Bayes' Theorem

${\displaystyle P(A\mid B)={\frac {P(B\mid A)\,P(A)}{P(B)}},}$

where ${\displaystyle A}$ and ${\displaystyle B}$ are events and ${\displaystyle P(B)\neq 0}$.

The basic notion is that if we believe that the expectation of an event of interest (A) is influenced by another event (B), we can improve pn the simple probability expectation of A by incorporating information about the B event. Thus we get an improved likelihood expectation of A.

${\displaystyle P(A\mid B)}$ is the likelihood of event ${\displaystyle A}$ occurring, given that ${\displaystyle B}$ is true. This is called a conditional probability.
${\displaystyle P(B\mid A)}$ is the likelihood of event ${\displaystyle B}$ occurring, given that ${\displaystyle A}$ is true. This is also a conditional probability.
${\displaystyle P(A)}$ and ${\displaystyle P(B)}$ are of observing ${\displaystyle A}$ and ${\displaystyle B}$, independently of each other. These are called marginal probabilities.

Example: Drug testing

    Suppose a blood test used to detect the presence of a particular banned sports drug is 99% sensitive and 99% specific. That is, the test will produce 99% true positive results for drug users and 99% true negative results for non-drug users. Suppose that 0.5% of athletes are users of the drug. What is the likelihood that a randomly selected athlete who tests positive is a user? Intuitively, this is the sensitivity of the test in the numerator, P = 0.99. However, we also know that the test is sometimes non-specific and returns a false positive, at a rate (1.00 - 0.99) = 0.01 in the denominator. Then:

${\displaystyle {\begin{aligned}P({\text{User}}\mid {\text{+}})&={\frac {P({\text{+}}\mid {\text{User}})P({\text{User}})}{P(+)}}\\&={\frac {P({\text{+}}\mid {\text{User}})P({\text{User}})}{P({\text{+}}\mid {\text{User}})P({\text{User}})+P({\text{+}}\mid {\text{Non-user}})P({\text{Non-user}})}}\\[8pt]&={\frac {0.99\times 0.005}{0.99\times 0.005+0.01\times 0.995}}\\[8pt]&\approx 33.2\%\end{aligned}}}$

    Even if an individual tests positive, it is more likely than not (1 - 33.2% = 66.8%) that they do not use the drug. Why? Even though the test appears to be highly accurate, the number of non-users is very large compared to the number of users. Then, the count of false positives will be greater than the count of true positives.

    To see this with actual numbers: In a test group of 1,000 individuals, we expect 995 non-users and 5 users. Among the 995 non-users, 0.01 × 995 ≃ 10 false positives are expected. Among the 5 users, 0.99 × 5 ≈ 5 true positives are expected. Out of 15 positive results, only 5 (~33%), are genuine.

    The importance of specificity in this example can be seen by calculating that even if sensitivity is improved to 100%, but specificity remains at 99%, then the probability that a person who tests positive is a drug user only rises only very slightly, from 33.2% to 33.4%. Alternatively, if sensitivity remains 99%, but specificity is improved to 99.5%, then the probability that a person who tests positive is a drug user rises to about 49.9%.

HOMEWORK

    1) Prove the statements in the last paragraph about the consequences of changing specificity & sensitivity.

    1) IOC decisions must allow for athletes who positive but protest their innocence. We hear in the news that "The test is being redone:"
            (A) What is the a priori probability of two false positives in a row?
            (B) What are the implications if all athletes are required to take two tests?

    2) Suppose that education and better screening in the home country reduces the fraction of users by one-half (0.25%). How will this modify the Bayes estimate that a positive test identifies a user?