Mixing
Will the following sequence ever repeat?
$begingroup$
I'm unsure if the notation used by the author is common, so I will define some terms before stating the problem.
{$0, 1$}$^infty$ is the set of all functions $f:mathbb{N} rightarrow ${$0, 1$}.
{$0, 1$}$^n$ is the set of all functions $f:$ {$1, ..., n$}$ rightarrow ${$0, 1$}.
{$0, 1$}$^*$ = $cup_{i=1}^infty${$0,1$}$^i$.
Simply put {$0, 1$}$^infty$ is the set of all infinite binary sequences, and {$0, 1$}$^n$ is the set of all binary sequences of length $n$, while {$0, 1$}$^*$ is the set of all finite binary sequences.
For any binary strings $X, Z$ we write $X preceq Z$ if $X$ is a prefix of $Z$, that is, if $exists Y$ such that $XY = Z$.
The problem:
Define an (infinite) binary sequence $S in$ {$0, 1$}$^ infty$ to be prefix-repetitive if there are infinitely many strings $w in ${$0, 1$}$^*$ such that $ww preceq S$
Prove: If the bits of the sequence $S in ${$0, 1$}$^infty$ are chosen by independent tosses of a fair coin, then
$Prob[S$ is prefix-repetitive$] =0$
My attempt.
Firstly, if $S$ is prefix-repetitive then $ forall N inmathbb{N}, exists n>N$ such that $aa preceq S$ for some $a in $ {$0, 1$}$^n$. We shall call this statement Lemma $1$.
Proof: choose $N in mathbb{N}$. Let $B$ be the set of all strings $b in ${$0, 1$}$^*$ such that $bb preceq S$. By the definition of prefix-repetitive $|B| = infty$. Suppose there does not exist a string $a in B$ such that $a in $ {$0, 1$}$^n$ for some $n>N$. Then $B subseteq cup_{i=1} ^N$ {$0,1$}$^i$. However $|cup_{i=1} ^N$ {$0,1$}$^i|$ is finite. A contradiction.
Secondly, let $Min mathbb{N}$, then
$Prob[exists a in ${$0, 1$}$^{m}$ such that $aa preceq S$ for some $m>M] =$
$sum_{i=M+1} ^infty Prob[exists a in ${$0, 1$}$^{i}$ such that $aa preceq S] = $
$sum_{i=M+1} ^infty frac{1}{2^i} = frac {1}{2^{M}}$
We shall call this result lemma $2$.
Lastly, we prove the main statement. Let $Nin mathbb{N}$. By succesive use of lemma $1$ $exists n_1, n_2, ...$ such that $N<n_1<n_2<...$ and $a_ka_k preceq S$ for some $a_k in ${$0,1$}$^{n_k}$. By lemma $2$ $Prob[a_k$ exists$] = frac{1}{2^{n_k}}$, and $Prob[a_k$ exists $forall k] = frac{1}{2^{n_1}} frac{1}{2^{n_2}} ... = 0$.
I'm not familiar with this area of mathematics and was wondering if my proof was valid. If not, how could I prove main statement?
I would really appreciate any help/thoughts!
probability sequences-and-series proof-verification binary
asked Jan 28 at 0:50
LeoLeo
788517
$endgroup$
add a comment |
$begingroup$
I'm unsure if the notation used by the author is common, so I will define some terms before stating the problem.
{$0, 1$}$^infty$ is the set of all functions $f:mathbb{N} rightarrow ${$0, 1$}.
{$0, 1$}$^n$ is the set of all functions $f:$ {$1, ..., n$}$ rightarrow ${$0, 1$}.
{$0, 1$}$^*$ = $cup_{i=1}^infty${$0,1$}$^i$.
Simply put {$0, 1$}$^infty$ is the set of all infinite binary sequences, and {$0, 1$}$^n$ is the set of all binary sequences of length $n$, while {$0, 1$}$^*$ is the set of all finite binary sequences.
