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I am trying to understand what's wrong with the following logic related to "multiple conditioning." Why is the probability of [(A given B) given C] not the same as the probability of [A given (B and C)] ? I know it's not true, but only because numbers disagree. I am having a hard time parsing what's wrong with the logic.

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Both are the same thing. What numbers disagree? –  leonbloy Dec 18 '11 at 0:32
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There is no such thing as [A given B]. There is such a thing as the probability given B, of A. –  Michael Hardy Dec 19 '11 at 22:53

3 Answers 3

Peter Milne's paper "Bruno de Finetti and the Logic of Conditional Events" is an account of attempts to base probability theory upon conditional events, with special regard to de Finetti's subjectivistic "coherent betting behavior" approach using a three-valued logic. The latter approach results in a conditional event algebra with the property that both $$(A|B)|C \ \equiv \ A|(B \And C)$$ and $$A|(B|C) \equiv A|(B \And C)$$ where $A, B, C$ are ordinary propositions, implying that the respective probabilities must be equal. (This is contrary to your expectation that they should differ. See p. 218 of the cited paper, where what you've called "multiple conditioning" is called "iterated conditioning".)

NB: In de Finetti's approach ...

To introduce the notion of conditional probability means extending the definition of $P(X)$ from the field of ordinary events $X$ to the field of conditional events.

This, of course, differs from standard presentations of probability theory, in which there are no such objects as "conditional events".

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There's no such thing as [A given B].

It is NOT "the probabilty of {A given B}".

Rather, it is "{the probability, given B} of A".

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What is wrong with saying "Given $C$, what is the probability of $A$ given $B$?" which becomes "Given $B$ and $C$, what is the the probability of $A$?" –  Henry Dec 18 '11 at 14:10
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+1 A lot of confusion is caused by the common notation $P(A|B)$, which to a newcomer may wrongly suggest (in standard presentations) that there is some object $A|B$ (i.e. a conditional event). In this sense, $P_B(A)$ is a much better notation for what is intended. –  r.e.s. Dec 18 '11 at 15:25
    
@Henry : It seems as if you're responding to my comment under your own answer rather than to what I wrote in my answer. –  Michael Hardy Dec 19 '11 at 6:49

$\Pr(X | Y) = \dfrac{\Pr(X \cap Y)}{\Pr(Y)}$ so $\Pr((A|B)|C) = \dfrac{\Pr(A \cap B|C)}{\Pr(B|C)} =\dfrac{\frac{\Pr(A \cap B\cap C)}{\Pr(C)}}{\frac{\Pr(B\cap C)}{\Pr(C)}} = \dfrac{\Pr(A \cap B\cap C)}{\Pr(B\cap C)}=\Pr(A | B \cap C)$ so they are the same.

Here is a way to check your numbers. If you have

  • $\Pr(A \cap B \cap C) = d/s$
  • $\Pr(A \cap B \cap C^c) = e/s$
  • $\Pr(A \cap B^c \cap C) = f/s$
  • $\Pr(A \cap B^c \cap C^c) = g/s$
  • $\Pr(A^c \cap B \cap C) = h/s$
  • $\Pr(A^c \cap B \cap C^c) = i/s$
  • $\Pr(A^c \cap B^c \cap C) = j/s$
  • $\Pr(A^c \cap B^c \cap C^c) = k/s$

where $s=d+e+f+g+h+i+j+k$, then $\Pr(B \cap C) = \dfrac{d+h}{s}$ and $\Pr(A | B \cap C) = \dfrac{d}{d+h}$.

Given $C$, we just need to look at

  • $\Pr(A \cap B | C) = d/t$
  • $\Pr(A \cap B^c | C) = f/t$
  • $\Pr(A^c \cap B | C) = h/t$
  • $\Pr(A^c \cap B^c | C) = j/t$

where $t=d+f+h+j$ (and $\Pr(C)=t/s$), then $\Pr(B|C) = \dfrac{d+h}{t}$ and $\Pr((A|B)|C) = \dfrac{d}{d+h}$, the same as before.

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How do you get $\Pr((A|B)|C) = \dfrac{\Pr(A \cap B|C)}{\Pr(B|C)}$? The expression before "${}={}$" makes no sense. It's almost as if you're taking $A|B$ to mean $A\cap B$. –  Michael Hardy Dec 18 '11 at 2:09
    
@Michael: I applied the initial expression $\Pr(X | Y) = \dfrac{\Pr(X \cap Y)}{\Pr(Y)}$. $\Pr((A|B)|C)$ is in the question "probability of [(A given B) given C]". I realise you think it makes no sense, but I attempted to give it an interpretation, namely that in the universe in which $C$ is true what is the probability of $A$ given $B$. I also gave the questioner a method of checking whether "numbers disagree". If I had intended $A|B$ to mean $A \cap B$ then I would have had $\Pr((A|B)|C) = {\Pr(A \cap B|C)}$ but clearly I did not. –  Henry Dec 18 '11 at 9:12
    
Whatever you meant, it's not a standard notation. –  Michael Hardy Dec 19 '11 at 1:34

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