Bayes' theorem (alternatively Bayes' law or Bayes' rule, after Thomas Bayes) gives a mathematical rule for inverting conditional probabilities, allowing us to find the probability of a cause given its effect. For example, if the risk of developing health problems is known to increase with age, Bayes' theorem allows the risk to an individual of a known age to be assessed more accurately by conditioning it relative to their age, rather than assuming that the individual is typical of the population as a whole. Based on Bayes law both the prevalence of a disease in a given population and the error rate of an infectious disease test have to be taken into account to evaluate the meaning of a positive test result correctly and avoid the base-rate fallacy.
One of the many applications of Bayes' theorem is Bayesian inference, a particular approach to statistical inference, where it is used to invert the probability of observations given a model configuration (i.e., the likelihood function) to obtain the probability of the model configuration given the observations (i.e., the posterior probability).
Bayes' theorem is named after the Reverend Thomas Bayes, also a statistician and philosopher. Bayes used conditional probability to provide an algorithm (his Proposition 9) that uses evidence to calculate limits on an unknown parameter. His work was published in 1763 as An Essay Towards Solving a Problem in the Doctrine of Chances. Bayes studied how to compute a distribution for the probability parameter of a binomial distribution (in modern terminology). On Bayes's death his family transferred his papers to a friend, the minister, philosopher, and mathematician Richard Price.
Over two years, Richard Price significantly edited the unpublished manuscript, before sending it to a friend who read it aloud at the Royal Society on 23 December 1763.[1] Price edited[2] Bayes's major work "An Essay Towards Solving a Problem in the Doctrine of Chances" (1763), which appeared in Philosophical Transactions,[3] and contains Bayes' theorem. Price wrote an introduction to the paper which provides some of the philosophical basis of Bayesian statistics and chose one of the two solutions offered by Bayes. In 1765, Price was elected a Fellow of the Royal Society in recognition of his work on the legacy of Bayes.[4] [5] On 27 April a letter sent to his friend Benjamin Franklin was read out at the Royal Society, and later published, where Price applies this work to population and computing 'life-annuities'.[6]
Independently of Bayes, Pierre-Simon Laplace in 1774, and later in his 1812 Théorie analytique des probabilités, used conditional probability to formulate the relation of an updated posterior probability from a prior probability, given evidence. He reproduced and extended Bayes's results in 1774, apparently unaware of Bayes's work.[7] The Bayesian interpretation of probability was developed mainly by Laplace.[8]
About 200 years later, Sir Harold Jeffreys put Bayes's algorithm and Laplace's formulation on an axiomatic basis, writing in a 1973 book that Bayes' theorem "is to the theory of probability what the Pythagorean theorem is to geometry".[9]
Stephen Stigler used a Bayesian argument to conclude that Bayes' theorem was discovered by Nicholas Saunderson, a blind English mathematician, some time before Bayes;[10] [11] that interpretation, however, has been disputed.[12] Martyn Hooper[13] and Sharon McGrayne[14] have argued that Richard Price's contribution was substantial:
Bayes' theorem is stated mathematically as the following equation:
where
A
B
P(B) ≠ 0
P(A\vertB)
A
B
A
B
P(B\vertA)
B
A
A
B
P(B\vertA)=L(A\vertB)
P(A)
P(B)
A
B
Bayes' theorem may be derived from the definition of conditional probability:
P(A\vertB)=
P(A\capB) | |
P(B) |
,ifP(B) ≠ 0,
where
P(A\capB)
P(B\vertA)=
P(A\capB) | |
P(A) |
,ifP(A) ≠ 0.
Solving for
P(A\capB)
P(A\vertB)
P(A\vertB)=
P(B\vertA)P(A) | |
P(B) |
,ifP(B) ≠ 0.
For two continuous random variables X and Y, Bayes' theorem may be analogously derived from the definition of conditional density:
fX(x)=
fX,Y(x,y) | |
fY(y) |
fY(y)=
fX,Y(x,y) | |
fX(x) |
Therefore,
fX(x)=
fY(y)fX(x) | |
fY(y) |
.