For any binary strings $X, Z$ we write $X preceq Z$ if $X$ is a prefix of $Z$, that is, if $exists Y$ such that $XY = Z$.
The problem:
Define an (infinite) binary sequence $S in$ {$0, 1$}$^ infty$ to be prefix-repetitive if there are infinitely many strings $w in ${$0, 1$}$^*$ such that $ww preceq S$
Prove: If the bits of the sequence $S in ${$0, 1$}$^infty$ are chosen by independent tosses of a fair coin, then
$Prob[S$ is prefix-repetitive$] =0$
My attempt.
Firstly, if $S$ is prefix-repetitive then $ forall N inmathbb{N}, exists n>N$ such that $aa preceq S$ for some $a in $ {$0, 1$}$^n$. We shall call this statement Lemma $1$.
Proof: choose $N in mathbb{N}$. Let $B$ be the set of all strings $b in ${$0, 1$}$^*$ such that $bb preceq S$. By the definition of prefix-repetitive $|B| = infty$. Suppose there does not exist a string $a in B$ such that $a in $ {$0, 1$}$^n$ for some $n>N$. Then $B subseteq cup_{i=1} ^N$ {$0,1$}$^i$. However $|cup_{i=1} ^N$ {$0,1$}$^i|$ is finite. A contradiction.
Secondly, let $Min mathbb{N}$, then
$Prob[exists a in ${$0, 1$}$^{m}$ such that $aa preceq S$ for some $m>M] =$
$sum_{i=M+1} ^infty Prob[exists a in ${$0, 1$}$^{i}$ such that $aa preceq S] = $
$sum_{i=M+1} ^infty frac{1}{2^i} = frac {1}{2^{M}}$
We shall call this result lemma $2$.
Lastly, we prove the main statement. Let $Nin mathbb{N}$. By succesive use of lemma $1$ $exists n_1, n_2, ...$ such that $N<n_1<n_2<...$ and $a_ka_k preceq S$ for some $a_k in ${$0,1$}$^{n_k}$. By lemma $2$ $Prob[a_k$ exists$] = frac{1}{2^{n_k}}$, and $Prob[a_k$ exists $forall k] = frac{1}{2^{n_1}} frac{1}{2^{n_2}} ... = 0$.
I'm not familiar with this area of mathematics and was wondering if my proof was valid. If not, how could I prove main statement?
I would really appreciate any help/thoughts!
probability sequences-and-series proof-verification binary
asked Jan 28 at 0:50
LeoLeo
788517
$endgroup$
add a comment |
$begingroup$
I'm unsure if the notation used by the author is common, so I will define some terms before stating the problem.
{$0, 1$}$^infty$ is the set of all functions $f:mathbb{N} rightarrow ${$0, 1$}.
{$0, 1$}$^n$ is the set of all functions $f:$ {$1, ..., n$}$ rightarrow ${$0, 1$}.
{$0, 1$}$^*$ = $cup_{i=1}^infty${$0,1$}$^i$.
Simply put {$0, 1$}$^infty$ is the set of all infinite binary sequences, and {$0, 1$}$^n$ is the set of all binary sequences of length $n$, while {$0, 1$}$^*$ is the set of all finite binary sequences.
For any binary strings $X, Z$ we write $X preceq Z$ if $X$ is a prefix of $Z$, that is, if $exists Y$ such that $XY = Z$.
The problem:
Define an (infinite) binary sequence $S in$ {$0, 1$}$^ infty$ to be prefix-repetitive if there are infinitely many strings $w in ${$0, 1$}$^*$ such that $ww preceq S$
Prove: If the bits of the sequence $S in ${$0, 1$}$^infty$ are chosen by independent tosses of a fair coin, then
$Prob[S$ is prefix-repetitive$] =0$
My attempt.
Firstly, if $S$ is prefix-repetitive then $ forall N inmathbb{N}, exists n>N$ such that $aa preceq S$ for some $a in $ {$0, 1$}$^n$. We shall call this statement Lemma $1$.
