An Erdős-Révész Type Law for the Length of the Longest Match of Two Coin-Tossing Sequences

Abstract

Consider a coin-tossing sequence, i.e., a sequence of independent variables, taking values 0 and 1 with probability $1/2$. The famous Erdős-Rényi law of large numbers implies that the longest run of ones in the first n observations has a length $R_n$ that behaves like $\log \left(n\right)$, as n tends to infinity (throughout this article, log denotes logarithm with base 2). Erdős and Révész refined this result by providing a description of the Lévy upper and lower classes of the process $R_n$. In another direction, Arratia and Waterman extended the Erdős-Rényi result to the longest matching subsequence (with shifts) of two coin-tossing sequences, finding that it behaves asymptotically like $2\log \left(n\right)$. The present paper provides some Erdős-Révész type results in this situation, obtaining a complete description of the upper classes and a partial result on the lower ones.

1. Introduction

Consider a coin-tossing sequence $\left(X_n\right)$, i.e., a sequence of independent random variables satisfying $\mathbb{P}(X_n=0)=\mathbb{P}(X_n=1)=1/2$. Let $R_n$ be the length of the longest head-run, i.e., the largest integer r for which there is an i, $0\le i\le n-r$, for which $X_{i+j}=1$ for $j=1,\dots r$. A result of Erdős and Rényi [1] implies that

$$\underset{n\to \infty }{lim}{\displaystyle \frac{R_n}{\log \left(n\right)}}=1$$

(1)

(throughout this paper, log will denote base 2 logarithms. The notation ${\log }_k$ will be used for its iterates: ${\log }_2\left(x\right)=\log \left(\log \left(x\right)\right)$, ${\log }_{k+1}\left(x\right)=\log \left({\log }_k\left(x\right)\right)$. Also, C and c, with or without an index, are used to denote generic constants that may have different values at each occurrence). The simple result (1) has seen a number of improvements. Erdős and Révész [2] provided a detailed description of the asymptotic behavior of $R_n$. In order to formulate their result, let us recall

Definition 1 

(Lévy classes). Let $\left(Y_n\right)$ be a sequence of random variables. We say that a sequence $\left(a_n\right)$ of real numbers belongs to

• The upper-upper class of $\left(Y_n\right)$ ($UUC\left(Y_n\right)$), if, with probability 1 as $n\to \infty$, $Y_n\le a_n$ eventually.
• The upper-lower class of $\left(Y_n\right)$ ($ULC\left(Y_n\right)$), if, with probability 1 as $n\to \infty$, $Y_n>a_n$ for infinitely many n.
• The lower-upper class of $\left(Y_n\right)$ ($LUC\left(Y_n\right)$), if, with probability 1 as $n\to \infty$, $Y_n<a_n$ for infinitely many n.
• The lower-lower class of $\left(Y_n\right)$ ($LLC\left(Y_n\right)$), if, with probability 1 as $n\to \infty$, $Y_n\ge a_n$ eventually.

Of course, these definitions work best if the sequence $\left(Y_n\right)$ obeys some zero-one law.

Their result is as follows:

Let $\left(a_n\right)$ be a nondecreasing integer sequence. Then

• $\left(a_n\right)\in UUC\left(R_n\right)$ if ${\sum }_n{2}^{-a_n}<\infty$,
• $\left(a_n\right)\in ULC\left(R_n\right)$ if ${\sum }_n{2}^{-a_n}=\infty$,
• for any $ϵ>0$, $a_n=\lfloor \log \left(n\right)-{\log }_3\left(n\right)+{\log }_2\left(e\right)-1+ϵ\rfloor \in LUC\left(R_n\right)$,
• for any $ϵ>0$, $a_n=\lfloor \log \left(n\right)-{\log }_3\left(n\right)+{\log }_2\left(e\right)-2-ϵ\rfloor \in LLC\left(R_n\right)$.

