Bernoulli measure is aperiodic
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Let $Sigma=s_1,dots,s_m$ be a finite list of symbols, and put $X=Sigma^mathbbZ$.
Consider the left two-sided shift $T:Xto X$ given by $T(x_n)=(x_n+1)$. Given an $m$-dimensional vector $vecp=(p_1,dots,p_m)$, we can construct a measure on $Sigma$ by $sum_i=1^m p_i delta_s_i$, which then generates an infinite product measure on $X$. Such a measure is called a Bernoulli shift, and is ergodic for $T$.
A measure is called aperiodic if set of periodic points has zero measure.
Question
Bernoulli measure is aperiodic?
probability-distributions dynamical-systems ergodic-theory
asked Aug 21 at 11:46
Adam
173
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up vote
2
down vote
favorite
Let $Sigma=s_1,dots,s_m$ be a finite list of symbols, and put $X=Sigma^mathbbZ$.
Consider the left two-sided shift $T:Xto X$ given by $T(x_n)=(x_n+1)$. Given an $m$-dimensional vector $vecp=(p_1,dots,p_m)$, we can construct a measure on $Sigma$ by $sum_i=1^m p_i delta_s_i$, which then generates an infinite product measure on $X$. Such a measure is called a Bernoulli shift, and is ergodic for $T$.
A measure is called aperiodic if set of periodic points has zero measure.
Question
Bernoulli measure is aperiodic?
probability-distributions dynamical-systems ergodic-theory
asked Aug 21 at 11:46
Adam
173
Hint: can you show that there are countably many periodic points in $X$? Now all you need to show is that the measure of any countable set is zero for the Bernoulli measure.
– Dan Rust
Aug 21 at 12:53
@DanRust any periodic sequence with a repeating an block of length n defines and periodic orbit. The set of all periodic sequences is countable.I can show that measure any countable set is zero but for Bernoulli measure,i do not know.
– Adam
Aug 21 at 15:04
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $Sigma=s_1,dots,s_m$ be a finite list of symbols, and put $X=Sigma^mathbbZ$.
Consider the left two-sided shift $T:Xto X$ given by $T(x_n)=(x_n+1)$. Given an $m$-dimensional vector $vecp=(p_1,dots,p_m)$, we can construct a measure on $Sigma$ by $sum_i=1^m p_i delta_s_i$, which then generates an infinite product measure on $X$. Such a measure is called a Bernoulli shift, and is ergodic for $T$.
A measure is called aperiodic if set of periodic points has zero measure.
Question
Bernoulli measure is aperiodic?
probability-distributions dynamical-systems ergodic-theory
asked Aug 21 at 11:46
Adam
173
Let $Sigma=s_1,dots,s_m$ be a finite list of symbols, and put $X=Sigma^mathbbZ$.
Consider the left two-sided shift $T:Xto X$ given by $T(x_n)=(x_n+1)$. Given an $m$-dimensional vector $vecp=(p_1,dots,p_m)$, we can construct a measure on $Sigma$ by $sum_i=1^m p_i delta_s_i$, which then generates an infinite product measure on $X$. Such a measure is called a Bernoulli shift, and is ergodic for $T$.
A measure is called aperiodic if set of periodic points has zero measure.
Question
Bernoulli measure is aperiodic?
probability-distributions dynamical-systems ergodic-theory
asked Aug 21 at 11:46
Adam
173
asked Aug 21 at 11:46
Adam
173
asked Aug 21 at 11:46
Adam
173
asked Aug 21 at 11:46
Adam
173
173
Hint: can you show that there are countably many periodic points in $X$? Now all you need to show is that the measure of any countable set is zero for the Bernoulli measure.
– Dan Rust
Aug 21 at 12:53
@DanRust any periodic sequence with a repeating an block of length n defines and periodic orbit. The set of all periodic sequences is countable.I can show that measure any countable set is zero but for Bernoulli measure,i do not know.
– Adam
Aug 21 at 15:04
add a comment |Â
Hint: can you show that there are countably many periodic points in $X$? Now all you need to show is that the measure of any countable set is zero for the Bernoulli measure.
– Dan Rust
Aug 21 at 12:53
@DanRust any periodic sequence with a repeating an block of length n defines and periodic orbit. The set of all periodic sequences is countable.I can show that measure any countable set is zero but for Bernoulli measure,i do not know.
– Adam
Aug 21 at 15:04
Hint: can you show that there are countably many periodic points in $X$? Now all you need to show is that the measure of any countable set is zero for the Bernoulli measure.
