- Bayes' theorem
In

probability theory ,**Bayes' theorem**(often called**Bayes' law**afterThomas Bayes ) relates the conditional and marginal probabilities of tworandom event s. It is often used to computeposterior probabilities given observations. For example, a patient may be observed to have certain symptoms. Bayes' theorem can be used to compute the probability that a proposed diagnosis is correct, given that observation. (See example 2)As a formal

theorem , Bayes' theorem is valid in all common interpretations of probability. However, it plays a central role in the debate around thefoundations of statistics :frequentist and Bayesian interpretations disagree about the ways in which probabilities should be assigned in applications. Frequentists assign probabilities to random events according to their frequencies of occurrence or to subsets of populations as proportions of the whole, while Bayesians describe probabilities in terms of beliefs and degrees of uncertainty. The articles onBayesian probability and frequentist probability discuss these debates at greater length.**tatement of Bayes' theorem**Bayes' theorem relates the conditional and marginal probabilities of events "A" and "B", where "B" has a non-vanishing probability:

:$P(A|B)\; =\; frac\{P(B\; |\; A),\; P(A)\}\{P(B)\}.$

Each term in Bayes' theorem has a conventional name:

* P("A") is theprior probability ormarginal probability of "A". It is "prior" in the sense that it does not take into account any information about "B".

* P("A"|"B") is theconditional probability of "A", given "B". It is also called theposterior probability because it is derived from or depends upon the specified value of "B".

* P("B"|"A") is the conditional probability of "B" given "A".

* P("B") is the prior or marginal probability of "B", and acts as anormalizing constant .Intuitively, Bayes' theorem in this form describes the way in which one's beliefs about observing 'A' are updated by having observed 'B'.

**An example**Suppose there is a co-ed school having 60% boys and 40% girls as students. The girl students wear trousers or skirts in equal numbers; the boys all wear trousers. An observer sees a (random) student from a distance; all they can see is that this student is wearing trousers. What is the probability this student is a girl?

It is clear that the probability is less than 40%, but by how much? Is it half that, since only half the girls are wearing trousers? The correct answer can be computed using Bayes' theorem.

The event "A" is that the student observed is a girl, and the event "B" is that the student observed is wearing trousers. To compute P("A"|"B"), we first need to know:

* P("A"), or the probability that the student is a girl regardless of any other information. Since the observers sees a random student, meaning that all students have the same probability of being observed, and the fraction of girls among the students is 40%, this probability equals 0.4.

* P("A"' ), or the probability that the student is a boy regardless of any other information ("A"' is the complementary event to "A"). This is 60%, or 0.6.

* P("B"|"A"), or the probability of the student wearing trousers given that the student is a girl. As they are as likely to wear skirts as trousers, this is 0.5.

* P("B"|"A"' ), or the probability of the student wearing trousers given that the student is a boy. This is given as 1.

* P("B"), or the probability of a (randomly selected) student wearing trousers regardless of any other information. Since P("B") = P("B"|"A")P("A") + P("B"|"A"' )P("A"' ), this is nowrap|1= 0.5×0.4 + 1×0.6 = 0.8.Given all this information, the probability of the observer having spotted a girl given that the observed student is wearing trousers can be computed by substituting these values in the formula:

:$P(A|B)\; =\; frac\{P(B|A)\; P(A)\}\{P(B)\}\; =\; frac\{0.5\; imes\; 0.4\}\{0.8\}\; =\; 0.25,.$

As expected, it is less than 40%, but more than half that.

Another, essentially equivalent way of obtaining the same result is as follows. Assume, for concreteness, that there are 100 students, 60 boys and 40 girls. Among these, 60 boys and 20 girls wear trousers. All together there are 80 trouser-wearers, of which 20 are girls. Therefore the chance that a random trouser-wearer is a girl equals nowrap|1=20/80 = 0.25.

It is often helpful when calculating conditional probabilities to create a simple table containing the number of occurrences of each outcome, or the relative frequencies of each outcome, for each of the independent variables. The table below illustrates the use of this method for the above girl-or-boy example.