Let
x | |
P | |
Y |
Y
X=x
PX
X
PX,Y(dx,dy)=
x | |
P | |
Y |
(dy)PX(dx)
y | |
P | |
X |
X
Y=y
Existence and uniqueness of the needed conditional expectation is a consequence of the Radon–Nikodym theorem. This was formulated by Kolmogorov in his famous book from 1933. Kolmogorov underlines the importance of conditional probability by writing "I wish to call attention to ... and especially the theory of conditional probabilities and conditional expectations ..." in the Preface.[15] The Bayes theorem determines the posterior distribution from the prior distribution. Uniqueness requires continuity assumptions.[16] Bayes' theorem can be generalized to include improper prior distributions such as the uniform distribution on the real line.[17] Modern Markov chain Monte Carlo methods have boosted the importance of Bayes' theorem including cases with improper priors.[18]
Bayes' rule and computing conditional probabilities provide a solution method for a number of popular puzzles, such as the Three Prisoners problem, the Monty Hall problem, the Two Child problem and the Two Envelopes problem.
Suppose, a particular test for whether someone has been using cannabis is 90% sensitive, meaning the true positive rate (TPR) = 0.90. Therefore, it leads to 90% true positive results (correct identification of drug use) for cannabis users.
The test is also 80% specific, meaning true negative rate (TNR) = 0.80. Therefore, the test correctly identifies 80% of non-use for non-users, but also generates 20% false positives, or false positive rate (FPR) = 0.20, for non-users.
Assuming 0.05 prevalence, meaning 5% of people use cannabis, what is the probability that a random person who tests positive is really a cannabis user?
The Positive predictive value (PPV) of a test is the proportion of persons who are actually positive out of all those testing positive, and can be calculated from a sample as:
PPV = True positive / Tested positive
If sensitivity, specificity, and prevalence are known, PPV can be calculated using Bayes theorem. Let
P(User\vertPositive)
\begin{align} P(User\vertPositive)&=
P(Positive\vertUser)P(User) | |
P(Positive) |
\\ &=
P(Positive\vertUser)P(User) | |
P(Positive\vertUser)P(User)+P(Positive\vertNon-user)P(Non-user) |
\\[8pt]&=
0.90 x 0.05 | |
0.90 x 0.05+0.20 x 0.95 |
=
0.045 | |
0.045+0.19 |
≈ 19\% \end{align}
The fact that
P(Positive)=P(Positive\vertUser)P(User)+P(Positive\vertNon-user)P(Non-user)
In other words, even if someone tests positive, the probability that they are a cannabis user is only 19%—this is because in this group, only 5% of people are users, and most positives are false positives coming from the remaining 95%.
If 1,000 people were tested:
The 1,000 people thus yields 235 positive tests, of which only 45 are genuine drug users, about 19%.
The importance of specificity can be seen by showing that even if sensitivity is raised to 100% and specificity remains at 80%, the probability of someone testing positive really being a cannabis user only rises from 19% to 21%, but if the sensitivity is held at 90% and the specificity is increased to 95%, the probability rises to 49%.
Positive | Negative | Total | |||
---|---|---|---|---|---|
User | 45 | 5 | 50 | ||
Non-user | 190 | 760 | 950 | ||
Total | 235 | 765 | 1000 | ||
90% sensitive, 80% specific, PPV=45/235 ≈ 19% |
Positive | Negative | Total | |||
---|---|---|---|---|---|
User | 50 | 0 | 50 | ||
Non-user | 190 | 760 | 950 | ||
Total | 240 | 760 | 1000 | ||
100% sensitive, 80% specific, PPV=50/240 ≈ 21% |
Positive | Negative | Total | |||
---|---|---|---|---|---|
User | 45 | 5 | 50 | ||
Non-user | 47 | 903 | 950 | ||
Total | 92 | 908 | 1000 | ||
90% sensitive, 95% specific, PPV=45/92 ≈ 49% |
Even if 100% of patients with pancreatic cancer have a certain symptom, when someone has the same symptom, it does not mean that this person has a 100% chance of getting pancreatic cancer. Assuming the incidence rate of pancreatic cancer is 1/100000, while 10/99999 healthy individuals have the same symptoms worldwide, the probability of having pancreatic cancer given the symptoms is only 9.1%, and the other 90.9% could be "false positives" (that is, falsely said to have cancer; "positive" is a confusing term when, as here, the test gives bad news).
