# Conditional Independence

Given mutually exclusive sets of random variables $$\mathbf{X},\mathbf{Y},\mathbf{Z}$$ we say $$\mathbf{X}$$ is conditionally independent of $$\mathbf{Y}$$ given $$\mathbf{Z}$$, denoted $$\mathbf{X} -||- \mathbf{Y}|\mathbf{Z}$$ or $$\mathbb{I}_\mathbb{P}(X,Y|Z)$$ if $$\mathbb{P}(\mathbf{X},\mathbf{Y}|\mathbf{Z}) = \mathbb{P}(\mathbf{X}|\mathbf{Z})\mathbb{P}(\mathbf{Y}|\mathbf{Z})$$.

Conditional independence (and dependence) can be graphically represented in Bayesian Networks (BNs). Let $$X,Y,Z$$ be random variables, we then have:

• The chains $$X \leftarrow Z \leftarrow Y$$, $$X\rightarrow Z \rightarrow Y$$ represent $$\mathbb{I}_\mathbb{P}(X,Y|Z)$$. To see this, without loss of generality take the chain $$X \rightarrow Z \rightarrow Y$$, then $$\mathbb{P}(X,Y|Z)=\frac{\mathbb{P}(X,Y,Z)}{\mathbb{P}(Z)}=\frac{\mathbb{P}(Y|Z)\mathbb{P}(X|Z)\mathbb{P}(Z)}{\mathbb{P}(Z)} =\mathbb{P}(X|Z)\mathbb{P}(Y|Z)$$. Here, we use Bayes theorem in the first equality and the Markov property in the second one.
• The fork/common cause $$X \leftarrow Z \rightarrow Y$$ represents $$\mathbb{I}(X,Y|Z)$$. The reasoning is exactly similar to above.
• The v-structure/immorality $$X \rightarrow Z \leftarrow Y$$ represents the conditional dependence relation $$\not \mathbb{I}_\mathbb{P}(X,Y|Z)$$. To see this, $$\mathbb{P}(X,Y|Z)=\frac{\mathbb{P}(X,Y,Z)}{\mathbb{P}(Z)}=\frac{\mathbb{P}(Y)\mathbb{P}(X)\mathbb{P}(Z|X,Y)}{\mathbb{P}(Z)} \neq \mathbb{P}(X|Z)\mathbb{P}(Y|Z)$$.

## Thoughts

• Find a way to write that independence symbol.

Created: 2022-04-04 Mon 23:39

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