Conditional independence

Let $A$, $B$, and $C$ be events. $A$ and $B$ are said to be conditionally independent given $C$ if and only if $P(C)\; >\; 0$ and:

:$P(A\; \backslash mid\; B,\; C)\; =\; P(A\; \backslash mid\; C)$

This property is often written: $(A\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; B\; \backslash mid\; C)$, which should be read $((A\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; B)\; \backslash vert\; C)$.

Equivalently, conditional independence may be stated as:

:$P(A,B|C)\; =\; P(A|C)P(B|C)$

where $P(A,B|C)$ is the joint probability of $A$ and $B$ given $C$. This alternate formulation states that $A$ and $B$ are independent events, given $C$.

It demonstrates that $(A\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; B\; \backslash mid\; C)$ is equivalent to $(B\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; A\; \backslash mid\; C)$.

:$P(A,\; B\; \backslash mid\; C)\; =\; P(A\backslash mid\; C)P(B\backslash mid\; C)$

:iff $\backslash frac\{P(A,\; B,\; C)\}\{P(C)\}\; =\; \backslash left(\backslash frac\{P(A,\; C)\}\{P(C)\}\backslash right)\; \backslash left(\backslash frac\{P(B,\; C)\}\{P(C)\}\; \backslash right)$      (definition of conditional probability)

:iff $P(A,\; B,\; C)\; =\; \backslash frac\{P(A,\; C)\; P(B,\; C)\}\{P(C)\}$       (multiply both sides by $P(C)$)

:iff $\backslash frac\{P(A,\; B,\; C)\}\{P(B,\; C)\}=\; \backslash frac\{P(A,\; C)\}\{P(C)\}$       (divide both sides by $P(B,\; C)$)

:iff $P(A\; \backslash mid\; B,\; C)\; =\; P(A\; \backslash mid\; C)$       (definition of conditional probability) $\backslash therefore$

Each cell represents a possible outcome. The events $\backslash color\{red\}R$, $\backslash color\{blue\}B$ and $\backslash color\{gold\}Y$ are represented by the areas shaded {{font color|red|red}}, {{font color|blue|blue}} and {{font color|gold|yellow}} respectively. The overlap between the events $\backslash color\{red\}R$ and $\backslash color\{blue\}B$ is shaded {{font color|purple|purple}}.

Image:Conditional independence.svg

The probabilities of these events are shaded areas with respect to the total area. In both examples $\backslash color\{red\}R$ and $\backslash color\{blue\}B$ are conditionally independent given $\backslash color\{gold\}Y$ because:

:$\backslash Pr(\{\backslash color\{red\}R\},\; \{\backslash color\{blue\}B\}\; \backslash mid\; \{\backslash color\{gold\}Y\})\; =\; \backslash Pr(\{\backslash color\{red\}R\}\; \backslash mid\; \{\backslash color\{gold\}Y\})\backslash Pr(\{\backslash color\{blue\}B\}\; \backslash mid\; \{\backslash color\{gold\}Y\})$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 (the purple shaded area) 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) = {{sfrac|2|12}} = {{sfrac|1|6}}. Similarly, Pr(R | Y) = {{sfrac|4|12}} = {{sfrac|1|3}} and Pr(B | Y) = {{sfrac|6|12}} = {{sfrac|1|2}}.

but not conditionally independent given $\backslash left[\; \backslash text\{not\; \}\{\backslash color\{gold\}Y\}\backslash right]$ because:

:$\backslash Pr(\{\backslash color\{red\}R\},\; \{\backslash color\{blue\}B\}\; \backslash mid\; \backslash text\{not\; \}\; \{\backslash color\{gold\}Y\})\; \backslash not=\; \backslash Pr(\{\backslash color\{red\}R\}\; \backslash mid\; \backslash text\{not\; \}\; \{\backslash color\{gold\}Y\})\backslash Pr(\{\backslash color\{blue\}B\}\; \backslash mid\; \backslash text\{not\; \}\; \{\backslash color\{gold\}Y\})$

Let events A and B be defined as the probability that person A and person B will be home in time for dinner where both people are randomly sampled from the entire world. Events A and B can be assumed to be independent i.e. knowledge that A is late has minimal to no change on the probability that B will be late. However, if a third event is introduced, person A and person B live in the same neighborhood, the two events are now considered not conditionally independent. Traffic conditions and weather-related events that might delay person A, might delay person B as well. Given the third event and knowledge that person A was late, the probability that person B will be late does meaningfully change.[https://math.stackexchange.com/q/23093 Could someone explain conditional independence?]

