I'm learning Bayesian networks and I have to "guess" if the following conditional independence are true or false using the following table:

enter image description here

And the conditional independence are:

  1. $I_p(A, B)$
  2. $I_p(A, C)$

p is the probability distribution.
A = a, B = b, and C = c.

I have no idea about how to know or which operation do I have to do to know if those conditional independence are true or false.

What do I have to do to know if those conditional independence are true or false?


closed as unclear what you're asking by Dilip Sarwate, mdewey, Michael Chernick, Juho Kokkala, Jeremy Miles Jan 8 at 0:57

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  • $\begingroup$ What does the subscript $p$ in $I_p(A,B)$ mean? In fact, what do $A,B,C$ mean and how are they related to the $a,b,c$ in your table? $\endgroup$ – Dilip Sarwate Jan 4 at 18:52
  • $\begingroup$ @DilipSarwate I have updated my question. $\endgroup$ – VansFannel Jan 4 at 18:54
  • $\begingroup$ In 1, are you asking about the independence of $A$ and $B$ given $C$? And in 2, $A$ and $C$ given $B$? $\endgroup$ – gunes Jan 4 at 19:03
  • $\begingroup$ @gunes I don't know. This is something that I don't understand and I'm trying to figure it out what it means. The book that I'm using has a very strange notation. $\endgroup$ – VansFannel Jan 4 at 19:11
  • $\begingroup$ which book are you using, and what is the page number? $\endgroup$ – gunes Jan 4 at 19:12

It appears that both from your book, and Learning Bayesian Networks by Richard Neapolitan uses the same notations. I've never come across with the notation $I_P(..)$ while dealing with Bayesian Networks, but it turns out your book is not the only one. This notation simply means Independence in P (Neapolitan page ~29). For example, if $P(A|B)=P(A)$, then $I_P(A,B)$, which means $A$ and $B$ are independent. If $P(A|B,C)=P(A|C)$ (for all $a,b,c$), then $I_P(A,B|C)$, which means $A$ and $B$ are independent conditioned on C.

In your question, you ask for the conditional independences $I_P(A,B)$ and $I_P(A,C)$, but according to the definitions there ask for usual independence, not conditional, which can still be referred as conditional independence on nothing.

Anyway, either it is conditional or not, we have the full joint distribution and we can calculate any probability we want using the table. For example, a marginal and a two-variable joint can be calculated as follows: $$P(A=a)=\sum_b\sum_c{P(A=a,B=b,C=c)}$$ $$P(A=a,B=b)=\sum_c{P(A=a,B=b,C=c)}$$ A conditional probability can be calculated using Bayes Law: $$P(A=a|B=b)=\frac{P(A=a,B=b)}{P(B=b)}$$ $$P(A=a,B=b|C=c)=\frac{P(A=a,B=b,C=c)}{P(C=c)}$$

You just need to plug in the numbers and verify that $P(A,B)=P(A)P(B)$ for $I_P(A,B)$ for the first case, assuming it questions independence of $A$ and $B$ conditioned on nothing, based on the definitions in both books.

  • $\begingroup$ The question seems to be about conditional independence and not about independence, which is what your answer discusses. $\endgroup$ – Dilip Sarwate Jan 4 at 18:54
  • $\begingroup$ You are right! I'll edit my answer as soon as I see what $I_p$ is. $\endgroup$ – gunes Jan 4 at 19:02
  • $\begingroup$ Thanks a lot for your help. Now the question is how to calculate any probability using the table. $\endgroup$ – VansFannel Jan 5 at 8:35
  • $\begingroup$ It's implicitly in the answer. For showing $P(A=a,B=b)=P(A=a)P(B=b)$ is true or not, you need to evaluate four possible situations: $(a,b)$ in ${(0,0),(0,1),(1,0),(1,1)}$. You can use the first two formula above. $\endgroup$ – gunes Jan 5 at 11:04
  • 1
    $\begingroup$ @gunes I've found how to use the table!!! I'm happy! $\endgroup$ – VansFannel Jan 7 at 17:59

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