A family of continuous distribution functions with a singular law to the Lebesgue measure











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Given a probability space $(Omega, mathcal{F}, mathbb{P})$, let ${X_n: n ge 1}$ be a sequence of i.i.d random variables with the common distribution
$$mathbb{P}(X_1 = 1) = p text{ and } mathbb{P}(X_1 = 0) = 1-p.$$
Define a mapping $D: Omega to [0,1]$ by $$D = sum_{n = 1} ^ {infty} frac{X_n}{2^n}.$$ Obviously $D$ is well-defined as a random variable because the series above is convergent almost surely. Denote by $mu_p$ and $F_p$ the law and the distribution function of $D$ respectively. (therefore, $mu_p$ is a probability measure on $([0, 1] , mathcal{B} ([0, 1])))$.



(i) Prove that $F_p$ is a continuous function.



(ii) Prove that $$mu_p left( left{x in [0, 1]: lim_{n to infty} frac{1}{n} sum_{j=1}^n E_n(x) = p right} right) = 1,$$
where, for each $x in [0,1]$ and $nge 1$
$$E_n(x) := begin{cases} 1, & text{ if } 2^{n-1}x - [2^{n-1}x] ge 0.5 \ 0, & text{ if } 2^{n-1}x - [2^{n-1}x] < 0.5end{cases}, $$
where $[s]$ denotes the integer part of $sge 0$.



From this argue that if $p_1 neq p_2$, then $mu_{p_1} perp mu_{p_2}$ , and in particular, if $p neq frac{1}
{2}$
, $mu_p perp lambda$ where $lambda$ is the Lebesgue measure on $mathbb{R}$.




For the first part, one can argue that $F_p$ is continuous iff $mu_p({x}) = 0, forall x in [0, 1]$. On the other hand, $mu_p$ for each interval $left[frac{k}{2^n}, frac{k+1}{2^n}right], k = 0, ldots 2^n$ can be readily shown to be $$mu_pleft(left[frac{k}{2^n}, frac{k+1}{2^n}right]right) = p^{n_1(n, k)}(1-p)^{n_0(n, k)}$$, where $n_1(n, k)$ and $n_0(n, k)$ are obtained by writing $k$ in the binary basis, padded with zeros on the left until it reaches
$n$ binary digits, and counting the number of 1s and 0s. For example $mu_p([0.25, 0.375]) = mu_p([frac{2}{2^3}, frac{2+1}{2^3}]) = mu_p([0.010, 0.011]) = p(1-p)^2$. Now for each $x in [0,1]$ there exists ${k_{x,n}, nge 1}$ such that ${x} = bigcap_{n=1}^{infty} [frac{k_{x,n}}{2^n}, frac{k_{x,n} +1}{2^n}]$, which implies $$mu_p({x}) = lim_{n to infty} p^{n_1(n, k_{x,n})}(1-p)^{n_0(n, k_{x,n})} = 0.$$



As for the second part, I understand that ${E_n(x)}$ gives the binary representation of $x in [0,1]$, but I cannot wrap my head around the claim of the problem. As a remark, it is stated that "the result in this problem offers a family of examples of continuous distribution functions that do not have probability density functions, i.e., if $p neq 1$, $F_p$ is a continuous distribution function
but $mu_p$ is singular to the Lebesgue measure."



Any hint is appreciated.










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  • All the statements here require independence of $X_n$'s.
    – Kavi Rama Murthy
    Nov 15 at 6:37










  • @KaviRamaMurthy Thank you. I edited the question.
    – math_enthuthiast
    Nov 15 at 11:50










  • I guess you want $X_1 in lbrace 0,1 rbrace$ a.s. and not $X_1 in lbrace -1,1 rbrace$
    – Stockfish
    Nov 15 at 12:01










  • @Stockfish right! Fixed it.
    – math_enthuthiast
    Nov 15 at 12:10










  • I don't know if this help. Have you thought of the measure of the set $$bigg{xin[0,1]:liminf_nfrac{1}{n}sum_{j=1}^nE_n(x)<alpha<beta<limsup_nfrac{1}{n}sum_{j=1}^nE_n(x)bigg}$$ where $alpha$, $beta$ are rationals such that $alpha<p<beta$?
    – ei2kpi
    Nov 15 at 13:47

















up vote
1
down vote

favorite













Given a probability space $(Omega, mathcal{F}, mathbb{P})$, let ${X_n: n ge 1}$ be a sequence of i.i.d random variables with the common distribution
$$mathbb{P}(X_1 = 1) = p text{ and } mathbb{P}(X_1 = 0) = 1-p.$$
Define a mapping $D: Omega to [0,1]$ by $$D = sum_{n = 1} ^ {infty} frac{X_n}{2^n}.$$ Obviously $D$ is well-defined as a random variable because the series above is convergent almost surely. Denote by $mu_p$ and $F_p$ the law and the distribution function of $D$ respectively. (therefore, $mu_p$ is a probability measure on $([0, 1] , mathcal{B} ([0, 1])))$.



