# Bayes' theorem

Bayes' theorem is a result in probability theory, which relates the conditional and marginal probability distributions of random variables. In some interpretations of probability, Bayes' theorem tells how to update or revise beliefs in light of new evidence.

The probability of an event A conditional on another event B is generally different from the probability of B conditional on A. However, there is a definite relationship between the two, and Bayes' theorem is the statement of that relationship.

As a formal theorem, Bayes' theorem is valid in all interpretations of probability. However, frequentist and Bayesian interpretations disagree about the kinds of variables for which the theorem holds. The articles on Bayesian probability and frequentist probability discuss these debates at greater length.

## Statement of Bayes' theorem

Bayes' theorem relates the conditional and marginal probabilities of stochastic events A and B:

$P(A|B) = \frac{P(B | A)\, P(A)}{P(B)} = \frac{L(A | B)\, P(A)}{P(B)} \!$

where

$L(A|B) = P(B|A) \,$

is the likelihood of A given B for a fixed value of B.

Each term in Bayes' theorem has a conventional name:

With this terminology, the theorem may be paraphrased as

$\mbox{posterior} = \frac{\mbox{likelihood} \times \mbox{prior}} {\mbox{normalizing constant}}$

In words: the posterior probability is proportional to the prior probability times the likelihood.

In addition, the ratio P(B|A)/P(B) is sometimes called the standardised likelihood, so the theorem may also be paraphrased as

$\mbox{posterior} = {\mbox{standardised likelihood} \times \mbox{prior} }\,$

## Derivation from conditional probabilities

To derive the theorem, we start from the definition of conditional probability. The probability of event A given event B is

$P(A|B) = \frac{P(A, B)}{P(B)} \!$

Likewise, the probability of event B given event A is

$P(B|A) = \frac{P(A, B)}{P(A)} \!$

Rearranging and combining these two equations, we find

$P(A|B)\, P(B) = P(A, B) = P(B|A)\, P(A) \!$

Dividing both sides by P(B), providing that it is non-zero, we obtain Bayes' theorem:

$P(A|B) = \frac{P(B | A)\,P(A)}{P(B)} \!$

## Alternative forms of Bayes' theorem

Bayes' theorem is often embellished by noting that

$\Pr(B) = \Pr(A, B) + \Pr(A^C, B) = \Pr(B|A) \Pr(A) + \Pr(B|A^C) \Pr(A^C)\,$

so the theorem can be restated as

$\Pr(A|B) = \frac{\Pr(B | A)\, \Pr(A)}{\Pr(B|A)\Pr(A) + \Pr(B|A^C)\Pr(A^C)} , \!$

where AC is the complementary event of A (often called "not A"). More generally, where {Ai} forms a partition of the event space,

$\Pr(A_i|B) = \frac{\Pr(B | A_i)\, \Pr(A_i)}{\sum_j \Pr(B|A_j)\,\Pr(A_j)} , \!$

for any Ai in the partition.

### Bayes' theorem in terms of odds and likelihood ratio

Bayes' theorem can also be written neatly in terms of a likelihood ratio Λ and odds O as

$O(A|B)=O(A) \cdot \Lambda (A|B)$

where

$O(A|B)=\frac{\Pr(A|B)}{\Pr(A^C|B)} \!$

are the odds of A given B,

$O(A)=\frac{\Pr(A)}{\Pr(A^C)} \!$

are the odds of A by itself, and

$\Lambda (A|B) = \frac{L(A|B)}{L(A^C|B)} = \frac{\Pr(B|A)}{\Pr(B|A^C)} \!$

is the likelihood ratio.

### Bayes' theorem for probability densities

There is also a version of Bayes' theorem for continuous distributions. It is somewhat harder to derive, since probability densities, strictly speaking, are not probabilities, so Bayes' theorem has to be established by a limit process; see Papoulis (citation below), Section 7.3 for an elementary derivation. Bayes' theorem for probability densities is formally similar to the theorem for probabilities:

$f(x|y) = \frac{f(x,y)}{f(y)} = \frac{f(y|x)\,f(x)}{f(y)} \!$

and there is an analogous statement of the law of total probability:

$f(x|y) = \frac{f(y|x)\,f(x)}{\int_{-\infty}^{\infty} f(y|x)\,f(x)\,dx}. \!$

As in the discrete case, the terms have standard names. f(x, y) is the joint distribution of X and Y, f(x|y) is the posterior distribution of X given Y=y, f(y|x) = L(x|y) is (as a function of x) the likelihood function of X given Y=y, and f(x) and f(y) are the marginal distributions of X and Y respectively, with f(x) being the prior distribution of X.

Here we have indulged in a conventional abuse of notation, using f for each one of these terms, although each one is really a different function; the functions are distinguished by the names of their arguments.

### Extensions of Bayes' theorem

Theorems analogous to Bayes' theorem hold in problems with more than two variables. For example:

$\Pr(A|B,C) = \frac{\Pr(A) \, \Pr(B|A) \, \Pr(C|A,B)}{\Pr(B) \, \Pr(C|B)} \!$

This can be derived in several steps from Bayes' theorem and the definition of conditional probability.

