# How can Radon-Nikodym and Borel-Cantelli be used to calculate Probability distribution?

Let $\zeta=\sum_{n=1}^{\infty}2^{-n}\,\omega_{n}$ , where $\omega_{n}$ are independent, $\mathbb{P}(\omega_{n}=1)=p\neq\frac{1}{2}$ . Also $\mathbb{P}(\omega_{n}=0)=1-p$ for all $n\geq1$ . I want to show that the distribution of $\zeta$ does not have a density, using Radon-Nikodym and Borel-Cantelli Lemma.

1. Radon-Nikodym establishes a connection between probability distribution and probability density function. A probability distribution exists if and only if the probability measure is absolutely continuous with respect to the Lebesgue measure. To show that the probability measure $Pr$ is absolutely continuous with respect to the Lebesgue measure $\mu$, we need to first realize that there is a one-to-one correspondence between probability distribution functions and probability measure on the real line. Therefore, for some subset $A$ on the real line it suffices for us to show that $Pr(A)$$\geq$ $\mu(A)$, so that absolute continuity does not hold and the density does not exist.

So, writing the probability distribution of partial sum $\zeta=\sum_{n=1}^{N}2^{-n}\,\omega_{n}$ $=$ $\sum_{m=1}^{N}(1-p)^{m}p^{N-1-m}\left(\sum_{\beta\in\{_{N-1}C_{m}\}}\Pr\left(2\omega_{1}\leq(j-\beta)\right)\right)$, where $\beta$ is just a different combination in the indicies of the sum $\sum 2^{-n}$.

I suspect Borel-Cantelli should come handy here to control the infinite sum perhaps? I am a bit lost, but I would appreciate any advice. Thanks!

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Why do you want to use Radon-Nikodym and Borel-Cantelli? This is not the usual method. (And the paragraph Radon-Nikodym establishes a connection... is a mystery to me.) –  Did Apr 15 '12 at 18:38
@ Didier- we want to deal with absolute continuity here. What would you suggest? –  Wei Apr 15 '12 at 19:05
By the law of large numbers for i.i.d. Bernoulli random variables, each distribution of $\zeta$ for a given $p$ is singular with respect to all the others. The case $p=\frac12$ is the Lebesgue measure on $[0,1]$, hence you are done. –  Did Apr 15 '12 at 19:33
Didier- I tried understand what you said, but I couldn't. Can you be more specific? –  Wei Apr 16 '12 at 0:13
I reverted your edit of the title: please do not use the title for what should have been done with a comment here! –  Mariano Suárez-Alvarez Apr 16 '12 at 1:07

This is to prove that the distribution of $\zeta$ has no density.

We start with a definition of the dyadic expansion of a real number in $S=[0,1)$. Let $s:S\to S$ and $b_1:S\to\{0,1\}$ be defined by $s(x)=2x-\lfloor 2x\rfloor$ and $b_1(x)=[x\geqslant\frac12]$ for every $x$ in $S$. For every $n\geqslant1$, let $b_{n+1}=b_n\circ s$.

For example $b_3$ is the indicator function of the set $[\frac18,\frac14)\cup[\frac38,\frac12)\cup[\frac58,\frac34)\cup[\frac78,1)$.

Then, one can check that $x=\sum\limits_{n\geqslant1}b_n(x)2^{-n}$ with $b_n(x)\in\{0,1\}$ for every $n\geqslant1$.

Next, for every $p$ in $(0,1)$, we introduce the set $B_p$ of the real numbers $x$ in $S$ such that the sequence $\frac1n\sum\limits_{k=1}^nb_k(x)$ converges when $n\to\infty$ and has limit $p$. Obviously, the sets $B_p$ are Borel measurable and disjoint.

We now apply this to our setting. For every $p$ in $(0,1)$, consider the distribution $\mu_p$ of the corresponding random variable $\zeta=\sum\limits_{n\geqslant1}\omega_n2^{-n}$. One sees that $\omega_n=b_n(\zeta)$ almost surely (the exception being when $\zeta$ is a dyadic rational, which is a null event). By the strong law of large numbers, since the sequence $(\omega_n)_n$ is i.i.d. Bernoulli with parameter $p$, $\frac1n\sum\limits_{k=1}^n\omega_k$ converges almost surely to $\mathrm E(\omega_1)=p$ when $n\to\infty$, that is, $\zeta$ is in $B_p$ almost surely. Finally:

For every $p$ in $(0,1)$, $\mu_p(B_p)=1$ and the sets $B_p$ are disjoint hence $(\mu_p)$ is a collection of mutually singular measures.

In particular, for every $p\ne\frac12$ in $(0,1)$, $\mu_p$ is singular with respect to $\mu_{1/2}$. Since $\mu_{1/2}$ is the Lebesgue measure on $S$, when $(\omega_n)_n$ is i.i.d. Bernoulli with parameter $p\ne\frac12$, the distribution of $\zeta$ has no density with respect to the Lebesgue measure (and one can show that $\mu_p$ has no discrete part either).

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How do you know $\mu_{1/2}$ is the Lebesgue measure? –  Wei Apr 17 '12 at 19:14
Because the interval $[k/2^n,(k+1)/2^n)$ corresponds to $\omega_i=a_i$ for every $1\leqslant i\leqslant n$, for some bits $(a_i)$, hence $\mu_{1/2}([k/2^n,(k+1)/2^n))=1/2^n$, for every $k$ and $n$, and because there is only one measure such that this holds, the Lebesgue measure. –  Did Apr 17 '12 at 20:51