(i) Prove that $F_p$ is a continuous function.



(ii) Prove that $$mu_p left( left{x in [0, 1]: lim_{n to infty} frac{1}{n} sum_{j=1}^n E_n(x) = p right} right) = 1,$$
where, for each $x in [0,1]$ and $nge 1$
$$E_n(x) := begin{cases} 1, & text{ if } 2^{n-1}x - [2^{n-1}x] ge 0.5 \ 0, & text{ if } 2^{n-1}x - [2^{n-1}x] < 0.5end{cases}, $$
where $[s]$ denotes the integer part of $sge 0$.



From this argue that if $p_1 neq p_2$, then $mu_{p_1} perp mu_{p_2}$ , and in particular, if $p neq frac{1}
{2}$
, $mu_p perp lambda$ where $lambda$ is the Lebesgue measure on $mathbb{R}$.




For the first part, one can argue that $F_p$ is continuous iff $mu_p({x}) = 0, forall x in [0, 1]$. On the other hand, $mu_p$ for each interval $left[frac{k}{2^n}, frac{k+1}{2^n}right], k = 0, ldots 2^n$ can be readily shown to be $$mu_pleft(left[frac{k}{2^n}, frac{k+1}{2^n}right]right) = p^{n_1(n, k)}(1-p)^{n_0(n, k)}$$, where $n_1(n, k)$ and $n_0(n, k)$ are obtained by writing $k$ in the binary basis, padded with zeros on the left until it reaches
$n$ binary digits, and counting the number of 1s and 0s. For example $mu_p([0.25, 0.375]) = mu_p([frac{2}{2^3}, frac{2+1}{2^3}]) = mu_p([0.010, 0.011]) = p(1-p)^2$. Now for each $x in [0,1]$ there exists ${k_{x,n}, nge 1}$ such that ${x} = bigcap_{n=1}^{infty} [frac{k_{x,n}}{2^n}, frac{k_{x,n} +1}{2^n}]$, which implies $$mu_p({x}) = lim_{n to infty} p^{n_1(n, k_{x,n})}(1-p)^{n_0(n, k_{x,n})} = 0.$$



As for the second part, I understand that ${E_n(x)}$ gives the binary representation of $x in [0,1]$, but I cannot wrap my head around the claim of the problem. As a remark, it is stated that "the result in this problem offers a family of examples of continuous distribution functions that do not have probability density functions, i.e., if $p neq 1$, $F_p$ is a continuous distribution function
but $mu_p$ is singular to the Lebesgue measure."



Any hint is appreciated.










share|cite|improve this question
























  • All the statements here require independence of $X_n$'s.
    – Kavi Rama Murthy
    Nov 15 at 6:37










  • @KaviRamaMurthy Thank you. I edited the question.
    – math_enthuthiast
    Nov 15 at 11:50










  • I guess you want $X_1 in lbrace 0,1 rbrace$ a.s. and not $X_1 in lbrace -1,1 rbrace$
    – Stockfish
    Nov 15 at 12:01










  • @Stockfish right! Fixed it.
    – math_enthuthiast
    Nov 15 at 12:10










  • I don't know if this help. Have you thought of the measure of the set $$bigg{xin[0,1]:liminf_nfrac{1}{n}sum_{j=1}^nE_n(x)<alpha<beta<limsup_nfrac{1}{n}sum_{j=1}^nE_n(x)bigg}$$ where $alpha$, $beta$ are rationals such that $alpha<p<beta$?
    – ei2kpi
    Nov 15 at 13:47















up vote
1
down vote

favorite









up vote
1
down vote

favorite












Given a probability space $(Omega, mathcal{F}, mathbb{P})$, let ${X_n: n ge 1}$ be a sequence of i.i.d random variables with the common distribution
$$mathbb{P}(X_1 = 1) = p text{ and } mathbb{P}(X_1 = 0) = 1-p.$$
Define a mapping $D: Omega to [0,1]$ by $$D = sum_{n = 1} ^ {infty} frac{X_n}{2^n}.$$ Obviously $D$ is well-defined as a random variable because the series above is convergent almost surely. Denote by $mu_p$ and $F_p$ the law and the distribution function of $D$ respectively. (therefore, $mu_p$ is a probability measure on $([0, 1] , mathcal{B} ([0, 1])))$.



(i) Prove that $F_p$ is a continuous function.



