# Thread: Derivation with Bayes theorem

1. ## Derivation with Bayes theorem

I'm afraid this is hopelessly trivial, but I'm not able to see why the following holds:

$Pr(X | Y \land Z) = Pr(X | Z) \Rightarrow Pr(Y | X \land Z) = Pr(Y | Z)$

Could anyone steer me in the right direction?

2. ## Re: Derivation with Bayes theorem

Hi,

We know that Pr(X/Y^Z) = Pr(X/Z)

Then:

Pr(X^Y^Z)/Pr(Y^Z) = Pr(X^Z)/Pr(Z)

Then:

Pr(X^Y^Z)/Pr(X^Z) = Pr(Y^Z)/Pr(Z)

Then:

Pr(Y/X^Z) = Pr(Y/Z)

Let me knw if you have any doubt

Thanks

Steve

3. ## Re: Derivation with Bayes theorem

Hey Lepzed.

So for the first statement you have P(X and Y and Z)/P(Y and Z) = P(X and Z)/P(Z) which implies that you should check that Z is a subset of Y. A subset S of a set Z is a set where P(S and Z) = P(S) for all valid subsets S without exception.

Now Bayes rule says P(B|A) = P(A)P(A|B)/P(B) So if you are looking at P(Y|Z) you need to consider P(Z|Y)*P(Y)/P(Z) which should equal P(X and Z|Y)*P(Y)/P(X and Z).

My guess is that if you show that Z is a subset of Y implies that Z is a subset of X then the implication will hold.

With conditional probabilities, basically what you are doing anyway is conditioning the probability with respect to some set so instead of your probability being with respect to the universal set omega, it's with respect to an arbitary set B where you find P(X|B): all you are doing is restricting your probability relative to a subset of events rather than omega.

Hopefully the above has outlined not only the math but the intuition behind what is actually going on.