Conditional independence

In probability theory, two events R and B are conditionally independent given a third event Y precisely if the occurrence or nonoccurrence of R and the occurrence or nonoccurrence of B are independent events in their conditional probability distribution given Y. In other words, R and B are conditionally independent if and only if, given knowledge of whether Y occurs, knowledge of whether R occurs provides no information on the likelihood of B occurring, and knowledge of whether B occurs provides no information on the likehood of R occurring.
In the standard notation of probability theory, R and B are conditionally independent given Y if and only if
or equivalently,
Two random variables X and Y are conditionally independent given a third random variable Z if and only if they are independent in their conditional probability distribution given Z. That is, X and Y are conditionally independent given Z if and only if, given any value of Z, the probability distribution of X is the same for all values of Y and the probability distribution of Y is the same for all values of X.
Two events R and B are conditionally independent given a σalgebra Σ if
where denotes the conditional expectation of the indicator function of the event A, χ_{A}, given the sigma algebra Σ. That is,
Two random variables X and Y are conditionally independent given a σalgebra Σ if the above equation holds for all R in σ(X) and B in σ(Y).
Two random variables X and Y are conditionally independent given a random variable W if they are independent given σ(W): the σalgebra generated by W. This is commonly written:
 or
This is read "X is independent of Y, given W"; the conditioning applies to the whole statement: "(X is independent of Y) given W".
If W assumes a countable set of values, this is equivalent to the conditional independence of X and Y for the events of the form [W = w]. Conditional independence of more than two events, or of more than two random variables, is defined analogously.
The following two examples show that X ⊥ Y neither implies nor is implied by X ⊥ Y  W. First, suppose W is 0 with probability 0.5 and is the value 1 otherwise. When W = 0 take X and Y to be independent, each having the value 0 with probability 0.99 and the value 1 otherwise. When W = 1, X and Y are again independent, but this time they take the value 1 with probability 0.99. Then X ⊥ Y  W. But X and Y are dependent, because Pr(X = 0) < Pr(X = 0Y = 0). This is because Pr(X = 0) = 0.5, but if Y = 0 then it's very likely that W = 0 and thus that X = 0 as well, so Pr(X = 0Y = 0) > 0.5. For the second example, suppose X ⊥ Y, each taking the values 0 and 1 with probability 0.5. Let W be the product X×Y. Then when W = 0, Pr(X = 0) = 2/3, but Pr(X = 0Y = 0) = 1/2, so X ⊥ Y  W is false. This also an example of Explaining Away. See Kevin Murphy's tutorial ^{[2]} where X and Y take the values "brainy" and "sporty".
Contents
Uses in Bayesian statistics
Let p be the proportion of voters who will vote "yes" in an upcoming referendum. In taking an opinion poll, one chooses n voters randomly from the population. For i = 1, ..., n, let X_{i} = 1 or 0 according as the ith chosen voter will or will not vote "yes".
In a frequentist approach to statistical inference one would not attribute any probability distribution to p (unless the probabilities could be somehow interpreted as relative frequencies of occurrence of some event or as proportions of some population) and one would say that X_{1}, ..., X_{n} are independent random variables.
By contrast, in a Bayesian approach to statistical inference, one would assign a probability distribution to p regardless of the nonexistence of any such "frequency" interpretation, and one would construe the probabilities as degrees of belief that p is in any interval to which a probability is assigned. In that model, the random variables X_{1}, ..., X_{n} are not independent, but they are conditionally independent given the value of p. In particular, if a large number of the Xs are observed to be equal to 1, that would imply a high conditional probability, given that observation, that p is near 1, and thus a high conditional probability, given that observation, that the next X to be observed will be equal to 1.
Rules of conditional independence
A set of rules governing statements of conditional independence have been derived from the basic definition.^{[3]}^{[4]}
Note: since these implications hold for any probability space, they will still hold if considers a subuniverse by conditioning everything on another variable, say K. For example, would also mean that .
Note: below, the comma can be read as an "AND".
Symmetry
Decomposition
Proof:
 p_{X,A,B}(x,a,b) = p_{X}(x)p_{A,B}(a,b) (meaning of )
 (ignore variable by integrating it out)
 p_{X,A}(x,a) = p_{X}(x)p_{A}(a)
repeat proof to show independence of X and B.
Weak union
Contraction
Contractionweakuniondecomposition
Putting the above three together, we have:
Intersection
If the probabilities of X, A, B are all positive^{[citation needed]}, then the following also holds:
References
 ^ To see that this is the case, one needs to realise that Pr(R ∩ B  Y) is the probability of an overlap of R and B in the Y area. Since, in the picture on the left, there are two squares where R and B overlap within the Y area, and the Y area has twelve squares, Pr(R ∩ B  Y) = = . Similarly, Pr(R  Y) = = and Pr(B  Y) = = .
 ^ http://people.cs.ubc.ca/~murphyk/Bayes/bnintro.html
 ^ Dawid, A. P. (1979). "Conditional Independence in Statistical Theory". Journal of the Royal Statistical Society Series B 41 (1): 1–31. JSTOR 2984718. MR0535541.
 ^ J Pearl, Causality: Models, Reasoning, and Inference, 2000, Cambridge University Press
See also
Categories: Probability theory
 Statistical dependence
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