A PROBABILISTIC ANALYSIS OF A PATTERN ... - Semantic Scholar
A PROBABILISTIC ANALYSIS OF A PATTERN MATCHING PROBLEM July 9, 1992 Mikhail J. Atallah Dept. of Computer Science Purdue University W. Lafayette, IN 47907 U.S.A.
Philippe Jacquety INRIA Rocquencourt 78153 Le Chesnay Cedex France
Wojciech Szpankowskiz Dept. of Computer Science Purdue University W. Lafayette, IN 47907 U.S.A.
Abstract The study and comparison of strings of symbols from a nite or an in nite alphabet is relevant to various areas of science, notably molecular biology, speech recognition, and computer science. In particular, the problem of nding the minimum "distance" between two strings (in general, two blocks of data) is of great importance. In this paper we investigate the (string) pattern matching problem in a probabilistic framework, namely, it is assumed that both strings form an independent sequences of i.i.d. symbols. Given a text string a of length n and a pattern string b of length m, let Mm;n be the maximum number of matches between p b and all m-substrings of a. Our main probabilistic result shows that Mm;n ? mP = ( 2m log n) in probability (pr.) provided m; n ! 1 such that log n = o(m), where P is the probability of a match between any two symbols of these strings. If the alphabet is symmetric (i.e., all symbols occur with the same probability), p a stronger result holds, namely (Mm;n ? mP ) 2m(1=V ? 1=V 2 ) log n (pr.), where V is the size of the alphabet. In either case, symmetric alphabet or not, we also prove that Mm;n =m ! P almost surely (a.s).
This author's research was supported by the Oce of Naval Research under Grants N0014-84-K-0502
and N0014-86-K-0689, and in part by AFOSR Grant 90-0107, and the NSF under Grant DCR-8451393, and in part by Grant R01 LM05118 from the National Library of Medicine. y This research was primary supported by NATO Collaborative Grant 0057/89. z This author's research was supported by AFOSR Grant 90-0107 and NATO Collaborative Grant 0057/89, and, in part by the NSF Grant CCR-8900305, and by Grant R01 LM05118 from the National Library of Medicine.
1
1. INTRODUCTION Pattern matching is one of the most fundamental problems in computer science. It is relevant to various areas of science, notably molecular biology, speech recognition, theory of languages, coding, and so forth. For example, in molecular biology often one compares two, say DNA, sequences to detect an unusual homology (similarity) between these strings. Comparisons of blocks of data representing, say time series of physical measurements, prices, etc., also lead to string comparisons, but with possibly in nite alphabets. In general, the problem of nding the minimum "distance" between two strings is of great importance. The version of the string matching problem we investigate here is the following one. Consider two strings, a text string a = a1 a2 :::an and a pattern string b = b1 b2 :::bm of lengths n and m respectively, such that symbols ai and bj belong to a V -ary alphabet = f1; 2; :::; V g. The alphabet may be nite or not. Let Ci be the number of positions at which the substring ai ai+1 :::ai+m?1 agrees with the pattern b (the index j that is out of P range is understood to stand for 1+(j mod n)). That is, Ci = mj=1 equal(ai+j ?1 ; bj ) where equal(x; y) is one if x = y, and zero otherwise. In other words, the "distance" between a and b is the Hamming distance. We are interested in the quantity Mm;n = max1in fCi g which represents the best matching between b and any m-substring of a. We analyze Mm;n under the following probabilistic assumption: symbols from the alphabet are generated independently, and symbol i 2 occurs with probability pi . This probabilistic model is known as the Bernoulli model [26]. An estimate of Mm;n can be used in various statistical tests, for example in the usual hypothesis-testing context in which one needs to detect a "signi cant" matching. In that context, a matching is considered signi cant when it is much larger than the one obtained for random strings, that is, than Mm;n ; indeed, if even random independent strings give rise to a match of Mm;n , it is reasonable not to attach particular signi cance to this amount of \similarity" between two strings. Furthermore, an estimate of Mm;n can be used to control some parameters of algorithms designed for pattern matching, as we did in Atallah et al. [8]. Our main probabilistic result concerning Mm;n can be summarized as follows. We prove p that with high probability Mm;n ?mP = ( 2m log n) for n; m ! 1 such that log n = o(m) where P is the probability that two symbols match. The restriction on n and m is important. We explicitly give the constants that are implicit in the () notation. Our simulation experp iments indicate that the second term in Mm;n , that is, ( 2m log n), spreads in probability between these two bounds, that is, we conjecture that the second term is rather a random variable than a constant. However, for the symmetric alphabet (i.e., all symbols occur with 2