Proof: choose $N in mathbb{N}$. Let $B$ be the set of all strings $b in ${$0, 1$}$^*$ such that $bb preceq S$. By the definition of prefix-repetitive $|B| = infty$. Suppose there does not exist a string $a in B$ such that $a in $ {$0, 1$}$^n$ for some $n>N$. Then $B subseteq cup_{i=1} ^N$ {$0,1$}$^i$. However $|cup_{i=1} ^N$ {$0,1$}$^i|$ is finite. A contradiction.
Secondly, let $Min mathbb{N}$, then
$Prob[exists a in ${$0, 1$}$^{m}$ such that $aa preceq S$ for some $m>M] =$
$sum_{i=M+1} ^infty Prob[exists a in ${$0, 1$}$^{i}$ such that $aa preceq S] = $
$sum_{i=M+1} ^infty frac{1}{2^i} = frac {1}{2^{M}}$
We shall call this result lemma $2$.
Lastly, we prove the main statement. Let $Nin mathbb{N}$. By succesive use of lemma $1$ $exists n_1, n_2, ...$ such that $N<n_1<n_2<...$ and $a_ka_k preceq S$ for some $a_k in ${$0,1$}$^{n_k}$. By lemma $2$ $Prob[a_k$ exists$] = frac{1}{2^{n_k}}$, and $Prob[a_k$ exists $forall k] = frac{1}{2^{n_1}} frac{1}{2^{n_2}} ... = 0$.
I'm not familiar with this area of mathematics and was wondering if my proof was valid. If not, how could I prove main statement?
I would really appreciate any help/thoughts!
probability sequences-and-series proof-verification binary
asked Jan 28 at 0:50
LeoLeo
788517
$endgroup$
I'm unsure if the notation used by the author is common, so I will define some terms before stating the problem.
{$0, 1$}$^infty$ is the set of all functions $f:mathbb{N} rightarrow ${$0, 1$}.
{$0, 1$}$^n$ is the set of all functions $f:$ {$1, ..., n$}$ rightarrow ${$0, 1$}.
{$0, 1$}$^*$ = $cup_{i=1}^infty${$0,1$}$^i$.
Simply put {$0, 1$}$^infty$ is the set of all infinite binary sequences, and {$0, 1$}$^n$ is the set of all binary sequences of length $n$, while {$0, 1$}$^*$ is the set of all finite binary sequences.
For any binary strings $X, Z$ we write $X preceq Z$ if $X$ is a prefix of $Z$, that is, if $exists Y$ such that $XY = Z$.
The problem:
Define an (infinite) binary sequence $S in$ {$0, 1$}$^ infty$ to be prefix-repetitive if there are infinitely many strings $w in ${$0, 1$}$^*$ such that $ww preceq S$
Prove: If the bits of the sequence $S in ${$0, 1$}$^infty$ are chosen by independent tosses of a fair coin, then
$Prob[S$ is prefix-repetitive$] =0$
My attempt.
Firstly, if $S$ is prefix-repetitive then $ forall N inmathbb{N}, exists n>N$ such that $aa preceq S$ for some $a in $ {$0, 1$}$^n$. We shall call this statement Lemma $1$.
Proof: choose $N in mathbb{N}$. Let $B$ be the set of all strings $b in ${$0, 1$}$^*$ such that $bb preceq S$. By the definition of prefix-repetitive $|B| = infty$. Suppose there does not exist a string $a in B$ such that $a in $ {$0, 1$}$^n$ for some $n>N$. Then $B subseteq cup_{i=1} ^N$ {$0,1$}$^i$. However $|cup_{i=1} ^N$ {$0,1$}$^i|$ is finite. A contradiction.
Secondly, let $Min mathbb{N}$, then
$Prob[exists a in ${$0, 1$}$^{m}$ such that $aa preceq S$ for some $m>M] =$
$sum_{i=M+1} ^infty Prob[exists a in ${$0, 1$}$^{i}$ such that $aa preceq S] = $
$sum_{i=M+1} ^infty frac{1}{2^i} = frac {1}{2^{M}}$
We shall call this result lemma $2$.
