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I think the statement $P(A|C) = \sum_{i} P(A|B_i)P(B_i|C)$ is true (where the events $B_i$ form a partition of the whole space). Is it, and why?

Thanks!

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    Have you tried writing $P(A \, | \, C) = P(A \cap C) / P(C)$ and so on for the other terms? Maybe something reaaally obvious will come out of it.2011-11-08
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    Yeah, but nothing good happened :/2011-11-08
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    Try to prove this instead: $P(A|C) = \sum_{i} P(A|B_i,C)P(B_i|C)$2011-11-08
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    @p.s. That works, thanks! Maybe I can reduce my problem to that problem instead.2011-11-08
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    @AndréNicolas: Okay, I tried that and I got $P(A|C) = \sum P(A \cap B_i | C)$ = $\sum P(A \cap B_i \cap C)/P(C)$ = $\sum P(A \cap C | B_i \cap C) P(B_i \cap C)/P(C)$ = $\sum P(A \cap C | B_i \cap C) P(B_i | C)$, which is kind of close but I'm beginning to suspect the statement is false.2011-11-08
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    @okay-at-math: I didn't push the calculation through.2011-11-08

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It’s false as stated. Suppose that the space is the sample space for two tosses of a fair coin,

  • $B_1$ is the event that the first toss is heads,
  • $B_2$ is the event that the first toss is tails,
  • $A$ is the event that both tosses are heads, and
  • $C$ is the event that at least one toss is tails (i.e, the complementary event to $A$).

Clearly $\mathbb{P}(A|C)=0$, but $$\mathbb{P}(A|B_1)\mathbb{P}(B_1|C)+\mathbb{P}(A|B_2)\mathbb{P}(B_2|C)=\frac12\cdot\frac13+0\cdot\frac23=\frac16.$$

Added: It never hurts to check some simple cases of a conjecture, if there are any. Those that are somehow atypical (like making $A$ and $C$ complementary events) are often especially useful.

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    Thank you! I guess I will have to solve my problem another way then.2011-11-08
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    We would be happy to take a look at your actual question if you are still bothered about it.2011-11-08