A general strategy is to work with a decomposition of the joint probability, and to marginalize (integrate) over the variables that are not of interest. Depending on the form of the decomposition, it may be possible to prove that some integrals must be 1, and thus they fall out of the decomposition; exploiting this property can reduce the computations very substantially. A Bayesian network, for example, specifies a factorization of a joint distribution of several variables in which the conditional probability of any one variable given the remaining ones takes a particularly simple form (see Markov blanket).

## Examples

### Example #1: False positives in a medical test

Suppose that a test for a particular disease has a very high success rate:

• if a tested patient has the disease, the test accurately reports this, a 'positive', 99% of the time (or, with probability 0.99), and
• if a tested patient does not have the disease, the test accurately reports that, a 'negative', 95% of the time (i.e. with probability 0.95).

Suppose also, however, that only 0.1% of the population have that disease (i.e. with probability 0.001). We now have all the information required to use Bayes' theorem to calculate the probability that, given the test was positive, that it is a false positive. This problem is discussed at greater length in Bayesian inference.

Let D be the event that the patient has the disease, and T be the event that the test returns a positive result. Then, using the second alternative form of Bayes' theorem (above), the probability of a true positive is

$P(D|T) = \frac{P(T|D)\,P(D)}{P(T|D)\,P(D) + P(T|D^C)\,P(D^C)} \!$
$P(D|T) = \frac{0.99\times 0.001}{0.99 \times 0.001 + 0.05\times 0.999} = 11/566 \approx 0.019, \!$

and hence the probability that a positive result is a false positive is about (1 − 0.019) = 0.981.

Despite the apparent high accuracy of the test, the incidence of the disease is so low (one in a thousand) that the vast majority of patients who test positive (98 in a hundred) do not have the disease.

### Example #2: Conditional probabilities

To illustrate, suppose there are two bowls full of cookies. Bowl #1 has 10 chocolate chip cookies and 30 plain cookies, while bowl #2 has 20 of each. Our friend Fred picks a bowl at random, and then picks a cookie at random. We may assume there is no reason to believe Fred treats one bowl differently from another, likewise for the cookies. The cookie turns out to be a plain one. How probable is it that Fred picked it out of bowl #1?

Intuitively, it seems clear that the answer should be more than a half, since there are more plain cookies in bowl #1. The precise answer is given by Bayes' theorem. But first, we can clarify the situation by rephrasing the question to "what’s the probability that Fred picked bowl #1, given that he has a plain cookie?” Thus, to relate to our previous explanation, the event A is that Fred picked bowl #1, and the event B is that Fred picked a plain cookie. To compute Pr(A|B), we first need to know:

• Pr(A), or the probability that Fred picked bowl #1 regardless of any other information. Since Fred is treating both bowls equally, it is 0.5.
• Pr(B), or the probability of getting a plain cookie regardless of any information on the bowls. In other words, this is the probability of getting a plain cookie from each of the bowls. It is computed as the sum of the probability of getting a plain cookie from a bowl multiplied by the probability of selecting this bowl. We know from the problem statement that the probability of getting a plain cookie from bowl #1 is 0.75, and the probability of getting one from bowl #2 is 0.5, and since Fred is treating both bowls equally the probability of selecting any one of them is 0.5. Thus, the probability of getting a plain cookie overall is 0.75×0.5 + 0.5×0.5 = 0.625.
• Pr(B|A), or the probability of getting a plain cookie given that Fred has selected bowl #1. From the problem statement, we know this is 0.75, since 30 out of 40 cookies in bowl #1 are plain.

Given all this information, we can compute the probability of Fred having selected bowl #1 given that he got a plain cookie, as such:

$\Pr(A|B) = \frac{\Pr(B | A) \Pr(A)}{\Pr(B)} = \frac{0.75 \times 0.5}{0.625} = 0.6.$

As we expected, it is more than half.

#### Tables of occurences and relative frequencies

It is often helpful when calculating conditional probabilities to create a simple table containing the number of occurences of each outcome, or the relative frequencies of each outcome, for each of the independent variables. The tables below illustrate the use of this method for the cookies:

Number of cookies in each bowl
Relative frequency of cookies in each bowl
Bowl #1 Bowl #2 Totals
Chocolate Chip
10
20
30
Plain
30
20
50
Totals
40
40
80
Bowl #1 Bowl #2 Totals
Chocolate Chip
0.125
0.250
0.375
Plain
0.375
0.250
0.625
Totals
0.500
0.500
1.000

The table on the right is derived from the table on the left by dividing each entry by the total number of cookies under consideration, or 80 cookies.

### Example #3: Bayesian inference

Applications of Bayes' theorem often assume the philosophy underlying Bayesian probability that uncertainty and degrees of belief can be measured as probabilities. One such example follows. For additional worked out examples, including simpler examples, please see the article on the examples of Bayesian inference.

We describe the marginal probability distribution of a variable A as the prior probability distribution or simply the prior. The conditional distribution of A given the "data" B is the posterior probability distribution or just the posterior.

