Given $A,B,C$ such that: $$ P(A\mid B),P(A\mid B^c),P(B\mid C),P(B\mid C^c) \text{ are known } $$ and that $A,C$ are conditionally independent given $B$, so that: $$ P(A\mid B\cap C)=P(A\mid B),P(A\mid B^c\cap C)=P(A\mid B^c) $$ (the equations above are not the definition of conditionally independent, but follow from it)

I want to obtain $P(A\mid C),P(A\mid C^c)$.

I've managed to show that: $P(A\mid C)=P(A\mid B)P(B\mid C)+P(A\mid B^c)P(B^c\mid C)$. But I'm stuck on what to do with $P(B^c\mid C)$. Any ideas?

Thanks for helping! :DDDD

  • $\begingroup$ P(Bc|C)=1-P(B|C). $\endgroup$ – Did Aug 27 '15 at 23:49
  • $\begingroup$ Indeed! Thanks! $\endgroup$ – Guilherme Salomé Aug 28 '15 at 0:55

In order to calculate $P(B^c\mid C)$ observe that: $$ P(B^c\mid C)+P(B|C)=\frac{P(B^c\cap C)}{P(C)}+\frac{P(B\cap C)}{P(C)}=\frac{P((B^c\cap C)\cup(B\cap C)}{P(C)}=\frac{P(C)}{P(C)}=1 $$ So that $P(B^c\mid C)=1-P(B\mid C)$ and $P(B\mid C)$ is known. The same applies to $P(B^c\mid C^c)=1-P(B\mid C^c)$, which is used to obtain $P(A|C^c)$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.