Based on incidence rate, the following table presents the corresponding numbers per 100,000 people.
Yes | No | Total | ||
---|---|---|---|---|
Yes | 1 | 0 | 1 | |
No | 10 | 99989 | 99999 | |
Total | 11 | 99989 | 100000 |
\begin{align} P(Cancer|Symptoms)&=
P(Symptoms|Cancer)P(Cancer) | |
P(Symptoms) |
\\ &=
P(Symptoms|Cancer)P(Cancer) | |
P(Symptoms|Cancer)P(Cancer)+P(Symptoms|Non-Cancer)P(Non-Cancer) |
\\[8pt] &=
1 x 0.00001 | |
1 x 0.00001+(10/99999) x 0.99999 |
=
1{11} | |
≈ |
9.1\% \end{align}
Defective | Flawless | Total | ||
---|---|---|---|---|
A | 10 | 190 | 200 | |
B | 9 | 291 | 300 | |
C | 5 | 495 | 500 | |
Total | 24 | 976 | 1000 |
A factory produces items using three machines—A, B, and C—which account for 20%, 30%, and 50% of its output respectively. Of the items produced by machine A, 5% are defective; similarly, 3% of machine B's items and 1% of machine C's are defective. If a randomly selected item is defective, what is the probability it was produced by machine C?
Once again, the answer can be reached without using the formula by applying the conditions to a hypothetical number of cases. For example, if the factory produces 1,000 items, 200 will be produced by Machine A, 300 by Machine B, and 500 by Machine C. Machine A will produce 5% × 200 = 10 defective items, Machine B 3% × 300 = 9, and Machine C 1% × 500 = 5, for a total of 24. Thus, the likelihood that a randomly selected defective item was produced by machine C is 5/24 (~20.83%).
This problem can also be solved using Bayes' theorem: Let Xi denote the event that a randomly chosen item was made by the i th machine (for i = A,B,C). Let Y denote the event that a randomly chosen item is defective. Then, we are given the following information:
P(XA)=0.2, P(XB)=0.3, P(XC)=0.5.
P(Y|XA)=0.05, P(Y|XB)=0.03, P(Y|XC)=0.01.
To answer the original question, we first find P(Y). That can be done in the following way:
P(Y)=\sumiP(Y|Xi)P(Xi)=(0.05)(0.2)+(0.03)(0.3)+(0.01)(0.5)=0.024.
Hence, 2.4% of the total output is defective.
We are given that Y has occurred, and we want to calculate the conditionalprobability of XC. By Bayes' theorem,
P(XC|Y)=
P(Y|XC)P(XC) | |
P(Y) |
=
0.01 ⋅ 0.50 | |
0.024 |
=
5 | |
24 |
Given that the item is defective, the probability that it was made by machine C is 5/24. Although machine C produces half of the total output, it produces a much smaller fraction of the defective items. Hence the knowledge that the item selected was defective enables us to replace the prior probability P(XC) = 1/2 by the smaller posterior probability P(XC | Y) = 5/24.
The interpretation of Bayes' rule depends on the interpretation of probability ascribed to the terms. The two predominant interpretations are described below.
In the Bayesian (or epistemological) interpretation, probability measures a "degree of belief". Bayes' theorem links the degree of belief in a proposition before and after accounting for evidence. For example, suppose it is believed with 50% certainty that a coin is twice as likely to land heads than tails. If the coin is flipped a number of times and the outcomes observed, that degree of belief will probably rise or fall, but might even remain the same, depending on the results. For proposition A and evidence B,
For more on the application of Bayes' theorem under the Bayesian interpretation of probability, see Bayesian inference.
In the frequentist interpretation, probability measures a "proportion of outcomes". For example, suppose an experiment is performed many times. P(A) is the proportion of outcomes with property A (the prior) and P(B) is the proportion with property B. P(B | A) is the proportion of outcomes with property B out of outcomes with property A, and P(A | B) is the proportion of those with A out of those with B (the posterior).
The role of Bayes' theorem is best visualized with tree diagrams. The two diagrams partition the same outcomes by A and B in opposite orders, to obtain the inverse probabilities. Bayes' theorem links the different partitionings.
