Conditional independence

See also: Conditional dependence

In probability theory, two events R and B are conditionally independent given a third event Y precisely if the occurrence or non-occurrence of R and the occurrence or non-occurrence of B are independent events in their conditional probability distribution given Y. In other words, R and B are conditionally independent given Y if and only if, given knowledge that 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 likelihood of R occurring.

Formal definition

These are two examples illustrating conditional independence. Each cell represents a possible outcome. The events R, B and Y are represented by the areas shaded red, blue and yellow respectively. The overlap between the events R and B is shaded purple. The probabilities of these events are shaded areas with respect to the total area. In both examples R and B are conditionally independent given Y because: $\Pr(R\cap B\mid Y)=\Pr(R\mid Y)\Pr(B\mid Y)\,$
but not conditionally independent given not Y because: : $\Pr(R\cap B\mid {\text{not }}Y)\not =\Pr(R\mid {\text{not }}Y)\Pr(B\mid {\text{not }}Y).\,$

In the standard notation of probability theory, R and B are conditionally independent given Y if and only if

$\Pr(R\cap B\mid Y)=\Pr(R\mid Y)\Pr(B\mid Y),\,$

or equivalently,

$\Pr(R\mid B\cap Y)=\Pr(R\mid Y).\,$

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 σ-algebra Σ if

$\Pr(R\cap B\mid \Sigma )=\Pr(R\mid \Sigma )\Pr(B\mid \Sigma )\ a.s.$

where $\Pr(A\mid \Sigma )$ denotes the conditional expectation of the indicator function of the event $A$, $\chi _{A}$, given the sigma algebra $\Sigma$. That is,

$\Pr(A\mid \Sigma ):=\operatorname {E} [\chi _{A}\mid \Sigma ].$

Two random variables X and Y are conditionally independent given a σ-algebra 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 σ-algebra generated by W. This is commonly written:

$X\perp \!\!\!\perp Y\mid W$ or

$X\perp Y\mid W$

This is read "X is independent of Y, given W"; the conditioning applies to the whole statement: "(X is independent of Y) given W".

$(X\perp \!\!\!\perp Y)\mid W$

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 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 = 0|Y = 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 = 0|Y = 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 = 0|Y = 0) = 1/2, so X ⊥ Y | W is false. This is also an example of Explaining Away. See Kevin Murphy's tutorial where X and Y take the values "brainy" and "sporty".

An example

The discussion on StackExchange provides a couple of useful examples.

1. Let the two events be the probabilities of persons A and B getting home in time for dinner, and the third event is the fact that a snow storm hit the city. While both A and B have a lower probability of getting home in time for dinner, the lower probabilities will still be independent of each other. That is, the knowledge that A is late does not tell you whether B will be late. (They may be living in different neighborhoods, traveling different distances, and using different modes of transportation.) However, if you have information that they live in the same neighborhood, use the same transportation, and work at the same place, then the two events are NOT conditionally independent.
2. Conditional independence depends on the nature of the third event. If you roll two dice, one may assume that the two dice behave independently of each other. Looking at the results of 1 die will not tell you about the result of the second die. (That is, the two dice are independent.) If, however, the 1st die's result is a 3, and someone tells you about a third event - that the sum of the two results is even - then this extra unit of information restricts the options for the 2nd result to an odd number. In other words, two events can be independent, but NOT conditionally independent.
3. Height and vocabulary are not independent; but they are conditionally independent if you add age.

Uses in Bayesian inference

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 corresponding, respectively, to whether or not 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 X₁, ..., 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 non-existence 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 X₁, ..., 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.

Note: below, the comma can be read as an "AND".

Symmetry

$X\perp \!\!\!\perp Y\quad \Rightarrow \quad Y\perp \!\!\!\perp X$

Decomposition

$X\perp \!\!\!\perp A,B\quad \Rightarrow \quad {\text{ and }}{\begin{cases}X\perp \!\!\!\perp A\\X\perp \!\!\!\perp B\end{cases}}$

Proof:

• $p_{X,A,B}(x,a,b)=p_{X}(x)p_{A,B}(a,b)$      (meaning of $X\perp \!\!\!\perp A,B$)
• $\int _{B}\!p_{X,A,B}(x,a,b)=\int _{B}\!p_{X}(x)p_{A,B}(a,b)$      (ignore variable B by integrating it out)
• $p_{X,A}(x,a)=p_{X}(x)p_{A}(a)$     

A similar proof shows the independence of X and B.

Weak union

$X\perp \!\!\!\perp A,B\quad \Rightarrow \quad {\text{ and }}{\begin{cases}X\perp \!\!\!\perp A\mid B\\X\perp \!\!\!\perp B\mid A\end{cases}}$

Proof:

• By definition, $\Pr(X)=\Pr(X\mid A,B)$.
• Due to the property of decomposition $X\perp \!\!\!\perp B$, $\Pr(X)=\Pr(X\mid B)$.
• Combining the above two equalities gives $\Pr(X\mid B)=\Pr(X\mid A,B)$, which establishes $X\perp \!\!\!\perp A\mid B$.

The second condition can be proved similarly.

Contraction

$\left.{\begin{aligned}X\perp \!\!\!\perp A\mid B\\X\perp \!\!\!\perp B\end{aligned}}\right\}{\text{ and }}\quad \Rightarrow \quad X\perp \!\!\!\perp A,B$

Proof:

This property can be proved by noticing $\Pr(X\mid A,B)=\Pr(X\mid B)=\Pr(X)$, each equality of which is asserted by $X\perp \!\!\!\perp A\mid B$ and $X\perp \!\!\!\perp B$, respectively.

Contraction-weak-union-decomposition

Putting the above three together, we have:

$\left.{\begin{aligned}X\perp \!\!\!\perp A\mid B\\X\perp \!\!\!\perp B\end{aligned}}\right\}{\text{ and }}\quad \iff \quad X\perp \!\!\!\perp A,B\quad \Rightarrow \quad {\text{ and }}{\begin{cases}X\perp \!\!\!\perp A\mid B\\X\perp \!\!\!\perp B\\X\perp \!\!\!\perp B\mid A\\X\perp \!\!\!\perp A\\\end{cases}}$

Intersection

For strictly positive probability distributions, the following also holds:

$\left.{\begin{aligned}X\perp \!\!\!\perp A\mid C,B\\X\perp \!\!\!\perp B\mid C,A\end{aligned}}\right\}{\text{ and }}\quad \Rightarrow \quad X\perp \!\!\!\perp B,A\mid C$

The five rules above were termed "Graphoid Axioms" by Pearl and Paz, because they hold in graphs, if $X\perp \!\!\!\perp A\mid B$ is interpreted to mean: "All paths from X to A are intercepted by the set B".

See also

• Graphoid
• Conditional dependence
• de Finetti's theorem
• Conditional expectation