3
$\begingroup$

enter image description here

Are A and B conditionally independent given the class label?

I calculated that

$$P(A=1) = \frac{1}{2}$$ $$P(B=1) = \frac{2}{5}$$ $$P(A=1,B=1)=\frac{1}{5}$$

My answer is yes. I do it by anding $(A\text{ and }B)$ which shows that when $A = 1$ and $B = 1$ it doesn't imply Class will be +. How can I actually prove this without this pseudo prove I have come up with?

$\endgroup$
0
$\begingroup$

A and B are conditionaly independent of C iff for every value of C, $P(A\cap B|C) = P(A|C)P(B|C)$

Lets try class = +.

$P(A\cap B|+)=\frac{P(A\cap\ B \cap +)}{P(+)} = \frac{1}{5} \neq P(A|+)P(B|+) = \frac{3}{5}\frac{2}{5} = \frac{6}{25}$

So no, they are not.

$\endgroup$
-1
$\begingroup$
  1. Marginalize to get the mass functions $p_{A \mid CL}(a \mid cl),p_{B \mid CL}(b \mid cl)$ and $p_{A,B \mid CL}(a,b \mid cl).$
  2. Check for every class label $CL=cl$ and $\forall a,b$ if the following holds (this is the definition of conditional independence of independence in general):
    $p_{A,B \mid CL}(A=a,B=b \mid CL=cl) = p_{A \mid CL}(A=a \mid CL=cl)p_{B \mid CL}(B=b \mid CL=cl)$
    If it holds, then $A,B$ are conditionally independent of the class label $CL$
$\endgroup$
  • $\begingroup$ Can you explain your answer a little more? I would appreciate it. $\endgroup$ – Mike John Nov 22 '13 at 2:52
  • $\begingroup$ Is there a reason for the downvote? $\endgroup$ – Sudarsan Nov 22 '13 at 23:23

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.