Lesson · Knowledge and Evidence · Credence

Conditional Credence and Base Rates

A conditional credence is a credence on the supposition that something else is true, and it is a ratio of two ordinary ones. Counting a million people shows that the chance of a positive result given the condition is a different number from the chance of the condition given a positive result, and that the gap is set by how rare the condition is.

Open the interactive lesson Practice

P(A | B) = P(A ∧ B) / P(B) P(+ | D) is not P(D | +) rare condition: most positives are false count a million people, then divide

Two conditionals, two questions

Core ideas

What this lesson establishes

Evidence arrives as a result of a test, a report or an observation. What it does to a credence is a conditional credence, and the first job is to keep its two directions apart.

A conditional credence is a ratio

The credence in A on the supposition that B is true is the credence in both, divided by the credence in B. It is defined only when B has credence above 0.

The two directions answer different questions

How likely a positive result is for someone who has the condition, and how likely the condition is for someone with a positive result, have different denominators. They are equal only by coincidence.

The base rate sets the gap

The proportion of people who have the condition before any test is the base rate. When it is small, the false positives from the many healthy people can outnumber the true positives from the few sick ones.

The material

A test, a condition and a million people

The same arithmetic done twice: as counts of people, then as a formula.

Almost every grandmother is a woman, and very few women are grandmothers. The two statements are about different groups, and nothing forces their proportions to match. A conditional credence makes the difference exact.

Definition · Conditional credence

The conditional credence in A given B is P(A | B) = P(A ∧ B) / P(B), defined when P(B) is above 0. Credences are written with P, like the probabilities whose rules they obey. It is the credence you would hold in A if you learned B and nothing else, and the direction of the bar matters: P(A | B) and P(B | A) are different numbers.

Now a test for a condition. One person in 1,000 has it. The test is positive for 99 in 100 of those who have it, and positive for 5 in 100 of those who do not. A person tests positive. The question is how likely it is that they have the condition, and the easiest way to answer it is to count a million people.

has it does not total ------------------------------------------- tests + 990 49950 50940 tests − 10 949050 949060 total 1000 999000 1000000

Of the 1,000 who have the condition, 99 in 100 test positive, which is 990, and the other 10 test negative. Of the 999,000 who do not, 5 in 100 test positive, which is 49,950, and the remaining 949,050 test negative. Everyone who tests positive is in the first row: 990 plus 49,950, which is 50,940 people.

P(D | +) = 990 / 50940 = 11/566, about 1.9%

Fewer than 2 in 100 of the people with a positive result have the condition. The number 99 in 100 is P(+ | D), the proportion of the 1,000 who test positive, and the question asked for P(D | +), the proportion of the 50,940 who have it. The two share a numerator and have different denominators.

The reason is the size of the second column. There are 999 healthy people for every sick one, so a false positive rate of 5 in 100 produces about fifty false positives for each true one. A very accurate test applied to a rare condition mostly produces false alarms, and the mistake of reading P(+ | D) as P(D | +) is called base-rate neglect.

The base rate is an input, not background

Hold the test fixed and change who is tested. If 30 in 100 of those tested have the condition, the same test gives P(D | +) = 297/332, about 89%. Nothing about the test has changed. The people have, and a positive result is worth what the population makes it worth.

A negative result is the other half of the table. Of the 949,060 who test negative, 10 have the condition, so P(D | −) is 1/94906. In this population a negative result is far more conclusive than a positive one, which is not what the phrase “99% accurate” suggests.

The lab states its limit plainly: it divides the counts it is given. Whether 1 in 1,000 is the right base rate for the person in front of you, who may have been tested because of symptoms, is a question about which group they belong to, and a later lesson takes it up.

Interactive

Bayes calculation

A million people, four whole-number cells

The test, counted

The lab opens on this lesson’s test: 1 in 1,000 have the condition, 99% of them test positive and 5% of the others do too. The table is the one above and the bar shows how small the true-positive share is. Choose “30 in 100” from the prevalence menu to see the same test in a different population, then move the false positive rate to 0 to see what it would take to remove the problem.

Prevalence P(D), the prior

sensitivity P(+ | D)99%

false positive rate P(+ | not D)5%

P(D), prior—

P(D | +), posterior—

false : true positives—

Method

From a test result to P(D | +)

The order that avoids the usual mistake: counts first, ratio last.

Take a round population

A million people, so that every cell is a whole number. The prior tells you how many have the condition.

Split it by the condition

Those who have it and those who do not. Do not skip the second group; it is the larger one.

Apply the test to each group separately

The sensitivity applies to the first group and the false positive rate to the second. Neither is applied to the whole population.

Collect the positives

Add the true positives and the false positives. This total, not the number with the condition, is the denominator.

Divide

True positives over all positives is P(D | +). Check it against the formula: the same two numbers appear.

Which conditional, and how many

Score 0 / 0

Choose the answer you can justify.

A positive result, counted

One person in 1,000 has the condition; sensitivity 99 in 100; false positives 5 in 100.

have it: 1,000 positive 990 do not: 999,000 positive 49,950 all positives: 990 + 49,950 = 50,940 P(D | +) = 990 / 50,940 = 11/566 P(+ | D) = 990 / 1,000 = 99/100

The last two lines have the same numerator and different denominators, and that is the whole difference between them. Fewer than 2 in 100 of the people with a positive result have the condition.

Common mistakes

Reading P(+ | D) as P(D | +)The two have the same numerator, 990, and different denominators: 1,000 people who have the condition for the first, 50,940 people who tested positive for the second. For this test one is 99/100 and the other is 11/566. A statement about how often the test finds the condition is not a statement about how often a positive result is right.

Treating the base rate as backgroundNothing about the test changes between the two populations, yet P(D | +) moves from 11/566 to 297/332. The base rate is as much an input as the sensitivity. Leaving it out is not a simplification; it replaces the question with a different one.

Applying the number to any person who tested positiveThe figure 11/566 is true of the population of 1,000,000 as described. A patient who was tested because of symptoms is not drawn from that population, and the lab has no way to know it. The calculation is exact about the group it is given and silent about which group a person belongs to.

Check your understanding

✓

Finish when you can fill the table before you divide.

You should be able to take a base rate, a sensitivity and a false positive rate, count a population into the four cells, and compute P(D | +) as a ratio of two of them. Stating the answer is not enough; say which two counts it is the ratio of and why the other direction has a different denominator.

Note: The condition and the test here are invented to make the arithmetic clean. Real screening programmes have numbers of exactly this shape, and the same counting is how they are evaluated.