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.
probability-theory measure-theory law-of-large-numbers
add a comment |
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.
probability-theory measure-theory law-of-large-numbers
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
add a comment |
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.
probability-theory measure-theory law-of-large-numbers
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
probability-theory measure-theory law-of-large-numbers
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
add a comment |
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
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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