Editing Conditional independence (section)

==Conditional independence of random variables==
Two discrete [[random variable]]s <math>X</math> and <math>Y</math> are conditionally independent given a third discrete random variable <math>Z</math> if and only if they are [[Independence (probability theory)|independent]] in their [[conditional probability distribution]] given <math>Z</math>. That is, <math>X</math> and <math>Y</math> are conditionally independent given <math>Z</math> if and only if, given any value of <math>Z</math>, the probability distribution of <math>X</math> is the same for all values of <math>Y</math> and the probability distribution of <math>Y</math> is the same for all values of <math>X</math>. Formally:

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|equation = {{NumBlk||<math>(X \perp\!\!\!\perp Y) \mid Z \quad \iff \quad F_{X,Y\,\mid\,Z\,=\,z}(x,y) = F_{X\,\mid\,Z\,=\,z}(x) \cdot F_{Y\,\mid\,Z\,=\,z}(y) \quad \text{for all } x,y,z</math>|{{EquationRef|Eq.2}}}}
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where <math>F_{X,Y\,\mid\,Z\,=\,z}(x,y)=\Pr(X \leq x, Y \leq y \mid Z=z)</math> is the conditional [[cumulative distribution function]] of <math>X</math> and <math>Y</math> given <math>Z</math>.

Two events <math>R</math> and <math>B</math> are conditionally independent given a [[sigma-algebra|&sigma;-algebra]] <math>\Sigma</math> if

:<math>\Pr(R, B \mid \Sigma) = \Pr(R \mid \Sigma)\Pr(B \mid \Sigma) \text{ a.s.}</math>

where <math>\Pr(A \mid \Sigma) </math> denotes the [[conditional expectation]] of the [[indicator function]] of the event <math>A</math>, <math>\chi_A</math>, given the sigma algebra <math>\Sigma</math>. That is,

:<math>\Pr(A \mid \Sigma) := \operatorname{E}[\chi_A\mid\Sigma].</math>

Two random variables <math>X</math> and <math>Y</math> are conditionally independent given a σ-algebra <math>\Sigma</math> if the above equation holds for all <math>R</math> in <math>\sigma(X)</math> and <math>B</math> in <math>\sigma(Y)</math>.

Two random variables <math>X</math> and <math>Y</math> are conditionally independent given a random variable <math>W</math> if they are independent given ''σ''(''W''): the σ-algebra generated by <math>W</math>. This is commonly written:

:<math>X \perp\!\!\!\perp Y \mid W </math> or
:<math>X \perp Y \mid W</math>

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

:<math>(X \perp\!\!\!\perp Y) \mid W</math>

This notation extends <math>X \perp\!\!\!\perp Y</math> for "<math>X</math> is [[Independence (probability theory)|independent]] of <math>Y</math>."

If <math>W</math> assumes a countable set of values, this is equivalent to the conditional independence of ''X'' and ''Y'' for the events of the form <math>[W=w]</math>.
Conditional independence of more than two events, or of more than two random variables, is defined analogously.

The following two examples show that <math>X \perp\!\!\!\perp Y</math> ''neither implies nor is implied by'' <math>(X \perp\!\!\!\perp Y) \mid W</math>.

First, suppose <math>W</math> is 0 with probability 0.5 and 1 otherwise.  When ''W''&nbsp;=&nbsp;0 take <math>X</math> and <math>Y</math> to be independent, each having the value 0 with probability 0.99 and the value 1 otherwise.  When <math>W=1</math>, <math>X</math> and <math>Y</math> are again independent, but this time they take the value 1 with probability 0.99.  Then <math>(X \perp\!\!\!\perp Y) \mid W</math>. But <math>X</math> and <math>Y</math> are dependent, because Pr(''X''&nbsp;=&nbsp;0) < Pr(''X''&nbsp;=&nbsp;0|''Y''&nbsp;=&nbsp;0).  This is because Pr(''X''&nbsp;=&nbsp;0) =&nbsp;0.5, but if ''Y''&nbsp;=&nbsp;0 then it's very likely that ''W''&nbsp;=&nbsp;0 and thus that ''X''&nbsp;=&nbsp;0 as well, so Pr(''X''&nbsp;=&nbsp;0|''Y''&nbsp;=&nbsp;0)&nbsp;>&nbsp;0.5.

For the second example, suppose <math>X \perp\!\!\!\perp Y</math>, each taking the values 0 and 1 with probability&nbsp;0.5. Let <math>W</math> be the product <math>X \cdot Y</math>.  Then when <math>W=0</math>, Pr(''X''&nbsp;=&nbsp;0)&nbsp;=&nbsp;2/3, but Pr(''X''&nbsp;=&nbsp;0|''Y''&nbsp;=&nbsp;0)&nbsp;=&nbsp;1/2, so <math>(X \perp\!\!\!\perp Y) \mid W</math> is false.
This is also an example of Explaining Away. See Kevin Murphy's tutorial <ref>{{Cite web|url=http://people.cs.ubc.ca/~murphyk/Bayes/bnintro.html|title=Graphical Models}}</ref> where <math>X</math> and <math>Y</math> take the values "brainy" and "sporty".