Bayes' theorem

Bayes' theorem may be derived from the definition of conditional probability:

${\displaystyle P(A\mid B)={\frac {P(A\cap B)}{P(B)}},{\text{ if }}P(B)\neq 0,}$

where ${\displaystyle P(A\cap B)}$ is the probability of both A and B being true. Similarly,

${\displaystyle P(B\mid A)={\frac {P(A\cap B)}{P(A)}},{\text{ if }}P(A)\neq 0,}$

Solving for ${\displaystyle P(A\cap B)}$ and substituting into the above expression for ${\displaystyle P(A\mid B)}$ yields Bayes' theorem:

${\displaystyle P(A\mid B)={\frac {P(B\mid A)P(A)}{P(B)}},{\text{ if }}P(B)\neq 0.}$

For two continuous random variables X and Y, Bayes' theorem may be analogously derived from the definition of conditional density:

${\displaystyle f_{X\mid Y=y}(x)={\frac {f_{X,Y}(x,y)}{f_{Y}(y)}}}$

${\displaystyle f_{Y\mid X=x}(y)={\frac {f_{X,Y}(x,y)}{f_{X}(x)}}}$

Therefore,

${\displaystyle f_{X\mid Y=y}(x)={\frac {f_{Y\mid X=x}(y)f_{X}(x)}{f_{Y}(y)}}.}$

Let ${\displaystyle P_{Y}^{x}}$ be the conditional distribution of ${\displaystyle Y}$ given ${\displaystyle X=x}$ and let ${\displaystyle P_{X}}$ be the distribution of ${\displaystyle X}$. The joint distribution is then ${\displaystyle P_{X,Y}(dx,dy)=P_{Y}^{x}(dy)P_{X}(dx)}$. The conditional distribution ${\displaystyle P_{X}^{y}}$ of ${\displaystyle X}$ given ${\displaystyle Y=y}$ is then determined by

$${\displaystyle P_{X}^{y}(A)=E(1_{A}(X)|Y=y)}$$

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.[4] The Bayes theorem determines the posterior distribution from the prior distribution. Bayes theorem can be generalized to include improper prior distributions such as the uniform distribution on the real line.[5] Modern Markov Chain Monte Carlo methods have boosted the importance of Bayes theorem including cases with improper priors.[6]

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.

Figure 1: Using a frequency box to show ${\displaystyle P({\text{User}}\mid {\text{Positive}})}$ visually by comparison of shaded areas

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 ${\displaystyle P({\text{User}}\mid {\text{Positive}})}$ mean "the probability that someone is a cannabis user given that they test positive," which is what is meant by PPV. We can write:

${\displaystyle {\begin{aligned}P({\text{User}}\mid {\text{Positive}})&={\frac {P({\text{Positive}}\mid {\text{User}})P({\text{User}})}{P({\text{Positive}})}}\\&={\frac {P({\text{Positive}}\mid {\text{User}})P({\text{User}})}{P({\text{Positive}}\mid {\text{User}})P({\text{User}})+P({\text{Positive}}\mid {\text{Non-user}})P({\text{Non-user}})}}\\[8pt]&={\frac {0.90\times 0.05}{0.90\times 0.05+0.20\times 0.95}}={\frac {0.045}{0.045+0.19}}\approx 19\%\end{aligned}}}$

The fact that ${\displaystyle P({\text{Positive}})=P({\text{Positive}}\mid {\text{User}})P({\text{User}})+P({\text{Positive}}\mid {\text{Non-user}})P({\text{Non-user}})}$ is a direct application of the Law of Total Probability. In this case, it says that the probability that someone tests positive is the probability that a user tests positive, times the probability of being a user, plus the probability that a non-user tests positive, times the probability of being a non-user. This is true because the classifications user and non-user form a partition of a set, namely the set of people who take the drug test. This combined with the definition of conditional probability results in the above statement.

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:

• 950 are non-users and 190 of them give false positive (0.20 × 950)
• 50 of them are users and 45 of them give true positive (0.90 × 50)

The 1,000 people thus yields 235 positive tests, of which only 45 are genuine drug users, about 19%. See Figure 1 for an illustration using a frequency box, and note how small the pink area of true positives is compared to the blue area of false positives.

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%.

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.

