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Using: $$ \begin{align} P(A =1|B=1) &= 0.9\\ P(A =1|C=1) &= 0.9\\ P(A=1) &= 0.5 \end{align} $$ I want to know the probability of $P(A=1|B=1,C=1)$

I also know that $B$ and $C$ and independent as well as conditionally independent given A.

My approach was to use the following: $$ \begin{align} P(A|BC) &= \frac{P(BC|A)P(A)}{P(BC)}\\ &= \frac{P(B|A)P(C|A)P(A)}{P(B)P(C)} &&\qquad \text{using independence}\\ &= \frac{\frac{P(A|B)P(B)}{P(A)}\frac{P(A|C)P(C)}{P(A)}P(A)}{P(B)P(C)}&&\qquad \text{using Bayes' rule}\\ &=\frac{P(A|B)P(B)P(A|C)P(C)}{P(A)P(B)P(C)}\\ &=\frac{P(A|B)P(A|C)}{P(A)} \end{align} $$

Using the known probabilities: $$ \begin{align} P(A=1|B=1,C=1) &= \frac{P(A=1|B=1)P(A=1|C=1)}{P(A=1)}\\ &=\frac{0.9*0.9}{0.5}\\ &=1.62 \end{align} $$

But this is clearly wrong, what step is wrong here?

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  • $\begingroup$ See math.stackexchange.com/questions/1288770/… $\endgroup$
    – Tim
    Mar 30, 2018 at 16:37
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    $\begingroup$ @Tim I am using exactly that in the second line, but that gives me this formula. $\endgroup$
    – trevog
    Mar 30, 2018 at 16:42
  • $\begingroup$ Are you sure about your assumptions? It seems correct to me. $\endgroup$
    – Ale
    Mar 30, 2018 at 17:46

1 Answer 1

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If $B$ and $C$ are independent and also conditionally independent given $A$, then your calculation is correct, and the conclusion is valid whenever $P(A)>0$: $$P(A\mid BC) = \frac{P(A\mid B)P(A\mid C)}{P(A)}.\tag1 $$ Since the LHS of (1) is at most $1$, you can deduce the following constraint: $$ P(A)\ge P(A\mid B)P(A\mid C), $$ which is not satisfied by your conditions. So the given assumptions are inconsistent.

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  • $\begingroup$ Let's assume $P(B) = 0.0001$ and $P(C) = 0.0001$. It seems quite possible that $P(A|B) = P(A|C) = 0.9$ while $P(A) = 0.5$ overall. $\endgroup$
    – jbowman
    Mar 30, 2018 at 19:01
  • $\begingroup$ @jbowman No matter what $P(B)$ and $P(C)$ are, the OP's assignment of probabilities is not compatible with the additional independence constraints. $\endgroup$
    – grand_chat
    Mar 30, 2018 at 19:08
  • $\begingroup$ Ah, yes, I hadn't worked it through in detail. $\endgroup$
    – jbowman
    Mar 30, 2018 at 19:17

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