# Bayes

advertisement ```Bayes Theorem
Motivating Example: Drug Tests
A drug test gives a false positive 2% of the time
(that is, 2% of those who test positive actually
are not drug users)
And the same test gives a false negative 1% of
the time (that is, 1% of those who test
negative actually are drug users)
If Joe tests positive, what are the odds Joe is a
drug user?
Insufficient information!
Suppose we know that 1% of the population
uses the drug?
Setting Up the Drug Test Problem
• Let T be the set of people who test positive
• Let D be the set of drug users
• These are events, and Pr(T|D) is the
probability that a drug user tests positive
• Pr(T|D) = .99 because the false negative rate
is 1%, that is, 99% of drug users test positive,
1% test negative
• We want to know: What is Pr(D|T)?
• This is a very different question: What is the
probability you are a drug user, given that you
test positive?
Bayes Theorem
Theorem: If Pr(A) and Pr(B) are both nonzero,
Pr(A | B) Pr(A)

Pr(B | A) Pr(B)
Proof. We know that
(*)Pr(A | B) Pr(B)  Pr(A  B)  Pr(B | A) Pr(A)
by the definition of conditional probability:
Pr(A and
B) similarly for Pr(B|A).
Pr(A | B) 
Pr(B)
,
Then divide the left and right sides of (*) by
Pr(B|A)∙Pr(B).
Bayes, v. 2
Pr(A) Pr(B | A)
Pr(A | B) 
Pr(B | A) Pr(A)  Pr(B | A) Pr(A)
This enables us to calculate Pr(A|B) using only the
absolute probability Pr(A) and the conditional
probabilities Pr(B|A) and Pr(B|&not;A).
Proof. We know that Pr(A | B) Pr(A)
Pr(B | A)

Pr(B)
Now multiply by Pr(B|A) and rewrite Pr(B) using the
law of total probability.
Drug Test again
• Suppose that a drug test has
– 2% false positives (that is, 2% of the people
who test positive are not drug users )
– 1% false negatives (1% of those who test
negative are drug users).
• Suppose 1% of the population uses drugs.
If you test positive, what are the odds you
are actually a drug user?
Drug test, cont’d
• Let D = “Uses drugs”
• Let T = “Tests positive”
Pr(D)  .01
Pr(T | D)  .02
Pr(T | D)  .99
• What is Pr(D|T)?
Pr(D)  .01
Pr(T | D)  .02
Pr(T | D)  .99
Pr(D) Pr(T | D)
Pr(D | T ) 
Pr(T | D) Pr(D)  Pr(T | D) Pr(D)
.01.99

 .33
.99 .01 .02 .99
• If you fail the drug test, there is only one
chance in three you are actually a drug user!
• How can this be? Think about it.
– Out of 1000 people there are 10 drug users and
990 non-users.
– Of those 990, 2% or almost 20 test positive.
– Almost all of the 10 users also test positive.
– So there are 2 non-users for every user, among
those who test positive!
FINIS
```