# What is a “strictly positive distribution”?

I am reading Judea Pearl's "Causality" (second edition 2009) and in section 1.1.5 Conditional Independence and Graphoids, he states:

The following is a (partial) list of properties satisfied by the conditional independence relation (X_||_Y | Z).

• Symmetry: (X_||_ Y | Z) ==> (Y_||_X | Z).
• Decomposition: (X_||_ YW | Z) ==> (X_||_Y | Z).
• Weak union: (X_||_ YW | Z) ==> (X_||_Y | ZW).
• Contraction: (X_||_ Y | Z) & (X_||_ W | ZY) ==> (X_||_ YW | Z).
• Intersection: (X_||_ W | ZY) & (X_||_ Y | ZW) (X_||_ YW | Z).

(Intersection is valid in strictly positive probability distributions.)

(formula (1.28) given earlier in the publicatiob: [(X_||_ Y | Z) iff P(X | Y,Z ) = P(X | Z) )

But what is an "strictly positive distribution" in general terms, and what distinguishes a "strictly positive distribution" form a distribution that is not strictly positive?

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Various properties of distributions and their manipulation tend to break as soon as you have a literal 0 probability of something. – Peteris Apr 6 '14 at 11:06
Can we see what it this "intersection" property ? – Stéphane Laurent Apr 8 '14 at 6:31
@StéphaneLaurent Done (enlarged the quote from Pearl's book – Willemien Apr 8 '14 at 8:39

A strictly positive distribution $D_{sp}$ has values $D_{sp}(x)>0$ for all $x$. This is different from a non-negative distribution $D_{nn}$ where $D_{nn}(x) \geq 0$.

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Aren't all distribution "nonnegative"? – Neil G Apr 7 '14 at 6:39
Very much not so. A lot of distributions can take negative values. Standard normal comes to mind as the most common example. – while Apr 7 '14 at 7:22
What is $x$, user11852 ? @while, you are talking about the support of the distribution. – Stéphane Laurent Apr 7 '14 at 8:16
Ah, sorry. You are absolutely right. Disregard my previous comment. – while Apr 7 '14 at 8:20
@StéphaneLaurent: I do not understand the point of your first comment as I never said something to that extend. Regarding your example with the $\Gamma$ whether or not you use $(0,\infty)$ or $[0,\infty)$ does not really matter in the sense that any function $g(x)$ that agrees with $f(x)$ everywhere except for a finite number of points is a member of the same equivalence class as $f(x)$ and to all intents and purposes is the same function. And as for support if you define as "the smallest closed set whose complement has probability zero" you alleviate any positivity concerns. – usεr11852 Apr 8 '14 at 18:09

The mass of each ball bearing in a population of ball bearings would be strictly positive because something with zero mass cannot be a ball bearing.

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As an example illustrating the definition of a strictly positive probability distribution in action (Courtesy of an old paper by Richard Holley on FKG Inequalities), imagine that we have $\Lambda$ which is a finite fixed set. Imagine also that we have $\Gamma$, which is a sublattice of the lattice of subsets of $\Lambda$. Let us then let $\mu$ be a strictly positive probability distribution on some finite distributed lattice $\Gamma$. For $\mu$ to be strictly positive, $\mu(A)>0$ for all $A\in\Gamma$ and $\sum_{A\in\Gamma}\mu(A)=1$

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