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Take a random $n \times n$ nonnegative symmetric matrix $D$ with zero diagonal. What is the probability that it is an abstract distance matrix, i.e. satisfies $D_{xy}+D_{yz} \geq D_{xz}$ for all index triples $x,y,z$?

In other words: what is the probability that a nonnegative function on $V \times V$, where $V$ is some finite set, defines a finite metric space?

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I added "pr." to probability, to use an existing tag rather than to create a new one, and added the random-matrices as it seems to fit. Other question, is it clear how you 'pick' such a matrix at random or are you interested in results for any way of doing this. –  quid Mar 4 '13 at 15:05
Do you want to impose $D_{xx}=0$ for each $x$, and that $D_{xy}=D_{yx}$? –  Anthony Quas Mar 4 '13 at 16:22
@quid: Any way is interesting. Thanks for the edit! –  Felix Goldberg Mar 5 '13 at 11:27
@AnthonyQuas: Yes, thanks. –  Felix Goldberg Mar 5 '13 at 11:27
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1 Answer

Since you haven't given a distribution, let me make an observation giving the right form of the answer in the case where the $D_{xy}$ are independent uniform $[0,1]$ random variables.

I want to claim that the probability, $p$, that the matrix defines a metric satisfies $\alpha^{n^2}\le p\le \beta^{n^2}$ for constants $\alpha$ and $\beta$.

First, notice that if all the $\binom n2$ edge lengths are in $[\frac12,1]$, then the triangle inequality is satisfied, so that $p\ge 2^{-\binom n2}$.

Conversely, consider the triples $(b,b+a,b+2a)$ where $a$ is an odd number in the range $1$ to $\frac n4$ and $1\le b\le a$. Notice that no two of these triples contain two elements in common. There are $\Theta(n^2)$ such triples. For such a triple, consider the event $E_{a,b}$ that $D_{b,b+2a}\le D_{b,b+a}+D_{b+a,b+2a}$. These events all have the same probability that is strictly less than 1. They are also independent, since no edge occurs in two events. Hence we obtain the upper bound.

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The lower bound argument works also if all distances are in $[\frac13,\frac23]$, or more generally in $[a,2a]$; perhaps one can find a better bound this way? –  Joel David Hamkins Mar 5 '13 at 11:56
For other distributions, the logarithm of the probability is either $0$ or $\Theta(n^2)$ by essentially the same argument. –  Douglas Zare Mar 5 '13 at 12:56
@Douglas: Just saw your answer (almost identical to the above) to a previous incarnation of this question (on which I commented). –  Anthony Quas Mar 5 '13 at 13:20
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