What is a strictly positive probability distribution?

Question

I'm reading about Markov Random Fields. In the wiki page it's written that

When the joint probability distribution of the random variables is strictly positive,...

I'm so confused! Because a probability distribution gives the probability of the random variable having a specific value and so a probability (per definition) can never be negative! So what do they mean by strictly positive?

Answer 1

At Markov random field we read:

When the joint probability distribution of the random variables is strictly positive, it is also referred to as a Gibbs random field, because, according to the Hammersley–Clifford theorem, it can then be represented by a Gibbs measure for an appropriate (locally defined) energy function.

The words Hammersley–Clifford theorem are a clickable link, and there we read about:

a probability distribution that has a positive mass or density

If it were to assign positive probability to every event, then it would assign positive probability to every point, and thus it would be a discrete distribution. Positive density, however means it assigns positive probability to every set whose "measure" is positive. Every probability density is a density with respect to some measure. For example, consider the density $$ f(x) = \begin{cases} 1/3 & \text{if }0<x<1, \\ 2/3 & \text{if } 1<x<2, \\ 0 & \text{otherwise}. \end{cases} $$ The probability that this assigns to the interval $(1.5,2)$ is the measure of that interval, which is $1/2$, i.e. just the length of the interval, times the density on that set, which is $2/3$. More generally, the probability of a subset $A$ is the integral over the set $A$ of the density with respect to the underlying measure.

So it appears that what is intended to be positive rather than zero at that point in that article is the density.

I have now edited the article thus.