From $P(A\mid B)$ and $P(B\mid C)$ can you compute $P(A\mid C)$?

Is this statement true? I couldn't find any formulae to that would relate these events. Bayes theorem certainly wasn't helpful here...

  • 3
    $\begingroup$ To see this is impossible, consider the case where $P(B) = 1$. Then $P(A|B) = P(A)$ and $P(B|C) = 1$, but $P(A|C)$ could be anything. $\endgroup$
    – fblundun
    Jan 15 '21 at 12:36

You can only if you assume $A$ is conditionally independent of $C$ given $B$, i.e. $P(A | B, C) = P(A | B)$. In other words, once you know the value of $B$, the outcome of $C$ tells you nothing further about $A$.

This conditional independence assumption allows: \begin{align} P(A | C ) & = \sum_{b\in\Omega_{B}} P( A | B = b, C) P(B = b | C) & \text{Law of total probability} \\ & = \sum_{b\in\Omega_{B}} P( A | B = b) P(B = b | C) & (A \perp\!\!\!\perp C)~ |~ B \end{align}

where $\Omega_B$ is the set of possible outcomes of $B$.

  • $\begingroup$ But this assumes that $P(A|B^c)$ is known which it isn't - we only know $P(A^c|B)=1-P(A|B)$. $\endgroup$ Jan 15 '21 at 11:54
  • $\begingroup$ I'm interpreting $A$, $B$, and $C$ from the original post to refer to random variables, not specific outcomes, in which case known $P(A | B)$ and $P(B | C)$ would be known conditional probability tables. $\endgroup$
    – jkpate
    Jan 15 '21 at 12:04

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