Which can then be used to calculate the probability of having cancer when you have the symptoms:

${\displaystyle {\begin{aligned}P({\text{Cancer}}|{\text{Symptoms}})&={\frac {P({\text{Symptoms}}|{\text{Cancer}})P({\text{Cancer}})}{P({\text{Symptoms}})}}\\&={\frac {P({\text{Symptoms}}|{\text{Cancer}})P({\text{Cancer}})}{P({\text{Symptoms}}|{\text{Cancer}})P({\text{Cancer}})+P({\text{Symptoms}}|{\text{Non-Cancer}})P({\text{Non-Cancer}})}}\\[8pt]&={\frac {1\times 0.00001}{1\times 0.00001+(10/99999)\times 0.99999}}={\frac {1}{11}}\approx 9.1\%\end{aligned}}}$

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 X_i 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:

${\displaystyle P(X_{A})=0.2,\quad P(X_{B})=0.3,\quad P(X_{C})=0.5.}$

If the item was made by the first machine, then the probability that it is defective is 0.05; that is, P(Y | X_A) = 0.05. Overall, we have

${\displaystyle P(Y|X_{A})=0.05,\quad P(Y|X_{B})=0.03,\quad P(Y|X_{C})=0.01.}$

To answer the original question, we first find P(Y). That can be done in the following way:

${\displaystyle P(Y)=\sum _{i}P(Y|X_{i})P(X_{i})=(0.05)(0.2)+(0.03)(0.3)+(0.01)(0.5)=0.024.}$

Hence, 2.4% of the total output is defective.

We are given that Y has occurred, and we want to calculate the conditional probability of X_C. By Bayes' theorem,

${\displaystyle P(X_{C}|Y)={\frac {P(Y|X_{C})P(X_{C})}{P(Y)}}={\frac {0.01\cdot 0.50}{0.024}}={\frac {5}{24}}}$

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(X_C) = 1/2 by the smaller posterior probability P(X_C | Y) = 5/24.

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,

• P (A), the prior, is the initial degree of belief in A.
• P (A | B), the posterior, is the degree of belief after incorporating news that B is true.
• the quotient P(B | A)/P(B) represents the support B provides for A.

For more on the application of Bayes' theorem under the Bayesian interpretation of probability, see Bayesian inference.

Figure 3: Illustration of frequentist interpretation with tree diagrams.

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 such as Figure 3. 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.

Figure 4: Tree diagram illustrating the beetle example. R, C, P and ${\displaystyle {\overline {P}}}$ are the events rare, common, pattern and no pattern. Percentages in parentheses are calculated. Three independent values are given, so it is possible to calculate the inverse tree.

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),

${\displaystyle {\begin{aligned}P({\text{Rare}}\mid {\text{Pattern}})&={\frac {P({\text{Pattern}}\mid {\text{Rare}})P({\text{Rare}})}{P({\text{Pattern}})}}\\[8pt]&={\frac {P({\text{Pattern}}\mid {\text{Rare}})P({\text{Rare}})}{P({\text{Pattern}}\mid {\text{Rare}})P({\text{Rare}})+P({\text{Pattern}}\mid {\text{Common}})P({\text{Common}})}}\\[8pt]&={\frac {0.98\times 0.001}{0.98\times 0.001+0.05\times 0.999}}\\[8pt]&\approx 1.9\%\end{aligned}}}$

For events A and B, provided that P(B) ≠ 0,

${\displaystyle P(A|B)={\frac {P(B|A)P(A)}{P(B)}}.}$

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:

${\displaystyle P(A|B)\propto P(A)\cdot P(B|A).}$

In words, the posterior is proportional to the prior times the likelihood.[7]

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

${\displaystyle P(A|B)=c\cdot P(A)\cdot P(B|A){\text{ and }}P(\neg A|B)=c\cdot P(\neg A)\cdot P(B|\neg A).}$

Adding these two formulas we deduce that

${\displaystyle 1=c\cdot (P(B|A)\cdot P(A)+P(B|\neg A)\cdot P(\neg A)),}$

or

${\displaystyle c={\frac {1}{P(B|A)\cdot P(A)+P(B|\neg A)\cdot P(\neg A)}}={\frac {1}{P(B)}}.}$