p
the same probability) stronger result holds, namely (Mm;n ?mP ) 2m(1=V ? 1=V 2 ) log n (pr.). In either case { symmetric alphabet or not { we demonstrate that almost surely (a.s.) Mm;n =m ! P as long as log n = o(m). A probabilistic analysis of any pattern matching is a "very complicated\ problem as asserted by Arratia, Gordon and Waterman [7]. In fact, the literature on probabilistic analysis of pattern matching is very scanty. To the best of our knowledge, it is restricted to three papers of Arratia, Gordon and Waterman [5, 6, 7], however, only [7] is relevant to our study. Papers [5, 6] investigate the Erdos-Renyi law for the longest contiguous run of matches. Only in [7] is the number of matches in a segment of length m investigated. This last problem resembles the one discussed in this paper, however the authors of [7] restrict themselves to log n = (m) while we analyze the case when log n = o(m). Finally, it should be mentioned that our formulation of the problem is an attempt to study a more important problem of deletion, insertion and substitution developed during the evolution of nucleotide sequences (cf. Apostolico et al. [4], Louchard and Szpankowski [23]). Algorithmic issues of pattern matching (with possible mismatches) are discussed in several papers. The problem of pattern matching with mismatches was posed by Galil [15] for the case of more than one mismatch. Of course, it has a linear time solution for the case of zero mismatch (i.e., k = m). For the case of a single mismatch (i.e., k = m ? 1) a linear time solution is also known (attributed to Vishkin in [15]). The best known time p bound for the general case of arbitrary k is O(n mpolylog(m)) and is due to Abrahamson [1]. The elegant probabilistic method of Chung and Lawler [9] led to an algorithm of O((n=m)k log m) time complexity on the average, assuming that k < m= log m + O(1) and that the alphabet size is O(1). The paper is organized as follows. The next section makes a more precise statement of our results. In particular, Theorem 2 contains our main probabilistic result. All proofs are delayed until Section 3, which is also of independent interest. It discusses a fairly general approach that can be used to analyze pattern matching in strings. In that section we apply extensively the saddle point method [18] to evaluate necessary asymptotic approximations.
2. MAIN RESULTS This section presents our main results derived under the Bernoulli model discussed P above. Note that, in such a model, P = Vi=1 p2i represents the probability of a match between a given position of the text string a and a given one in the pattern string b. All our asymptotic results are valid in the so called right domain asymptotic, that is, for 3
log n = o(m). In fact, the most interesting range of values for m and n is such that m = (n ) for some 0 1. However, to cover the whole right domain asymptotics, we assume that may also slowly converge to zero, that is, 1 ?1 o(m= log m) (for simplicity of notation we write instead of n ). Finally, for simplicity of the presentation, it helps to imagine that a is written on a cycle of size n 2m. Then, we call i (b) the version of b written on the same cycle, and cyclically shifted by i positions relative to a. Note that Ci is the number of matches between a and b when b is shifted by i positions. In our cyclic representation, Ci can alternatively be thought of as the number of places on the cycle in which a and i (b) agree. It is easy to see that the distribution of Ci is binomial, hence for any i ! m PrfCi = `g = ` P ` (1 ? P )m?` : (1) Naturally, the average number of matches EC1 is equal to mP . Furthermore, Ci tends almost surely to its mean mP (by the Strong Law of Large Numbers [12]). Now we can present our rst result regarding Mm;n = max1in fCi g which can be interpreted as the amount of similarity between a and b.
Theorem 1 If m and n are in the right domain asymptotic, that is , log n = o(m), then Mm;n = P lim m!1 m
(a:s:)
(2)
Proof. A lower bound on Mm;n follows from the fact that, by its de nition, Mm;n must be greater than C1 which tends { by the Strong Law of Large Numbers { almost surely to mP . So, now we concentrate on an upper bound. We rst establish the bound in probability (pr.) sense. From Boole's inequality we have PrfMm;n > rg = PrfC1 > r or C2 > r or ::: Cn > rg nPrfC1 > rg :
(3)
It suces to show that for r mP the above probability becomes o(1), that is, nPrfC1 > (1 + )mP g = o(1). For this we need an estimate of the tail for the binomial distribution (1). Such an estimate is computed in Section 2 by the saddle point method. A simpler (and more general) approach, however, is necessary for the purpose of this proof. We note that C1 can be represented as a sum of m independent Bernoulli distributed random variables Xi , where Xi is equal to one when there is a match at the ith position, and zero p otherwise. From the Central Limit Theorem we know that (C1 ? EC1 )=( mP (1 ? P )) ! 4
N (0; 1), where N (0; 1) is the standard normal distribution. Let Gm(x) be the distribution of p P ( mi=1 Xi ? mEX1 )= m varX1 , and let (x) be the standard normal distribution. Then, from Feller [12]
?x2 =2 p e p o( m) ; Gm(x) = (x) + 2 2 p where (x) ex?px2=2 . In (3), set r = mP + (1 + ") m2P (1 ? P ) log n. Then q PrfMm;n > mP + (1 + ") 2mP (1 ? P ) log ng n1" :
(4)
The above rate of convergence does not yet warrant the application of the Borel-Cantelli lemma to prove (2), but clearly (4) proves Mm;n =m ! P (pr.) in the right domain asymptotic. To establish the stronger (a.s.) convergence, we proceed as follows. De ne a subsep f = (M quence nr = s2r for some integers r and s. Let M n m;n ? mP )= 2m. Then, by (4) p p f = log n P ? P 2 (a.s.). To nish the proof we need to extend this to all n. Note M nr r that for a xed s we can always nd an r such that s2r n (s + 1)2r . Furthermore, since f is a stochastically increasing sequence, we have M n p f f M r log( s + 1)2r = pP ? P 2 (a:s:) : M ( s +1)2 n p p p lim sup lim sup r n!1 log n r!1 log s2r log(s + 1)2 This proves (2). Theorem 1 does not provide much useful information, and an estimate of Mm;n based on it would be a very poor one. From the proof of Theorem 1 we learn, however, that p Mm;n ? EC1 = O( m log n), hence the next term in the asymptotics of Mm;n plays a very signi cant role, and de nitely cannot be omitted in any reasonable computation. The next theorem { our main result { provides an extension of Theorem 1, and shows how much the maximum Mm;n di ers from the average EC1 . The proof of the above theorem will be given in Section 3.
Theorem 2 Under the hypothesis of Theorem 1 we also have for every " > 0 Mpm;n ? mP < (1 + ")g = 1 ; lim Pr f (1 ? " ) n!1 2m log n
where
q
q
= maxf (1 ? )(P ? T ); minf(P ? T ); (1 ? )1 gg p
= minf P ? P 2 ;
V X
j =1
q
pj 1 ? pj g : 5
(5) (6) (7)
In the above, P
where T = Vi=1 p3i .
)(P ? 3P + 2T ) ; 1 = (P ? T6( T ? P 2) 2
Remark 1. The lower bound (6) and the upper bound (7) are composed of two di erent
expressions. None of the expressions is better than the other for the entire range of the probabilities 0 < pi < 1. With respect to the lower bound, we can show that there exist values of the probabilities fpi gVi=1 such that either 2 = (1 ? )(P ? T ) or 2 = minf(P ? T ); (1 ? )1 g. The latter case clearly can occur. Surprisingly enough, the former case may happen too. Indeed, let one of the probability pi dominates all the others, e.g., p1 = 1 ? ". Then, we have either 2 = (1 ? )" + O("2 ) or 2 = minf2=3 (1 ? )"; "g + O("2 ). For close to one the former bound is tighter. With respect to the upper bound, we have a p P similar situation. In most cases, 2 = P ? P 2 . However, sometimes = Vi=1 pi 1 ? pi . Indeed, this occurs for example for a binary alphabet with p1 < 0:1. 2 Note that, when all the pi ! 1=V (the symmetric case) we have 1 ! 1. Thus, for all p < 1 we have = = 1=V ? 1=V 2 . Thus, we obtain the following corollary.
Corollary 1 If the alphabet is symmetric and < 1, then for every " > 0 the following holds
Mm;n ? mP p lim Pr f 1 ? " < < 1 + "g = 1 ; n!1 2m(1=V ? 1=V 2 ) log n p that is, Mm;n ? mP 2m(1=V ? 1=V 2 ) log n (pr.).
(8)
In the last subsection we will extend the condition of the corollary to the limiting case = 1. In order to verify our bounds for Mm;n we have performed some simulation experiments, which are presented in Figures 1, 2, and 3. In these gures we plot the experimental distrip bution function Prf(Mm;n ? mP )= mP (1 ? P ) xg versus x. Our simulation experiments presented in Figure 1 con rm our theoretical result of Corollary 1. Namely, for the symp metric alphabet, (Mm;n ? mP )= m log n) converges to a constant. But { surprisingly { this is not true for an asymmetric alphabet, as indicated by Figures 2 and 3. In fact, based on p these simulations we conjecture, that (Mm;n ? mP )= m log n) converges in probability to a random variable, say Z , which is not a degenerate one, except for the symmetric alphabet. A study of the limiting distribution of Z seems to be very dicult. Our Theorem 2 provides some information regarding the behavior of the tail of the distribution of Z . In particular, 6
1.0
m;n ?mP < xg Prf pMmP (1?P ) 0.8
0.6
0.4
0.2
0 0.0
x
Lower=Upper bound 2
4
6
8
10
Figure 1: Distribution function of (Mm;n ? mP )(mP (1 ? P ))?1=2 for V = 2 with p1 = 0:5 (symmetric case), via simulations. we know that Z degenerates for the symmetric alphabet, that is, possibly the variance of Z becomes asymptotically zero in this case.
3. ANALYSIS In this section, we prove our main result, namely Theorem 2. In the course of deriving it, we establish some interesting combinatorial properties of pattern matching that have some similarities with the work of Guibas and Odlyzko [16, 17] (see also [19]). The proof itself consists of two parts: the upper bound (cf. Section 3.2) and the more dicult lower bound (cf. Section 3.3). Section 3.1 provides necessary mathematical background, and presents some preliminary results. Before we plunge into technicalities, we present an overview of our approach. An upper bound on Mm;n is obtained from (3), or a modi cation of it, and the real challenge is in establishing a tight lower bound. The idea was to apply the second moment method, in the form of Chung and Erdos [10] which states that for events fCi > rg, the following holds P n [ > rg)2 ( P PrfMm;n > rg = Prf (Ci > r)g PrfC > rg + iPPrfCi Pr : (9) i (i6=j ) fCi > r & Cj > rg i i=1 7
1.0
m;n ?mP < xg Prf pMmP (1?P ) 0.8
0.6
0.4
0.2
0 0.0
Lower bound 2
x
Upper bound 4
6
8
10
Figure 2: Distribution function of (Mm;n ? mP )(mP (1 ? P ))?1=2 for V = 2 when n = 50; 000 and m = 600, when n = 50; 000 and m = 600, with p1 = 0:2, via simulations. Thus, one needs to estimate the joint distribution PrfCi > r & Cj > rg. Ideally, we would p expect that the right-hand side of (9) goes to one when r = mP +(1 ? ") m2(P ? P 2 ) log n for any " > 0, which would match the upper bound. Provided the above works, one needs an estimate of PrfCi > r & Cj > rg. This is a dicult task since C1 and Cj are strictly positively correlated random variables for all values of 1 < j n. In fact, there are two types of dependency. For j > m, the patterns aligned at positions 1 and j do not overlap, but nevertheless C1 and Cj are correlated since the same pattern is aligned. Although this correlation is not dicult to estimate, it is strong enough to "spoil" the second moment method. In addition, an even stronger correlation between C1 and Cj occurs for j < m since in addition overlapping takes place. We treat this type of dependency in Section 3.4. To resurrect the second moment method, we introduce a conditional second moment method. The idea is to x a string of length m built over the same alphabet as the pattern b, and to estimate C1 and Cj under the condition b = . Then, C1 and Cj are conditionally independent for all j > m, and the dependency occurs only for j m, that is, when C1 and Cj overlap. This idea we shall develop below. 8
1.0
m;n ?mP < xg Prf pMmP (1?P ) 0.8
0.6
0.4
0.2
0 0.0
Lower bound
x
Upper bound
2
4
6
8
10
Figure 3: Distribution function of (Mm;n ? mP )(mP (1 ? P ))?1=2 for V = 2 when n = 100; 000 and m = 1000, with p1 = 0:1, via simulations. We introduce some notation. For a given , we de ne
G(r; ) = PrfC1 > rjb = g ; m X F (r; ) = PrfC1 > r & Ci > r jb = g i=2
p() = Prfb = g :
(10) (11) (12)
=V p j where the denotes the number of Considering our Bernoulli model, p() = jj =1 j j occurrence of the symbol j in the string . When varies, we may consider the conditional probabilities G(r; ) and F (r; ) as random variables with respect to . We adopt this point of view. Q
3.1 Preliminary Results In this subsection, we put together several results concerning the tail of a binomially distributed random variable, that are used in the proof of our main results. If X is a binomially distributed random variable with parameters m and p, then we write B (m; p; r) = PrfX > rg for the tail of X . We need the following result concerning the tail of C1 . 9
Lemma 1 When m and both tend to in nity such that = O(log m), then p p ? 1 exp 2(P ? P 2 ) : (13) B (m; P; mP + m ) = PrfC1 mP + m g p 2(P ? P 2 ) Proof: According to (1), C1 is binomially distributed, that is, PrfC1 = rg = ?mrP r (1 ? P )m?r . Introducing the generating function C (u) = Pmr=1 PrfC1 = rgur for u complex, we easily get the formula C (u) = (1 + P (u ? 1))m . Then, by Cauchy's celebrated formula [18] I 1 (14) PrfC1 rg = 2i (1 + P (u ? 1))m ur (u1? 1) du ; where the integration is along a path around the unit disk for u complex. The problem is how to evaluate this integral for large m. In this case the best suited method seems to be the saddle point method [18, 13]. This method applies to integrals of the following form
I (m) =
Z
P
(x)e?mh(x) dx ;
(15)
where P is a closed curve, and (x) and h(x) are analytical functions inside P . We evaluate I (m) for large m. The value of the integral does not depend on the shape of the curve P . The idea is to run the path of integration through the saddle point, which is de ned to be a place where the derivative of h(x) is zero. To apply this idea to our integral (14) we represent it in the form of (15) and nd the minimum of the exponent. De ne u = 1 + h, and then the exponent in our integral can be expanded as log((1 + P (u ? 1))m =ur ) = m log(1 + hP ) ? r log(1 + h) = (Pm ? r)h ? 1=2(P 2 m ? r)h2 + O(m + r)h3 ) 2 = ? 12 (r ? mP )2 (r ? mP 2 )?1 + (r ? mP 2 ) (h?2h0 ) + O(m + r)h3 )
p
with h0 = (r ? mP )=(r ? mP 2 ). Let r = mP + mx, with x > 0. Changing the scale of p the variable under the integrand: h = h0 + it= m(P ? P 2 ), we obtain !
Z log m 2 p exp(?t2 =2) dt ?1 + O(1=pm) : 1 x p PrfC1 mP + mxg = exp 2(P ? P 2 + x= m) 2 ? log m pP x?P 2 + it p Therefore, when x ! 1 (e.g., as log m) we get p p Z log m exp(?t2 =2) dt P ? P 2 Z 1 exp(?t2 =2)dt = 2(P ? P 2 )
? log m pP x?P 2 + it
x
?1
x
where the last integral can be computed from the error function [2]. This completes the proof of our lemma. 10
To estimate the conditional probability G(r; ), we need some additional asymptotics for the binomial distribution, which are summarized in the lemma below.
Lemma 2 Let rm = mp + pm m where m = O(log m). Then for all > 1 there exists a sequence m ! +1 such that the following hold pm; p; r ) exp( m ) 1 (16) lim inf B ( m ?
m m m!1 2(p ? p2 ) and
p
lim sup B (m + m m; p; rm ) exp( 2(p m? p2 ) ) 1 : m!1
(17)
Proof: These are computational consequences of Lemma 1. Details are deferred to the Appendix.
Now, we use Lemma 2 to estimate a typical behavior of the probability G(r; ) (recall that we treat G(r; ) as a random variable with respect to ). For a string 2 m , we de ne j as the number of occurrences of the symbol j 2 in . If varies, then j can be treated as a random variable, too. Let also C1 (j ) be the number of matches that involve P symbol of j 2 between b = and a1 a2 am . Clearly, C1 = Vj=1 C1 (j ), and C1 (j ) is binomially distributed with parameters j and pj . We have the following simple estimate of G(r; ).
Lemma 3 For all rj such that PVj=1 rj = r, the following holds V Y j =1
B ( j ; pj ; rj ) G(r; )
V X j =1
B ( j ; pj ; rj ) :
(18)
Proof: This follows immediately from the next two trivial bounds PrfC1 > rg Prf8j : C1 (j ) > rj g ; and
PrfC1 > rg Prf9j : C1 (j ) > rj g :
This completes the proof. Our main result of this subsection is contained in the next theorem. It provides (a.s.) bounds for the probability G(r; ). 11
Theorem 3 Let rm = mP + pm m with m = O(log m), and P and T as de ned in Theorem 2. For all > 1, the following estimates hold: (i) m mlim !1 Prf : G(rm ; ) exp ? 2(P ? T ) g = 1 ; (ii) m mlim !1 Prf : G(rm ; ) exp ? 2S 2 g = 1 ; p P where S = Vj=1 pj 1 ? pj .
(19) (20)
Proof: We rst deal with part (i). For convenience of presentation we drop the subscript m from rm and m. Let rj = mp2j + j pm with PVj=1 j = 1. By (16) of Lemma 2, for any > 1 there exists a sequence m ! 1 such that the following holds for m large enough V V Y Y 2 p B (mpj ? m m; pj ; rj ) exp(? 2p2 (1j ? p ) ) : j j j =1 j =1 P Taking j = p2j (1 ? pj )=( p2i (1 ? pi )), after some algebra involving an optimization of the right-hand side of the above, we obtain (? =2(P ? T )) in the exponent of the right-hand p Q side (RHS) of the above. But, by Lemma 3, G(r; ) Vj=1 B (mpj ? m m; pj ; rj ) where is such that for all j : j mpj ? m pm. Thus, p Prf : G(r; ) exp(? 2(P ? T ) )g Prf8j : j mpj ? m mg :
By the Central Limit Theorem, every random variable
j tends to the Gaussian distribution q with mean mpj and with the standard deviation m(pj ? p2j ). So, it is clear that mlim !1 Prf8j :
p
j mpj ? m mg = 1
as m ! 1. The proof of part (ii) goes along the same lines. For completeness, we provide the details p P below. Let rj = mp2j + j m with Vj=1 j = 1. By Lemma 2, for every > 1 we can nd a sequence m ! 1 such that the following holds for m large enough: V X
p
B (mpj + m m; pj ; rj ) V ?1
V X
2
exp(? 2p2 (1j ? p ) )
j j j =1 j =1 p Taking j = S ?1 pj 1 ? pj , we obtain (? =2S 2 ) in the exponent of the RHS of the above. p P But, by Lemma 3, G(r; ) Vj=1 B (mpj + m m; pj ; rj ) where is such that for all j : p
j mpj + m m. Thus,
p
Prf : G(r; ) exp ? 2 S 2 g Prf8j : j mpj + m mg : 12
By the Central Limit Theorem, every
j tends to the Gaussian random variable with mean q mpj and with standard deviation m(pj ? p2j ). Hence, mlim !1 Prf8j :
p
j mpj + m mg = 1
for m ! 1.
Remark 2. If we denote by Gm the set of 2 m such that G(r; ) exp ? 2(P ?T ) , then (19) asserts that PrfGm g ! 1. As above, if we denote by Gm0 a set of 2 m such that G(r; ) exp(? 2 S 2 ) ), then (20) implies that PrfGm0 g ! 1. 3.2 Upper Bounds We establish here the two upper bounds claimed in Theorem 2. We prove the following theorem.
Theorem 4 Let " > 0 be an arbitrary non-negative real number. Then (i) If = 2(1 + ")(P ? P 2 ) log n, then the following holds limn;m!1 PrfMm;n < rg = 1 p with r = mP + m .
p P (ii) If = 2(1 + ")S 2 log n, with S = Vi=1 pi 1 ? pi, then limn;m!1 PrfMm;n < rg = 1 p with r = mP + m .
Proof: Part (i) was in fact already proved in Theorem 1, and it really follows directly from PrfMm;n < rg nB (m; P; r). We concentrate on proving part (ii). We have PrfMm;n > rg
X
2Gm0
n p()G(r; ) + Prf 62 Gm0 g
max fnG(r; )g + Prf 62 Gm0 g ; 2Gm0 p where the set Gm0 is de ned in Theorem 3 (ii) with = 1 + ". Note that Prf 2= Gm0 g ! 0, by Theorem 3. Then by (20) G(r; ) exp(? =(2S 2 )) for 2 Gm0 , thus p PrfMm;n > rg n1? 1+" + Prf 62 Gm g :
But both terms of the above tend to zero, and this establishes the second upper bound in Theorem 2.
3.3 Lower Bounds 13
The lower bounds are much more intricate to prove. We start with a simple one that does not take into account overlapping. Then, we extend this bound to include the overlapping. This extension is of interest to us since it leads to the exact constant in the symmetric case. Mm;n where In the rst lower bound we ignore overlapping by considering Mm;n = max1ibn=mc fC1+im g, that is, the maximum is taken over all nonoverlapping Mm;n positions of a. Note also that C1 and C1+m are conditionally independent (i.e., under b = ). Our rst lower bound is contained in the theorem below.
Theorem 5 Let 0 < " < 1. If = 2(1 ? ")(P ? T )(1 ? ) log n and r = mP + pm , then limn;m!1 PrfMm;n < rg ! 0. Proof. Due to conditional independence we can write < rg = PrfMm;n
X
2m
p()(1 ? G(r; ))n=m
max f(1 ? G(r; ))n=m g + Prf 2= Gm g ; 2Gm where Gm is de ned in remark 2, that is, for = (1?")?1=2 we have G(r; ) > exp(? =(2(P ? T ))) for all 2 Gm . Note also that Prf 2= Gmg ! 0. We concentrate now on the rst term of the above. Using log(1 ? x) ?x, we obtain (1 ? G(r; ))n=m e?(n=m)G(r;) : Now, it suces to show that (n=m)G(r; ) ! 1 for 2 Gm . By Theorem 3, we have for
2 Gm
n G(r; ) n exp ? (n=m)1?p1?" ! 1 ; m m 2(P ? T ) where the convergence is a consequence of our restrictions log n = o(m). This completes the proof of the "easier" lower bound in Theorem 2.
The second lower bound is more elaborate. It requires an estimate of PrfC1 > r & Ci > = rg also for i < m. Let Fm;n = Pmi=2 PrfC1 > r & Ci > rg. Note that Fm;n Fm;n Pm i=2 PrfC1 + Ci > 2rg. In the next subsection we prove the following theorem.
