Tutorial 12. Logical learning
Exercise sheetFor the last workgroup, we just have one exercise that illustrate an important concepts about Bayesian learning. The rest of the workgroup you can use to ask questions for the exam.
Baserate fallacy
The baserate fallacy is about a common misestimation concerning conditional probabilities.
A simple example concerns diagnostic tests for rare diseases.
Suppose that a certain disease has a prevalence of 1% in the population. This is already quite high, most “serious” infectious diseases have a way lower prevalence in the general population.
Now suppose that we have a test for the disease, which is 90% accurate in the following sense:

If you have the disease, the test is positive with 0.90 probability. This is called the test sensitivity .

If you don’t have the disease, the test is negative with 0.90 probability. This is called the test specificity .
Note that test sensitivity and specificity can differ, but we assume both are the same for simplicity.
a)
Using a language with only the propositional variables $$\mathsf{DISEASE},\mathsf{POSITIVE},$$ determine the probabilities given by the above description (it’s sufficient to determine the ones you need for b).
Hint: You can estimate the unconditional probability of the person having the disease as the prevalence of the disease in the population.
b)
Suppose we’re administering the test to a randomly selected person from the population. The test results positive. What is the probability that the person has the disease given this outcome?
Hint: Use the alternative formulation of Bayes theorem.
c)
Let’s take the Netherlands with around 18 million people as sample population? Interpreting probabilities as frequencies and work out what the numbers from before mean. This may help understand the situation.
Solution
a)
The information we’re given can be represented as follows:
 $Pr(\mathsf{DISEASE})=0.01$
 $Pr(\mathsf{POSITIVE}\mid\mathsf{DISEASE})=0.9$
 $Pr(\neg\mathsf{POSITIVE}\mid\neg\mathsf{DISEASE})=0.9$
From this we can derive a few probabilities that will be useful for the following question:
 $Pr(\neg\mathsf{DISEASE})=0.99$
 $Pr(\mathsf{POSITIVE}\mid\neg\mathsf{DISEASE})=0.1$
b)
We can use the following instance of the alternative Bayes theorem:
$$Pr(\mathsf{DISEASE}\mid \mathsf{POSITIVE})=$$ $$\frac{Pr(\mathsf{POSITIVE}\mid \mathsf{DISEASE})\times Pr(\mathsf{DISEASE})}{Pr(\mathsf{POSITIVE}\mid \mathsf{DISEASE})\times Pr(\mathsf{DISEASE})+Pr(\mathsf{POSITIVE}\mid\neg \mathsf{DISEASE})\times Pr(\neg \mathsf{DISEASE})}$$Plugging in the values from above, we get:
$$Pr(\mathsf{DISEASE}\mid \mathsf{POSITIVE})=$$ $$\frac{0.9\times 0.01}{(0.9\times 0.01)+(0.1\times 0.99)}=0.08\bar{3}$$
That is, a positive test for a randomly picked person should raise your credence in that person having the disease from 1% to about 8%. This may be surprising in is nowhere near the 90% reliability of the test.