I suddenly remembered that there is an English version of Hertz' method in a review by Chandrasekhar (Rev. Mod. Phys, 15, (1943), 2) of which I attached the relevant appendix. Chandrasekhar only treats 3 D, but it is easy to use this method for any dimension, which is in  fact what Hertz does. Hertz' formula (7) is Chandrasekhar's formula (669). In (7) k(r) is the volume of a N dimensional sphere. Hertz solves the resulting differential equation for any dimension. It is relatively simple to do the whole thing yourself, and even though the paper is in German, there is enough math that you should be able to  follow without translating the text. If you need help, let me know.

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some elements here, but not the answer you're looking for :

http://math.stackexchange.com/questions/64028/expectation-of-pairwise-differences-between-uniform-random-points-in-hypercube

also (but for k = 3 only !) :

https://people.kth.se/~johanph/habc.pdf

and some references in the attached paper

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An additional element: I'm looking into this as a way to determine whether two discrete probability distributions (vectors X and Y) can be considered significantly different.

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then, i suppose you know this one :

http://www.mii.lt/informatica/pdf/INFO558.pdf

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I think the answer to your question is given in an old (1908) paper by Hertz: P. Hertz, Uber den gegenseitigen durchschnittlichen Abstand von Punkten, die mit bekannter mittlerer Dichte in Raume angeordnet sind. (Average distance between points in a system of points of known average density). His derivation is for all dimensions, but he gives explicit expressions for the first four on p. 397.

I suddenly remembered that there is an English version of Hertz' method in a review by Chandrasekhar (Rev. Mod. Phys, 15, (1943), 2) of which I attached the relevant appendix. Chandrasekhar only treats 3 D, but it is easy to use this method for any dimension, which is in  fact what Hertz does. Hertz' formula (7) is Chandrasekhar's formula (669). In (7) k(r) is the volume of a N dimensional sphere. Hertz solves the resulting differential equation for any dimension. It is relatively simple to do the whole thing yourself, and even though the paper is in German, there is enough math that you should be able to  follow without translating the text. If you need help, let me know.

@Gert

Very interesting reference, thank you.

f.

Sorry, but I have some problem with my browser and it posts an empty message, so I retry to write it. From your second message it seems that you have two populations X and Y and you have to test if they have the same discrete distribution. Is it the case?

Hi Andrea,

Well, actually I have the distributions, not the populations.

f.

Perhaps the following will help.

PROPOSITION. If a d+1 dimensional random vector X= (X0,X1,...,Xd) possesses a uniform pdf concentrated on the simplex {xi>0 for every i, x0+x1+...+xd=1}, and if this random vector is independent of the random variable S of standard gamma distribution with exponent d+1, then the components of the random vector Y =S X are iid with standard exponential pdf.

Francesco> ... the constraint that x1 + ... + xn = 1 and y1 + ... + yn = 1.

Let also the coordinates of the points be uniformly distributed in the [0,1] interval, i.e.: xi,yi ~ U[0,1].

What do you mean? For n≠2, the constraint is not consistent with the requirement of uniform distribution of each coordinate on [0, 1]. Do you perhaps mean the probability distribution

P(x1, x2, ..., xn) = (n-1)! δ(x1 + x2 + ... + xn -1) when all 0

Dear Kåre Olaussen,

I have assumed your supposition about the uniform didtribution over the simplex, since according to my experience - the way Francesco Bianconi has presented the problem is very popular among applications. And the marginal density (the same for each coordinate) equals as you have written.

BTW: isn't it interesting that the same is the pdf of the min{U_1,U_2,...,U_{n-1}} whenever the U-s are random variables independent identically distributed (iid) uniformly over [0,1]? I explain this as follows: If one posed at random n-1 points uniformly within [0,1], then the distances between consecutive order statistics possess the joint pd as the n coordinates x1,x2,...xn in your answer. In particular their marginals have the same pd as the first order statistics min{U_1,U_2,...,U_{n-1} }, or that P{ min{U_1,U_2,...,U_{n-1} } >x } = (1-x)^{n-1}.