Arratia and Waterman [3] extend Erdős and Rényi’s result in another direction: they consider two independent coin-tossing sequences, $\left(X_n\right)$ and $\left(Y_n\right)$, and look for the longest matching subsequences when shifting is allowed. Formally, let $M_n$ be the the largest integer m for which there are $i,j$ with $0\le i,j\le n-m$ and $X_{i+k}=Y_{j+k}$ for all $k=1,\dots ,m$. They prove that, with probability 1

$$\underset{n\to \infty }{lim}{\displaystyle \frac{M_n}{\log \left(n\right)}}=2.$$

(2)

In the present paper, we will make this more precise by providing a description for the upper classes of $\left(M_n\right)$ and also some results on its lower classes:

Theorem 1. 

Let $\left(a_n\right)$ be a nondecreasing integer sequence. We have

• $\left(a_n\right)\in UUC\left(M_n\right)$ if ${\sum }_{n}n{2}^{-a_n}<\infty$.
• $\left(a_n\right)\in ULC\left(M_n\right)$ if ${\sum }_{n}n{2}^{-a_n}=\infty$.
• for some c, $a_n=\lfloor 2\log \left(n\right)-{\log }_3\left(n\right)+c\rfloor \in LUC\left(M_n\right)$.
• for some c, $a_n=\lfloor 2\log \left(n\right)-2{\log }_2\left(n\right)-{\log }_3\left(n\right)+c\rfloor \in LLC\left(M_n\right)$.

2. Discussion

We leave the proof of Theorem 1 for later and rather discuss some of the concepts that are connected to this problem. One of them is the so-called independence principle: in many, although not all, situations, one may pretend that the waiting times until a given pattern of length l is observed have an exponential distribution with parameter $2^{-l}$, and that the waiting times for different patterns are independent. Móri [4] and Móri and Székely [5] provide an account of this principle and its limitations. In our case, all results but the lower-lower class one are more or less in tune with this principle.

Another question that is closely related is that of the number $N(n,l)$ of different length l subsequences of $(X_1,\dots ,X_n)$. This question does not seem to have been considered by literature very much; there is one remarkable result by Móri [6]: in the remark following the statement of Theorem 3 in that paper, he mentions that with probability one for large n, the largest l for which all $2^l$ possible patterns occur as subsequences of $(X_1,\dots ,X_n)$ is either $\lfloor \log \left(n\right)-{\log }_2\left(n\right)-ϵ\rfloor$ or $\lfloor \log \left(n\right)-{\log }_2\left(n\right)+ϵ\rfloor$ for any $ϵ>0$. The independence principle would suggest that $N(n,\log (n\left)\right)/n$ is bounded away from 0 with probability one, and this or even the less stringent $N(n,\log \left(n\right))\ge n{\left({\log }_2\left(n\right)\right)}^{-c}$ for large n would be an important step towards removing the double log term from the $LLC$ result, as we conjecture that, for some $c>0$, $\log \left(n\right)-c{\log }_3\left(n\right)\in LLC\left(M_n\right)$. Unfortunately, we are only able to obtain $N(n,\log (n\left)\right)\ge cn/\log \left(n\right)$, which is also implied by Móri’s result.

3. Proofs

Proof of the upper-upper class result. 

Both upper class statements are fairly easy to prove. First, observe that under our assumptions, the convergence of

$$\sum _{n=1}^{\infty }n{2}^{-a_n}$$

(3)

is equivalent to that of

$$\sum _{k=1}^{\infty }{n}_k^2{2}^{-a_{n_k}}$$

(4)

with $n_k=2^k$.

Now, define events

$$A_k=[M_{n_k}\ge a_{n_{k-1}}].$$

(5)

$A_k$ occurs if in one of the ${(n_k+1-a_{n_{k-1}})}^2$ pairs of sequences

$$((X_{i+1},\dots ,X_{i+a_{k-1}}),\left((Y_{j+1},\dots ,Y_{j+a_{k-1}})\right)$$

(6)

both sequences agree. That provides the trivial upper bound

$$\mathbb{P}\left(A_k\right)\le n_k^2{2}^{-a_{n_{k-1}}},$$

(7)

so, by our assumptions, ${\sum }_n\mathbb{P}\left(A_k\right)<\infty$, and the Borel-Cantelli Lemma implies that, with probability 1, only finitely many events $A_k$ occur. Thus, for sufficiently large k, $M_{n_k}\le a_{n_{k-1}}$, and for $n_{k-1}\le n\le n_k$, we have

$$M_n\le M_{n_k}\le a_{n_{k-1}}\le a_n.$$

(8)

This shows that $\left(a_n\right)\in UUC\left(M_n\right)$, as claimed. □

Proof of the upper-lower class result. 

