Conditional independence  (Page 2/4)

An examination of the sixteen equivalent conditions for independence, with probability measure P replaced by probability measure P _C , shows that we have independence of the pair $\{A,B\}$ with respect to the conditional probability measure $P_C(·)=P(·|C)$ . Thus, $P\left(AB\right|C)=P(A\left|C\right)P\left(B\right|C)$ . For this example, we should also expect that $P\left(A\right|C^c)=P\left(A\right|B{C}^c)$ , so that there is independence with respect to the conditional probability measure $P(·|C^c)$ . Does this make the pair $\{A,B\}$ independent (with respect to the prior probability measure P )? Some numerical examples make it plain that only in the most unusual cases would thepair be independent. Without calculations, we can see why this should be so. If the first customer buys an umbrella, this indicates a higher than normal likelihoodthat the weather is rainy, in which case the second customer is likely to buy. The condition leads to $P\left(B\right|A)>P(B)$ . Consider the following numerical case. Suppose $P\left(AB\right|C)=P(A\left|C\right)P\left(B\right|C)$ and $P\left(AB\right|C^c)=P\left(A\right|C^c)P\left(B\right|C^c)$ and

Then

As a result,

The product rule fails, so that the pair is not independent. An examination of the pattern of computation shows that independence would require very specialprobabilities which are not likely to be encountered.

Sixteen equivalent conditions

Using the facts on repeated conditioning and the equivalent conditions for independence , we may produce a similar table of equivalent conditions for conditional independence. In the hybrid notation we use for repeatedconditioning, we write