Suppose we wish to know about the proportion r of voters in a large population who will vote "yes" in a referendum. Let n be the number of voters in a random sample (chosen with replacement, so that we have statistical independence) and let m be the number of voters in that random sample who will vote "yes". Suppose that we observe n = 10 voters and m = 7 say they will vote yes. From Bayes' theorem we can calculate the probability distribution function for r using

$f(r | n=10, m=7) = \frac {f(m=7 | r, n=10) \, f(r)} {\int_0^1 f(m=7|r, n=10) \, f(r) \, dr}. \!$

From this we see that from the prior probability density function f(r) and the likelihood function L(r) = f(m = 7|r, n = 10), we can compute the posterior probability density function f(r|n = 10, m = 7).

The prior probability density function f(r) summarizes what we know about the distribution of r in the absence of any observation. We provisionally assume in this case that the prior distribution of r is uniform over the interval [0, 1]. That is, f(r) = 1. If some additional background information is found, we should modify the prior accordingly. However before we have any observations, all outcomes are equally likely.

Under the assumption of random sampling, choosing voters is just like choosing balls from an urn. The likelihood function L(r) = P(m = 7|r, n = 10,) for such a problem is just the probability of 7 successes in 10 trials for a binomial distribution.

$\Pr( m=7 | r, n=10) = {10 \choose 7} \, r^7 \, (1-r)^3.$

As with the prior, the likelihood is open to revision -- more complex assumptions will yield more complex likelihood functions. Maintaining the current assumptions, we compute the normalizing factor,

$\int_0^1 \Pr( m=7|r, n=10) \, f(r) \, dr = \int_0^1 {10 \choose 7} \, r^7 \, (1-r)^3 \, 1 \, dr = {10 \choose 7} \, \frac{1}{1320} \!$

and the posterior distribution for r is then

$f(r | n=10, m=7) = \frac{{10 \choose 7} \, r^7 \, (1-r)^3 \, 1} {{10 \choose 7} \, \frac{1}{1320}} = 1320 \, r^7 \, (1-r)^3$

for r between 0 and 1, inclusive.

One may be interested in the probability that more than half the voters will vote "yes". The prior probability that more than half the voters will vote "yes" is 1/2, by the symmetry of the uniform distribution. In comparison, the posterior probability that more than half the voters will vote "yes", i.e., the conditional probability given the outcome of the opinion poll – that seven of the 10 voters questioned will vote "yes" – is

$1320\int_{1/2}^1 r^7(1-r)^3\,dr \approx 0.887, \!$

which is about an "89% chance".

## Historical remarks

Bayes' theorem is named after the Reverend Thomas Bayes (17021761), who studied how to compute a distribution for the parameter of a binomial distribution (to use modern terminology). His friend, Richard Price, edited and presented the work in 1763, after Bayes' death, as An Essay towards solving a Problem in the Doctrine of Chances. Pierre-Simon Laplace replicated and extended these results in an essay of 1774, apparently unaware of Bayes' work.

One of Bayes' results (Proposition 5) gives a simple description of conditional probability, and shows that it does not depend on the order in which things occur:

If there be two subsequent events, the probability of the second b/N and the probability of both together P/N, and it being first discovered that the second event has also happened, the probability I am right [i.e., the conditional probability of the first event being true given that the second has happened] is P/b.

Bayes' main result (Proposition 9 in the essay) is the following: assuming a uniform distribution for the prior distribution of the binomial parameter p, the probability that p is between two values a and b is

$\frac {\int_a^b {n+m \choose m} p^m (1-p)^n\,dp} {\int_0^1 {n+m \choose m} p^m (1-p)^n\,dp} \!$

where m is the number of observed successes and n the number of observed failures. His preliminary results, in particular Propositions 3, 4, and 5, imply the result now called Bayes' Theorem (as described below), but it does not appear that Bayes himself emphasized or focused on that result.

What is "Bayesian" about Proposition 9 is that Bayes presented it as a probability for the parameter p. So, one can compute probability for an experimental outcome, but also for the parameter which governs it, and the same algebra is used to make inferences of either kind.

Bayes states his question in a way that might make the idea of assigning a probability distribution to a parameter palatable to a frequentist. He supposes that a billiard ball is thrown at random onto a billiard table, and that the probabilities p and q are the probabilities that subsequent billiard balls will fall above or below the first ball.

## References

### Commentaries

• G. A. Barnard (1958) "Studies in the History of Probability and Statistics: IX. Thomas Bayes's Essay Towards Solving a Problem in the Doctrine of Chances", Biometrika 45:293–295. (biographical remarks)
• Daniel Covarrubias. "An Essay Towards Solving a Problem in the Doctrine of Chances". (an outline and exposition of Bayes's essay)
• Stephen M. Stigler (1982). "Thomas Bayes' Bayesian Inference," Journal of the Royal Statistical Society, Series A, 145:250–258. (Stigler argues for a revised interpretation of the essay; recommended)
• Isaac Todhunter (1865). A History of the Mathematical Theory of Probability from the time of Pascal to that of Laplace, Macmillan. Reprinted 1949, 1956 by Chelsea and 2001 by Thoemmes.