We may assume without loss of generality that $n^2{2}^{-a_n}\le 1/4$.

Again, let $n_k=2^k$. We want to use the second Borel-Cantelli Lemma, so we are defining independent events

$$A_k=[\exists i,j:n_{k-1}<i,j\le n_k:X_{i+s}=Y_{j+s},s=0,\dots ,a_{n_k}-1]$$

(9)

This is the union of the events

$$B_{ij}=[X_{i+s}=Y_{j+s},s=0,\dots ,a_{n_k}-1]$$

(10)

with $n_{k-1}<i,j\le n_k$. We endow the set of pairs $(i,j)$ with the lexicographic order. For a subset I of the integers, Bonferroni’s inequality provides

$$\mathbb{P}(\bigcup _{(i,j)\in I\times I}{B}_{ij})\ge \sum _{(i,j)\in I\times I}\mathbb{P}\left(B_{ij}\right)-\sum _{(i,j),(i^{\prime },j^{\prime })\in I\times I,(i,j)<(i^{\prime },j^{\prime })}\mathbb{P}(B_{ij}\cap B_{i^{\prime }{j}^{\prime }}).$$

(11)

Let $d((i,j),(i^{\prime },j^{\prime }))=max\left(\right|i-i^{\prime }|,|j-j^{\prime }\left|\right)$. If $d((i,j),(i^{\prime },j^{\prime }))\ge a_{n_k}$, then $\mathbb{P}(B_{ij}\cap B_{i^{\prime }{j}^{\prime }})=2^{-2a_{n_k}}$, otherwise $\mathbb{P}(B_{ij}\cap B_{i^{\prime }{j}^{\prime }})=2^{-(a_{n_k}+d((i,j),(i^{\prime },j^{\prime }))}$.

Let $I=\{i:n_{k-1}<i\le n_k:4|i\}$. Using this in (11) yields the value ${\left|I\right|}^2{2}^{-a_{n_k}}$ for the first sum. In the second sum, for given $(i,j)$ and $d<a_{n_k}/4$, there are no more than $2(2d+1)$ pairs $(i^{\prime },j^{\prime })$ with $d((i,j),(i^{\prime },j^{\prime }))=4d$. The number of pairs of pairs $((i,j),(i^{\prime },j^{\prime })$ with $d((i,j),(i^{\prime },j^{\prime }))\ge a_{n_k}$ is trivially bounded by ${\left|I\right|}^4/2$. Putting these together, we arrive at the upper estimate

$${\left|I\right|}^2{2}^{-a_{n_k}}(\sum _{d=1}^{\infty }(2d+1)2^{1-4d}+{\left|I\right|}^2{2}^{-1-a_{n_k}})<{\displaystyle \frac{1}{2}}{\left|I\right|}^2{2}^{-a_{n_k}}.$$

(12)

In total,

$$\mathbb{P}\left(A_k\right)\ge {\displaystyle \frac{1}{2}}{\left|I\right|}^2{e}^{-a_{n_k}}={\displaystyle \frac{1}{128}}{n}_k^2{2}^{-a_{n_k}},$$

(13)

keeping in mind that $\left|I\right|=n_k/8$.

and ${\sum }_k\mathbb{P}\left(A_k\right)=\infty$. Borel-Cantelli implies that, with probability 1, infinitely many events $A_k$ occur. Thus, for infinitely many k, $M_{n_k}\ge a_{n_k}$, so $\left(a_n\right)\in ULC\left(M_n\right)$. □

For the lower class results, we first prove some lemmas:

Lemma 1. 

With probability 1 for n sufficiently large

$${\displaystyle \frac{n}{4\log \left(n\right)}}\le N(n,\lfloor \log \left(n\right)\rfloor )\le n$$

(14)

Proof of Lemma 1. 

