Hi!

I don't know exact answer, but simple observation can be made.

This problem can be formulated as non-interacting fermions problem:

numbers 1, 2, 3, ... 2N are energy levels of one particle;

N is number of fermions;

A is energy of N fermions.

Than the main question is: what are the statistical weights of energy levels A?

Answer to this question is probably treated in stat.mech. books.

Dear Gennadiy, thanks for the kind reminder. Indeed the problem was derived from a model that fits the general framework of stat mech. Unfortunately to my limited knowledge the canonical examples are often put in the setting of grand canonical ensembles, where the particle numbers are not given a priori, and here we have a fixed, given number of particles.

Dear Gennadiy, thanks for the kind reminder. Indeed the problem was derived from a model that fits the general framework of stat mech. Unfortunately to my limited knowledge the canonical examples are often put in the setting of grand canonical ensembles, where the particle numbers are not given a priori, and here we have a fixed, given number of particles.

Dear Yen Ting Lin,

Are you interested to know the full distribution of A? I think this distribution is concentrated in vicinity of mean value. It is possible to find large N asymptotics of moments of A.

Besides, there is trick based on auxiliary integration. It is possible to write canonical sum as integral of grand canonical sum due to this trick. Then you can find asymptotics of distribution for any A. I do not remember appropriate reference, sorry.

I would agree with Gennadly that A is going to follow an approximately normal distribution. The only difference this system has from a normal distribution is that it has a limited degree of freedom. That means the distribution will not spread out as much as a normal curve does.

Dear Gennadiy & Haochun, I agree both of you that the as infinite N there could be some calculations we can do for the distribution. On the other hand, I am *not* interested in this limit since it is already well-known and not that exciting. I am interested in the distribution of $A$, for any positive number $N>0$; but thank you both still :)

El problema es "we randomly draw N number" fuera del conjunto S. ¿How you choice number 234345654, for example? Or the another way: ¿Which probability we use to choice "one" number of "Natural - {1,2,..., 2N}"? One trivial probability or probability like Poisson? (for example)?

Eliseo: just imagine drawing N balls exclusively out of 2N balls, each on which has a number from 1 to 2N.

It wouldn't be one of your common distributions, e.g.,

Binomial or Poisson. This is a special case of sampling

w/o replacement from a finite population. No doubt

the distribution has been characterixzed. Check the

internet, and/or Johnson and Katz. It would be easy to

compute the mean and standard deviation. I would guess

that the limiting distribution as N -> infinity is

normal.

Dear Prof. Zurakowski, thanks for the input. I agree with your opinion regarding sampling w/o replacement from a finite population. Could you elaborate on the reference of Jonson & Katz? From my limited knowledge I do not recall any textbook / paper out of the names. I did try a preliminary internet search before posting here, but there was no luck. Once again thanks very much for your input!

My english is very poor. Well, you choice N ball between 2N Balls? OK. Then you compute sum of this numbers, is OK.? Then, I think you get a limit thermodinamic called Law zero-one. I recommend the article "Phase transition in the countdown problem" from Lucas Lacasa and Luque, Physical Review. And Scientific American May 2013, number ? (I read the Spanish version of Scientific American, the article is called All or Nothing, author: Bartolo Luque)

Yen Ting Lin, I still thinking about that.. Very interesting problem. Will write here when will have some results. I think that a program can calculate it using formulas from there http://en.wikipedia.org/wiki/Partition_(number_theory)

If you want derive explicit formula you should try to use them too, I guess :)

David refers to a handbook of probability distributions published around 1970 in several volumes. The title is "Distributions in Statistics" and the authors are Johnson and Kotz.

You can try the 2005 edition published by Wiley (you can preview it at Google Books):

Univariate Discrete Distributions (by Norman L. Johnson, Adrienne W. Kemp, Samuel Kotz).

The preface mentions further similar books.

Another possibility is that maybe that problem has been studied in number theory. Some number theory problems involve the number of ways a whole number can be written as a sum of other numbers. I'm sure someone already came across the case in which you need a sum of *different* numbers, thus maybe also your situation (different numbers

What you're describing is essentially the sampling distribution of the sample mean, except where you're "stopping one step short". That is, if you were to add one more operation of calculating the average value of each of your samples (dividing each sum of N picks by the sample size N itself), you have precisely this scenario. Your statement implies N is constant from one trial to the next, so your problem is a proxy of the standard problem. Classical texts on mathematical statistics discuss the problem thoroughly (e.g. any edition of Paul Hoel's "Introduction to Mathematical Statistics"). Khan Academy also has a nice screencast illustrating the problem.

https://www.khanacademy.org/math/probability/statistics-inferential/sampling_distribution/v/sampling-distribution-of-the-sample-mean

Yen, please see attached information on Katz distribution.

Dear Pedro Terán: Thanks for the clarification!

Dear Michael Wendl: I humbly disagree. The original problem is to sample the set *without replacement* while the normal textbook examples are always sampling *with* replacement. The no-replacement approach inevitably introduces dependency of different samples, and it won't be as straightforward as one applies CLT's to the traditional problems.

Dear Prof. Zurakowski: Looks like a very interesting work. I will certainly read it, thanks very much!!

The distinction may be one without a real difference. That is, it will only matter if N is very small. However, if N is indeed small for your particular application, then why not simply enumerate all "2N choose N" combinations to obtain the actual distribution of the sample mean for your specific problem? For example, for N=20 there are roughly 1e11 combinations, which can be computed in a fairly short time on modern hardware.

Out of curiosity, what actually is the problem for which you are trying to find this distribution?

Dear Michael, sorry for the late reply. Things had been quite crazy these couple of days.

I agree with you (and other answers) that as $N$ goes to large number it should be normally distributed. However, out of mathematical curiosity, I would like to ask if any mathematicians had invested this case when $N$ is *finite*.

I have worked out the large $N$ case in the grand canonical ensemble (GCE) settings, however, it does not solve the original problem since in the grand canonical ensemble the temperature is introduced, while in the original problem it is more related to the idea of microcanonical ensemble.

Indeed, I also coded up a very easy computer program, which shows that when $N=30$ the results is already very good with the GCE approximation.

Meanwhile, I am still very interested in the case when $N$ is finite (as a pure mathematical inquiry), and how this finite distribution converges to the limiting distribution.

As for the motivation of the problem, I am thinking about $N$ particles connected by rigid rods with free joints, in a uniform external field with the constraint that one of the particle is held fixed in the space. The original motivation is to consider similar polymer structures in oscillatory field (the motivation of the model came from a biological phenomenon---chromosome oscillation). Our group (http://cdc.pks.mpg.de) is actively studying this problem as of this moment.