[1] "4.5.1-3, 4.5.9, 4.5.15"Lecture 10
Bayes’ Rule
Section 4.5 Bayes’ Rule
- Remember the experiment on color blindness:
| Men(\(B\)) | Women(\(B^c\)) | Total | |
|---|---|---|---|
| Colorblind (\(A\)) | .04 | .002 | .042 |
| Not Colorblind (\(A^c\)) | .47 | .488 | .958 |
| Total | .51 | .49 | 1 |
- Event \(B\): person is a man
- Event \(B^c\): person is a woman
Taken together it’s the full sample space and are mutually exclusive.
What if we want to describe all color blind people? We have two types, those in \(B\) and those in \(B^c\)
How do we represent color blind people in \(B\)?
- \(A \cap B\)
How do we present color blind people in \(B^c\)?
- \(A \cap B^c\)
How do we represent all color blind people:
\((A \cap B) \cup (A \cap B^c)\)
Using the addition rule:
\[ P(A) = P(A \cap B) + P(A \cap B^c) \]
Why is:
\[ P((A \cap B) \cap (A \cap B^c)) = 0 \]
Let’s now assume there are more than 2 mutually exclusive and exhaustive categories - we can extend the rule using k populations \(S_1, S_2, S_3, \dots, S_k\).
\[ A = (A \cap S_1) \cup (A \cap S_2) \cup (A \cap S_3) \cup \dots \cup (A \cap S_k) \]
Then \[ P(A) = P(A \cap S_1) + P(A \cap S_2) + P(A \cap S_3) + \dots + P(A \cap S_k) \]
- Show venn diagram that is on page 159
Remember also that:
\[ P(A \cap S_i) = P(S_i)P(A|S_i) \]
The Law of Total Probability
\[ P(A) = P(S_1)P(A|S_1) + P(S_2)P(A|S_2) + P(S_3)P(A|S_3) + \dots + P(S_k)P(A|S_k) \]
Bayes’ Rule
One useful use of this is when you need to find the conditional probability that A occurred given B \(P(A|B)\), but you know \(P(B|A)\). This is common with screening tests.
- Event \(D\) is that you have the disease
- Event \(T\) is that the test is positive
A test is often characterized by the true positive rate \(P(T|D)\), true negative rate \(P(T^c|D^c)\), false positive rate \(P(T|D^c)\) and the false negative rate \(P(T^c|D)\). You can see all these values depend on a state of disease. This is because when they are creating a new test, they can find people with the disease and without the disease - i.e. they are known to have the disease.
- However, what we really care about is the probability you have the disease given a positive test \(P(D|T)\).
Let’s imagine a scenario where you have a test that is pretty good when it detects \(P(T|D) = .99\) and doesn’t have a ton of false positives \(P(T|D^c) = .01\). Does this seem like a good test?
The answer is that it depends. Let’s see why:
Recall that:
\[ P(D)(P(T|D)) = P(D\cap T) = P(T)P(D|T) \]
Then:
\[ P(D|T) = \frac{P(T|D)P(D)}{P(T)} \]
Where
\[ P(T) = P(D)P(T|D) + P(D^c)P(T|D^c) \]
Plugging in for \(P(T)\):
\[ P(D|T) = \frac{P(T|D)P(D)}{P(D)P(T|D) + P(D^c)P(T|D^c)} \]
What can we fill in?
\(P(T|D) = .99\) and \(P(T|D^c) = .01\)
\[ P(D|T) = \frac{.99P(D)}{.99P(D) + .01P(D^c)} \]
Let’s say we are trying to detect the common cold. What if it’s the winter? There is a lot of cold out there, assuming you are feeling sick then there is a 40% chance it’s the cold.
Therefore: \(P(D) = .4\) and \(P(D^c) = 1 - .4 = .6\) So what’s the probability that you have the cold given there is a positive test?
\[ P(D|T) = \frac{.99 \times .4}{.99 \times .4 + .01 \times .6} = 0.985 = 98.5\% \]
However, what if it’s the summer - there isn’t a lot of cold going around out there. Not a lot of people have the cold so \(P(D) = .05\) and \(P(D^c) = .95\).
\[ P(D|T) = \frac{.99 \times .05}{.99 \times .05 + .01 \times .95} = 0.839 = 83.9\% \]
The posterior probability \(P(D|T)\) changes even though the test information is the same. It changes because the prior (\(P(D)\)) matters.
This can be generalized to:
\[ P(S_i|A) = \frac{P(S_i)P(A|S_i)}{\sum_{j=1}^kP(S_j)P(A|S_j)} \]
for \(i = 1, 2, \dots, k\)
Homework
Answers: Chapter 4 - Section 5
Midterm Review
- 1.4.21, 1.4.22, 1.4.25
- 2.1.4, 2.1.9, 2.1.21
- 2.2.10
- 2.3.1, 2.3.6-14, 2.3.27
- 4.1.13-15
- 4.2.11-15, 4.2.38, 4.2.40,
- 4.3.20, 4.3.25
- 4.4.16-18, 4.4.30, 4.4.36
- 4.5.7, 4.5.15
Answers: Midterm 1 Answers