Is P(A|B) equal to P(A) if and only if P(B|A) is equal to P(B)?

[x^2]

Independent Probability...

Hello,

Could someone point me in the right direction of how to prove that P(A|B) = P(A) if and only if P(B|A) = P(B)? I think I understand the program and I can't formulate any contradictions, but I'm having difficulty showing this property with a formal proof.

Thanks,
x^2

 

[lanedance]

can you start with the definition

P(A|B) = \frac{P(A \cap B)}{P(B)}

 

[x^2]

P(A|B) = \frac{P(A \cup B)}{P(B)} = P(A)

P(B|A) = \frac{P(A \cup B)}{P(A)} = P(B)

P(B|A) = \frac{P(A \cup B)}{P(B)} = P(A)

P(A|B) = \frac{P(A \cup B)}{P(A)} = \frac{P(A \cup B)}{P(A)} = P(B|A)

I think that is right... Thank you for the hint!
x^2

 

[lanedance]

i think you mean intersections, not unions

a better way to show it would be to start with
P(A|B) = P(A)

and use the definition and simplify to show P(B|A) = P(B)

 

[x^2]

i think you mean intersections, not unions

a better way to show it would be to start with
P(A|B) = P(A|B)

and use the definition and simplify to show P(A) = P(B)

Yes, sorry, you are correct; they should be intersections and not unions.

Thank you for the help!
x^2

 

[honestrosewater]

P(A|B) = \frac{P(A \cup B)}{P(B)} = P(A)

P(B|A) = \frac{P(A \cup B)}{P(A)} = P(B)

P(B|A) = \frac{P(A \cup B)}{P(B)} = P(A)

P(A|B) = \frac{P(A \cup B)}{P(A)} = \frac{P(A \cup B)}{P(A)} = P(B|A)

I think that is right... Thank you for the hint!
x^2

What is going on in this proof? It looks like you are assuming P(A|B) = P(A) and P(B|A) = P(B) = P(A), and then you don't prove what you set out to prove.

What does P(A|B) = P(A) mean? It means that B's occurring has no effect on the probability of A occurring, i.e., A is independent of B, yes? You need to show that A being independent of B also makes B independent of A. So there is a hint: Do you have a rule that has a different version for independent events?

To prove an equivalence, you can prove the implication both ways: Assume P(A|B) = P(A) and use this to derive P(B|A) = P(B). Then assume P(B|A) = P(B) and use this to derive P(A|B) = P(A).

So the first line in your proof should be

1] P(A|B) = P(A)​

What can you say given also

1] P(A|B) = P(A)
2] P(A ∩ B) = P(B) * P(A|B)​

Remember that you are trying to get to P(B|A) = P(B). So, as a general rule, P(B|A) = ??

Note that A and B are arbitrary events, so the proof in the other direction will be the same.

 

[lanedance]

sorry yeah corected post