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 one 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.

Height and vocabulary are dependent since very small people tend to be children, known for their more basic vocabularies. But knowing that two people are 19 years old (i.e., conditional on age) there is no reason to think that one person's vocabulary is larger if we are told that they are taller.

Two discrete random variables $X$ and $Y$ are conditionally independent given a third discrete 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 $X$. Formally:

{{Equation box 1

|indent =

|title=

|equation = {{NumBlk||$(X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y)\; \backslash mid\; Z\; \backslash quad\; \backslash iff\; \backslash quad\; F\_\{X,Y\backslash ,\backslash mid\backslash ,Z\backslash ,=\backslash ,z\}(x,y)\; =\; F\_\{X\backslash ,\backslash mid\backslash ,Z\backslash ,=\backslash ,z\}(x)\; \backslash cdot\; F\_\{Y\backslash ,\backslash mid\backslash ,Z\backslash ,=\backslash ,z\}(y)\; \backslash quad\; \backslash text\{for\; all\; \}\; x,y,z$|{{EquationRef|Eq.2}}}}

|cellpadding= 6

|border

|border colour = #0073CF

|background colour=#F5FFFA}}

where $F\_\{X,Y\backslash ,\backslash mid\backslash ,Z\backslash ,=\backslash ,z\}(x,y)=\backslash Pr(X\; \backslash leq\; x,\; Y\; \backslash leq\; y\; \backslash mid\; Z=z)$ is the conditional cumulative distribution function of $X$ and $Y$ given $Z$.

Two events $R$ and $B$ are conditionally independent given a σ-algebra $\backslash Sigma$ if

:$\backslash Pr(R,\; B\; \backslash mid\; \backslash Sigma)\; =\; \backslash Pr(R\; \backslash mid\; \backslash Sigma)\backslash Pr(B\; \backslash mid\; \backslash Sigma)\; \backslash text\{\; a.s.\}$

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

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

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

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:

:$X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y\; \backslash mid\; W$ or

:$X\; \backslash perp\; Y\; \backslash mid\; W$

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

:$(X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y)\; \backslash mid\; W$

This notation extends $X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y$ for "$X$ is independent of $Y$."

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 $X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y$ neither implies nor is implied by $(X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y)\; \backslash mid\; W$.

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 $(X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y)\; \backslash mid\; W$. 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 $X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y$, each taking the values 0 and 1 with probability 0.5. Let $W$ be the product $X\; \backslash cdot\; Y$. Then when $W=0$, Pr(X = 0) = 2/3, but Pr(X = 0|Y = 0) = 1/2, so $(X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y)\; \backslash mid\; W$ is false.

This is also an example of Explaining Away. See Kevin Murphy's tutorial {{Cite web|url=http://people.cs.ubc.ca/~murphyk/Bayes/bnintro.html|title=Graphical Models}} where $X$ and $Y$ take the values "brainy" and "sporty".

Two random vectors $\backslash mathbf\{X\}=(X\_1,\backslash ldots,X\_l)^\{\backslash mathrm\; T\}$ and $\backslash mathbf\{Y\}=(Y\_1,\backslash ldots,Y\_m)^\{\backslash mathrm\; T\}$ are conditionally independent given a third random vector $\backslash mathbf\{Z\}=(Z\_1,\backslash ldots,Z\_n)^\{\backslash mathrm\; T\}$ if and only if they are independent in their conditional cumulative distribution given $\backslash mathbf\{Z\}$. Formally:

