$\newcommand{\ci}{\!\perp\!\!\!\perp\!}$On page 11 of the book in the title, Pearl introduces the Dawid notation for conditional independence: $(X\ci Y|Z)_P$ if and only if $P(x|y,z)=P(x|z)$ for all values $x,y,z$ such that $P(y,z)>0.$ A little later on on the same page, Pearl introduces the graphoid axioms: \begin{align*} \text{Symmetry: } (X\ci Y|Z)&\implies(Y\ci X|Z)\\ \text{Decomposition: } (X\ci YW|Z)&\implies(X\ci Y|Z)\\ \text{Weak union: } (X\ci YW|Z)&\implies(X\ci Y|ZW)\\ \text{Contraction: } (X\ci Y|Z)\land(X\ci W|ZY)&\implies(X\ci YW|Z)\\ \text{Intersection: } (X\ci W|ZY)\land(X\ci Y|ZW)&\implies(X\ci YW|Z). \end{align*}

My question is this: what does the notation $YW$ stand for in Decomposition? Or what does $ZW$ stand for in Weak union? The author never explains that notation. Is it set union?

I have looked at this thread, but none of the answers appear to be certain of themselves!

Many thanks for your time!


1 Answer 1


$\newcommand{\ci}{\!\perp\!\!\!\perp\!}$The notation $YW$ here stands for the set of variables $\{Y, W\}$.

Thus, for instance, $(X\ci YW|Z)$ means$P(X, Y, W |Z) = P(X|Z)P(Y, W|Z)$, for all instantiations of the variables.

  • $\begingroup$ What would be the difference, if any, between $YW=\{Y,W\}$ and $YW=Y\cup W?$ $\endgroup$ Apr 27, 2020 at 19:20
  • 1
    $\begingroup$ @AdrianKeister say $Y$ and $W$ are real numbers. The random variable $YW = \{Y, W\}$ is a vector valued random variable. I'm not sure what your union notation would mean, since the union makes sense with respect to events. But also the union of events would not work here, what we are computing is P(X = x, Y=y, Z = z) not P(X = x, Y = y or Z =z). $\endgroup$ Apr 29, 2020 at 1:50
  • $\begingroup$ In the errata, on page 11, Pearl adds this: (We use $YW$ to abbreviate $Y\cup W.$) $\endgroup$ May 15, 2020 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.