**Bayes' theorem in terms of likelihood**Bayes' theorem can also be interpreted in terms of likelihood:

:$P(A|B)\; propto\; L(A\; |\; b),\; P(A).$

Here "L"("A"|"b") is the likelihood of "A" given fixed "b". The rule is then an immediate consequence of the relationship $P(B\; |\; A)\; propto\; L(A\; |\; B)$.

With this terminology, the theorem may be paraphrased as

:$mbox\{posterior\}\; =\; frac\{mbox\{likelihood\}\; imes\; mbox\{prior\; \{alpha\}$(where $alpha$ is a normalising constant equal to "P"("B")).

In words: the posterior probability is proportional to the product of the prior probability and the likelihood.

**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:$P(A|B)=frac\{P(A\; cap\; B)\}\{P(B)\}.$

Equivalently, the probability of event "B" given event "A" is

:$P(B|A)\; =\; frac\{P(A\; cap\; B)\}\{P(A)\}.\; !$

Rearranging and combining these two equations, we find

:$P(A|B),\; P(B)\; =\; P(A\; cap\; B)\; =\; P(B|A),\; P(A).\; !$

This lemma is sometimes called the product rule for probabilities. Dividing both sides by P("B"), providing that it is non-zero, we obtain Bayes' theorem:

:$P(A|B)\; =\; frac\{P(A\; cap\; B)\}\{P(B)\}\; =\; frac\{P(B|A),P(A)\}\{P(B)\}.\; !$

**Alternative forms of Bayes' theorem**Bayes' theorem is often embellished by noting that

:$P(B)\; =\; P(Acap\; B)\; +\; P(A^Ccap\; B)\; =\; P(B|A)\; P(A)\; +\; P(B|A^C)\; P(A^C),$

where "A"

^{"C"}is the complementary event of "A" (often called "not A"). So the theorem can be restated as the following formula:$P(A|B)\; =\; frac\{P(B\; |\; A),\; P(A)\}\{P(B|A)\; P(A)\; +\; P(B|A^C)\; P(A^C)\}.\; !$

More generally, where {"A"

_{"i"}} forms a partition of the event space,:$P(A\_i|B)\; =\; frac\{P(B\; |\; A\_i),\; P(A\_i)\}\{sum\_j\; P(B|A\_j),P(A\_j)\}\; ,\; !$

for any "A"

_{"i"}in the partition.See also the

law of total probability .**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 $O(A|B)=frac\{P(A|B)\}\{P(A^C|B)\}\; !$ are the "odds" of "A" given "B",

and $O(A)=frac\{P(A)\}\{P(A^C)\}\; !$ are the odds of "A" by itself,

while $Lambda\; (A|B)\; =\; frac\{L(A|B)\}\{L(A^C|B)\}\; =\; frac\{P(B|A)\}\{P(B|A^C)\}\; !$ is the likelihood ratio.

**Bayes' theorem for probability densities**There is also a version of Bayes' theorem for

continuous distribution s.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::$f\_X(x|Y=y)\; =\; frac\{f\_\{X,Y\}(x,y)\}\{f\_Y(y)\}\; =\; frac\{f\_Y(y|X=x),f\_X(x)\}\{f\_Y(y)\}\; =\; frac\{f\_Y(y|X=x),f\_X(x)\}\{int\_\{-infty\}^\{infty\}\; f\_Y(y|X=xi\; ),f\_X(xi\; ),dxi\; \}.!$

There is an analogous statement of the

law of total probability , which is used in the denominator::$f\_Y(y)\; =\; int\_\{-infty\}^\{infty\}\; f\_Y(y|X=x\; ),f\_X(x),dx\; .!$As in the discrete case, the terms have standard names. :$f\_\{X,Y\}(x,y),$ is the joint distribution of "X" and "Y",:$f\_X(x|Y=y),$ is the posterior distribution of "X" given "Y"="y",:$f\_Y(y|X=x)\; =\; L(x|y),$ is (as a function of "x") the likelihood function of "X" given "Y"="y",and :$f\_X(x),$and :$f\_Y(y)!$

are the marginal distributions of "X" and "Y" respectively, with $f\_X(x),$ being the prior distribution of "X".

