Mixing
Cantor distribution extended
$begingroup$
Let $A_{m,n}$ be the $m$-th removed interval (from left to right) in step $n$ of constructing the Cantor set $C$. For example $A_{1,2} = (1/9,2/9)$, $A_{2,2} = (3/9,4/9)$, $A_{3,2} = (7/9,8/9)$. Define $F$ on the complement of $C$
begin{equation*}
F(x) =
begin{cases}
0 hfill& x leq 0 \
frac{m}{2^n} hfill& text{if $x in A_{m,n}$} \
1 hfill& x geq 1
end{cases}
end{equation*}
It's easy to see that $F$ is well defined. Furthermore, $F$ is uniformly continuous so there exists a function $tilde{F}$ on $(-infty,infty)$ such that $tilde{F} = F$ on $text{dom}(F)$. We know any $x in C$ can be written
$$x = sum_{n=1}^infty frac{a_n}{3^n}$$
where $a_n in {0,2}$. I'm trying to show for $x in C$ we have
$$tilde{F}(x) = sum_{n=1}^infty frac{a_n}{2^{n+1}}$$
I'm confused as to how we can use the definition of $F$ to help prove this.
real-analysis probability-theory cantor-set
asked Jan 30 at 4:36
user20354139user20354139
498211
$endgroup$
add a comment |
$begingroup$
Let $A_{m,n}$ be the $m$-th removed interval (from left to right) in step $n$ of constructing the Cantor set $C$. For example $A_{1,2} = (1/9,2/9)$, $A_{2,2} = (3/9,4/9)$, $A_{3,2} = (7/9,8/9)$. Define $F$ on the complement of $C$
begin{equation*}
F(x) =
begin{cases}
0 hfill& x leq 0 \
frac{m}{2^n} hfill& text{if $x in A_{m,n}$} \
1 hfill& x geq 1
end{cases}
end{equation*}
It's easy to see that $F$ is well defined. Furthermore, $F$ is uniformly continuous so there exists a function $tilde{F}$ on $(-infty,infty)$ such that $tilde{F} = F$ on $text{dom}(F)$. We know any $x in C$ can be written
$$x = sum_{n=1}^infty frac{a_n}{3^n}$$
where $a_n in {0,2}$. I'm trying to show for $x in C$ we have
$$tilde{F}(x) = sum_{n=1}^infty frac{a_n}{2^{n+1}}$$
I'm confused as to how we can use the definition of $F$ to help prove this.
real-analysis probability-theory cantor-set
asked Jan 30 at 4:36
user20354139user20354139
498211
$endgroup$
add a comment |
$begingroup$
Let $A_{m,n}$ be the $m$-th removed interval (from left to right) in step $n$ of constructing the Cantor set $C$. For example $A_{1,2} = (1/9,2/9)$, $A_{2,2} = (3/9,4/9)$, $A_{3,2} = (7/9,8/9)$. Define $F$ on the complement of $C$
begin{equation*}
F(x) =
begin{cases}
0 hfill& x leq 0 \
frac{m}{2^n} hfill& text{if $x in A_{m,n}$} \
1 hfill& x geq 1
end{cases}
end{equation*}
It's easy to see that $F$ is well defined. Furthermore, $F$ is uniformly continuous so there exists a function $tilde{F}$ on $(-infty,infty)$ such that $tilde{F} = F$ on $text{dom}(F)$. We know any $x in C$ can be written
$$x = sum_{n=1}^infty frac{a_n}{3^n}$$
where $a_n in {0,2}$. I'm trying to show for $x in C$ we have
$$tilde{F}(x) = sum_{n=1}^infty frac{a_n}{2^{n+1}}$$
I'm confused as to how we can use the definition of $F$ to help prove this.
real-analysis probability-theory cantor-set
asked Jan 30 at 4:36
user20354139user20354139
498211
$endgroup$
Let $A_{m,n}$ be the $m$-th removed interval (from left to right) in step $n$ of constructing the Cantor set $C$. For example $A_{1,2} = (1/9,2/9)$, $A_{2,2} = (3/9,4/9)$, $A_{3,2} = (7/9,8/9)$. Define $F$ on the complement of $C$
begin{equation*}
F(x) =
begin{cases}
0 hfill& x leq 0 \
frac{m}{2^n} hfill& text{if $x in A_{m,n}$} \
1 hfill& x geq 1
end{cases}
end{equation*}
It's easy to see that $F$ is well defined. Furthermore, $F$ is uniformly continuous so there exists a function $tilde{F}$ on $(-infty,infty)$ such that $tilde{F} = F$ on $text{dom}(F)$. We know any $x in C$ can be written
$$x = sum_{n=1}^infty frac{a_n}{3^n}$$
where $a_n in {0,2}$. I'm trying to show for $x in C$ we have
$$tilde{F}(x) = sum_{n=1}^infty frac{a_n}{2^{n+1}}$$
I'm confused as to how we can use the definition of $F$ to help prove this.
real-analysis probability-theory cantor-set
real-analysis probability-theory cantor-set
asked Jan 30 at 4:36
user20354139user20354139
498211
asked Jan 30 at 4:36
user20354139user20354139
498211
asked Jan 30 at 4:36
user20354139user20354139
498211
asked Jan 30 at 4:36
user20354139user20354139
498211
asked Jan 30 at 4:36
user20354139user20354139
498211
498211
add a comment |
add a comment |
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