Another form of Bayes' theorem for two competing statements or hypotheses is:

${\displaystyle P(A|B)={\frac {P(B|A)P(A)}{P(B|A)P(A)+P(B|\neg A)P(\neg A)}}.}$

For an epistemological interpretation:

For proposition A and evidence or background B,[8]

• ${\displaystyle P(A)}$ is the prior probability, the initial degree of belief in A.
• ${\displaystyle P(\neg A)}$ is the corresponding initial degree of belief in not-A, that A is false, where ${\displaystyle P(\neg A)=1-P(A)}$
• ${\displaystyle P(B|A)}$ is the conditional probability or likelihood, the degree of belief in B given that proposition A is true.
• ${\displaystyle P(B|\neg A)}$ is the conditional probability or likelihood, the degree of belief in B given that proposition A is false.
• ${\displaystyle P(A|B)}$ is the posterior probability, the probability of A after taking into account B.

Often, for some partition {A_j} of the sample space, the event space is given in terms of P(A_j) and P(B | A_j). It is then useful to compute P(B) using the law of total probability:

${\displaystyle P(B)={\sum _{j}P(B|A_{j})P(A_{j})},}$

${\displaystyle \Rightarrow P(A_{i}|B)={\frac {P(B|A_{i})P(A_{i})}{\sum \limits _{j}P(B|A_{j})P(A_{j})}}\cdot }$

In the special case where A is a binary variable:

${\displaystyle P(A|B)={\frac {P(B|A)P(A)}{P(B|A)P(A)+P(B|\neg A)P(\neg A)}}\cdot }$

Figure 5: Bayes' theorem applied to an event space generated by continuous random variables X and Y. There exists an instance of Bayes' theorem for each point in the domain. In practice, these instances might be parametrized by writing the specified probability densities as a function of x and y.

Consider a sample space Ω generated by two random variables X and Y. In principle, Bayes' theorem applies to the events A = {X = x} and B = {Y = y}.

${\displaystyle P(X{=}x|Y{=}y)={\frac {P(Y{=}y|X{=}x)P(X{=}x)}{P(Y{=}y)}}}$

However, terms become 0 at points where either variable has finite probability density. To remain useful, Bayes' theorem must be formulated in terms of the relevant densities (see Derivation).

If X is continuous and Y is discrete,

${\displaystyle f_{X|Y{=}y}(x)={\frac {P(Y{=}y|X{=}x)f_{X}(x)}{P(Y{=}y)}}}$

where each ${\displaystyle f}$ is a density function.

If X is discrete and Y is continuous,

${\displaystyle P(X{=}x|Y{=}y)={\frac {f_{Y|X{=}x}(y)P(X{=}x)}{f_{Y}(y)}}.}$

If both X and Y are continuous,

${\displaystyle f_{X|Y{=}y}(x)={\frac {f_{Y|X{=}x}(y)f_{X}(x)}{f_{Y}(y)}}.}$

Figure 6: A way to conceptualize event spaces generated by continuous random variables X and 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 f_Y(y), this becomes an integral:

${\displaystyle f_{Y}(y)=\int _{-\infty }^{\infty }f_{Y|X=\xi }(y)f_{X}(\xi )\,d\xi .}$

Bayes' theorem in odds form is:

${\displaystyle O(A_{1}:A_{2}\mid B)=O(A_{1}:A_{2})\cdot \Lambda (A_{1}:A_{2}\mid B)}$

where

${\displaystyle \Lambda (A_{1}:A_{2}\mid B)={\frac {P(B\mid A_{1})}{P(B\mid A_{2})}}}$

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

${\displaystyle O(A_{1}:A_{2})={\frac {P(A_{1})}{P(A_{2})}},}$

${\displaystyle O(A_{1}:A_{2}\mid B)={\frac {P(A_{1}\mid B)}{P(A_{2}\mid B)}},}$

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 ${\displaystyle A_{1}=A}$ and ${\displaystyle A_{2}=\neg A}$, one writes ${\displaystyle O(A)=O(A:\neg A)=P(A)/(1-P(A))}$, and uses a similar abbreviation for the Bayes factor and for the conditional odds. The odds on ${\displaystyle A}$ is by definition the odds for and against ${\displaystyle A}$. Bayes' rule can then be written in the abbreviated form

${\displaystyle O(A\mid B)=O(A)\cdot \Lambda (A\mid B),}$

or, in words, the posterior odds on ${\displaystyle A}$ equals the prior odds on ${\displaystyle A}$ times the likelihood ratio for ${\displaystyle A}$ given information ${\displaystyle B}$. In short, posterior odds equals prior odds times likelihood ratio.

