# Probability of a proposition conditional on a disjunction

I am reading through E.T. Jaynes' Probability Theory: The Logic of Science, and I am stuck on one of the exercises in Chapter 2. The exercise is

Is it possible to find a general formula for $p(C|A + B)$, analogous to (2.66), from the product and sum rules? If so, derive it; if not, explain why this cannot be done.

where (2.66) is $p(A + B|C) = p(A|C) + p(B|C) − p(AB|C)$.

My intuition is that $p(C|A + B) = p(C|A) + p(C|B) - p(C|AB)$. But I can't see how to prove or disprove this from the product and sum rules. I don't see how to expand the expression $p(C|A + B)$, since there is only a single proposition on the left side of the conditional symbol. And I don't know of any identities that can be used for the expression on the right side of the conditional symbol.

For context, Jaynes gives the product and sum rules as:

Product Rule: $P(AB|C) = P(A|C)P(B|AC) = P(B|C)P(A|BC)$

Sum Rule: $P(A|C) + P(\neg{A}|C) = 1$

I have seen this previous posting which seems to be about the same question. But it doesn't look like there was any resolution.

This is my first post, so any meta comments on my question are appreciated.

By "analogous to (2.66)", I think Jaynes intends a formula that expresses $p(C∣A+B)$ in terms of conditional probabilities involving $A,B,C$ only -- which (due to the form of the sum and product rules) is evidently impossible without special assumptions about the propositions $A,B,C$. (The very next exercise in the book concerns such a special case.)

In your question you say ...

My intuition is that $p(C|A+B)=p(C|A)+p(C|B)−p(C|AB)$.

... and in a comment ...

But I’m not sure how to use [the linked counterexample] to demonstrate that the intuition is invalid. My understanding is that $\Omega$ is a distribution over three propositions, each having 1/3 probability. My attempt to plug that into the intuitive formula seems wrong: $P(C|A+B)=P(c|a,c)+P(c|a,b)−P(c|a,c,a,b).$

Your commas appear to stand for logical OR (otherwise denoted by $+$), so there's some confusion about the meaning of $AB$ (the logical conjunction of propositions $A$ and $B$).

NB: The counterexample at your link applies when your equation is (mis-)interpreted in terms of standard probability-as-measure, whereas the context here is Jaynes' propositional theory of probability. (In this theory, there is no such object as an unconditional probability; in particular, the notation $P(A\mid B)$ does not mean $P(AB)/P(B)$ as it would in the standard theory.) However, it's easy to reformulate it as a counterexample in the present context, as follows:

Suppose $P(a\mid X)=P(b\mid X)=P(c\mid X)=1/3$, where proposition $X$ implies that the three propositions $a,b,c$ are mutually exclusive and exhaustive (i.e., exactly one of the three is true and the other two are false). Now consider the propositions $A=a+c,\ B=a+b,\ C=c$. We have $(A+B)X=(a+b+c)X$, and $ABX=(a+c)(a+b)X=(aa+ab+ca+cb)X=aX$ because $X$ implies that $ab,ca$, and $cb$ are false. We then find the following: \begin{align} P(C\mid AX+BX)&=P(c\mid (A+B)X)\\ &=P(c\mid (a+b+c)X)\\ &=P(c\mid X)\\ &=1/3 \\ \\ P(C\mid AX)&=P(c\mid (a+c)X)\\ &={P(c(a+c)\mid X)\over P(a+c)\mid X)}\\ &={P(c\mid X)\over P(a+c\mid X)}\\ &={1/3\over 2/3}\\ &=1/2\\ \\ P(C\mid BX)&=P(c\mid (a+b)X)\\ &=0\\ \\ P(C\mid ABX)&=P(c\mid aX)\\ &=0\\ \end{align} Now if your intuitive formula were correct, then $$\underbrace{P(C\mid AX+BX)}_{1/3}=\underbrace{P(C\mid AX)}_{1/2}+\underbrace{P(C\mid BX)}_{0}-\underbrace{P(C\mid ABX)}_{0}$$ hence the formula is not correct.

In your cited post, there is already a comment answering your question that the intuition is wrong. However you can always expand it, if you like:

\begin{align} P(C|A \cup B) &= \frac {P(C \cap (A \cup B))} {P(A \cup B)} \\ &= \frac {P((C \cap A) \cup (C \cap B))} {P(A) + P(B) - P(A \cap B)} \\ &= \frac {P(C \cap A) + P(C \cap B) - P(C \cap A \cap B)} {P(A) + P(B) - P(A \cap B)} \end{align}

Not sure if that is what you want.

• By "analogous to (2.66)", I think Jaynes intends a formula that expresses $p(C\mid A+B)$ in terms of conditional probabilities involving $A,B,C$ only -- which I suspect is impossible unless some special assumptions are made about $A,B,C$. (For example, in this answer the unconditional probabilities can be written as conditional on a sample space, say $S$, but then that introduces an additional symbol $S$ for the sample space. – r.e.s. Feb 2 '18 at 1:02
• @BGM Ah, I see the comment giving a counter example. But I’m not sure how to use it to demonstrate that the intuition is invalid. My understanding is that $\Omega$ is a distribution over three propositions, each having 1/3 probability. My attempt to plug that into the intuitive formula seems wrong: $P(C|A+B) = P(c|a, c) +P(c|a, b) - P(c|a, c, a, b)$ – Matt R. Feb 3 '18 at 17:13