Bayes' theorem

In probability theory, Bayes' theorem (often called Bayes' law and named after Rev Thomas Bayes; IPA:/'beɪz/) shows how one conditional probability (such as the probability of a hypothesis given observed evidence) depends on its inverse (in this case, the probability of that evidence given the hypothesis). The theorem expresses the posterior probability (i.e. after evidence E is observed) of a hypothesis H in terms of the prior probabilities of H and E, and the probability of E given H. It implies that evidence has a stronger confirming effect if it was more unlikely before being observed.

As a formal theorem, Bayes' theorem is valid in all common interpretations of probability. However, frequentist and Bayesian interpretations disagree on how (and to what) probabilities are assigned. In the Bayesian interpretation, probabilities are rationally coherent degrees of belief, or a degree of belief in a proposition given a body of well-specified information. Bayes' theorem can then be understood as specifying how an ideally rational person responds to evidence. In the frequentist interpretation, probabilities are the frequencies of occurrence of random events as proportions of a whole. Though his name has become associated with subjective probability, Bayes himself interpreted the theorem in an objective sense.

The theorem was given extra prominence by a theorem by physicist R. T. Cox which showed that any system of inference fitting certain requirements could be mapped onto probability. Bayes' Theorem has since found a wide variety of applications in science and engineering.

Statement
Bayes gave a special case involving continuous prior and posterior probability distributions and discrete probability distributions of data.

Discrete case
In its simplest setting involving only discrete distributions, 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 the prior probability or marginal probability of A. It is "prior" in the sense that it does not take into account any information about B.
 * P(A|B) is the conditional probability of A, given B. It is also called the posterior 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 a normalizing constant.

Intuitively, Bayes' theorem in this form describes the way in which one's beliefs about observing A are updated by having observed B.

Likelihood functions and continuous prior and posterior distributions
Suppose a continuous probability distribution with probability density function &fnof;&Theta; is assigned to an uncertain quantity &Theta;. (In the conventional language of mathematical probability theory &Theta; would be a "random variable", but in certain kinds of scientific applications such language may be considered objectionable.) The probability that the event B will be the outcome of an experiment depends on &Theta;; it is P(B | &Theta;). As a function of &Theta; this is the likelihood function:


 * $$ L(\theta) = P(B \mid \Theta = \theta) \, $$

Then the posterior probability distribution of &Theta;, i.e. the conditional probability distribution of &Theta; given the observed data B, has probability density function


 * $$ f_\Theta(\theta \mid B) = \text{constant}\cdot f_\Theta(\theta) L(B \mid \theta)\,$$,

where the "constant" is a normalizing constant so chosen as to make the integral of the function equal to 1, so that it is indeed a probability density function. This is the form of Bayes' theorem actually considered by Thomas Bayes.

In other words, Bayes' theorem says:


 * To get the posterior probability distribution, multiply the prior probability distribution by the likelihood function and then normalize.

More generally still, the new data B may be the value of an observed continuously distributed random variable X. The probability that it has any particular value is therefore 0. In such a case, the likelihood function is the value of a probability density function of X given &Theta;, rather than a probability of B given &Theta;:


 * $$ L(\theta) = f_X(x \mid \Theta = \theta)\, $$

The theorem still prescribes multiplying the prior distribution by the likelihood function and then normalizing, to get the posterior distribution.

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


 * $$P(A|B) = \frac{P(B|A) P(A)}{P(B)} = \frac{0.5 \times 0.4}{0.8} = 0.25$$

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 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

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), provided 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
Bayes' theorem is often completed by noting that, according to the Law of total probability,


 * $$P(B) = P(A\cap B) + P(A^C\cap B) = P(B|A) P(A) + P(B|A^C) P(A^C)$$.

where AC 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 {Ai} forms a partition of the event space,


 * $$P(A_i|B) = \frac{P(B | A_i)\, P(A_i)}{P(B)} = \frac{P(B | A_i)\, P(A_i)}{\sum_j P(B\cap A_j)} = \frac{P(B | A_i)\, P(A_i)}{\sum_j P(B|A_j)\,P(A_j)} \!$$

for any Ai in the partition.

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) are the (posterior) odds of A given B,


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

O(A) are the (prior) odds of A by itself


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

and Λ(A|B) is the likelihood ratio.


 * $$\Lambda (A|B) = \frac{P(B|A)}{P(B|A^C)} \!$$

For probability densities
There is also a version of Bayes' theorem for continuous distributions. It is somewhat harder to derive, since probability densities 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 originally used the proposition to find a continuous posterior distribution given discrete observations.

Bayes' theorem for probability densities is formally similar to the theorem for probabilities:


 * $$ f_X(x|Y=y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} = \frac{f_Y(y|X=x)\,f_X(x)}{f_Y(y)} = \frac{f_Y(y|X=x)\,f_X(x)}{\int_{-\infty}^{\infty} f_Y(y|X=\xi )\,f_X(\xi )\,d\xi }\!$$.

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 density function of X and Y,


 * $$ f_X(x|Y=y)\,$$

is the posterior probability density function 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 probability density functions of X and Y respectively, where ƒX(x) is the prior probability density function of X.

Abstract form
Given two absolutely continuous probability measures P ~ Q on the probability 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


 * $$E_P \left[ X|\mathcal{G} \right] = \frac{E_Q \left[ \left.\frac{dP}{dQ} X \right|\mathcal{G} \right] }{E_Q \left[\left. \frac{dP}{dQ}\right|\mathcal{G} \right] } $$.