For example, if a medical test has a sensitivity of 90% and a specificity of 91%, then the positive Bayes factor is ${\displaystyle \Lambda _{+}=P({\text{True Positive}})/P({\text{False Positive}})=90\%/(100\%-91\%)=10}$. Now, if the prevalence of this disease is 9.09%, and if we take that as the prior probability, then the prior odds is about 1:10. So after receiving a positive test result, the posterior odds of actually having the disease becomes 1:1; In other words, the posterior probability of actually having the disease is 50%. If a second test is performed in serial testing, and that also turns out to be positive, then the posterior odds of actually having the disease becomes 10:1, which means a posterior probability of about 90.91%. The negative Bayes factor can be calculated to be 91%/(100%-90%)=9.1, so if the second test turns out to be negative, then the posterior odds of actually having the disease is 1:9.1, which means a posterior probability of about 9.9%.

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

${\displaystyle {\frac {91}{909}}\times {\frac {90.1\%}{9.0\%}}={\frac {91\times 90.1\%}{909\times 9.0\%}}=1:1}$

which is consistent with the fact that there are 82 true positives and 82 false positives in the group of 1000 people.

Using ${\displaystyle P(\neg B\mid A)=1-P(B\mid A)}$ twice, one may use Bayes' theorem to also express ${\displaystyle P(\neg B\mid \neg A)}$ in terms of ${\displaystyle P(A\mid B)}$ and without negations:

${\displaystyle P(\neg B\mid \neg A)=1-\left(1-P(A\mid B)\right){\frac {P(B)}{P(\neg A)}}}$,

when ${\displaystyle P(\neg A)=1-P(A)\neq 0}$. From this we can read off the inference

${\displaystyle P(A\mid B)=1\implies P(\neg B\mid \neg A)=1}$.

In words: If certainly ${\displaystyle B}$ implies ${\displaystyle A}$, we infer that certainly ${\displaystyle \neg A}$ implies ${\displaystyle \neg B}$. Where ${\displaystyle P(B)\neq 0}$, the two implications being certain are equivalent statements. In the probability formulas, the conditional probability ${\displaystyle P(A\mid B)}$ generalizes the logical implication ${\displaystyle B\implies A}$, where now beyond assigning true or false, we assign probability values to statements. The assertion of ${\displaystyle B\implies A}$ is captured by certainty of the conditional, the assertion of ${\displaystyle P(A\mid B)=1}$. Relating the directions of implication, Bayes' theorem represents a generalization of the contraposition law, which in classical propositional logic can be expressed as:

${\displaystyle (B\implies A)\iff (\neg A\implies \neg B)}$.

Note that in this relation between implications, the positions of ${\displaystyle A}$ resp. ${\displaystyle B}$ get flipped.

The corresponding formula in terms of probability calculus is Bayes' theorem, which in its expanded form involving the prior probability/base rate ${\displaystyle a}$ of only ${\displaystyle A}$, is expressed as:[9]

${\displaystyle P(A\mid B)=P(B\mid A){\frac {a(A)}{P(B\mid A)\,a(A)+P(B\mid \neg A)\,a(\neg A)}}}$.

Bayes' theorem represents a special case of deriving inverted conditional opinions in subjective logic expressed as:

${\displaystyle (\omega _{A{\tilde {|}}B}^{S},\omega _{A{\tilde {|}}\lnot B}^{S})=(\omega _{B\mid A}^{S},\omega _{B\mid \lnot A}^{S}){\widetilde {\phi }}a_{A},}$