{{Equation box 1

|indent =

|title=

|equation = {{NumBlk||$(\backslash mathbf\{X\}\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; \backslash mathbf\{Y\})\; \backslash mid\; \backslash mathbf\{Z\}\; \backslash quad\; \backslash iff\; \backslash quad\; F\_\{\backslash mathbf\{X\},\backslash mathbf\{Y\}|\backslash mathbf\{Z\}=\backslash mathbf\{z\}\}(\backslash mathbf\{x\},\backslash mathbf\{y\})\; =\; F\_\{\backslash mathbf\{X\}\backslash ,\backslash mid\backslash ,\backslash mathbf\{Z\}\backslash ,=\backslash ,\backslash mathbf\{z\}\}(\backslash mathbf\{x\})\; \backslash cdot\; F\_\{\backslash mathbf\{Y\}\backslash ,\backslash mid\backslash ,\backslash mathbf\{Z\}\backslash ,=\backslash ,\backslash mathbf\{z\}\}(\backslash mathbf\{y\})\; \backslash quad\; \backslash text\{for\; all\; \}\; \backslash mathbf\{x\},\backslash mathbf\{y\},\backslash mathbf\{z\}$|{{EquationRef|Eq.3}}}}

|cellpadding= 6

|border

|border colour = #0073CF

|background colour=#F5FFFA}}

where $\backslash mathbf\{x\}=(x\_1,\backslash ldots,x\_l)^\{\backslash mathrm\; T\}$, $\backslash mathbf\{y\}=(y\_1,\backslash ldots,y\_m)^\{\backslash mathrm\; T\}$ and $\backslash mathbf\{z\}=(z\_1,\backslash ldots,z\_n)^\{\backslash mathrm\; T\}$ and the conditional cumulative distributions are defined as follows.

: $\backslash begin\{align\}$

F_{\mathbf{X},\mathbf{Y}\,\mid\,\mathbf{Z}\,=\,\mathbf{z}}(\mathbf{x},\mathbf{y}) &= \Pr(X_1 \leq x_1,\ldots,X_l \leq x_l, Y_1 \leq y_1,\ldots,Y_m \leq y_m \mid Z_1=z_1,\ldots,Z_n=z_n) \\[6pt]

F_{\mathbf{X}\,\mid\,\mathbf{Z}\,=\,\mathbf{z}}(\mathbf{x}) &= \Pr(X_1 \leq x_1,\ldots,X_l \leq x_l \mid Z_1=z_1,\ldots,Z_n=z_n) \\[6pt]

F_{\mathbf{Y}\,\mid\,\mathbf{Z}\,=\,\mathbf{z}}(\mathbf{y}) &= \Pr(Y_1 \leq y_1,\ldots,Y_m \leq y_m \mid Z_1=z_1,\ldots,Z_n=z_n)

\end{align}

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.

A set of rules governing statements of conditional independence have been derived from the basic definition.{{cite journal

|first=A. P. |last=Dawid |authorlink=Philip Dawid

|title=Conditional Independence in Statistical Theory

|journal=Journal of the Royal Statistical Society, Series B

|year=1979

|volume=41 |issue=1 |pages=1–31

|mr=0535541

|jstor=2984718

}}J Pearl, Causality: Models, Reasoning, and Inference, 2000, Cambridge University Press

These rules were termed "Graphoid Axioms"

by Pearl and Paz,{{cite conference

| last1 = Pearl | first1 = Judea | author1-link = Judea Pearl

| last2 = Paz | first2 = Azaria

| editor1-last = du Boulay | editor1-first = Benedict

| editor2-last = Hogg | editor2-first = David C.

| editor3-last = Steels | editor3-first = Luc

| contribution = Graphoids: Graph-Based Logic for Reasoning about Relevance Relations or When would x tell you more about y if you already know z?

| pages = 357–363

| publisher = North-Holland

| title = Advances in Artificial Intelligence II, Seventh European Conference on Artificial Intelligence, ECAI 1986, Brighton, UK, July 20–25, 1986, Proceedings

| url = https://ftp.cs.ucla.edu/pub/stat_ser/r53-L.pdf

| year = 1986}} because they hold in graphs, where $X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; A\backslash mid\; B$ is interpreted to mean: "All paths from X to A are intercepted by the set B".{{cite book|last1=Pearl|first1=Judea|title=Probabilistic reasoning in intelligent systems: networks of plausible inference|url=https://archive.org/details/probabilisticrea00pear|url-access=registration|date=1988|publisher=Morgan Kaufmann|isbn=9780934613736}}

