1
$\begingroup$

I have the following Bayes' Net: {D}->{C}<-{A}->{B}

I need to find P(A|B,C) with B and C both being true. I have calculated the probability of A given B using the formula

$$ P(A|B)=\frac{P(B|A)P(A)}{P(B)} $$

but I am unsure how to add C into this equation since it links to D.

Would I use the same method as described in Extended Bayes' theorem: p(A | B, C, D) - Constructing a Bayesian Network ?

The nodes in my diagram have the following values:

P(A)=0.01

P(D)=0.02

+---+---+----------+
| A | D | P(C|A,D) |
+---+---+----------+
| T | T | 0.9      |
| F | T | 0.5      |
| T | F | 0.8      |
| F | F | 0.05     |
+---+---+----------+
+---+---+--------+
| A | B | P(B|A) |
+---+---+--------+
| T | T | 0.6    |
| F | T | 0.02   |
+---+---+--------+
$\endgroup$
  • $\begingroup$ In general you can use: $$ P[A|B,C]=\frac{P[A,B,C]}{P[B,C]} =\frac{P[B,C|A]P[A]}{P[B,C]}$$ However, it seems like more information about the joint probabilities is needed to solve. In your Baye's Net, are you assuming the arrows represent conditional independence given all other events that are not associated with the arrows? For example, are you assuming that $P[B|A]=P[B|A,C]=P[B|A,C,D]=P[B|A,D]$? $\endgroup$ – Michael Mar 28 '16 at 4:19
  • $\begingroup$ This represents a disease network. So diseases A and D cause symptoms. Disease A causes B and C. Disease D causes symptom C but has no link to symptom B. A and D are conditionally dependent because in the problem C is known which makes them no longer independent. $\endgroup$ – Evilsithgirl Mar 28 '16 at 10:43
  • $\begingroup$ I am not sure what you are asking then. I am following the burglary example given in Ch.14 of AI: A Modern Approach 3rd ed (images.slideplayer.com/20/6042848/slides/slide_42.jpg ) . I have guessed based on all the information given (which I have provided here), that arrows represent conditional independence given all other events that are not associated with the arrows as you stated. In the book example, children of a parent node are independent of each other and this is the example the professor used in class for learning this material so I am assuming the same. $\endgroup$ – Evilsithgirl Mar 28 '16 at 16:27
0
$\begingroup$

You have $P(C)=0.02$ but I assume you mean $P(D)=0.02$. We can proceed as follows:

\begin{align} P(A\mid B,C) &= \dfrac{P(A,B,C)}{P(B,C)} = \dfrac{P(A,B,C)}{P(A,B,C) + P(A^c,B,C)}. \\ \end{align}

\begin{align} & \\ P(A,B,C) &= P(B,C\mid A)\cdot P(A) \\ &= P(B\mid A)\cdot P(C\mid A)\cdot P(A) \qquad\text{since $B,C$ are conditionally independent given $A$} \\ &= P(B\mid A)\cdot P(A)\cdot \left[P(C\mid D,A)\cdot P(D\mid A) + P(C\mid D^c,A)\cdot P(D^c\mid A)\right] \\ &= P(B\mid A)\cdot P(A)\cdot \left[P(C\mid D,A)\cdot P(D) + P(C\mid D^c,A)\cdot P(D^c)\right] \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{since $A,D$ are independent} \\ &= 0.6\times 0.01\times [0.9\times 0.02 + 0.8\times 0.98] \\ &= 0.004812. \\ \end{align}

Similarly, \begin{align} P(A^c,B,C) &= P(B,C\mid A^c)\cdot P(A^c) \\ &= P(B\mid A^c)\cdot P(C\mid A^c)\cdot P(A^c) \qquad\text{since $B,C$ are conditionally independent given $A^c$} \\ &= P(B\mid A^c)\cdot P(A^c)\cdot \left[P(C\mid D,A^c)\cdot P(D\mid A^c) + P(C\mid D^c,A^c)\cdot P(D^c\mid A^c)\right] \\ &= P(B\mid A^c)\cdot P(A^c)\cdot \left[P(C\mid D,A^c)\cdot P(D) + P(C\mid D^c,A^c)\cdot P(D^c)\right] \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{since $A^c,D$ are independent} \\ &= 0.02\times 0.99\times [0.5\times 0.02 + 0.05\times 0.98] \\ &= 0.0011682. \\ \end{align}

Therefore,

\begin{align} P(A\mid B,C) &= \dfrac{0.004812}{0.004812 + 0.0011682} \approx 0.80466. \\ \end{align}

$\endgroup$
  • $\begingroup$ Now, how do you know $B,C$ are conditionally independent given $A$? That was never specified. $\endgroup$ – Michael Mar 28 '16 at 16:00
  • $\begingroup$ @Michael Because $A$ is the parent of both $B$ and $C$. $\endgroup$ – Mick A Mar 28 '16 at 16:08
  • $\begingroup$ You seem to be assuming some structure associated with "parent" for such problems. For example, one might have a parent relationship with $B=C$, in which case $P[B,C|A]\neq P[B|A]P[C|A]$. Or we might have $B$ and $C$ being disjoint, in which case $P[B,C|A]=0 \neq P[B|A]P[C|A]$. $\endgroup$ – Michael Mar 28 '16 at 16:23
  • $\begingroup$ @Michael This is a Bayesian network. Do you know how they work? $\endgroup$ – Mick A Mar 28 '16 at 16:25
  • $\begingroup$ No, I do not know the assumed structure of such problems. My suspicion is that "we assume conditional independence for such problems." Else, they cannot be solved like this. It would be good to emphasize this as an assumption in the model, since this is not true in general. $\endgroup$ – Michael Mar 28 '16 at 16:26

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.