– Dan Rust
Aug 21 at 12:53
Hint: can you show that there are countably many periodic points in $X$? Now all you need to show is that the measure of any countable set is zero for the Bernoulli measure.
– Dan Rust
Aug 21 at 12:53
@DanRust any periodic sequence with a repeating an block of length n defines and periodic orbit. The set of all periodic sequences is countable.I can show that measure any countable set is zero but for Bernoulli measure,i do not know.
– Adam
Aug 21 at 15:04
@DanRust any periodic sequence with a repeating an block of length n defines and periodic orbit. The set of all periodic sequences is countable.I can show that measure any countable set is zero but for Bernoulli measure,i do not know.
– Adam
Aug 21 at 15:04
add a comment |Â
1 Answer
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Let $(x_i)_i geq 1$ be a countable subset of $X$ and let $$B_i,k = x in X mid (x)_[-k,k] = (x_i)_[-k,k]$$ be the cylinder set of length $k$ which contains $x_i$.
Show that $$B(n) := bigcup_i=0^infty B_i,n2^i$$ is a nested sequences of subsets that contains every $x_i$ for any value of $n$ and then show that the measure of $B(n)$ converges to $0$ as $n$ increases. Hence, what must the measure of $x_i mid i geq 0$ be?
answered Aug 21 at 15:30
Dan Rust
22.3k114784
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $(x_i)_i geq 1$ be a countable subset of $X$ and let $$B_i,k = x in X mid (x)_[-k,k] = (x_i)_[-k,k]$$ be the cylinder set of length $k$ which contains $x_i$.
Show that $$B(n) := bigcup_i=0^infty B_i,n2^i$$ is a nested sequences of subsets that contains every $x_i$ for any value of $n$ and then show that the measure of $B(n)$ converges to $0$ as $n$ increases. Hence, what must the measure of $x_i mid i geq 0$ be?
answered Aug 21 at 15:30
Dan Rust
22.3k114784
add a comment |Â
up vote
0
down vote
Let $(x_i)_i geq 1$ be a countable subset of $X$ and let $$B_i,k = x in X mid (x)_[-k,k] = (x_i)_[-k,k]$$ be the cylinder set of length $k$ which contains $x_i$.
Show that $$B(n) := bigcup_i=0^infty B_i,n2^i$$ is a nested sequences of subsets that contains every $x_i$ for any value of $n$ and then show that the measure of $B(n)$ converges to $0$ as $n$ increases. Hence, what must the measure of $x_i mid i geq 0$ be?
answered Aug 21 at 15:30
Dan Rust
22.3k114784
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $(x_i)_i geq 1$ be a countable subset of $X$ and let $$B_i,k = x in X mid (x)_[-k,k] = (x_i)_[-k,k]$$ be the cylinder set of length $k$ which contains $x_i$.
Show that $$B(n) := bigcup_i=0^infty B_i,n2^i$$ is a nested sequences of subsets that contains every $x_i$ for any value of $n$ and then show that the measure of $B(n)$ converges to $0$ as $n$ increases. Hence, what must the measure of $x_i mid i geq 0$ be?
answered Aug 21 at 15:30
Dan Rust
22.3k114784
Let $(x_i)_i geq 1$ be a countable subset of $X$ and let $$B_i,k = x in X mid (x)_[-k,k] = (x_i)_[-k,k]$$ be the cylinder set of length $k$ which contains $x_i$.
Show that $$B(n) := bigcup_i=0^infty B_i,n2^i$$ is a nested sequences of subsets that contains every $x_i$ for any value of $n$ and then show that the measure of $B(n)$ converges to $0$ as $n$ increases. Hence, what must the measure of $x_i mid i geq 0$ be?
answered Aug 21 at 15:30
Dan Rust
22.3k114784
answered Aug 21 at 15:30
Dan Rust
22.3k114784
answered Aug 21 at 15:30
Dan Rust
22.3k114784
answered Aug 21 at 15:30
Dan Rust
22.3k114784
22.3k114784
add a comment |Â
add a comment |Â
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Hint: can you show that there are countably many periodic points in $X$? Now all you need to show is that the measure of any countable set is zero for the Bernoulli measure.
– Dan Rust
Aug 21 at 12:53
@DanRust any periodic sequence with a repeating an block of length n defines and periodic orbit. The set of all periodic sequences is countable.I can show that measure any countable set is zero but for Bernoulli measure,i do not know.
– Adam
Aug 21 at 15:04