Proof :

by definition of conditional probability,


 * $$ E_P \left[ X|\mathcal{G} \right] = \frac{E_P \left[ X 1_\mathcal{G} \right] }{P \left[ \mathcal{G} \right] }  = \frac{E_P \left[ X 1_\mathcal{G} \right] }{E_P \left[ 1_\mathcal{G} \right] } $$

We further have that


 * $$ E_Q \left[ \frac{dP}{dQ} X\right] = \int \frac{dP}{dQ} X \,dQ   = \int X \, dP = E_P \left[ X \right] $$.

Hence,


 * $$ E_Q \left[ \left.\frac{dP}{dQ} X\right|\mathcal{G} \right] = \frac{E_Q \left[ \frac{dP}{dQ} X 1_\mathcal{G} \right] }{E_Q \left[ 1_\mathcal{G} \right] } = \frac{E_P \left[ X 1_\mathcal{G} \right] }{E_Q \left[ 1_\mathcal{G} \right] }$$


 * $$ E_Q \left[ \left.\frac{dP}{dQ}\right|\mathcal{G} \right] = \frac{E_Q \left[ \frac{dP}{dQ} 1_\mathcal{G} \right] }{E_Q \left[ 1_\mathcal{G} \right] } = \frac{E_P \left[ 1_\mathcal{G} \right] }{E_Q \left[ 1_\mathcal{G} \right] }$$.

Summarizing, we obtain the desired result.

This formulation is used in Kalman filtering to find Zakai equations. It is also used in financial mathematics for change of numeraire techniques.

Extensions
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. 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).

Example 1: Drug testing
An example of the use of Bayes' theorem is the evaluation of drug test results. 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 can be used to demonstrate the relatively high likelihood of miss-classifying non-users as users. Let's assume a corporation decides to test its employees for drug use, and that only 0.5% of the employees actually use the drug. What is the probability that, given a positive drug test, an employee is actually a drug user? Let "D" stand for 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: Given this information, we can compute the posterior probability P(D|+) of an employee who tested positive actually being a drug user:
 * 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 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 a false 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 +.


 * $$\begin{align}P(D|+) & = \frac{P(+ | D) P(D)}{P(+)} \\

& = \frac{P(+ | D) P(D)}{P(+ | D) P(D) + P(+ | N) P(N)} \\ & = \frac{0.99 \times 0.005}{0.99 \times 0.005 + 0.01 \times 0.995} \\ & = 0.3322.\end{align}$$

Despite the specificity and sensitivity of the test, the low base-rate of use renders the accuracy of the test low: the probability that an employee who tests positive actually using drugs is only about 33%, so it is in fact 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 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.


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

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,



\begin{align} \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} \left / {11 \choose 3,7,1} \right. = {10 \choose 7} \, \frac{1}{1320} \end{align} $$

and the posterior distribution for r is then


 * $$ f(r | n=10, m=7) =

\frac = 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".

Example 3: The Monty Hall problem
We are presented with three doors—red, green, and blue—from which to choose, one of which has a prize hidden behind it. Suppose we choose the red door. The presenter, who knows where the prize is (and will not choose that door to open), opens the blue door and reveals that there is no prize behind it. He then asks if we wish to change our choice from our initial selection of red. Will changing our mind at this point improve our chances of winning the prize?

You might think that, with two doors left unopened, you have a 50:50 chance with either door, and so there is no point in changing doors. However, this is not the case. Let us call the situation that the prize is behind a given door Ar, Ag, and Ab.

To start with, $$P(A_r) = P(A_g) = P(A_b) = \tfrac 1 3$$, 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 presenter 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 presenter must pick the blue door. Thus, $$P(B|A_g) = 1$$
 * In the situation where the prize is behind the blue door, the presenter must pick the green door. Thus, $$P(B|A_b) = 0$$

Thus,



\begin{align} 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} = \tfrac 1 3 \\ P(A_g|B) & = \frac{P(B | A_g) P(A_g)}{P(B)} =   \frac{1 \cdot \frac 1 3}{\frac 1 2} = \tfrac 2 3 \\ 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{align} $$

So, we should always choose the green door. This assumes that the presenter chooses at random with equal probability. If the presenter ALWAYS chooses the blue door, it makes no difference what we do.

Note how this depends on the value of P(B). Another way of looking at the apparent inconsistency is that when you chose the first door, you had a 1/3 chance of being right. When the second door was removed from the list of possibilities, the conditional probability for the chosen door to be the winning door is also 1/3 and hence this left the last door with a 2/3 chance of being right.

Historical remarks
Bayes' theorem is named after the Reverend Thomas Bayes (1702–1761), 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 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 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.

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 describes the evolution of the field from "inverse probability" at the time of Bayes and Laplace, and even of Harold Jeffreys (1939) to "Bayesian" in the 1950s. The irony is that this label was introduced by R.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.

An investigation by Stigler. suggests that Bayes' theorem was discovered by Nicholas Saunderson some time before Bayes. However, this interpretation was later argued against by Edwards.

Versions of the essay

 * Thomas Bayes "An essay towards solving a Problem in the Doctrine of Chances". (Bayes' essay in the original notation)
 * Thomas Bayes "An essay towards solving a Problem in the Doctrine of Chances". (Bayes' essay in the original notation)
 * Thomas Bayes "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. "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.
 * 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.
 * Jeff Miller, et al., 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.
 * The on-line textbook: Information Theory, Inference, and Learning Algorithms, by David 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.
 * Stanford Encyclopedia of Philosophy: Inductive Logic provides a comprehensive Bayesian treatment of Inductive Logic and Confirmation Theory.
 * Weisstein, Eric W. Bayes' Theorem. From MathWorld--A Wolfram Web Resource. Accessed 15 June 2010.
 * * Eliezer S. Yudkowsky (2003), "An Intuitive Explanation of Bayesian Reasoning"
 * A tutorial on probability and Bayes’ theorem devised for Oxford University psychology students
 * Confirmation Theory An extensive presentation of Bayesian Confirmation Theory