# Notational confusion about conditional independence in Pearl 2009

First I read this definition which introduces $$X$$, $$Y$$ and $$Z$$ as sets of random variables.

Definition (Pearl 2009)

Let $$V = \{V_1, V_2, \ldots \}$$ be a finite set of variables. Let $$P(\cdot)$$ be a joint probability function over the variables in $$V$$, and let $$X$$, $$Y$$, $$Z$$ stand for any three subsets of the variables in $$V$$. The sets $$X$$ and $$Y$$ are said to be conditionally independent given $$Z$$ if

$$P(x | y, z) = P(x|z)$$ whenever $$P(y,z) > 0$$.

Pearl then introduces Dawid's notation:

$$(X \mathrel{\unicode{x2AEB}} Y | Z)_P \iff P(x | y, z) = P(x | z)$$

That's all fine as far as I understand, but I got confused by the decomposition property of conditional independence:

$$(X \mathrel{\unicode{x2AEB}} YW | Z)_P \implies (X \mathrel{\unicode{x2AEB}} Y | Z)_P$$

The symbol $$W$$ is not defined, nor is putting it to the right of $$Y$$ in the form $$YW$$. I'm not sure if there is a set operation implied here. What does this notation mean?

• All of this is on page 11, for those that have Pearl's 2009 textbook. Commented Feb 19, 2023 at 23:28
• Commented May 22, 2023 at 5:30

I think $$X, Y, Z, W$$ should all be interpreted as random variables, and "$$YW$$" is not the product of $$Y$$ and $$W$$, but the random vector $$(Y, W)$$. Therefore, the expression means: \begin{align} (X \mathrel{\unicode{x2AEB}} (Y, W)|Z)_P \implies (X \mathrel{\unicode{x2AEB}} Y|Z)_P, \end{align} i.e., if $$X$$ and $$(Y, W)$$ are conditionally independent given $$Z$$, then $$X$$ and $$Y$$ are conditionally independent given $$Z$$. This interpretation is supported by the following statement which appears in the second paragraph after the expression:
The decomposition axiom asserts that if two combined items of information are judged irrelevant to $$X$$, then each separate item is irrelevant as well.
The mathematical proof for this property should also be straightforward: $$(X \mathrel{\unicode{x2AEB}} (Y, W)|Z)_P$$ means that for any $$A_1 \in \sigma(X), A_2 \in \sigma(Y, W)$$, it holds that (see this thread for a rigorous definition of conditional independence): \begin{align} P(A_1A_2|\sigma(Z)) = P(A_1|\sigma(Z))P(A_2|\sigma(Z)). \tag{1} \end{align}
Since $$\sigma(Y) \subset \sigma(Y, W)$$, $$(1)$$ holds for any $$A_2 \in \sigma(Y)$$, hence $$(X \mathrel{\unicode{x2AEB}} Y | Z)_P$$.