(ii) Prove that $$mu_p left( left{x in [0, 1]: lim_{n to infty} frac{1}{n} sum_{j=1}^n E_n(x) = p right} right) = 1,$$
where, for each $x in [0,1]$ and $nge 1$
$$E_n(x) := begin{cases} 1, & text{ if } 2^{n-1}x - [2^{n-1}x] ge 0.5 \ 0, & text{ if } 2^{n-1}x - [2^{n-1}x] < 0.5end{cases}, $$
where $[s]$ denotes the integer part of $sge 0$.



From this argue that if $p_1 neq p_2$, then $mu_{p_1} perp mu_{p_2}$ , and in particular, if $p neq frac{1}
{2}$
, $mu_p perp lambda$ where $lambda$ is the Lebesgue measure on $mathbb{R}$.




For the first part, one can argue that $F_p$ is continuous iff $mu_p({x}) = 0, forall x in [0, 1]$. On the other hand, $mu_p$ for each interval $left[frac{k}{2^n}, frac{k+1}{2^n}right], k = 0, ldots 2^n$ can be readily shown to be $$mu_pleft(left[frac{k}{2^n}, frac{k+1}{2^n}right]right) = p^{n_1(n, k)}(1-p)^{n_0(n, k)}$$, where $n_1(n, k)$ and $n_0(n, k)$ are obtained by writing $k$ in the binary basis, padded with zeros on the left until it reaches
$n$ binary digits, and counting the number of 1s and 0s. For example $mu_p([0.25, 0.375]) = mu_p([frac{2}{2^3}, frac{2+1}{2^3}]) = mu_p([0.010, 0.011]) = p(1-p)^2$. Now for each $x in [0,1]$ there exists ${k_{x,n}, nge 1}$ such that ${x} = bigcap_{n=1}^{infty} [frac{k_{x,n}}{2^n}, frac{k_{x,n} +1}{2^n}]$, which implies $$mu_p({x}) = lim_{n to infty} p^{n_1(n, k_{x,n})}(1-p)^{n_0(n, k_{x,n})} = 0.$$



As for the second part, I understand that ${E_n(x)}$ gives the binary representation of $x in [0,1]$, but I cannot wrap my head around the claim of the problem. As a remark, it is stated that "the result in this problem offers a family of examples of continuous distribution functions that do not have probability density functions, i.e., if $p neq 1$, $F_p$ is a continuous distribution function
but $mu_p$ is singular to the Lebesgue measure."



Any hint is appreciated.










share|cite|improve this question
















Given a probability space $(Omega, mathcal{F}, mathbb{P})$, let ${X_n: n ge 1}$ be a sequence of i.i.d random variables with the common distribution
$$mathbb{P}(X_1 = 1) = p text{ and } mathbb{P}(X_1 = 0) = 1-p.$$
Define a mapping $D: Omega to [0,1]$ by $$D = sum_{n = 1} ^ {infty} frac{X_n}{2^n}.$$ Obviously $D$ is well-defined as a random variable because the series above is convergent almost surely. Denote by $mu_p$ and $F_p$ the law and the distribution function of $D$ respectively. (therefore, $mu_p$ is a probability measure on $([0, 1] , mathcal{B} ([0, 1])))$.



(i) Prove that $F_p$ is a continuous function.



(ii) Prove that $$mu_p left( left{x in [0, 1]: lim_{n to infty} frac{1}{n} sum_{j=1}^n E_n(x) = p right} right) = 1,$$
where, for each $x in [0,1]$ and $nge 1$
$$E_n(x) := begin{cases} 1, & text{ if } 2^{n-1}x - [2^{n-1}x] ge 0.5 \ 0, & text{ if } 2^{n-1}x - [2^{n-1}x] < 0.5end{cases}, $$
where $[s]$ denotes the integer part of $sge 0$.



From this argue that if $p_1 neq p_2$, then $mu_{p_1} perp mu_{p_2}$ , and in particular, if $p neq frac{1}
{2}$
, $mu_p perp lambda$ where $lambda$ is the Lebesgue measure on $mathbb{R}$.




For the first part, one can argue that $F_p$ is continuous iff $mu_p({x}) = 0, forall x in [0, 1]$. On the other hand, $mu_p$ for each interval $left[frac{k}{2^n}, frac{k+1}{2^n}right], k = 0, ldots 2^n$ can be readily shown to be $$mu_pleft(left[frac{k}{2^n}, frac{k+1}{2^n}right]right) = p^{n_1(n, k)}(1-p)^{n_0(n, k)}$$, where $n_1(n, k)$ and $n_0(n, k)$ are obtained by writing $k$ in the binary basis, padded with zeros on the left until it reaches
$n$ binary digits, and counting the number of 1s and 0s. For example $mu_p([0.25, 0.375]) = mu_p([frac{2}{2^3}, frac{2+1}{2^3}]) = mu_p([0.010, 0.011]) = p(1-p)^2$. Now for each $x in [0,1]$ there exists ${k_{x,n}, nge 1}$ such that ${x} = bigcap_{n=1}^{infty} [frac{k_{x,n}}{2^n}, frac{k_{x,n} +1}{2^n}]$, which implies $$mu_p({x}) = lim_{n to infty} p^{n_1(n, k_{x,n})}(1-p)^{n_0(n, k_{x,n})} = 0.$$