Best, Joachim

Regarding the question of the ``distance'' between probability distributions, the usual approach is provided by the so-called Kullback-Leibler divergence, described, for instance, here: http://hanj.cs.illinois.edu/cs412/bk3/KL-divergence.pdf

Of course this can't settle the question, whether this distance is ``significant'' or not, anymore than that of any distance-that depends on what quantity sets the scale of distances in the problem in question (and whether such a scale even exists, which goes under the study of whether invariance under conformal transformations is a useful concept for the problem studied).

Dear Stam Nicolis,

Thanks for interesting and valuable reference. However the question is not about the distance between distributions but the rndom vectors treated as a random variable (thus having the probability distribution following from the probability distributions of the components.

Best, Joachim Domsta

In the case n = 1, if we exclude the proposed constraints x1 = 1 and y1 =1 (which imply a related constant zero distribution), then the determination of cumulative distributtion function (CDF) of the distance |x1 - y1| reduces to the well known problem in Geometric Probability “Probability that two persons will meet”. Namely, the probability of the event |x1 - y1| < d with 0 ≤ d ≤ 1 is actually area equal to 1 – 2(1 -d)2/2 = 2 d – d2. Hence, CDF F1 of the random variable |x1 - y1| is defined as: F1(x) = 0 for x < 0, F1(x) = 2x – x2 for 0 ≤ x ≤ 1 and F1(x) = 1 for x > 1.

Notice also that a related problem (without constraints) was solved for the cases n = 2 and n = 3 in J. Philip’s paper “THE PROBABILITY DISTRIBUTION OF THE DISTANCE BETWEEN TWO RANDOM POINTS IN A BOX” available at https://people.kth.se/~johanph/habc.pdf.

Notice that for n = 2, under the proposed constraints x1 + x2 = 1 and y1 + y2 = 1, the condition that the Euclidean distance betweeen points X(x1, y1) and Y(x2, y2) is less than d (0 ≤ d ≤ √2) reduces to the inequality |x1 - y1| < d/√2. Hence, the detrrmination of CDF F2(x) of the desired random variable reduced to the known problem in Geometric Probability “Probability that two persons will meet” (presented in my previous answer, with x/√2 instead of x and √2 instead of 1).

I am now trying to exploit the formula presented above by Kåre Olaussen, that the pdf for a single Xi as well Yi is equal to

(n-1) (1-x)^{n-2} i=1,2,..,n.

Now, since the pair consistes of independent variables, hence the expectation of square of the difference Xi-Yi equals twive the varians of the given pdf, which equals 1/[n(n+1)]. By additivity of the expactation, w arrive at the formula for the expectation of the square of the euclidean distance:

E{ dist(X,Y)^2 } = 1/(n+1).

But this is not much wrt the to the question. From this we even do not have the expectation of the distance. Indeed note that

E{ dist(X,Y) } < \sqrt{ E{ dist(X,Y)^2 } }.

The answer I think you're looking for is p-value tied to the dimension value, N, tied to a (euclidean) distance between X and Y.

https://math.stackexchange.com/questions/80848/calculate-p-value

The euclidean distance in N dimensions expands as sqrt(sqrt(a^2 + b^2)^2 + c^2)... but since you've constrained it to the case where a + b + c + ... = 1

This problem looks to have an excess when you make each value complex (as the euclidean distance becomes expanding and geometric). The t statistic scales with root N, so whatever you have left over is just the p-value from the distribution, and the standard deviation value of the observed data (X) rather than that of the difference data (Y), however, you could include both as metrics of significance, but that is not kosher.

Although Gert's (Hertz') answer shows it comes out as proportional to a 3rd root of an inverse of N, showing an inherent mathematical relationship between infinite dimensions and 3 dimensions, that to move some distance of significance would take a cubing of total dimensions.

But I think we should not resolve to Calculus for this solution, but Alcalculus, particularly if you beg of it to be continuous.

But as Joachim Domsta is so eloquently stating, if you assume the probabilities sum to 1, then the next measurement must lower all previous probabilities - retrocausality.