**Abstract Bayes' theorem**Given two

absolutely continuous probability measures $P\; sim\; Q$ on theprobability space $(Omega,\; mathcal\{F\})$ and a sigma-algebra $mathcal\{G\}\; subset\; mathcal\{F\}$, the abstract Bayes theorem for a $mathcal\{F\}$-measurable random variable $X$ becomes:$E\_P\; [X|mathcal\{G\}]\; =\; frac\{E\_Q\; [frac\{dP\}\{dQ\}\; X\; |mathcal\{G\}]\; \}\{E\_Q\; [frac\{dP\}\{dQ\}|mathcal\{G\}]\; \}$.

Proof :

by definition of conditional probability,

$E\_P\; [X|mathcal\{G\}]\; =\; frac\{E\_P\; [X\; 1\_mathcal\{G\}]\; \}\{E\_P\; [mathcal\{G\}]\; \}$

and we have also

$E\_Q\; [frac\{dP\}\{dQ\}\; X|mathcal\{G\}]\; =\; frac\{E\_Q\; [frac\{dP\}\{dQ\}\; X\; 1\_mathcal\{G\}]\; \}\{E\_Q\; [mathcal\{G\}]\; \}\; =\; frac\{E\_P\; [X\; 1\_mathcal\{G\}]\; \}\{E\_Q\; [mathcal\{G\}]\; \}$

$E\_Q\; [frac\{dP\}\{dQ\}|mathcal\{G\}]\; =\; frac\{E\_Q\; [frac\{dP\}\{dQ\}\; 1\_mathcal\{G\}]\; \}\{E\_Q\; [mathcal\{G\}]\; \}\; =\; frac\{E\_P\; [1\_mathcal\{G\}]\; \}\{E\_Q\; [mathcal\{G\}]\; \}$

This formulation is used in

Kalman filtering to findZakai equation s. It is also used infinancial mathematics for change ofnumeraire techniques.**Extensions of Bayes' theorem**Theorems analogous to Bayes' theorem hold in problems with more than two variables. For example:

:$P(A|B\; cap\; C)\; =\; frac\{P(A)\; ,\; P(B|A)\; ,\; P(C|A\; cap\; B)\}\{P(B)\; ,\; P(C|B)\},.$

This can be derived in a few steps from Bayes' theorem and the definition of conditional probability:

:$P(A|B\; cap\; C)\; =\; frac\{P(A\; cap\; B\; cap\; C)\}\{P(B\; cap\; C)\}\; =\; frac\{P(C|A\; cap\; B)\; ,\; P(A\; cap\; B)\}\{P(B)\; ,\; P(C|B)\}\; =\; frac\{P(A)\; ,\; P(B|A)\; ,\; P(C|A\; cap\; B)\}\{P(B)\; ,\; P(C|B)\},.$

Similarly,

:$P(A|B\; cap\; C)\; =\; frac\{P(B|A\; cap\; C)\; ,\; P(A|C)\}\{P(B|C)\},,$

which can be regarded as a conditional Bayes' Theorem, and can be derived by as follows:

:$P(A|B\; cap\; C)\; =\; frac\{P(A\; cap\; B\; cap\; C)\}\{P(B\; cap\; C)\}\; =\; frac\{P(B|A\; cap\; C)\; ,\; P(A|C)\; ,\; P(C)\}\{P(C)\; ,\; P(B|C)\}\; =\; frac\{P(B|A\; cap\; C)\; ,\; P(A|C)\}\{P(B|C)\},.$

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.ABayesian network , for example, specifies a factorization of ajoint distribution of several variables in which the conditional probability of any one variable given the remaining ones takes a particularly simple form (seeMarkov blanket ).**Further examples****Example 1: Drug testing**Bayes' theorem is useful in evaluating the result of

drug test s. Suppose a certain drug test is 99% sensitive and 99% specific, that is, the test will correctly identify a drug user as testing positive 99% of the time, and will correctly identify a non-user as testing negative 99% of the time. This would seem to be a relatively accurate test, but Bayes' theorem will reveal a potential flaw. Let's assume a corporation decides to test its employees foropium use, and 0.5% of the employees use the drug. We want to know theprobability that, given a positive drug test, an employee is actually a drug user. Let "D" be the event of being a drug user and "N" indicate being a non-user. Let "+" be the event of a positive drug test. We need to know the following:

* "P"("D"), or the probability that the employee is a drug user, regardless of any other information. This is 0.005, since 0.5% of the employees are drug users. This is the "prior probability" of D.

