Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Sounds like infinite divisibility. Edit: sorry, more like decomposability, since the binomial is not infinitely divisible. But infinite divisibility is decomposability to the max. ;) –  Raskolnikov Dec 5 '10 at 18:10
    
What about the uniform distribution? A random uniform variable is bounded and decomposable. –  Raskolnikov Dec 5 '10 at 18:13
    
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. ;) –  Raskolnikov Dec 5 '10 at 18:33
    
The uniform distribution seems to correspond to a sum of two independent but not identically distributed rv's. –  Shai Covo Dec 5 '10 at 18:36
    
Yep, just read it before you edited that message. So it's no good either. You got an interesting question there. +1 –  Raskolnikov Dec 5 '10 at 18:58

2 Answers 2

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.

share|improve this answer
    
It seems that the "free independence" concept is quite complicated. –  Shai Covo Dec 5 '10 at 19:21
    
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. –  Raskolnikov Dec 5 '10 at 19:26

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...

share|improve this answer
    
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. –  Raskolnikov Dec 6 '10 at 9:11
    
Forgot to add that since these singular continuous measures are constructed out of a Bernoulli distribution, they are sometimes called Bernoulli measures. –  Raskolnikov Dec 6 '10 at 9:33
    
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. –  Shai Covo Dec 6 '10 at 15:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.