An entomologist spots what might, due to the pattern on its back, be a rare subspecies of beetle. A full 98% of the members of the rare subspecies have the pattern, so P(Pattern | Rare) = 98%. Only 5% of members of the common subspecies have the pattern. The rare subspecies is 0.1% of the total population. How likely is the beetle having the pattern to be rare: what is P(Rare | Pattern)?
From the extended form of Bayes' theorem (since any beetle is either rare or common),
\begin{align} P(Rare\vertPattern)&=
P(Pattern\vertRare)P(Rare) | |
P(Pattern) |
\\ [8pt]&=
P(Pattern\vertRare)P(Rare) | |
P(Pattern\vertRare)P(Rare)+P(Pattern\vertCommon)P(Common) |
\\ [8pt]&=
0.98 x 0.001 | |
0.98 x 0.001+0.05 x 0.999 |
\\ [8pt]& ≈ 1.9\% \end{align}
For events A and B, provided that P(B) ≠ 0,
P(A|B)=
P(B|A)P(A) | |
P(B) |
.
In many applications, for instance in Bayesian inference, the event B is fixed in the discussion, and we wish to consider the impact of its having been observed on our belief in various possible events A. In such a situation the denominator of the last expression, the probability of the given evidence B, is fixed; what we want to vary is A. Bayes' theorem then shows that the posterior probabilities are proportional to the numerator, so the last equation becomes:
P(A|B)\proptoP(A) ⋅ P(B|A).
In words, the posterior is proportional to the prior times the likelihood.[19]
If events A1, A2, ..., are mutually exclusive and exhaustive, i.e., one of them is certain to occur but no two can occur together, we can determine the proportionality constant by using the fact that their probabilities must add up to one. For instance, for a given event A, the event A itself and its complement ¬A are exclusive and exhaustive. Denoting the constant of proportionality by c we have
P(A|B)=c ⋅ P(A) ⋅ P(B|A)andP(\negA|B)=c ⋅ P(\negA) ⋅ P(B|\negA).
Adding these two formulas we deduce that
1=c ⋅ (P(B|A) ⋅ P(A)+P(B|\negA) ⋅ P(\negA)),
or
c=
1 | |
P(B|A) ⋅ P(A)+P(B|\negA) ⋅ P(\negA) |
=
1 | |
P(B) |
.
A | P(B | A)⋅P(A) = P(A | B)⋅P(B) | P(¬B | A)⋅P(A) = P(A | ¬B)⋅P(¬B) | P(A) | ||
---|---|---|---|---|---|---|---|---|---|
¬A (not A) | P(B | ¬A)⋅P(¬A) = P(¬A | B)⋅P(B) | P(¬B | ¬A)⋅P(¬A) = P(¬A | ¬B)⋅P(¬B) | P(¬A) = 1-P(A) | ||
Total | P(B) | P(¬B) = 1-P(B) | 1 |
P(A|B)=
P(B|A)P(A) | |
P(B|A)P(A)+P(B|\negA)P(\negA) |
.
For an epistemological interpretation:
For proposition A and evidence or background B,[20]
P(A)
P(\negA)
P(\negA)=1-P(A)
P(B|A)
P(B|\negA)
P(A|B)
Often, for some partition of the sample space, the event space is given in terms of P(Aj) and P(B | Aj). It is then useful to compute P(B) using the law of total probability:
P(B)=\sumjP(B\capAj),
Or (using the multiplication rule for conditional probability),[21]
P(B)={\sumjP(B|Aj)P(Aj)},
⇒ P(Ai|B)=
P(B|Ai)P(Ai) | |
\sum\limitsjP(B|Aj)P(Aj) |
⋅
In the special case where A is a binary variable:
P(A|B)=
P(B|A)P(A) | |
P(B|A)P(A)+P(B|\negA)P(\negA) |
⋅
Consider a sample space Ω generated by two random variables X and Y with known probability distributions. In principle, Bayes' theorem applies to the events A = and B = .
P(X{=}x|Y{=}y)=
P(Y{= | |
y |
|X{=}x)P(X{=}x)}{P(Y{=}y)}
However, terms become 0 at points where either variable has finite probability density. To remain useful, Bayes' theorem can be formulated in terms of the relevant densities (see Derivation).