* "P"("N"), or the probability that the employee is not a drug user. This is nowrap|1 − "P"("D"), or 0.995.

* "P"(+|"D"), or the probability that the test is positive, given that the employee is a drug user. This is 0.99, since the test is 99% accurate.

* "P"(+|"N"), or the probability that the test is positive, given that the employee is not a drug user. This is 0.01, since the test will produce afalse positive for 1% of non-users.

* "P"(+), or the probability of a positive test event, regardless of other information. This is 0.0149 or 1.49%, which is found by adding the probability that a true positive result will appear (= 99% x 0.5% = 0.495%) plus the probability that a false positive will appear (= 1% x 99.5% = 0.995%). This is the prior probability of +.Given this information, we can compute the posterior probability "P"("D"|+) of an employee who tested positive actually being a drug user::$egin\{align\}P(D|+)\; =\; frac\{P(+\; |\; D)\; P(D)\}\{P(+)\}\; \backslash \; =\; frac\{P(+\; |\; D)\; P(D)\}\{P(+\; |\; D)\; P(D)\; +\; P(+\; |\; N)\; P(N)\}\; \backslash \; =\; frac\{0.99\; imes\; 0.005\}\{0.99\; imes\; 0.005\; +\; 0.01\; imes\; 0.995\}\; \backslash \; =\; 0.3322.end\{align\}$

Despite the apparently high accuracy of the test, the probability that an employee who tested positive actually did use drugs is only about 33%, so it is "actually more likely" that the employee is not a drug user. The rarer the condition for which we are testing, the greater the percentage of positive tests that will be false positives.

**Example 2: 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 theposterior 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 havestatistical 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 .:$P(\; m=7\; |\; r,\; n=10)\; =\; \{10\; choose\; 7\}\; ,\; r^7\; ,\; (1-r)^3.$

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,

:$int\_0^1\; P(\; m=7|r,\; n=10)\; ,\; f(r)\; ,\; dr\; =\; int\_0^1\; \{10\; choose\; 7\}\; ,\; r^7\; ,\; (1-r)^3\; ,\; 1\; ,\; dr\; =\; \{10\; choose\; 7\}\; ,\; frac\{1\}\{1320\}\; !$

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

which is about an "89% chance".

**Example 3: The Monty Hall problem**We are presented with three doors - red, green, and blue - one of which has a prize. We choose the

**red**door, which is not opened until the presenter performs an action. The presenter "who knows what door the prize is behind, and who must open a door, but is not permitted to open the door we have picked or the door with the prize", opens the "blue" door and reveals that there is no prize behind it and subsequently asks if we wish to change our mind about our initial selection of**red**. What is the probability that the prize is behind each of the green and red doors?Let us call the situation that the prize is behind a given door

**A**,_{r}**A**, and_{g}**A**._{b}To start with, $P(A\_r)\; =\; P(A\_g)\; =\; P(A\_b)\; =\; frac\; 1\; 3$, and to make things simpler we shall assume that we have already picked the red door.

Let us call

**B**"the presenter opens the blue door". Without any prior knowledge, we would assign this a probability of 50%.* In the situation where the prize is behind the red door, the host is free to pick between the green or the blue door at random. Thus, $P(B|A\_r)\; =\; 1/2$

* In the situation where the prize is behind the green door, the host must pick the blue door. Thus, $P(B|A\_g)\; =\; 1$

* In the situation where the prize is behind the blue door, the host must pick the green door. Thus, $P(B|A\_b)\; =\; 0$Thus,

$egin\{matrix\}\; P(A\_r|B)\; =\; frac\{P(B\; |\; A\_r)\; P(A\_r)\}\{P(B)\}\; =\; frac\{frac\; 1\; 2\; cdot\; frac\; 1\; 3\}\{frac\; 1\; 2\}\; =\; frac\; 1\; 3\backslash \; P(A\_g|B)\; =\; frac\{P(B\; |\; A\_g)\; P(A\_g)\}\{P(B)\}\; =\; frac\{1\; cdot\; frac\; 1\; 3\}\{frac\; 1\; 2\}\; =\; frac\; 2\; 3\backslash \; P(A\_b|B)\; =\; frac\{P(B\; |\; A\_b)\; P(A\_b)\}\{P(B)\}\; =\; frac\{0\; cdot\; frac\; 1\; 3\}\{frac\; 1\; 2\}\; =\; 0.end\{matrix\}$

So, we should always choose the green door.

