# Bayes' Theorem

## Theorem

Let $\Pr$ be a probability measure on a probability space $\struct {\Omega, \Sigma, \Pr}$.

Let $\condprob A B$ denote the conditional probability of $A$ given $B$.

Let $\map \Pr A > 0$ and $\map \Pr B > 0$.

Then:

- $\condprob B A = \dfrac {\condprob A B \, \map \Pr B} {\map \Pr A}$

### General Result

There are other more or less complicated ways of saying very much the same thing, all of which can be derived from the basic version with the help of other fairly elementary results.

For example:

Let $\set {B_1, B_2, \ldots}$ be a partition of the event space $\Sigma$.

Then, for any $B_i$ in the partition:

- $\condprob {B_i} A = \dfrac {\condprob A {B_i} \map \Pr {B_i} } {\map \Pr A} = \dfrac {\condprob A {B_i} \map \Pr {B_i} } {\sum_j \condprob A {B_j} \map \Pr {B_j} }$

where $\ds \sum_j$ denotes the sum over $j$.

## Proof

From the definition of conditional probabilities, we have:

- $\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$

- $\condprob B A = \dfrac {\map \Pr {A \cap B} } {\map \Pr A}$

After some algebra:

- $\condprob A B \, \map \Pr B = \map \Pr {A \cap B} = \condprob B A \, \map \Pr A$

Dividing both sides by $\map \Pr A$ (we are told that it is non-zero), the result follows:

- $\condprob B A = \dfrac {\condprob A B \, \map \Pr B} {\map \Pr A}$

$\blacksquare$

## Also known as

This result is also known as **Bayes' Formula**.

The formula:

- $\condprob A B \, \map \Pr B = \map \Pr {A \cap B} = \condprob B A \, \map \Pr A$

is sometimes called the **product rule for probabilities**.

## Examples

### Arbitrary Example $1$

Let **box $1$** contain $10$ good screws and $2$ unslotted screws.

Let **box $2$** contain $8$ good screws and $4$ unslotted screws.

Let a box be selected at random.

Let a screw chosen from that box prove to be unslotted.

What is the probability that it came from **box $2$**?

### Arbitrary Example $2$

Suppose that, in a population, $6$ out of every $1000$ people has an illness $X$.

It is known that:

- if a person has $X$, there is a $92 \%$ probability that a blood test will be positive for $X$
- if a person does not have $X$, there is a $0 \cdotp 5 \%$ probability that a blood test will be positive for $X$.

Let a person selected at random test positive for $X$.

What is the probability that this person actually has $X$?

## Source of Name

This entry was named for Thomas Bayes.

## Historical Note

**Bayes' Theorem** was published posthumously in $1763$.

## Sources

- 1986: Geoffrey Grimmett and Dominic Welsh:
*Probability: An Introduction*... (previous) ... (next): $\S 1.6$: Exercise $18$ - 1991: Roger B. Myerson:
*Game Theory*... (previous) ... (next): $1.2$ Basic Concepts of Decision Theory