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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.)2012-04-15
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    @ Didier- we want to deal with absolute continuity here. What would you suggest?2012-04-15
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    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.2012-04-15
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    Didier- I tried understand what you said, but I couldn't. Can you be more specific?2012-04-16
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    I reverted your edit of the title: please do not use the title for what should have been done with a comment here!2012-04-16
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    The paragraph on the Radon-Nikodym connection makes sense to me. After all, that's the precise way in which the Cantor function (which is a valid CDF) is shown to have no PDF.2012-04-16
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    @EMS No. (But *precise* is sweet.)2013-02-28
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    @Did I'm sorry that you don't understand the paragraph. It is a very nice way of explaining absolute continuity in the context of probability distributions here. Perhaps if you explain why you don't understand it, we can work together to provide a better explanation for you.2013-02-28
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    @Did For instance, have you considered the [Wikipedia article on the Radon-Nikodym Theorem](http://en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem)? There it says: "... Specifically, the probability density function of a random variable is the Radon–Nikodym derivative of the induced measure with respect to some base measure (usually the Lebesgue measure for continuous random variables)."2013-02-28
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    @EMS Soooo sweet again. As regards the paragraph in the OP, I salivate at the idea of seeing *A probability distribution exists if and only if the probability measure is absolutely continuous with respect to the Lebesgue measure* explained to me (second sentence of the paragraph, the third and fourth are still more mysterious).2013-02-28
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    @Did The random variable induces a probability measure via is cumulative distribution function. This is standard in any probability theory textbook. If that measure is absolutely continuous w.r.t. the Lebesgue measure, then the random variable admits a probability density function (unfortunately often just called a 'distribution' function).2013-03-01
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    @EMS Your entire last comment is true, trivial, and (unfortunately) quite unrelated to the paragraph in the OP. Unless you have something to say related to the actual paragraph, please let me sleep.2013-03-01
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    @Did I apologize for trying to be helpful. Since you would rather win the argument than understand the paragraph, I will not bother you about it further.2013-03-01
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    @EMS Sure... :-)2013-03-01

1 Answers 1

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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?2012-04-17
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    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.2012-04-17