Note how this depends on the value of P(B).

**Historical remarks**An investigation by a statistics professor (Stigler 1983) suggests that Bayes' theorem was discovered by

Nicholas Saunderson some time before Bayes.Bayes' theorem is named after the Reverend

Thomas Bayes (1702–1761), who studied how to compute a distribution for the parameter of abinomial 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 can be expressed independently of 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, from hence I guess that the first 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 also happened] is P/b.

Note that the expression says nothing about the "order" in which the events occurred; it measures correlation, not causation. His preliminary results, in particular Propositions 3, 4, and 5, imply the result now called Bayes' Theorem (as described above), but it does not appear that Bayes himself emphasized or focused on that result.

Bayes' main result (Proposition 9 in the essay) is the following: assuming a

uniform distribution for theprior distribution of thebinomial 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.

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.

Stephen Fienberg [*http://ba.stat.cmu.edu/journal/2006/vol01/issue01/fienberg.pdf*] describes the evolution of the field from "inverse probability" at the time of Bayes and Laplace, and even ofHarold Jeffreys (1939) to "Bayesian" in the 1950's. The irony is that this label was introduced byR.A. Fisher in a derogatory sense. So, historically, Bayes was not a "Bayesian". It is actually unclear whether or not he was a Bayesian in the modern sense of the term, i.e. whether or not he was interested in inference or merely in probability: the 1763 essay is more of a probability paper.**See also**

*Bayesian inference

*Bayesian network

*Bayesian probability

*Bayesian spam filtering

*Thomas Bayes

*Bogofilter

*Conjugate prior

*Empirical Bayes method

*Monty Hall problem

*Occam's razor

*Prosecutor's fallacy

*Raven paradox

*Recursive Bayesian estimation

*Revising opinions in statistics

*Sequential bayesian filtering

*Borel's paradox

*Naive Bayes classifier **References****Versions of the essay*** Thomas Bayes (1763), "An Essay towards solving a Problem in the Doctrine of Chances. By the late Rev. Mr. Bayes, F. R. S. communicated by Mr. Price, in a letter to John Canton, A. M. F. R. S.", "

Philosophical Transactions , Giving Some Account of the Present Undertakings, Studies and Labours of the Ingenious in Many Considerable Parts of the World" 53:370–418.

* Thomas Bayes (1763/1958) "Studies in the History of Probability and Statistics: IX. Thomas Bayes' Essay Towards Solving a Problem in the Doctrine of Chances", "Biometrika " 45:296–315. "(Bayes' essay in modernized notation)"

* Thomas Bayes [*http://www.stat.ucla.edu/history/essay.pdf "An essay towards solving a Problem in the Doctrine of Chances"*] . "(Bayes' essay in the original notation)"**Commentaries*** G. A. Barnard (1958) "Studies in the History of Probability and Statistics: IX. Thomas Bayes' Essay Towards Solving a Problem in the Doctrine of Chances", "Biometrika" 45:293–295. "(biographical remarks)"

* Daniel Covarrubias. [*http://www.stat.rice.edu/~blairc/seminar/Files/danTalk.pdf "An Essay Towards Solving a Problem in the Doctrine of Chances"*] . "(an outline and exposition of Bayes' 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.

* [*http://yudkowsky.net/bayes/bayes.html An Intuitive Explanation of Bayesian Reasoning*] (includes biography)**Additional material*** Pierre-Simon Laplace (1774/1986), "Memoir on the Probability of the Causes of Events", "Statistical Science" 1(3):364–378.

* Stephen M. Stigler (1986), "Laplace's 1774 memoir on inverse probability", "Statistical Science" 1(3):359–378.

* Stephen M. Stigler (1983), "Who Discovered Bayes' Theorem?" "The American Statistician" 37(4):290–296.