where ${\displaystyle {\widetilde {\phi }}}$ denotes the operator for inverting conditional opinions. The argument ${\displaystyle (\omega _{B\mid A}^{S},\omega _{B\mid \lnot A}^{S})}$ denotes a pair of binomial conditional opinions given by source ${\displaystyle S}$, and the argument ${\displaystyle a_{A}}$ denotes the prior probability (aka. the base rate) of ${\displaystyle A}$. The pair of derivative inverted conditional opinions is denoted ${\displaystyle (\omega _{A{\tilde {|}}B}^{S},\omega _{A{\tilde {|}}\lnot B}^{S})}$. The conditional opinion ${\displaystyle \omega _{A\mid B}^{S}}$ generalizes the probabilistic conditional ${\displaystyle P(A\mid B)}$, i.e. in addition to assigning a probability the source ${\displaystyle S}$ can assign any subjective opinion to the conditional statement ${\displaystyle (A\mid B)}$. A binomial subjective opinion ${\displaystyle \omega _{A}^{S}}$ is the belief in the truth of statement ${\displaystyle A}$ with degrees of epistemic uncertainty, as expressed by source ${\displaystyle S}$. Every subjective opinion has a corresponding projected probability ${\displaystyle P(\omega _{A}^{S})}$. The application of Bayes' theorem to projected probabilities of opinions is a homomorphism, meaning that Bayes' theorem can be expressed in terms of projected probabilities of opinions:

${\displaystyle P(\omega _{A{\tilde {|}}B}^{S})={\frac {P(\omega _{B\mid A}^{S})a(A)}{P(\omega _{B\mid A}^{S})a(A)+P(\omega _{B\mid \lnot A}^{S})a(\lnot A)}}.}$

Hence, the subjective Bayes' theorem represents a generalization of Bayes' theorem.[10]

A conditioned version of the Bayes' theorem[11] results from the addition of a third event ${\displaystyle C}$ on which all probabilities are conditioned:

${\displaystyle P(A\mid B\cap C)={\frac {P(B\mid A\cap C)\,P(A\mid C)}{P(B\mid C)}}}$

Using the chain rule

${\displaystyle P(A\cap B\cap C)=P(A\mid B\cap C)\,P(B\mid C)\,P(C)}$

And, on the other hand

${\displaystyle P(A\cap B\cap C)=P(B\cap A\cap C)=P(B\mid A\cap C)\,P(A\mid C)\,P(C)}$

The desired result is obtained by identifying both expressions and solving for ${\displaystyle P(A\mid B\cap C)}$.

In the case of 3 events - A, B, and C - it can be shown that:

$${\displaystyle P(A\mid B,C)={\frac {P(B\mid A,C)\;P(A\mid C)}{P(B\mid C)}}}$$

Proof[12]

$${\displaystyle {\begin{aligned}P(A\mid B,C)&={\frac {P(A,B,C)}{P(B,C)}}\\[1ex]&={\frac {P(B\mid A,C)\,P(A,C)}{P(B,C)}}\\[1ex]&={\frac {P(B\mid A,C)\,P(A\mid C)\,P(C)}{P(B,C)}}\\[1ex]&={\frac {P(B\mid A,C)\,P(A\mid C)P(C)}{P(B\mid C)P(C)}}\\[1ex]&={\frac {P(B\mid A,C)\;P(A\mid C)}{P(B\mid C)}}\end{aligned}}}$$

Example of a Bayesian analysis table for a female individual's risk for a disease based on the knowledge that the disease is present in her siblings but not in her parents or any of her four children. Based solely on the status of the subject's siblings and parents, she is equally likely to be a carrier as to be a non-carrier (this likelihood is denoted by the Prior Hypothesis). However, the probability that the subject's four sons would all be unaffected is 1/16 (1⁄2·1⁄2·1⁄2·1⁄2) if she is a carrier, about 1 if she is a non-carrier (this is the Conditional Probability). The Joint Probability reconciles these two predictions by multiplying them together. The last line (the Posterior Probability) is calculated by dividing the Joint Probability for each hypothesis by the sum of both joint probabilities.[28]

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,[29] located on the q arm of chromosome 7.[30]

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:

Given that the patient is unaffected, there are only three possibilities. Within these three, there are two scenarios in which the patient carries the mutant allele. Thus the prior probabilities are 2⁄3 and 1⁄3.

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.

After carrying out the same analysis on the patient's male partner (with a negative test result), the chances of their child being affected is equal to the product of the parents' respective posterior probabilities for being carriers times the chances that two carriers will produce an affected offspring (1⁄4).

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 scan2. 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).[28]

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.