Editing Conditional independence (section)

===Weak union===

: <math>
X \perp\!\!\!\perp Y \mid Z
\quad \Rightarrow \quad
X \perp\!\!\!\perp Y \mid (Z, h(X))
</math>

'''Proof:'''

Given <math> X \perp\!\!\!\perp Y \mid Z </math>, we aim to show
: <math>
\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}
</math>
. We begin with the left side of the equation
: <math>
\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}
</math>
. From the given condition
: <math>
\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}
</math>
. Thus <math> P(Y \mid X, Z, h(X)) = P(Y \mid Z, h(X))
</math>, so we have shown that <math>
X \perp\!\!\!\perp Y \mid (Z, h(X))
</math>.

'''Special Cases:'''

Some textbooks present the property as
* <math> X \perp\!\!\!\perp (Y, W) \mid Z
\quad \Rightarrow \quad
X \perp\!\!\!\perp Y \mid (Z, W) </math> <ref name="Koller">{{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}}</ref>.

* <math> (X,W)  \perp\!\!\!\perp Y \mid Z
\quad \Rightarrow \quad
X \perp\!\!\!\perp Y \mid (Z,W) </math>.
Both versions can be shown to follow from the weak union property given initially via the same method as in the decomposition section above.