* Jeff Miller, "et al.", [*http://members.aol.com/jeff570/mathword.html Earliest Known Uses of Some of the Words of Mathematics (B)*] . ("very informative; recommended")

*Athanasios Papoulis (1984), "Probability, Random Variables, and Stochastic Processes", second edition. New York: McGraw-Hill.

* [*http://www.inference.phy.cam.ac.uk/mackay/itila/ The on-line textbook: Information Theory, Inference, and Learning Algorithms*] , byDavid J. C. MacKay provides an up to date overview of the use of Bayes' theorem in information theory and machine learning.

* Provides a comprehensive introduction to Bayes' theorem.

* [*http://plato.stanford.edu/entries/logic-inductive/ Stanford Encyclopedia of Philosophy: Inductive Logic*] provides a comprehensive Bayesian treatment of Inductive Logic and Confirmation Theory.

*

*

*Eliezer S. Yudkowsky (2003), " [*http://yudkowsky.net/bayes/bayes.html An Intuitive Explanation of Bayesian Reasoning*] "

* [*http://www.celiagreen.com/charlesmccreery/statistics/bayestutorial.pdf A tutorial on probability and Bayes’ theorem devised for Oxford University psychology students*]

* [*http://faculty-staff.ou.edu/H/James.A.Hawthorne-1/Hawthorne%20--%20Confirmation%20Theory.pdf Confirmation Theory*] An extensive presentation of Bayesian Confirmation Theory

*Wikimedia Foundation.
2010.*

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**Bayes theorem**— A probability principle set forth by the English mathematician Thomas Bayes (1702 1761). Bayes theorem is of value in medical decision making and some of the biomedical sciences. Bayes theorem is employed in clinical epidemiology to determine the … Medical dictionary**Bayes-Theorem**— Das Bayestheorem (auch Satz von Bayes) ist ein Ergebnis der Wahrscheinlichkeitstheorie, benannt nach dem Mathematiker Thomas Bayes. Es gibt an, wie man mit bedingten Wahrscheinlichkeiten rechnet. Inhaltsverzeichnis 1 Formel 2 Interpretation 3… … Deutsch Wikipedia**Bayes Theorem**— Das Bayestheorem (auch Satz von Bayes) ist ein Ergebnis der Wahrscheinlichkeitstheorie, benannt nach dem Mathematiker Thomas Bayes. Es gibt an, wie man mit bedingten Wahrscheinlichkeiten rechnet. Inhaltsverzeichnis 1 Formel 2 Interpretation 3… … Deutsch Wikipedia**Bayes' Theorem**— A formula for determining conditional probability named after 18th century British mathematician Thomas Bayes. The theorem provides a way to revise existing predictions or theories given new or additional evidence. In finance, Bayes’… … Investment dictionary**Bayes' theorem**— /beɪz ˈθɪərəm/ (say bayz thearruhm) noun Statistics a method of measuring the degree of belief that can be attached to a proposition, based on the relevance of ascertainable facts. Also, Bayes law. {named posthumously after Thomas Bayes, 1701–61 … Australian English dictionary**Bayes' theorem**— noun Etymology: Thomas Bayes died 1761 English mathematician Date: 1865 a theorem about conditional probabilities: the probability that an event A occurs given that another event B has already occurred is equal to the probability that the event B … New Collegiate Dictionary**Bayes' theorem**— noun Statistics a theorem expressing the conditional probability of each of a set of possible causes for a given observed outcome in terms of the known probability of each cause and the conditional probability of the outcome of each cause.… … English new terms dictionary**Bayes' theorem**— noun (statistics) a theorem describing how the conditional probability of a set of possible causes for a given observed event can be computed from knowledge of the probability of each cause and the conditional probability of the outcome of each… … Useful english dictionary**Bayes' theorem**— /bayz, bay ziz/, Statistics. a theorem describing how the conditional probability of each of a set of possible causes, given an observed outcome, can be computed from knowledge of the probability of each cause and of the conditional probability… … Universalium**Bayes-Theorem**— Es sei f(y | δ) die bedingte ⇡ Dichtefunktion von y. Gemäß der baysianischen Sichtweise ist der Parameter δ die Realisation einer Zufallsvariablen mit einer Dichtefunktion f (δ), dem sog. „Prior“. Der Prior kann durch eine meist subjektive… … Lexikon der Economics