A Weird but Interesting Distribution pt II
by 3ch03s, Mar 27, 2016, 12:47 AM
In the first part I define an interesting distribution. Let 
a bernoulli iid sequence with parameter ![$p \in ]0,1[$](//latex.artofproblemsolving.com/2/7/c/27cb262a70c5720fc44eed93acef192f69070f54.png)
, 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 ![$x \in [0,1]$](//latex.artofproblemsolving.com/d/3/4/d3423a3200d0ad7660fc4265d59d5598a53d551d.png)
, 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:![$x \in [0,1], x=\sum x_n/2^n|\frac 1n \sum x_i \to 1/2$](//latex.artofproblemsolving.com/9/c/7/9c71081a732517b2c25462caf98f2edf733f9e0e.png)
.
So we define:
and hence![$[0,1]=\bigcup_{p \in [0,1]} A_p$](//latex.artofproblemsolving.com/a/5/3/a538d5c2e9d8bf814d22c59055451aa9322af74f.png)
.
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, ![$X \sim U[0,1]$](//latex.artofproblemsolving.com/2/a/5/2a59b696376fef15b6ec53da5e0f354dfbc30b1d.png)
, 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:
![$[0,1]=A_{1/2} \cup \bigcup_{p \neq 1/2} A_p$](//latex.artofproblemsolving.com/7/f/a/7fad08e83f703c2fae17b0ddb370835fe45b9521.png)
Clearly
and the rest of the union are disjoint.
so:
![$\lambda([0,1])=1=\lambda (A_{1/2})+\lambda (\bigcup_{p \neq 1/2} A_p)$](//latex.artofproblemsolving.com/3/c/f/3cf39ce31b9c0e07431d77a95e127e35eb9a9faa.png)
a bernoulli iid sequence with parameter ![$p \in ]0,1[$](http://latex.artofproblemsolving.com/2/7/c/27cb262a70c5720fc44eed93acef192f69070f54.png)
, 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 ![$x \in [0,1]$](http://latex.artofproblemsolving.com/d/3/4/d3423a3200d0ad7660fc4265d59d5598a53d551d.png)
, 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:
![$x \in [0,1], x=\sum x_n/2^n|\frac 1n \sum x_i \to 1/2$](http://latex.artofproblemsolving.com/9/c/7/9c71081a732517b2c25462caf98f2edf733f9e0e.png)
.So we define:
and hence
![$[0,1]=\bigcup_{p \in [0,1]} A_p$](http://latex.artofproblemsolving.com/a/5/3/a538d5c2e9d8bf814d22c59055451aa9322af74f.png)
.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, ![$X \sim U[0,1]$](http://latex.artofproblemsolving.com/2/a/5/2a59b696376fef15b6ec53da5e0f354dfbc30b1d.png)
, 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:![$[0,1]=A_{1/2} \cup \bigcup_{p \neq 1/2} A_p$](http://latex.artofproblemsolving.com/7/f/a/7fad08e83f703c2fae17b0ddb370835fe45b9521.png)
Clearly
and the rest of the union are disjoint.so:
![$\lambda([0,1])=1=\lambda (A_{1/2})+\lambda (\bigcup_{p \neq 1/2} A_p)$](http://latex.artofproblemsolving.com/3/c/f/3cf39ce31b9c0e07431d77a95e127e35eb9a9faa.png)
March 2016
August 2015