\(\displaystyle P(A\vert B) = P(A)\cdot\frac{P(B\vert A)}{P(B)}\qquad\qquad(1)\)
If event \(A\) means “have a disease”, and event \(B\) means “have a positive test”, then, knowing that a person has a positive test (”given \(B\)“), how to calculate the probability \(P(A\vert B)\) of this person having a disease (”probability of \(A\) given \(B\)”) ?
Well, from the Bayes equation \((1)\), it’s the probability \(P(A)\) of a random person (from the set of all people) having a disease (a.k.a. the “prior” probability), multiplied by \(\displaystyle \frac{P(B\vert A)}{P(B)}\).
In practice, calculating \(\displaystyle \frac{P(B\vert A)}{P(B)}\) is not straightforward — one needs to know \(P(B)\), i. e. the total probability of getting a positive test.
Instead, if for the given test, we know two values: the sensitivity (true positive rate, or TPR) and specificity (true negative rate, or TNR), then it’s easier to use the equation \((2)\), which operates in terms of odds:
\(\displaystyle O(A\vert B) = O(A)\cdot \frac{P(B\vert A)}{P(B\vert \neg A)}\qquad\qquad(2)\)
If there’s an event \(A\) in the universe \(U\), then the probability of \(A\) is given by: \(\displaystyle P(A) = \frac{\vert A\vert}{\vert U\vert}\)
For another event \(B\), the probability is: \(\displaystyle P(B) = \frac{\vert B\vert}{\vert U\vert}\)
If events \(A\) and \(B\) intersect, e. g. we know we’re in the region \(B\), what is the probability of being in the intersection (\(A\) “given \(B\)”)? Well obviously it’s the size of the intersection divided by the size of \(B\): \(\displaystyle P(A\vert B) = \frac{\vert A\cap B\vert}{\vert B\vert}\)
The same goes for probability of \(B\) “given \(A\)”: \(\displaystyle P(B\vert A) = \frac{\vert A\cap B\vert}{\vert A\vert}\)
The numerator is the same, so this follows: \(\displaystyle P(A\vert B)\cdot\vert B\vert = P(B\vert A)\cdot\vert A\vert\)
Using the first two equations to substitute \(\vert A\vert\) and \(\vert B\vert\) with their probabilities in \(U\), we get the Bayes equation \((1)\): \(\displaystyle P(A\vert B)\cdot P(B) = P(B\vert A)\cdot P(A)\)
The equation \((2)\) directly follows from:
Odds of having a disease are: \(\displaystyle O(A) = \frac{\vert A\vert }{\vert \neg A\vert }\) (e. g. when \(P(A) = 0.1\), the odds are “1 to 9”, or \(\displaystyle \frac{1}{9}\)). Here, \(A\) is the set of people having the disease and \(\neg A\) is the set of people not having the disease.
Odds of having a disease given a positive test is then (decreasing sizes of the sets in both numerator and denominator by the corresponding probabilities of getting a positive test result):
\(\displaystyle O(A\vert B) = \frac{\vert A\vert \cdot P(B\vert A)}{\vert \neg A\vert \cdot P(B\vert \neg A)} = O(A)\cdot \frac{P(B\vert A)}{P(B\vert \neg A)}\)
The second term is known as the Bayes factor, where \(P(B\vert A)\) is the test sensitivity, and \(P(B\vert \neg A)\) is the test’s false positive rate (FPR, or 1-specificity). The first term \(O(A)\) is known as the “prior” odds.
Say the prior odds of having a disease are 1 to 99 (i. e. the probability is 1%).
If a test has 90% sensitivity (TPR = 90%) and 90% specificity (FPR = 10%), then its Bayes factor is 9.
Meaning that having a positive test result updates the prior odds by the factor of 9, and the odds of having a disease given a positive test become 9 to 99.
Sensitivity, recall, true positive rate (TPR), probability of detection, hit rate, power
\(\displaystyle =\frac {TP}{P}\)
(the number of true positives divided by number of positives)
Specificity, true negative rate (TNR), selectivity
\(\displaystyle =\frac {TN}{N}\)
(the number of true negatives divided by number of negatives)
Precision, positive predictive value (PPV)
\(\displaystyle =\frac {TP}{PP}\)
(the number of true positives divided by predicted positives)
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