If X is continuous and Y is discrete,
fXy}(x)=
P(Y{= | |
y| |
X{=}x)fX(x)}{P(Y{=}y)}
where each
f
If X is discrete and Y is continuous,
P(X{=}x|Y{=}y)=
fYx | |
(y) |
P(X{=}x)}{fY(y)}.
If both X and Y are continuous,
fX|y}(x)=
fYx | |
(y) |
fX(x)}{fY(y)}.
A continuous event space is often conceptualized in terms of the numerator terms. It is then useful to eliminate the denominator using the law of total probability. For fY(y), this becomes an integral:
fY(y)=
infty | |
\int | |
-infty |
fY|(y)fX(\xi)d\xi.
Bayes' theorem in odds form is:
O(A1:A2\vertB)=O(A1:A2) ⋅ Λ(A1:A2\vertB)
where
Λ(A1:A2\vertB)=
P(B\vertA1) | |
P(B\vertA2) |
is called the Bayes factor or likelihood ratio. The odds between two events is simply the ratio of the probabilities of the two events. Thus
O(A1:A2)=
P(A1) | |
P(A2) |
,
O(A1:A2\vertB)=
P(A1\vertB) | |
P(A2\vertB) |
,
Thus, the rule says that the posterior odds are the prior odds times the Bayes factor, or in other words, the posterior is proportional to the prior times the likelihood.
In the special case that
A1=A
A2=\negA
O(A)=O(A:\negA)=P(A)/(1-P(A))
A
A
O(A\vertB)=O(A) ⋅ Λ(A\vertB),
or, in words, the posterior odds on
A
A
A
B
For example, if a medical test has a sensitivity of 90% and a specificity of 91%, then the positive Bayes factor is
Λ+=P(TruePositive)/P(FalsePositive)=90\%/(100\%-91\%)=10
The example above can also be understood with more solid numbers: Assume the patient taking the test is from a group of 1000 people, where 91 of them actually have the disease (prevalence of 9.1%). If all these 1000 people take the medical test, 82 of those with the disease will get a true positive result (sensitivity of 90.1%), 9 of those with the disease will get a false negative result (false negative rate of 9.9%), 827 of those without the disease will get a true negative result (specificity of 91.0%), and 82 of those without the disease will get a false positive result (false positive rate of 9.0%). Before taking any test, the patient's odds for having the disease is 91:909. After receiving a positive result, the patient's odds for having the disease is
91 | x | |
909 |
90.1\% | = | |
9.0\% |
91 x 90.1\% | |
909 x 9.0\% |
=1:1
Using
P(\negB\vertA)=1-P(B\vertA)
P(\negB\vert\negA)
P(A\vertB)
P(\negB\vert\negA)=1-\left(1-P(A\vertB)\right)
P(B) | |
P(\negA) |
,
P(\negA)=1-P(A) ≠ 0
P(A\vertB)=1\impliesP(\negB\vert\negA)=1
B
A
\negA
\negB
P(B) ≠ 0
P(A\vertB)
B\impliesA
B\impliesA
P(A\vertB)=1
(B\impliesA)\iff(\negA\implies\negB)
In this relation between implications, the positions of
A
B
a
A
P(A\vertB)=P(B\vertA)
a(A) | |
P(B\vertA)a(A)+P(B\vert\negA)a(\negA) |
.
Bayes' theorem represents a special case of deriving inverted conditional opinions in subjective logic expressed as:
S | |
(\omega | |
A\tilde{| |
S | |
B},\omega | |
A\tilde{| |
lnotB})=
S | |
(\omega | |
B\vertA |
,
S | |
\omega | |
B\vertlnotA |
)\widetilde{\phi}aA,
where
\widetilde{\phi}
S | |
(\omega | |
B\vertA |
S | |
,\omega | |
B\vertlnotA |
)
S
aA
A
S | |
(\omega | |
A\tilde{| |
S | |
B},\omega | |
A\tilde{| |
lnotB})
S | |
\omega | |
A\vertB |
P(A\vertB)
S
(A\vertB)
S | |
\omega | |
A |
A
S
S | |
P(\omega | |
A |
)
S | |
P(\omega | |
A\tilde{| |
B})=
| |||||||||||||||
|
.
