A Weird but Interesting Distribution

by 3ch03s, Nov 18, 2014, 5:51 AM

Let $X_i$ an iid bernoulli sequence variables with success parameter $p$, and let compute the (random) sum:
$R_n=\sum_{i=1}^{\infty} X_i/2^i$
then, when $n$ tends to infinity, the variable $R_n$ converges to a weird distribution $R$, presumibly without radon-nikodym derivative for $p \neq 1/2$ and for $p=1/2$ $R_n$ tends to an uniform distribution in [0,1]. in both cases we have to take care about the non-numerability of R, because of $R_n$, and diadic arguments, combined by density.

Let's call $Rm(p)$ to this distribution. this distribution has a several properties like:

a) $R \sim Rm(p) \Rightarrow \ 1-R \sim Rm(1-p)$
b) if $X$ is bernoulli with parameter $p$ independent of $R \sim Rm(p)$ then:

$\frac{X+R}2 \sim Rm(p)$

c) From (b) we can compute the mean and variance of $R$, as $R$ has the same distribution as $\frac{X+R}2$ then
$E[R]=\frac{E[X]+E[R]}2=\frac{p+E[R]}2$
Obtaining $E[R]=p$
For the variance we have
$V[R]=\frac 14 V[X+R]$, and by independence:
$V[R]=\frac 14 (V[X]+V[R])=\frac 14(p(1-p)+V[R])$
Hence $V[R]=\frac 13p(1-p)$

d) Also from (b) we have an interesting property about the cdf of R:

$P[R \leq r]=P[X+R \leq 2r]=P[R\leq 2r](1-p)+pP[R\leq 2r+1]$

$F(r)=(1-p)F(2r)+pF(2r+1)$

With this formula, we can compute by recursion the cdf evaluated in any $r \in [0,1]$, and also, give a closed form when $r$ is rational.

From this, the cdf has an interesting property of self-similarity, and a fractal-like effect when we plot the cdf.

i'll be continue.....
This post has been edited 1 time. Last edited by 3ch03s, Mar 24, 2016, 5:27 PM
Reason: 2^i

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    also dead blog

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  • omg so pro

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  • puedes hablar espanol conmigo?

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  • tengo shouts? yaaaa.....

    by 3ch03s, Aug 11, 2015, 7:38 PM

  • hola. Mi nombre es claudio, soy chileno (y la ctm) Ingeniero civil Matematico con menci??n en estadistica aceptado en el doctorado de Estadpistica de la puc. Mucho Floyd Buki TB-303 y Lacie Baskerville y disco house.

    by 3ch03s, Nov 18, 2014, 4:29 AM

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