# Probability of consequent if dependent on two independent antecedents

Suppose

P(Z|X) = A

P(Z|Y) = B

where X and Y are independent. How does one determine P(Z|X,Y)?

I haven't studied maths in years, so what I post below may be entirely off base, but I don't like to ask a question without some attempt, so:

From what I read, Baye's theorem is used to calculate update a probability based on new information. If the prior probability is P(Z|X), and the posterior P(Z|X,Y):

P(Z|X,Y) = (P(Y|Z,X)P(Z|X))/P(Y)

(Some of the above was worked out on intuition, and I'd love for someone to give me an explanation of how to properly apply Baye's theorem to update an already dependent result, every discussion I found dealt with the prior probability as a simple probability, not a conditional). Since Y is independent of X, this is equal to:

P(Z|X,Y) = (P(Y|Z)P(Z|X))/P(Y)

To digress:

P(Y|Z) = P(Z|Y)P(Y)/P(Z)

(By Baye's theorem, again), so:

P(Z|X,Y) = ((P(Z|Y)P(Y)/P(Z))P(Z|X))/P(Y)

P(Z|X,Y) = (P(Z|Y)P(Y)P(Z/X))/P(Y)P(Z)

Ignoring the degenerate case P(Y) = 0:

P(Z|X,Y) = (P(Z|X)P(Z|Y))/P(Z)

Could someone first confirm whether or not this is correct? Secondly, can this be extended to more than two dependencies? It seems to work, but honestly my brain started melting when I tried to figure it.

Also, if possible, explain simply why it works? I always like to know intuitively to reason for a result. I assume it's dependent on P(Z) because its necessary to find the overlap between P(X) and P(Y), but aside from that it's a bit hazy.

Finally, if my guess is correct and the term P(Z) is necessary to determine the overlap between P(X) and P(Y), what extra information would I need and what equation would I have to solve in order to calculate P(Z|X,Y) without knowledge of P(Z)? This one I haven't been able to make any progress on, because I can't see where to start.

I realise this is an awfully large question, and thank you all very much in advance for any contributions. If any parts of the question are too large (the last part, particularly, seems a candidate), please feel free to ignore them. This is all out of curiousity, so I don't specifically need an answer to any individual part.

-
add comment

## 1 Answer

Unfortunately your result is not correct. Consider two coin tosses (with a fair coin). Let X be: "You get heads in the first toss" and Y be "You get heads in the second toss". Then X and Y are independent events. Let Z be "You get at least one head". Then P(Z)=3/4, but P(Z|X,Y)=P(Z|X)=P(Z|Y)=1.

The problem in your argument is that P(Y|Z,X) is not necessary equal to P(Y|Z), or in other words, when constrained to Z, X and Y need no longer be independent.

-
yes indeed, and if I'm not mistaken, in your example, $p(XY|Z)$ would be 1/3, but $p(X|Z)=p(Y|Z)=2/3$, so $p(XY|Z)\neq p(X|Z)p(Y|Z)$ showing that independence is not inherited into the conditional probability space. Gross mistake on my side, thank you. –  Weltschmerz Oct 3 '10 at 15:49
add comment