I have to calculate/approximate the following multiple conditional probability:


with only $P(X|C_1)$, $P(X|C_2)$ and $P(X|C_3)$ given. Is it possible to express the above in terms of these single conditional probabilities only? What independence assumptions would I need to make?

Edit: Concretely, I have an algorithm that can calculate single conditional probabilities $P(X|C_i)$ directly, and I want to use these to calculate multiple conditional probabilities. I'm fairly certain it is not possible to calculate this exactly, but perhaps I can make some independence assumptions so some terms can be dropped?

  • $\begingroup$ The question doesn't seem to be clear $\endgroup$ – freak_warrior Nov 15 '13 at 11:03
  • $\begingroup$ What part is unclear? I will clarify where needed... $\endgroup$ – niefpaarschoenen Nov 15 '13 at 11:05
  • $\begingroup$ I will try to add a bit more detail. $\endgroup$ – niefpaarschoenen Nov 15 '13 at 11:10
  • $\begingroup$ I don't quite understand why you get an upvote for a double comment without mentioning what is wrong with the question. In the meantime I have already edited the question... $\endgroup$ – niefpaarschoenen Nov 18 '13 at 10:17
  • $\begingroup$ I think I had accidentally reviewed your question twice. Anyway below is my answer! $\endgroup$ – freak_warrior Nov 18 '13 at 10:37

Assume $C_1,C_2,C_3$ are independent of each other. Then

$$P(X|C_1)P(X|C_2)P(X|C_3)=\frac{P(XC_1)}{P(C_1)}\cdot\frac{P(XC_2)}{P(C_2)}\cdot\frac{P(XC_3)}{P(C_3)}\\ =\frac{P(XC_1C_2C_3)}{P(C_1C_2C_3)}=P(X|C_1C_2C_3)$$

  • $\begingroup$ Seems to not match with this answer : math.stackexchange.com/questions/1288770/… $\endgroup$ – DollarAkshay Oct 30 '18 at 9:36
  • 3
    $\begingroup$ @AkshayLAradhya Indeed, this is wrong. Consider $C_1,C_2,C_3$ to be the events of throwing a head on, respectively, the first, second, and third of three fair coins, and $X$ to be the event of throwing exactly two heads among those three coins. So $C_1,C_2,C_3$ are mutually independent events (although clearly not conditionally independent). Now $\mathsf P(X\mid C_1,C_2,C_3)=0$ and $\mathsf P(X\mid C_i)=1/2$ for each $i\in\{1,2,3\}$. As $0\neq 1/8$, we see that we do not have $\mathsf P(X\mid C_1,C_2,C_3) = \mathsf P(X\mid C_1)~\mathsf P(X\mid C_2)~\mathsf P(X\mid C_3)$. $\endgroup$ – Graham Kemp Oct 30 '18 at 12:03

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