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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    $\begingroup$ Various properties of distributions and their manipulation tend to break as soon as you have a literal 0 probability of something. $\endgroup$ – Peteris Apr 6 '14 at 11:06
  • $\begingroup$ Can we see what it this "intersection" property ? $\endgroup$ – Stéphane Laurent Apr 8 '14 at 6:31
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    $\begingroup$ @StéphaneLaurent Done (enlarged the quote from Pearl's book $\endgroup$ – 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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    $\begingroup$ Aren't all distribution "nonnegative"? $\endgroup$ – Neil G Apr 7 '14 at 6:39
  • $\begingroup$ Very much not so. A lot of distributions can take negative values. Standard normal comes to mind as the most common example. $\endgroup$ – while Apr 7 '14 at 7:22
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    $\begingroup$ What is $x$, user11852 ? @while, you are talking about the support of the distribution. $\endgroup$ – Stéphane Laurent Apr 7 '14 at 8:16
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    $\begingroup$ Modifying a denumerable number of values of a density does not change the distribution, so I would really be surprised that such a positivity condition could be relevant. $\endgroup$ – Stéphane Laurent Apr 8 '14 at 6:31
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    $\begingroup$ @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. $\endgroup$ – 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.


A strictly positive probability distribution over a state space simply means that all states are possible, ie no state has a probability of zero. All states have a probability greater than zero. "Strictly positive" means greater than zero.

Strictly positive does not imply that the probability of any state could be negative. There is no such thing as negative probability.

  • $\begingroup$ For continuous distributions you would have to say positive probability density everywhere. Never 0 for any finite value. $\endgroup$ – Michael Chernick Aug 13 '18 at 13:18
  • $\begingroup$ Allan, could you supply a reference for this notion of "strictly positive"? It conflicts with other answers in this thread, so we need to come to some resolution of the difference. @Michael Consider the distribution of $Y=UX$ where $U$ is a Rademacher variable and independently $X$ has a Gamma$(k)$ distribution with $k\gt 1.$ $Y$ has a density function defined everywhere. Would you exclude this example because its density at $0$ is zero? $\endgroup$ – whuber Aug 13 '18 at 13:24
  • $\begingroup$ I am not sure of what the definition is either but the way I interpret it the answer to your question would be yes. $\endgroup$ – Michael Chernick Aug 13 '18 at 13:28

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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