Hence, the subjective Bayes' theorem represents a generalization of Bayes' theorem.[23]
A version of Bayes' theorem for 3 events[24] results from the addition of a third event
C
P(C)>0,
P(A\vertB\capC)=
P(B\vertA\capC)P(A\vertC) | |
P(B\vertC) |
Using the chain rule
P(A\capB\capC)=P(A\vertB\capC)P(B\vertC)P(C)
And, on the other hand
P(A\capB\capC)=P(B\capA\capC)=P(B\vertA\capC)P(A\vertC)P(C)
The desired result is obtained by identifying both expressions and solving for
P(A\vertB\capC)
In genetics, Bayes' rule can be used to estimate the probability of an individual having a specific genotype. Many people seek to approximate their chances of being affected by a genetic disease or their likelihood of being a carrier for a recessive gene of interest. A Bayesian analysis can be done based on family history or genetic testing, in order to predict whether an individual will develop a disease or pass one on to their children. Genetic testing and prediction is a common practice among couples who plan to have children but are concerned that they may both be carriers for a disease, especially within communities with low genetic variance.[25]
Hypothesis | Hypothesis 1: Patient is a carrier | Hypothesis 2: Patient is not a carrier | |
---|---|---|---|
Prior Probability | 1/2 | 1/2 | |
Conditional Probability that all four offspring will be unaffected | (1/2) ⋅ (1/2) ⋅ (1/2) ⋅ (1/2) = 1/16 | About 1 | |
Joint Probability | (1/2) ⋅ (1/16) = 1/32 | (1/2) ⋅ 1 = 1/2 | |
Posterior Probability | (1/32) / (1/32 + 1/2) = 1/17 | (1/2) / (1/32 + 1/2) = 16/17 |
Parental genetic testing can detect around 90% of known disease alleles in parents that can lead to carrier or affected status in their child. Cystic fibrosis is a heritable disease caused by an autosomal recessive mutation on the CFTR gene,[27] located on the q arm of chromosome 7.[28]
Bayesian analysis of a female patient with a family history of cystic fibrosis (CF), who has tested negative for CF, demonstrating how this method was used to determine her risk of having a child born with CF:
Because the patient is unaffected, she is either homozygous for the wild-type allele, or heterozygous. To establish prior probabilities, a Punnett square is used, based on the knowledge that neither parent was affected by the disease but both could have been carriers:
WHomozygous for the wild- type allele (a non-carrier) | MHeterozygous (a CF carrier) | ||
---|---|---|---|
WHomozygous for the wild- type allele (a non-carrier) | WW | MW | |
MHeterozygous (a CF carrier) | MW | MM (affected by cystic fibrosis) |
Next, the patient undergoes genetic testing and tests negative for cystic fibrosis. This test has a 90% detection rate, so the conditional probabilities of a negative test are 1/10 and 1. Finally, the joint and posterior probabilities are calculated as before.
Hypothesis | Hypothesis 1: Patient is a carrier | Hypothesis 2: Patient is not a carrier | |
---|---|---|---|
Prior Probability | 2/3 | 1/3 | |
Conditional Probability of a negative test | 1/10 | 1 | |
Joint Probability | 1/15 | 1/3 | |
Posterior Probability | 1/6 | 5/6 |
Bayesian analysis can be done using phenotypic information associated with a genetic condition, and when combined with genetic testing this analysis becomes much more complicated. Cystic fibrosis, for example, can be identified in a fetus through an ultrasound looking for an echogenic bowel, meaning one that appears brighter than normal on a scan. This is not a foolproof test, as an echogenic bowel can be present in a perfectly healthy fetus. Parental genetic testing is very influential in this case, where a phenotypic facet can be overly influential in probability calculation. In the case of a fetus with an echogenic bowel, with a mother who has been tested and is known to be a CF carrier, the posterior probability that the fetus actually has the disease is very high (0.64). However, once the father has tested negative for CF, the posterior probability drops significantly (to 0.16).[26]
Risk factor calculation is a powerful tool in genetic counseling and reproductive planning, but it cannot be treated as the only important factor to consider. As above, incomplete testing can yield falsely high probability of carrier status, and testing can be financially inaccessible or unfeasible when a parent is not present.