Lastly, we prove the main statement. Let $Nin mathbb{N}$. By succesive use of lemma $1$ $exists n_1, n_2, ...$ such that $N<n_1<n_2<...$ and $a_ka_k preceq S$ for some $a_k in ${$0,1$}$^{n_k}$. By lemma $2$ $Prob[a_k$ exists$] = frac{1}{2^{n_k}}$, and $Prob[a_k$ exists $forall k] = frac{1}{2^{n_1}} frac{1}{2^{n_2}} ... = 0$.
I'm not familiar with this area of mathematics and was wondering if my proof was valid. If not, how could I prove main statement?
I would really appreciate any help/thoughts!
probability sequences-and-series proof-verification binary
probability sequences-and-series proof-verification binary
asked Jan 28 at 0:50
LeoLeo
788517
asked Jan 28 at 0:50
LeoLeo
788517
edited Jan 28 at 0:57
Leo
asked Jan 28 at 0:50
LeoLeo
788517
asked Jan 28 at 0:50
LeoLeo
788517
asked Jan 28 at 0:50
LeoLeo
788517
788517
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Let $R_n$ be the event that the second block of $n$ bits matches the first block of $n$ bits, that is, the event that $wwprec S$ for some $win{0,1}^n$, and let $N=N(S)=sum_{nge1}mathbb 1_{R_n}(S)$ be the number of $n$ such that $Sin R_n$, that is, the number of $n$ such that $wwprec S$ for $w$ of length $n$. It might be finite or it might be infinite. If $N(S)<infty$ then $S$ is not prefix repetitive, but if $N(S)=infty$ then $S$ is prefix repetitive. Now note that $P(R_n)=2^{-n}$, so $EN = EN(S) = sum_{nge1} 2^{-n} < infty$. Hence $P(N(S)<infty) = 1$. (By the Borel-Cantelli lemma, or the fact that random variables with finite expectations are finite with probability 1.) So with probability 1, $S$ is not prefix repetitive.
Where does this come up? What are you reading?
answered Jan 28 at 2:11
kimchi loverkimchi lover
11.5k31229
$endgroup$
$begingroup$
Would $EN$ be the probability that there exists some $r≥n$ such that the second block of $r$ bits matches the first block of $r$?
$endgroup$
– Leo
Jan 28 at 9:19
$begingroup$
Some guy shared the problem in a math groupchat on kik. I have never studied anything similar myself.
$endgroup$
– Leo
Jan 28 at 9:21
$begingroup$
Sorry not to have seen your notes earlier. About your $EN$ query: No. The events I call $R_n$ have some complicated relationship with each other, and there is no simple formula for $P(bigcup_{rge n}R_r)$; it is certainly not $EN$. But the expected number of events $R_n$ that do occur is finite, which implies that with prob 1, only finitely many of them do occur.
$endgroup$
– kimchi lover
Jan 29 at 3:32
add a comment |
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$begingroup$
Let $R_n$ be the event that the second block of $n$ bits matches the first block of $n$ bits, that is, the event that $wwprec S$ for some $win{0,1}^n$, and let $N=N(S)=sum_{nge1}mathbb 1_{R_n}(S)$ be the number of $n$ such that $Sin R_n$, that is, the number of $n$ such that $wwprec S$ for $w$ of length $n$. It might be finite or it might be infinite. If $N(S)<infty$ then $S$ is not prefix repetitive, but if $N(S)=infty$ then $S$ is prefix repetitive. Now note that $P(R_n)=2^{-n}$, so $EN = EN(S) = sum_{nge1} 2^{-n} < infty$. Hence $P(N(S)<infty) = 1$. (By the Borel-Cantelli lemma, or the fact that random variables with finite expectations are finite with probability 1.) So with probability 1, $S$ is not prefix repetitive.
Where does this come up? What are you reading?
answered Jan 28 at 2:11
kimchi loverkimchi lover
11.5k31229
$endgroup$
$begingroup$
Would $EN$ be the probability that there exists some $r≥n$ such that the second block of $r$ bits matches the first block of $r$?