The lower part is a direct consequence of Móri’s result: as for sufficiently large n $N(n,\lfloor \log \left(n\right)-{\log }_2\left(n\right)-1\rfloor )=2^{\lfloor \log \left(n\right)-{\log }_2\left(n\right)-1\rfloor }\ge {\displaystyle \frac{n}{4\log \left(n\right)}}$ and, obviously $N(n,\lfloor \log \left(n\right)\rfloor )\ge N(n,\lfloor \log \left(n\right)-{\log }_2\left(n\right)-1\rfloor )-{\log }_2\left(n\right)-2$, as extending two different sequences from length $\lfloor \log \left(n\right)-{\log }_2\left(n\right)\rfloor$ to $\lfloor \log \left(n\right)\rfloor$ keeps them different; it can only happen that some of them are extended beyond index n, but this can affect at most ${\log }_2\left(n\right)+2$ of them. □

Lemma 2. 

Let S be a set of $m<2^l$ sequences of length $l<n$, and let A be the event that none of the sequences in S occurs as a subsequence of $(X_1,\dots ,X_n)$. For $l\le n^{\prime }<n$, let B be the event that some sequence from S is a subsequence of $X_1,\dots ,X_{n^{\prime }}$. Then

$$({(1-\mathbb{P}\left(B\right))}^2-\lfloor n/n^{\prime }\rfloor m{2}^{-l}){(1-\mathbb{P}\left(B\right))}^{\lfloor n/n^{\prime }\rfloor }\le \mathbb{P}\left(A\right)\le {(1-\mathbb{P}\left(B\right))}^{\lfloor n/n^{\prime }\rfloor }.$$

(15)

Proof of Lemma 2, upper part. 

Consider the $\lfloor n/n^{\prime }\rfloor$ sequences $X_{k{n}^{\prime }+1},\dots ,X_{(k+1)n^{\prime }}$ for $k=0,\dots \lfloor n/n^{\prime }\rfloor -1$. Each of these has probability $1-\mathbb{P}\left(B\right)$ that it does not contain a subsequence that lies in S, and by independence

$$\mathbb{P}\left(A\right)\le {(1-\mathbb{P}\left(B\right))}^{\lfloor n/n^{\prime }\rfloor }.$$

(16)

□

Proof of Lemma 2, lower part. 

Assume for simplicity that n is a multiple of $n^{\prime }$, say $n=N{n}^{\prime }$. Again, we split $(X_1,\dots ,X_n)$ into N blocks of length $n^{\prime }$, and the probability that none of those contains a sequence from S is

$${(1-\mathbb{P}\left(B\right))}^N.$$

(17)

It can still happen that there is a sequence from S that crosses one of the boundaries between the blocks. There are $N-1$ boundaries, and for each of those, there are $l-1$ possible subsequences of length l crossing it. The probability that this one is from S but none of the N blocks contains one can be estimated above by the probability that it is from S and none of the $N-2$ blocks not adjacent to it contains one from S. This provides the upper bound

$$(N-1)(l-1)m{2}^{-l}{(1-\mathbb{P}\left(B\right))}^{N-2}$$

(18)

for the probability that there is a subsequence from S in $(X_1,\dots ,X_n)$ but none in any of the N blocks. Subtracting this from (17), we obtain the lower bound

$$({(1-\mathbb{P}\left(B\right))}^2-(N-1)(l-1)m{2}^{-l}){(1-\mathbb{P}\left(B\right))}^{N-2},$$

(19)

and the general case is obtained by observing that the probability $\mathbb{P}\left(A\right)$ for a given n is bounded below by the one that we get for $n^{\prime }\lceil n/n^{\prime }\rceil$. □

In this lemma, the lower and upper bounds are rather close. Its applicability, however, depends on the availability of good estimates for the probability $\mathbb{P}\left(B\right)$. There is the trivial upper bound $n^{\prime }m{2}^{-l}$ and an almost as simple lower bound $\mathrm{const}.n^{\prime }m{l}^{-1}{2}^{-l}$ (given $n^{\prime }m{l}^{-1}{2}^{-l}<1$). Bridging the gap between these requires deeper insight into the structure of S.

Proof of the lower-lower class result. 

