9
$\begingroup$

Below is the scheme of conditional dependence and the probabilities of events:

enter image description here

P(A=1) = 0.01
P(A=0) = 0.99
P(B=1) = 0.1
P(B=0) = 0.9
P(C=1|A=0,B=0) = 0.1
P(C=1|A=0,B=1) = 0.5
P(C=1|A=1,B=0) = 0.6
P(C=1|A=1,B=1) = 0.9

Given the probabilities above I wanted to calculate P(B=1|C=1) and P(B=1|C=1,A=1) but didn't get the correct result.

I wrote the probabilistic function the following way:

P(A, B, C) = P(A)P(B)P(C|A, B)

and then set the variables

P(B=1, C=1) = P(A=0, B=1, C=1) + P(A=1, B=1, C=1)=
=P(A=0)P(B=1)P(C=1|A=0, B=1) + P(A=1)P(B=1)P(C=1|A=1, B=1)=
=0.99*0.1*0.5 + 0.01*0.1*0.9 = 0.0495

The result however is not correct and don't know where is the error. I would be very thankful if anyone could correct/explain what's wrong.

$\endgroup$
4
  • $\begingroup$ You said you were asked P(B=1|C=1) but you computed P(B=1,C=1) hence you are off by the factor P(C=1) (which you will need to compute). $\endgroup$
    – Did
    Commented Mar 30, 2012 at 11:06
  • 1
    $\begingroup$ Is P(C=1) = ( P(C=1|A=0,B=0)+P(C=1|A=0,B=1)+P(C=1|A=1,B=0)+P(C=1|A=1,B=1) )/4 ? $\endgroup$ Commented Apr 1, 2012 at 11:53
  • $\begingroup$ Maybe because 0.99*0.1*0.5 + 0.01*0.1*0.9 = 0.0504 (and not 0.0495) ? $\endgroup$ Commented Oct 9, 2016 at 12:10
  • $\begingroup$ math.stackexchange.com/questions/884728/… $\endgroup$ Commented Apr 1, 2017 at 7:03

1 Answer 1

6
$\begingroup$

The typical way I do inter-causal reasoning is to flip the conditional probabilities around --

$$ \begin{align} P(B = 1 \vert C = 1) & = \frac{P(B = 1, C = 1)}{P(C = 1)} \\ & = \frac{P(C = 1 \vert B = 1)P(B = 1)}{P(C = 1)} \\ \\ P(B = 1 \vert C = 1, A = 1) & = \frac{P(B = 1, C = 1, A = 1)}{P(C = 1, A = 1)} \\ & = \frac{P(C = 1 \vert B = 1, A = 1)P(B = 1)P(A = 1)}{P(C = 1, A = 1)} \\ & = \frac{P(C = 1 \vert B = 1, A = 1) P(B = 1) P(A = 1)}{P(C = 1 \vert A = 1)P(A = 1)} \\ & = \frac{P(C = 1 \vert B = 1, A = 1) P(B = 1)}{P(C = 1 \vert A = 1)} \end{align} $$

Does that help?

$\endgroup$
3
  • 1
    $\begingroup$ And how do you calculate P(C=1) ? Is it ( P(C=1|A=0,B=0) + P(C=1|A=0,B=1) ) * ( P(C=1|A=1,B=0) + P(C=1|A=1,B=1) ) ? $\endgroup$
    – jaor
    Commented Jul 30, 2014 at 13:02
  • $\begingroup$ No. That would give you P(C) > 1 $\endgroup$ Commented Oct 10, 2016 at 10:04
  • $\begingroup$ This is very old, but how is it that $P(B = 1, C = 1, A = 1) = P(C = 1 \vert B = 1, A = 1)P(B = 1)P(A = 1)$ in the second half? More specifically, why are we assuming that $A$ and $B$ are independent, when it's the case that variable $C$ is observed? As far as I know, in a $v$-structure, the middle variable being observed now means that the two parents are dependent. $\endgroup$
    – Sean
    Commented Dec 12, 2020 at 12:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .