In the first part I define an interesting distribution. Let a bernoulli iid sequence with parameter , and let


We called it .
We also showed some interesting properties of his probability distribution (cumulated) function, and we give some clues about the behavior of the distribution when p=1/2.

I claim when p=1/2, has the uniform distribution (which is in fact, the lebesgue measure in [0,1]). Also we give this property of cdf:



Then we can compute any value of this distribution using dyadic aproximations of any number in [0,1]. Is pretty clear the sequence of aproximations has the form . this implies that for any number of the form , with natural, F(r) has a "closed" form, because we are dealing with a finite number of bernoulli variables in that case.

we know every number in [0,1] has a unique (in the sense of unique sequence of 0's and 1's without getting finite 0's) binary representation, with this, we can compute if , which using the recurssion formula leads to:



Another aproach to the formula is thinking in the secuence of 0 and 1 for the finite dyadic representation.
So for every aproximation we can compute using this recurssion:



¿What happens when ?

the formula for dyadic numbers just reduces to:

, hence, any dyadic number has the propierty , then, the recursion for any dyadic approximation gives , So in the limit, by density arguments that i will justificate soon, we have for any , so, for p=1/2, the distribution is in fact, the uniform distribution and hence has a pdf.

But... what about the image of X? are [0,1]?... in fact, because the representation, the uniform distribution does not go trough the whole interval, just, a select set of numbers in [0,1] with the property: .

So we define:



and hence .

if then the image of X is .
its clear that and are disjoint if . We will use this property using the fact if p=1/2, , and hence the probability measure is in fact, the Lebesgue measure in [0,1], so, exists a pdf by radon-nykodym theorem. Let the Lebesgue measure in [0,1], write:

Clearly and the rest of the union are disjoint.
so:

In this year, I found something interesting. this post in AOPS

What about a random variable with support in [0,1]?

well, we can think, there is so many distributions, some well known (like the beta family, bernoulli trials, and a large etc), some very strange (like the rm distribution in my first entry blog). all with their mean and variance, could we think a distribution given a arbitrary mean and variance? the answer is no. in this entry, we find a several interesting things for the mean and variance of any distribution in [0,1].

First, ¿Why not all the pair mean-variance are possible in this kind of distributions?.

The answer is pretty simple. for X random variable distributed in [0,1] we can check that all the moments exists, and his mgf also exists always, in particular for first and second moment, who determines the mean and variance.

as for we have for p<q naturals, is easy to show that for this kind of distributions, specially setting p=1 and q=2: . We know that the variance is defined by: , so how to use the preceding fact?. just note this simply operation:


then

This simple fact shows for a given mean in [0,1] and a distribution in [0,1] the variance is bounded by the term (in fact is a parabola) and most important: as (a well-known inequality) we conclude that the variance can't be greater than 1/4. this is his maximal variance, so thinking a distribution with an arbitrary pair of mean-variance is not posible unless this pair lies on the "factible set" given by:

more interestingly, if we think a pair mean-standard.deviation, the factible set is a semicircle of radius 1/2 and center in x=1/2.

The main questions are: ¿what's happening about the distribution for a given pair in the factible set? can the proximity of the pair respect the boundary of this set tell us how is the distribution behavior?, what distribution reaches is maximal variance? is unique?, what about the moments? how does affect for a more general distributions with bounded support? what's happening if the variance is very near to the maximal line?.

We know that the distribution who reaches is maximal variance is the bernoulli distribution, this distribution is often regarded a "discrete" distribution, but by the pointwiew of measuring theory, can be also regarded a concentrated distribution at the points 0 and 1 in some proportion given by the mean in [0,1], in fact the relation are for the point 1 and for the point 0 (that is, a simple trial giving 1 with prob ). the uniqueness of the "maximal variance" distribution can be proved using chebyshev inequalities. setting for fixed h<1/2 we have using some algebra that so putting the function and the chebyshev inequalities we have:

But

the preceding inequality tell us for a bernoulli distribution, for all h<1/2 the probability is (in fact is an equivalency), so, if there is another distribution different of bernoulli there exists "h" such that , if also, this distribution reaches his maximal variance so we have for the same h: which is a contradiction.
Again this inequality tell us something more about how the distribution is because the probability of X taking values around 1/2, or even, every set A (medible) in ]0,1[ it's bounded by a quantity who depends of the diference of the maximal variance and the variance and the maximal distance respect of x=1/2, then, if the variances comes closer to his maximal, the probability begins to be more smaller, and evidently, more concentrated in the extremes, so, the more the variance approaches is maximum, more concentrated in 0 and 1 is the distribution.
Then, converges into a bernoulli.

It will continue.....