In both the lower-lower and lower-upper parts, we consider the asymptotics of the longest match found between $(X_1,\dots ,X_n)$ and $(Y_1,\dots ,Y_n)$ under the condition that the sequence $(Y_n,n\in \mathbb{N})$ is given. Doing so, we may assume that $\mathbf{Y}=(Y_n,n\in \mathbb{N})$ is a “typical” coin-tossing sequence, in the sense that it possesses some property that holds with probability 1. In the sequel, all probabilities are understood as conditional with respect to such a typical sequence $\mathbf{Y}$. For the lower-lower class result, we let $n=n_l=\lceil C{2}^{l/2}l\sqrt{\log \left(l\right)}\rceil$, $n^{\prime }=n_l^{\prime }=l$, and $m=m_l=\lceil {\displaystyle \frac{n}{4\log \left(n\right)}}\rceil$ in Lemma 2. Clearly, as $(X_1,\dots ,X_l)$ only has one length l subsequence, $\mathbb{P}\left(B\right)$ equals $\tilde{m}{2}^{-l}$, where $\tilde{m}$ is the number of different sequences of length l in $Y_1,\dots ,Y_{n_l}$. For sufficiently large l, $\tilde{m}\ge m_l$ by Lemma 1, and we obtain an upper estimate

$$p_l=exp(-m\lfloor n/l\rfloor 2^{-l})=exp\left(-{\displaystyle \frac{1}{2}}{C}^2\log \left(l\right)(1+o\left(1\right))\right)$$

(20)

for the probability (conditional on $\mathbf{Y}$) that there is no match of length l between $(Y_1,\dots ,Y_{n_l})$ and $(X_1,\dots ,X_{n_l})$. For $C>\sqrt{2/\log \left(e\right)}$, the series ${\sum }_l{p}_l$ converges, so with probability 1, we have $M\left(n_l\right)\ge l$ for all but finitely many l. Thus, for $n_{l-1}\le n<n_l$, $M_n\ge M_{n_{l-1}}\ge l-1$. Inverting the relationship between $n_l$ and l yields $l=2\log \left(n_l\right)-2{\log }_2\left(n_l\right)-{\log }_3\left(n_l\right)+O\left(1\right)$, so, for some constant c and l large enough, we obtain $M_n\ge l-1\ge 2\log \left(n\right)-2{\log }_2\left(n\right)-{\log }_3\left(n\right)-c$, which proves the lower-lower class result. □

Proof of the lower-upper class result. 

This time, we need to make our choice of the parameters in Lemma 2 with a little more sophistication. We start with $l=l_k=k^2$ for $k\in \mathbb{N}$. Then, $n=n_k=\lfloor C{2}^{l_k/2}\sqrt{\log \left(l_k\right)}\rfloor \prime$. As the set S, we choose the set $S_k$ of all sequences of length $l_k$ contained in $(Y_1,\dots ,Y_{n_k})$. $n^{\prime }=n_k^{\prime }$ is chosen in such a way that $m_k{n}_k^{\prime }{2}^{-l_k}\to 0$ and $m_k{n}_k{l}_k{2}^{-l_k}/n_k^{\prime }\to 0$, $n_k^{\prime }=\lfloor 2^{l_k/4}\rfloor$ is a possible choice.

We define the events

$$A_k=\left\{\mathrm{There\; is\; no\; sequence\; from} S_k \mathrm{in}(X_1,\dots X_{n_k})\right\},$$

(21)

$${\tilde{A}}_k=\left\{\mathrm{There\; is\; no\; sequence\; from} S_k \mathrm{in}(X_{n_{k-1}+1},\dots X_{n_k})\right\}$$

(22)

$$B_k=\left\{\mathrm{There\; is\; a\; sequence\; from} S_k \mathrm{in}(X_1,\dots X_{n_k^{\prime }})\right\}$$

(23)

(this last is just the event B from Lemma 2).

Lemma 2 gives us

$$\mathbb{P}\left(A_k\right)={(1-\mathbb{P}\left(B_k\right))}^{\lfloor n_k/n_k^{\prime }\rfloor }(1+o\left(1\right))$$

(24)

and

$$\mathbb{P}\left({\tilde{A}}_k\right)={(1-\mathbb{P}\left(B_k\right))}^{\lfloor (n_k-n_{k-1})/n_k^{\prime }\rfloor }(1+o\left(1\right)).$$

(25)

The trivial estimate $\mathbb{P}\left(B_k\right)\le m_k{n}_k^{\prime }{2}^{-l_k}\le n_k^2{2}^{-l_k}$ yields

$$\mathbb{P}\left(A_k\right)\ge e^{-2C^2\log \left(k\right)(1+o\left(1\right))},$$

(26)

which diverges if we choose $C<1/\sqrt{2\log \left(e\right)}$.

We are going to use the Borel-Cantelli Lemma in the usual form for dependent events:

Lemma 3 

(Borel-Cantelli II). If the sequence $\left(A_n\right)$ satisfies

$$\sum _{n\in \mathbb{N}}\mathbb{P}\left(A_n\right)=\infty$$

(27)

and

$$\underset{n\to \infty }{lim}{\displaystyle \frac{{\sum }_{i=1}^n{\sum }_{j=1}^n\mathbb{P}(A_i\cap A_j)}{{\left({\sum }_{i=1}^n\mathbb{P}\left(A_i\right)\right)}^2}}=1,$$

(28)

then

$$\mathbb{P}(\underset{n}{lim\; sup}{A}_n)=1,$$

(29)

To this end, we need an upper bound for $\mathbb{P}(A_i\cap A_j)$ for $i<j$. We have

$$\mathbb{P}(A_i\cap A_j)\le \mathbb{P}(A_i\cap {\tilde{A}}_j)=\mathbb{P}\left(A_i\right)\mathbb{P}\left({\tilde{A}}_j\right).$$

(30)

By our Equations (24) and (25) from above

$$\mathbb{P}\left({\tilde{A}}_j\right)=\mathbb{P}\left(A_j\right){(1-\mathbb{P}\left(B_j\right))}^{-n_{j-1}/n_j^{\prime }}(1+o\left(1\right))=\mathbb{P}\left(A_j\right)e^{m_j{n}_{j-1}{2}^{-l_j}(1+o\left(1\right))}=$$

$$\mathbb{P}\left(A_j\right)(1+o\left(1\right)),$$

(31)

as $n_{j-1}\le n_j{e}^{1-2j}$ and $m_j{n}_j{2}^{-l_j}\le n_j^2{2}^{-l_j}=O\left(\log j\right)$.

This means that for any $ϵ>0$, there is a number $j_0$ such that, for $j>j_0$ and $i<j$, the inequality

$$\mathbb{P}(A_i\cap A_j)\le (1+ϵ)\mathbb{P}\left(A_i\right)\mathbb{P}\left(A_j\right)$$

(32)

holds. Plugging this into

$$\sum _{j=1}^n\sum _{i=1}^n\mathbb{P}(A_i\cap A_j)=\sum _{i=1}^n\mathbb{P}\left(A_i\right)+2\sum _{j=2}^n\sum _{i=1}^{j-1}\mathbb{P}(A_i\cap A_j)$$

(33)

yields the estimate

$$\sum _{j=1}^n\sum _{i=1}^n\mathbb{P}(A_i\cap A_j)\le \sum _{i=1}^n\mathbb{P}\left(A_i\right)+j_0^2+(1+ϵ){(\sum _{i=1}^n\mathbb{P}\left(A_i\right))}^2.$$

(34)

As ${\sum }_{i=1}^{\infty }\mathbb{P}\left(A_i\right)=\infty$, we get

$$\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{{\sum }_{i=1}^n{\sum }_{j=1}^n\mathbb{P}(A_i\cap A_j)}{{\left({\sum }_{i=1}^n\mathbb{P}\left(A_i\right)\right)}^2}}\le 1+ϵ.$$

(35)

As $ϵ>0$ is arbitrary, the sequence $\left(A_k\right)$ satisfies the assumptions of Lemma 3, so, with probability 1, infinitely many of the events $\left(A_k\right)$ occur. Thus, with probability 1 for infinitely many k, $M_{n_k}<l_k$. Observing that $l_k=2\log \left(n_k\right)-{\log }_3\left(n_k\right)+O\left(1\right)$, we obtain our lower-upper class result. □