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











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













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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked Nov 15 at 5:04









math_enthuthiast

14910




14910












  • 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

















active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














 

draft saved


draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999244%2fa-family-of-continuous-distribution-functions-with-a-singular-law-to-the-lebesgu%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown






























active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes
















 

draft saved


draft discarded



















































 


draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999244%2fa-family-of-continuous-distribution-functions-with-a-singular-law-to-the-lebesgu%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

AnyDesk - Fatal Program Failure

How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

QoS: MAC-Priority for clients behind a repeater