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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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