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For any positive integer $N$, the binomial$(N!,p)$ distribution has the following property: for any $1 \leq n \leq N$, there exist i.i.d. random variables $X_1,\ldots,X_n$ such that $X_1 + \cdots + X_n \sim {\rm binomial}(N!,p)$ (specifically, we take $X_1,\ldots,X_n$ to be i.i.d. binomial$(N!/n,p)$ rv's). It may be interesting to consider the following question: given $N \geq 3$, arbitrary but fixed, is there a continuous bounded distribution $\mu = \mu_N$ having the same property? (I stress: continuous and bounded.)

EDIT: Well, it turns out this is a really trivial problem, but worth remembering...; see my answer below.

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    Sounds like [infinite divisibility](http://en.wikipedia.org/wiki/Infinite_divisibility_%28probability%29). Edit: sorry, more like [decomposability](http://en.wikipedia.org/wiki/Decomposable_distributions), since the binomial is not infinitely divisible. But infinite divisibility is decomposability to the max. ;)2010-12-05
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    What about the uniform distribution? A random uniform variable is bounded and decomposable.2010-12-05
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    By the way, by non-constant, do you mean non-constant density function? Or you apply it to the random variable as such? Which is not really interesting. Maybe you should rewrite your OP so as to reflect you already know about infinite divisibility and decomposable distributions and the exampls of wikipedia. This will spare people a lot of second-guessing. ;)2010-12-05
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    The uniform distribution seems to correspond to a sum of two independent but not identically distributed rv's.2010-12-05
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    Yep, just read it before you edited that message. So it's no good either. You got an interesting question there. +12010-12-05

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EDIT: NEW ANSWER

It is convenient to express the solution in terms of convolutions: for any distribution $\mu$, positive integer $N$, and $1 \leq n \leq N$, we have $$ \mu ^{*N!} = \big(\mu ^{*(N!/n)} \big)^{*n}. $$ So simple, yet instructive...

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    Hello Shai! I've thought about your construction, and I realized that for $p=1/2$ the building blocks of your distributions are just uniform distributions over $[0,1]$. So in that case $\mu(N!,1/2)$ is the result of convoluting $N!$ uniform distributions. When $p\neq 1/2$ your basic building blocks are singular continuous distributions. Besides, yours is a generalization of the construction on the wikipage.2010-12-06
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    Forgot to add that since these singular continuous measures are constructed out of a Bernoulli distribution, they are sometimes called Bernoulli measures.2010-12-06
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    The fact concerning the singular distributions is interesting. Actually, thanks to your observation regarding the convolution of $N!$ uniform distributions, I noticed that the problem is trivial.2010-12-06
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OK, I think I've got one for you, but I have to admit it's a bit of a cheat. If you replace independence in your requirements by free independence then there exists a continuous bounded and infinitely divisible distribution, namely the Wigner semi-circle distribution.

That's the only example I can think of for now.

I also give this reference for details.

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    It seems that the "free independence" concept is quite complicated.2010-12-05
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    Well, understanding the basics requires some knowledge of functional analysis, especially Hilbert spaces, von Neumann algebras and C*-algebras. It has been developed by Dan Voiculescu as an alternative statistics theory, a bit like there are non-Euclidean geometries. I think there's also a third type of statistics, based on another kind of independence. But I forgot how it's called.2010-12-05