A. The dogs owner wants to be certain about a diagnosis so he takes a series of n identical medical tests. He wants multiple tests to reduce his uncertinity of his dogs disease. Events: D: Owners dog has the disease being tested for. Tj: Owners dog tests positive on the jth test for j = 1, 2, 3, …, n. Let p = P(D) be the prior probability that he has the disease. Assume for this part the test results are conditionally independent given the owners dog disease status. Let a0 of the n tests. Since a0 does not depend on j (the index of a test result), the algebra simplifies a lot since P(Ti | D) P(Tj | D)= a0^2 for any test indices I and j. The same holds for b0. B. Suppose some doges have a gene that makes them always test positive on this type of medical test. Let G the event that Omer has the gene. Assume that P(G) = 0.5 and that D and G are independent — that is, the gene does not make you more or less susceptible to the disease. If Omer has the gene, he’ll test positive on all n tests. If the owners dog does not have the gene, then the test results are conditionally independent given his disease status. Let a1 =P(Tj | D, G^c) and b1=P(Tj | D^c, G^c) where a1 and b1 do notdepend on j. Note: D^c=D complement, G^c=G complement. C. Now, suppose that the owners dog tests positive on all tests and find the posterior probability that the owners dog has the disease. 3. Using the same setup as in part B, find the posterior probability that the owners dog has the gene given that he tests positive on all n of the tests Show all calculations in steps (hand written solutions)
A.
The dogs owner wants to be certain about a diagnosis so he takes a series of n identical medical tests. He wants multiple tests to reduce his uncertinity of his dogs disease.
Events:
D: Owners dog has the disease being tested for.
Tj: Owners dog tests positive on the jth test for j = 1, 2, 3, …, n.
Let p = P(D) be the prior probability that he has the disease.
Assume for this part the test results are conditionally independent given the owners dog disease status. Let a0 of the n tests. Since a0 does not depend on j (the index of a test result), the algebra simplifies a lot since P(Ti | D) P(Tj | D)= a0^2 for any test indices I and j. The same holds for b0.
B.
Suppose some doges have a gene that makes them always test positive on this type of medical test. Let G the event that Omer has the gene. Assume that P(G) = 0.5 and that D and G are independent — that is, the gene does not make you more or less susceptible to the disease. If Omer has the gene, he’ll test positive on all n tests.
If the owners dog does not have the gene, then the test results are conditionally independent given his disease status. Let a1 =P(Tj | D, G^c) and b1=P(Tj | D^c, G^c) where a1 and b1 do notdepend on j. Note: D^c=D complement, G^c=G complement.
C.
Now, suppose that the owners dog tests positive on all tests and find the posterior probability that the owners dog has the disease. 3. Using the same setup as in part B, find the posterior probability that the owners dog has the gene given that he tests positive on all n of the tests
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