: $$

X \perp\!\!\!\perp Y \mid Z \quad

\Leftrightarrow

\quad

Y \perp\!\!\!\perp X \mid Z

Proof:

From the definition of conditional independence,

: $$

X \perp\!\!\!\perp Y \mid Z \quad

\Leftrightarrow

\quad P(X, Y \mid Z) = P(X \mid Z) P(Y \mid Z) \quad

\Leftrightarrow

\quad

Y \perp\!\!\!\perp X \mid Z

: $$

X \perp\!\!\!\perp Y \mid Z

\quad \Rightarrow \quad

h(X) \perp\!\!\!\perp Y \mid Z

Proof

From the definition of conditional independence, we seek to show that:

: $$

X \perp\!\!\!\perp Y \mid Z

\quad \Rightarrow \quad

P(h(X), Y \mid Z) = P(h(X) \mid Z) P(Y \mid Z)

. The left side of this equality is:

: $$

P(h(X)=a, Y=y \mid Z=z) = \sum_{X \colon h(X)=a} P(X=x, Y=y \mid Z=z)

, where the expression on the right side of this equality is the summation over $X$ such that $h(X)=a$ of the conditional probability of $X,\; Y$ on $Z$.

Further decomposing,

: $$

\begin{align}

\sum_{X \colon h(X)=a} P(X=x, Y=y \mid Z=z) =& \sum_{X \colon h(X)=a} P(X=x \mid Z=z) P(Y=y \mid Z=z) \\

=& P(Y=y \mid Z=z) \sum_{X \colon h(X)=a} P(X=x \mid Z=z) \\

=& P(Y \mid Z) P (h(X) \mid Z)

\end{align}

. Special cases of this property include

(X, W) \perp\!\!\!\perp Y \mid Z

\quad \Rightarrow \quad

X \perp\!\!\!\perp Y \mid Z

• Proof: Let us define $A\; =\; (X,W)$ and $h(\backslash cdot)$ be an 'extraction' function $h(X,W)\; =\; X$. Then:

: $$

\begin{align}

(X,W) \perp\!\!\!\perp Y \mid Z

\quad &\Leftrightarrow \quad

A \perp\!\!\!\perp Y \mid Z \\

&\Rightarrow \quad

h(A) \perp\!\!\!\perp Y \mid Z \quad &\text{Decomposition} \\

&\Leftrightarrow \quad

X \perp\!\!\!\perp Y \mid Z

\end{align}

X \perp\!\!\!\perp (Y, W) \mid Z

\quad \Rightarrow \quad

X \perp\!\!\!\perp Y \mid Z

• Proof: Let us define $V\; =\; (Y,W)$ and $h(\backslash cdot)$ be again an 'extraction' function $h(Y,W)\; =\; Y$. Then:

: $$

\begin{align}

X \perp\!\!\!\perp (Y,W) \mid Z

\quad &\Leftrightarrow \quad

X \perp\!\!\!\perp V \mid Z \\

&\Leftrightarrow \quad

V \perp\!\!\!\perp X \mid Z \quad &\text{Symmetry} \\

&\Rightarrow \quad

h(V) \perp\!\!\!\perp X \mid Z \quad &\text{Decomposition} \\

&\Leftrightarrow \quad

Y \perp\!\!\!\perp X \mid Z \\

&\Leftrightarrow \quad

X \perp\!\!\!\perp Y \mid Z \quad &\text{Symmetry}

\end{align}

: $$

X \perp\!\!\!\perp Y \mid Z

\quad \Rightarrow \quad

X \perp\!\!\!\perp Y \mid (Z, h(X))

Proof:

Given $X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y\; \backslash mid\; Z$, we aim to show

: $$

\begin{align}

X \perp\!\!\!\perp Y \mid (Z, h(X))

\quad &\Leftrightarrow \quad

X \perp\!\!\!\perp Y \mid U \quad &\text{where} \quad U = (Z, h(X)) \\

&\Leftrightarrow \quad

Y \perp\!\!\!\perp X \mid U \quad &\text{Symmetry} \\

&\Leftrightarrow \quad

P(Y\mid X, U) = P(Y\mid U) \\

&\Leftrightarrow \quad

P(Y \mid X, Z, h(X)) = P(Y \mid Z, h(X))

