0
$\begingroup$

I am brushing up on graphical models, and doing the following problem 3.3 from the book PGM by Kophler.

An alarm A can be set off by either Burglary B or earthquake E. Prove that if $P(a^1| b^1, e^1) = P(a^1 | b^0, e^1) = 1$ then $P(b^1| a^1, e^1) = P(b^1)$.

From the result I can only derive the equivalent formula that $P(e^1) = P(a^1, e^1)$. How to go from here?

The derivation:

$P(b^1| a^1, e^1) = P(b^1)$

$P(b^1, a^1, e^1) = P(b^1) P(a^1, e^1)$

$P(a^1 | b^1, e^1) P(b^1) P(e^1) = P(b^1) P(a^1, e^1)$

$P(e^1) = P(a^1, e^1)$

$P(e^1) = P(a^1| b^1, e^1)P(b^1) + P(a^1 | b^0, e^1) P(b^0)$

$P(e^1) = P(b^1) + P(b^0) = 1$ wrong???

$\endgroup$
2
  • $\begingroup$ Your "derivation" is puzzling because it begins with what you are attempting to show, whereas logically (if not always narratively) derivations always begin with what you assume and end with what you want to show. There also do not seem to be clear connections between successive statements in your work. What reasons justify each of the equalities you have written? $\endgroup$
    – whuber
    Commented Jun 3, 2014 at 22:14
  • 1
    $\begingroup$ It is equivalent derivation, if and only if. I'm trying to get some correct result, then based on iff the proof is correct. $\endgroup$ Commented Jun 4, 2014 at 18:44

1 Answer 1

1
$\begingroup$

We need to assume that earth quakes are independent from burglaries, otherwise it won't follow. I'm not willing to grant this highly dubious assumption. Smart burglars would clearly use the chaos caused by an earthquake to make their move!

Anyway, let's start with Bayes:

$P(B=1|A=1, E=1) = [P(A=1,E=1|B=1)/P(A=1,E=1)]*P(B=1)$

You want to show that the term in square brackets is 1.

The assumption says that conditional on there being an earthquake, burglary is independent from the alarm. Hence $P(A=1|E=1)=1$

Then $P(A=1,E=1)=P(A=1|E=1)*P(E=1)=P(E=1)$... if the earthquake sets the alarm of with certainty the probability of both happening is just the probability an earthquake happens..makes sense.

Now the numerator: $P(A=1,E=1|B=1)=P(E=1|B=1)*P(A=1|E=1,B=1)$. That's the same formula that was used above the find the probability of an intersection, except everything's conditional on $B=1$.

The second term in the product is 1. The first term needs to be $P(E=1)$ which requires burglaries and earthquakes to be independent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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