$endgroup$
– Leo
Jan 28 at 9:19
$begingroup$
Some guy shared the problem in a math groupchat on kik. I have never studied anything similar myself.
$endgroup$
– Leo
Jan 28 at 9:21
$begingroup$
Sorry not to have seen your notes earlier. About your $EN$ query: No. The events I call $R_n$ have some complicated relationship with each other, and there is no simple formula for $P(bigcup_{rge n}R_r)$; it is certainly not $EN$. But the expected number of events $R_n$ that do occur is finite, which implies that with prob 1, only finitely many of them do occur.
$endgroup$
– kimchi lover
Jan 29 at 3:32
add a comment |
$begingroup$
Let $R_n$ be the event that the second block of $n$ bits matches the first block of $n$ bits, that is, the event that $wwprec S$ for some $win{0,1}^n$, and let $N=N(S)=sum_{nge1}mathbb 1_{R_n}(S)$ be the number of $n$ such that $Sin R_n$, that is, the number of $n$ such that $wwprec S$ for $w$ of length $n$. It might be finite or it might be infinite. If $N(S)<infty$ then $S$ is not prefix repetitive, but if $N(S)=infty$ then $S$ is prefix repetitive. Now note that $P(R_n)=2^{-n}$, so $EN = EN(S) = sum_{nge1} 2^{-n} < infty$. Hence $P(N(S)<infty) = 1$. (By the Borel-Cantelli lemma, or the fact that random variables with finite expectations are finite with probability 1.) So with probability 1, $S$ is not prefix repetitive.
Where does this come up? What are you reading?
answered Jan 28 at 2:11
kimchi loverkimchi lover
11.5k31229
$endgroup$
$begingroup$
Would $EN$ be the probability that there exists some $r≥n$ such that the second block of $r$ bits matches the first block of $r$?
$endgroup$
– Leo
Jan 28 at 9:19
$begingroup$
Some guy shared the problem in a math groupchat on kik. I have never studied anything similar myself.
$endgroup$
– Leo
Jan 28 at 9:21
$begingroup$
Sorry not to have seen your notes earlier. About your $EN$ query: No. The events I call $R_n$ have some complicated relationship with each other, and there is no simple formula for $P(bigcup_{rge n}R_r)$; it is certainly not $EN$. But the expected number of events $R_n$ that do occur is finite, which implies that with prob 1, only finitely many of them do occur.
$endgroup$
– kimchi lover
Jan 29 at 3:32
add a comment |
$begingroup$
Let $R_n$ be the event that the second block of $n$ bits matches the first block of $n$ bits, that is, the event that $wwprec S$ for some $win{0,1}^n$, and let $N=N(S)=sum_{nge1}mathbb 1_{R_n}(S)$ be the number of $n$ such that $Sin R_n$, that is, the number of $n$ such that $wwprec S$ for $w$ of length $n$. It might be finite or it might be infinite. If $N(S)<infty$ then $S$ is not prefix repetitive, but if $N(S)=infty$ then $S$ is prefix repetitive. Now note that $P(R_n)=2^{-n}$, so $EN = EN(S) = sum_{nge1} 2^{-n} < infty$. Hence $P(N(S)<infty) = 1$. (By the Borel-Cantelli lemma, or the fact that random variables with finite expectations are finite with probability 1.) So with probability 1, $S$ is not prefix repetitive.
Where does this come up? What are you reading?
answered Jan 28 at 2:11
kimchi loverkimchi lover
11.5k31229
$endgroup$
Let $R_n$ be the event that the second block of $n$ bits matches the first block of $n$ bits, that is, the event that $wwprec S$ for some $win{0,1}^n$, and let $N=N(S)=sum_{nge1}mathbb 1_{R_n}(S)$ be the number of $n$ such that $Sin R_n$, that is, the number of $n$ such that $wwprec S$ for $w$ of length $n$. It might be finite or it might be infinite. If $N(S)<infty$ then $S$ is not prefix repetitive, but if $N(S)=infty$ then $S$ is prefix repetitive. Now note that $P(R_n)=2^{-n}$, so $EN = EN(S) = sum_{nge1} 2^{-n} < infty$. Hence $P(N(S)<infty) = 1$. (By the Borel-Cantelli lemma, or the fact that random variables with finite expectations are finite with probability 1.) So with probability 1, $S$ is not prefix repetitive.