\end{align}

. We begin with the left side of the equation

: $$

\begin{align}

P(Y \mid X, Z, h(X)) &= P(Y \mid X, Z) \\

&= P(Y \mid Z) &\text{Since by symmetry } Y \perp\!\!\!\perp X \mid Z

\end{align}

. From the given condition

: $$

\begin{align}

X \perp\!\!\!\perp Y \mid Z

\quad &\Rightarrow \quad

h(X) \perp\!\!\!\perp Y \mid Z

\quad &\text{Decomposition} \\

&\Leftrightarrow \quad

Y \perp\!\!\!\perp h(X) \mid Z

\quad &\text{Symmetry} \\

&\Rightarrow \quad

P(Y \mid Z, h(X)) = P(Y \mid Z)

\end{align}

. Thus $P(Y\; \backslash mid\; X,\; Z,\; h(X))\; =\; P(Y\; \backslash mid\; Z,\; h(X))$

, so we have shown that $$

X \perp\!\!\!\perp Y \mid (Z, h(X))

.

Special Cases:

Some textbooks present the property as

• $X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; (Y,\; W)\; \backslash mid\; Z$

\quad \Rightarrow \quad

X \perp\!\!\!\perp Y \mid (Z, W) {{cite book |last1=Koller |first1=Daphne |last2=Friedman |first2=Nir |title=Probabilistic Graphical Models |date=2009 |publisher=The MIT Press |location=Cambridge, MA |isbn=9780262013192}}.

• $(X,W)\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y\; \backslash mid\; Z$

\quad \Rightarrow \quad

X \perp\!\!\!\perp Y \mid (Z,W) .

Both versions can be shown to follow from the weak union property given initially via the same method as in the decomposition section above.

: $$

\left.\begin{align}

X \perp\!\!\!\perp A \mid B \\

X \perp\!\!\!\perp B

\end{align}\right\}\text{ and }

\quad \Rightarrow \quad

X \perp\!\!\!\perp A,B

Proof

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

For strictly positive probability distributions, the following also holds:

: $$

\left.\begin{align}

X \perp\!\!\!\perp Y \mid Z, W\\

X \perp\!\!\!\perp W \mid Z, Y

\end{align}\right\}\text{ and }

\quad \Rightarrow \quad

X \perp\!\!\!\perp W, Y \mid Z

Proof

By assumption:

: $P(X|Z,\; W,\; Y)\; =\; P(X|Z,\; W)\; \backslash land\; P(X|Z,\; W,\; Y)\; =\; P(X|Z,\; Y)\; \backslash implies\; P(X|Z,\; Y)\; =\; P(X|Z,\; W)$

Using this equality, together with the Law of total probability applied to $P(X|Z)$:

: $\backslash begin\{align\}$

P(X|Z) &= \sum_{w \in W} P(X|Z, W=w)P(W=w|Z) \\[4pt]

&= \sum_{w \in W} P(X|Y, Z)P(W=w|Z) \\[4pt]

&= P(X|Z, Y) \sum_{w \in W} P(W=w|Z) \\[4pt]

&= P(X|Z, Y)

\end{align}

Since $P(X|Z,\; W,\; Y)\; =\; P(X|Z,\; Y)$ and $P(X|Z,\; Y)\; =\; P(X|Z)$, it follows that $P(X|Z,\; W,\; Y)\; =\; P(X|Z)\; \backslash iff\; X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y,W\; |\; Z$.

Technical note: since these implications hold for any probability space, they will still hold if one considers a sub-universe by conditioning everything on another variable, say K. For example, $X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y\; \backslash Rightarrow\; Y\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; X$ would also mean that $X\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; Y\; \backslash mid\; K\; \backslash Rightarrow\; Y\; \backslash perp\backslash !\backslash !\backslash !\backslash perp\; X\; \backslash mid\; K$.

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

{{Reflist}}

• {{Commons category-inline}}

{{DEFAULTSORT:Conditional Independence}}

Category:Independence (probability theory)