As for the second part, I understand that ${E_n(x)}$ gives the binary representation of $x in [0,1]$, but I cannot wrap my head around the claim of the problem. As a remark, it is stated that "the result in this problem offers a family of examples of continuous distribution functions that do not have probability density functions, i.e., if $p neq 1$, $F_p$ is a continuous distribution function
but $mu_p$ is singular to the Lebesgue measure."



Any hint is appreciated.







probability-theory measure-theory law-of-large-numbers






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edited 2 days ago

























asked Nov 15 at 5:04









math_enthuthiast

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  • All the statements here require independence of $X_n$'s.
    – Kavi Rama Murthy
    Nov 15 at 6:37










  • @KaviRamaMurthy Thank you. I edited the question.
    – math_enthuthiast
    Nov 15 at 11:50










  • I guess you want $X_1 in lbrace 0,1 rbrace$ a.s. and not $X_1 in lbrace -1,1 rbrace$
    – Stockfish
    Nov 15 at 12:01










  • @Stockfish right! Fixed it.
    – math_enthuthiast
    Nov 15 at 12:10










  • I don't know if this help. Have you thought of the measure of the set $$bigg{xin[0,1]:liminf_nfrac{1}{n}sum_{j=1}^nE_n(x)<alpha<beta<limsup_nfrac{1}{n}sum_{j=1}^nE_n(x)bigg}$$ where $alpha$, $beta$ are rationals such that $alpha<p<beta$?
    – ei2kpi
    Nov 15 at 13:47




















  • All the statements here require independence of $X_n$'s.
    – Kavi Rama Murthy
    Nov 15 at 6:37










  • @KaviRamaMurthy Thank you. I edited the question.
    – math_enthuthiast
    Nov 15 at 11:50










  • I guess you want $X_1 in lbrace 0,1 rbrace$ a.s. and not $X_1 in lbrace -1,1 rbrace$
    – Stockfish
    Nov 15 at 12:01










  • @Stockfish right! Fixed it.
    – math_enthuthiast
    Nov 15 at 12:10










  • I don't know if this help. Have you thought of the measure of the set $$bigg{xin[0,1]:liminf_nfrac{1}{n}sum_{j=1}^nE_n(x)<alpha<beta<limsup_nfrac{1}{n}sum_{j=1}^nE_n(x)bigg}$$ where $alpha$, $beta$ are rationals such that $alpha<p<beta$?
    – ei2kpi
    Nov 15 at 13:47


















All the statements here require independence of $X_n$'s.
– Kavi Rama Murthy
Nov 15 at 6:37




All the statements here require independence of $X_n$'s.
– Kavi Rama Murthy
Nov 15 at 6:37












@KaviRamaMurthy Thank you. I edited the question.
– math_enthuthiast
Nov 15 at 11:50




@KaviRamaMurthy Thank you. I edited the question.
– math_enthuthiast
Nov 15 at 11:50












I guess you want $X_1 in lbrace 0,1 rbrace$ a.s. and not $X_1 in lbrace -1,1 rbrace$
– Stockfish
Nov 15 at 12:01




I guess you want $X_1 in lbrace 0,1 rbrace$ a.s. and not $X_1 in lbrace -1,1 rbrace$
– Stockfish
Nov 15 at 12:01












@Stockfish right! Fixed it.
– math_enthuthiast
Nov 15 at 12:10




@Stockfish right! Fixed it.
– math_enthuthiast
Nov 15 at 12:10












I don't know if this help. Have you thought of the measure of the set $$bigg{xin[0,1]:liminf_nfrac{1}{n}sum_{j=1}^nE_n(x)<alpha<beta<limsup_nfrac{1}{n}sum_{j=1}^nE_n(x)bigg}$$ where $alpha$, $beta$ are rationals such that $alpha<p<beta$?
– ei2kpi
Nov 15 at 13:47






I don't know if this help. Have you thought of the measure of the set $$bigg{xin[0,1]:liminf_nfrac{1}{n}sum_{j=1}^nE_n(x)<alpha<beta<limsup_nfrac{1}{n}sum_{j=1}^nE_n(x)bigg}$$ where $alpha$, $beta$ are rationals such that $alpha<p<beta$?
– ei2kpi
Nov 15 at 13:47

















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