Where does this come up? What are you reading?
answered Jan 28 at 2:11
kimchi loverkimchi lover
11.5k31229
edited Jan 28 at 12:31
answered Jan 28 at 2:11
kimchi loverkimchi lover
11.5k31229
answered Jan 28 at 2:11
kimchi loverkimchi lover
11.5k31229
answered Jan 28 at 2:11
kimchi loverkimchi lover
11.5k31229
11.5k31229
$begingroup$
Would $EN$ be the probability that there exists some $r≥n$ such that the second block of $r$ bits matches the first block of $r$?
$endgroup$
– Leo
Jan 28 at 9:19
$begingroup$
Some guy shared the problem in a math groupchat on kik. I have never studied anything similar myself.
$endgroup$
– Leo
Jan 28 at 9:21
$begingroup$
Sorry not to have seen your notes earlier. About your $EN$ query: No. The events I call $R_n$ have some complicated relationship with each other, and there is no simple formula for $P(bigcup_{rge n}R_r)$; it is certainly not $EN$. But the expected number of events $R_n$ that do occur is finite, which implies that with prob 1, only finitely many of them do occur.
$endgroup$
– kimchi lover
Jan 29 at 3:32
add a comment |
$begingroup$
Would $EN$ be the probability that there exists some $r≥n$ such that the second block of $r$ bits matches the first block of $r$?
$endgroup$
– Leo
Jan 28 at 9:19
$begingroup$
Some guy shared the problem in a math groupchat on kik. I have never studied anything similar myself.
$endgroup$
– Leo
Jan 28 at 9:21
$begingroup$
Sorry not to have seen your notes earlier. About your $EN$ query: No. The events I call $R_n$ have some complicated relationship with each other, and there is no simple formula for $P(bigcup_{rge n}R_r)$; it is certainly not $EN$. But the expected number of events $R_n$ that do occur is finite, which implies that with prob 1, only finitely many of them do occur.
$endgroup$
– kimchi lover
Jan 29 at 3:32
$begingroup$
Would $EN$ be the probability that there exists some $r≥n$ such that the second block of $r$ bits matches the first block of $r$?
$endgroup$
– Leo
Jan 28 at 9:19
$begingroup$
Would $EN$ be the probability that there exists some $r≥n$ such that the second block of $r$ bits matches the first block of $r$?
$endgroup$
– Leo
Jan 28 at 9:19
$begingroup$
Some guy shared the problem in a math groupchat on kik. I have never studied anything similar myself.
$endgroup$
– Leo
Jan 28 at 9:21
$begingroup$
Some guy shared the problem in a math groupchat on kik. I have never studied anything similar myself.
$endgroup$
– Leo
Jan 28 at 9:21
$begingroup$
Sorry not to have seen your notes earlier. About your $EN$ query: No. The events I call $R_n$ have some complicated relationship with each other, and there is no simple formula for $P(bigcup_{rge n}R_r)$; it is certainly not $EN$. But the expected number of events $R_n$ that do occur is finite, which implies that with prob 1, only finitely many of them do occur.
$endgroup$
– kimchi lover
Jan 29 at 3:32
$begingroup$
Sorry not to have seen your notes earlier. About your $EN$ query: No. The events I call $R_n$ have some complicated relationship with each other, and there is no simple formula for $P(bigcup_{rge n}R_r)$; it is certainly not $EN$. But the expected number of events $R_n$ that do occur is finite, which implies that with prob 1, only finitely many of them do occur.
$endgroup$
– kimchi lover
Jan 29 at 3:32
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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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
