Show that for a given random event $A$, the function $R: \alpha \rightarrow \mathbb{R}$ defined as $R (B) = P (A | B)$ does not satisfy the axioms: https://en.wikipedia.org/wiki/Probability_axioms

$ \alpha $ is the sigma-algebra

This is what I tried:

It is a simple counterexample: I have $A_{i}´s$ that belongs to $ \alpha $. If I take the $\emptyset$ set. So $R (B) = P (A | B)$ it is not well defined.

I am not sure. Any help? Am I right?

Also, If I define $P(B)>0$ of course my counterexmple doesnt work. How I could proof this?

Answer 1

We must have $P(B)>0$ for $P(A|B)$ to be defined (your textbook may vary).

1. Axiom 1: $P(A|B) := \frac{P(A \cap B)}{P(B)} \ge 0$. There's no problem here.

2. Axiom 2: $P(\Omega) = P(A | \Omega) = \frac{P(A)}{P(\Omega)} = P(A) \ne 1$. There is a problem here if $P(A) \ne 1$. Of course there is no problem here if $P(A) = 1$ (even if $A \ne \Omega$, which is indeed possible).

3. Axiom 3: $P(A|\cup_{\text{countable} \ i} B_i) = \sum_{\text{countable} \ i} P(A|B_i)$. I think we don't have to check anymore because Axiom 2 is violated, but I think Axiom 3 is violated as well.

The correct probability function involving conditional probability should be $R(A)=P(A|B)$, I think:

1. Axiom 1: $P(A|B) := \frac{P(A \cap B)}{P(B)} \ge 0$. There's no problem here.

2. Axiom 2: $P(\Omega) = P(\Omega | B) = \frac{P(B)}{P(B)} = 1$. Now, there's no problem here.

3. Axiom 3: $P(\cup_{\text{countable} \ i} A_i | B) = \sum_{\text{countable} \ i} P(A_I |B)$. I think Axiom 3 is satisfied here. I'll leave it to you.