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For example, for Bayes formula:
P(A|B)=P(B|A)*P(A)/P(B)

If we add condition on C for each term, we get:
P(A|B,C)=P(B|A,C)*P(A|C)/P(B|C)

Is it still correct? Will every equation remain correct after such operation?
Can you prove it, or provide intuition for it?

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4 Answers 4

up vote 1 down vote accepted

As joriki notes, for any event $C$ with $P(C)>0$ we can define a new probability $$P_C(A):=P(A|C)=P(AC)/P(C).$$ Applying Bayes rule to the new probability gives $$P_C(A|B)=P_C(B|A) P_C(A)/P_C(B)$$ and translating back to the original probability this reads: $$P(A|BC)=P(B|AC) P(A|C)/P(B|C).$$

Note that the conditioning events in this equation are $BC$, $AC$, and $C$. I'm not quite sure what you mean by the comma in $P(A|B,C)$ but it should be the intersection of $B$ and $C$.

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If by a "true probability equation" you mean a tautological one, that is, one that doesn't depend on the concrete probabilities and expresses a theorem on probabilities in general, then the answer is "yes". This is a consequence of the fact that the conditional probabilities are, as the name indicates, also probabilities; they satisfy the probability axioms and therefore all theorems that are derived from them. Intuitively, you can imagine the event $C$ actually occuring; then what were conditional probabilities under the condition $C$ will turn into actual probabilities.

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Can't grasp the intuition.. Can you mathematically prove that for all tautological probability formulas, adding same conditional to all terms yields another tautological formula? –  Sunny88 Oct 19 '11 at 18:32

Yes, it's true. Informally: Conditioning a set of variables (events) $(A,B, \ldots)$ on the ocurrence of some event $Z$ can be thought as restricting the original universe to a "smaller universe", so that we have get new variables (events) :$(A|Z, B|Z, \ldots ) \equiv (A',B', \ldots)$

(More precisely, the new universe, and hence the new variables, depende on the value of the conditioning variable).

An example: let our universe be the french people (selected by some random procedure), we could have the events $A\equiv$french person speaks english $B\equiv$french person is married, etc. $Z\equiv$person is male If we condition on $Z$ being true, we have $A|Z \equiv $ french person speaks english, given he is male. But this could equivalently be defined as a new event, if we restrict our universe to the males french people: calling this event ("male french speaks english") $A'$ , we see that $A | Z \equiv A'$

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All probabilities are conditional: events are subsets of a "probability space" or "sample space" $\Omega$. If $A$ is such an event, then $\Pr(A)$ is the same thing as $\Pr(A\mid \Omega)$. If something is true of every probability space $\Omega$, then if one puts in place of $\Omega$ some subset $C$, so that for example $\Pr(A)$ is replaced by $\Pr(A \mid C)$, then the word "every" implies it's true with $C$ in that place.

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