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In the examples considered so far, it has been reasonable to assume conditional independence, given an event C , and conditional independence, given the complementary event. But there are cases in which the effect ofthe conditioning event is asymmetric. We consider several examples.
Since conditional independence is ordinary independence with respect to a conditional probability measure, it should be clear how to extend the concept to larger classes of sets.
Definition . A class , where J is an arbitrary index set, is conditionally independent, given event C , denoted , iff the product rule holds for every finite subclass of two or more.
As in the case of simple independence, the replacement rule extends.
The replacement rule
If the class , then any or all of the events A i may be replaced by their complements and still have a conditionally independent class.
Since conditional independence is independence, we may use independence techniques in the solution of problems. We consider two types of problems: an inference problemand a conditional Bernoulli sequence.
Sharon is investigating a business venture which she thinks has probability 0.7 of being successful. She checks with five “independent” advisers. If theprospects are sound, the probabilities are 0.8, 0.75, 0.6, 0.9, and 0.8 that the advisers will advise her to proceed; if the venture is not sound, the respective probabilitiesare 0.75, 0.85, 0.7, 0.9, and 0.7 that the advice will be negative. Given the quality of the project, the advisers are independent of one another in thesense that no one is affected by the others. Of course, they are not independent, for they are all related to the soundness of the venture. We mayreasonably assume conditional independence of the advice, given that the venture is sound and also given that the venture is not sound. If Sharon goes with the majorityof advisers, what is the probability she will make the right decision?
Let E i be the event the i th adviser is positive. Then the sum of probabilities of the form , where M k are minterms generated by the class . Because of the assumed conditional independence,
with similar expressions for each and . This means that if we want the probability of three or more successes, given H , we can use ckn with the matrix of conditional probabilities. The following MATLAB solution of theinvestment problem is indicated.
P1 = 0.01*[80 75 60 90 80];P2 = 0.01*[75 85 70 90 70];PH = 0.7;
PE = ckn(P1,3)*PH + ckn(P2,3)*(1 - PH)PE = 0.9255 Often a Bernoulli sequence is related to some conditioning event H . In this case it is reasonable to assume the sequence and . We consider a simple example.
A race track regular claims he can pick the winning horse in any race 90 percent of the time. In order to test his claim, he picks a horse to win in each of ten races.There are five horses in each race. If he is simply guessing, the probability of success on each race is 0.2. Consider the trials to constitute a Bernoulli sequence.Let H be the event he is correct in his claim. If S is the number of successes in picking the winners in the ten races, determine for various numbers k of correct picks. Suppose it is equally likely that his claim is valid or that he is merely guessing. We assume two conditional Bernoulli trials:
Claim is valid: Ten trials, probability .
Guessing at random: Ten trials, probability .
Let number of correct picks in ten trials. Then
Giving him the benefit of the doubt, we suppose and calculate the conditional odds.
k = 0:10;
Pk1 = ibinom(10,0.9,k); % Probability of k successes, given HPk2 = ibinom(10,0.2,k); % Probability of k successes, given H^c
OH = Pk1./Pk2; % Conditional odds-- Assumes P(H)/P(H^c) = 1e = OH > 1; % Selects favorable odds
disp(round([k(e);OH(e)]'))
6 2 % Needs at least six to have creditability 7 73 % Seven would be creditable,
8 2627 % even if P(H)/P(H^c) = 0.1 9 94585
10 3405063
Under these assumptions, he would have to pick at least seven correctly to give reasonable validation of his claim.
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