# Conditional independence  (Page 2/4)

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$P\left(A|C\right)=P\left(A|BC\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{or,}\phantom{\rule{4.pt}{0ex}}\text{in}\phantom{\rule{4.pt}{0ex}}\text{another}\phantom{\rule{4.pt}{0ex}}\text{notation,}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{P}_{C}\left(A\right)={P}_{C}\left(A|B\right)$

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 $\left\{A,\phantom{\rule{0.166667em}{0ex}}B\right\}$ with respect to the conditional probability measure ${P}_{C}\left(·\right)=P\left(·|C\right)$ . Thus, $P\left(AB|C\right)=P\left(A|C\right)P\left(B|C\right)$ . For this example, we should also expect that $P\left(A|{C}^{c}\right)=P\left(A|B{C}^{c}\right)$ , so that there is independence with respect to the conditional probability measure $P\left(·|{C}^{c}\right)$ . Does this make the pair $\left\{A,\phantom{\rule{0.166667em}{0ex}}B\right\}$ 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 $P\left(B|A\right)>P\left(B\right)$ . Consider the following numerical case. Suppose $P\left(AB|C\right)=P\left(A|C\right)P\left(B|C\right)$ and $P\left(AB|{C}^{c}\right)=P\left(A|{C}^{c}\right)P\left(B|{C}^{c}\right)$ and

$P\left(A|C\right)=0.60,P\left(A|{C}^{c}\right)=0.20,P\left(B|C\right)=0.50,\phantom{\rule{0.277778em}{0ex}}P\left(B|{C}^{c}\right)=0.15,\text{with}P\left(C\right)=0.30.$

Then

$P\left(A\right)=P\left(A|C\right)P\left(C\right)+P\left(A|{C}^{c}\right)P\left({C}^{c}\right)=0.3200\phantom{\rule{5pt}{0ex}}P\left(B\right)=P\left(B|C\right)P\left(C\right)+P\left(B|{C}^{c}\right)P\left({C}^{c}\right)=0.2550$
$P\left(AB\right)=P\left(AB|C\right)P\left(C\right)+P\left(AB|{C}^{c}\right)P\left({C}^{c}\right)\phantom{\rule{21.68121pt}{0ex}}=P\left(A|C\right)P\left(B|C\right)P\left(C\right)+P\left(A|{C}^{c}\right)P\left(B|{C}^{c}\right)P\left({C}^{c}\right)=0.1110$

As a result,

$P\left(A\right)P\left(B\right)=0.0816\ne 0.1110=P\left(AB\right)$

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.

## Students and exams

Two students take exams in different courses, Under normal circumstances, one would suppose their performances form an independent pair. Let A be the event the first student makes grade 80 or better and B be the event the second has a grade of 80 or better. The exam is given on Monday morning. It is the fall semester. Thereis a probability 0.30 that there was a football game on Saturday, and both students are enthusiastic fans. Let C be the event of a game on the previous Saturday. Now it is reasonable to suppose

$P\left(A|C\right)=P\left(A|BC\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(A|{C}^{c}\right)=P\left(A|B{C}^{c}\right)$

If we know that there was a Saturday game, additional knowledge that B has occurred does not affect the lielihood that A occurs. Again, use of equivalent conditions shows that the situation may be expressed

$P\left(AB|C\right)=P\left(A|C\right)P\left(B|C\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(AB|{C}^{c}\right)=P\left(A|{C}^{c}\right)P\left(B|{C}^{c}\right)$

Under these conditions, we should suppose that $P\left(A|C\right) and $P\left(B|C\right) . If we knew that one did poorly on the exam, this would increase the likelihoood there was a Saturday game and hence increase thelikelihood that the other did poorly. The failure to be independent arises from a common chance factor that affects both. Although their performances are “operationally”independent, they are not independent in the probability sense. As a numerical example, suppose

$P\left(A|C\right)=0.7\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(A|{C}^{c}\right)=0.9\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(B|C\right)=0.6\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(B|{C}^{c}\right)=0.8\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(C\right)=0.3$

Straightforward calculations show $P\left(A\right)=0.8400,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(B\right)=0.7400,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(AB\right)=0.6300$ . Note that $P\left(A|B\right)=0.8514>P\left(A\right)$ as would be expected.

## 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

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