Theorem 6 For m = O(log m) ! 1 we have 2 2T )5=2 (mP + pm m ) m(P ? 3P + p exp(? P ? 3P m2 + 2T ) : Fm;n 2 3 2(T ? P ) m for m ! 1. 14
(21)
Assuming that Theorem 6 is available, we proceed as follows. Using the second moment method we have X PrfMm;n > rg p()S (r; ) ; (22) 2m
where following (9)
(r; ))2 S (r; ) = nG(r; ) + nF (r; )(nG + (n2 ? n(2m + 1))(G(r; ))2 : P In the above, the probability F (r; ) is de ned in (12) (to recall, F (r; ) = mi=2 PrfC1 > r & Ci > r jb = g). We rewrite S (r; ) as ?1 S (r; ) = nG(1r; ) + n(FG((r;r; )))2 + 1 ? 2mn+ 1 :
(23)
We also have the following identity X
2m
p()F (r; ) = Fm;n (r) ;
(r) is estimated in Theorem 6. The almost sure where the probability Fm;n (r) Fm;n behavior of F (r; ) { as a random function of { is discussed in the next lemma.
Lemma 4 If ! 1, then for any constant we have
mlim !1 Prf : F (r; ) m exp ? P ? 3P 2 + 2T g = 1 :
(24)
Proof: It is a simple consequence of Markov's inequality: Since P2m p()F (r; ) = Fm (r), it is clear that PrfF (r; ) > m Fm (r)g 1= m . Remark 3. We denote by Gm00 the set of 2 m such that F (r; ) m exp(? P ?3P 2+2T ) holds. The lemma shows that limm!1 Prf 2 Gm00 g = 1. 2 Now we are ready to prove the main result of this subsection that provides the most elaborate lower bound. Let 1 be de ned as in Theorem 2, that is, )(P ? 3P + 2T ) : 1 = (P ? T3( T ? P 2) 2
Let also 2 = minf(1 ? )1 ; 2(P ? T )g. 15
Theorem 7 Let 0 < " < 1. If = (1 ? ")2 log n, then limm;n!1 PrfMm;n > rg = 1 for r = mP + pm . Proof: Note that it suces to prove the theorem for any positive and arbitrarily small ".
By (22) and (23) we need to show that
nG(r; ) ! 1 (25) F (r; ) ! 0 : (26) nG2(r; ) The rst identity is easy to prove. For 2 Gm by Theorem 3 with (1 ? ")?1=2 we have p nG(r; ) n1?1=2 1?"2 =(P ?T ) ! 1
since 2 =(2(P ? T )) 1. Now, we deal with (26). Note that 1 > 0 since P 2 < T . For 2 Gm \ Gm00 where Gm and Gm00 are de ned as in (respectively) Theorem 3 and Lemma 4 (with = 1), we have F (r; ) m exp( ? 2 nG (r; ) n P ? T P ? 3P 2 + 2P ) 1? ?(11?")2 =1 exp(( ? 1) P ? T )
n
We know that 1 ? ? 2 =1 (1 ? ") "2 =1 . Choosing ? 1 = O("2 ) in the above, we nally obtain 1 F (r; ) !0; 2 1 ? ? = 2 1 (1?")+O ("2 ) nG (r; ) n since " can be arbitrary small. Putting everything together, we have just proved that S (r; ) ! 1 for all 2 Gm \ Gm00 . But, by (22), Theorem 3 and Lemma 4 PrfMm;n > rg
X
2m
p()S (r; )
Prf 2 Gm \ Gm00 g 2Gmin fS (r; )g ! 1 ; m \G 00 m
which completes the proof.
3.4 Proof of Theorem 6 In this subsection, we prove Theorem 6 concerning the asymptotic behavior of the probability PrfC1 > r; Ci > rg for i < m. In fact, we evaluate the generating function Hm;` (u; v) = Pr1 ;r2 PrfC1 = r1 & C` = r2 gur1 vr2 , and then compute the probability PrfC1 = r1 & C` = r2 g through the Cauchy integral as it was done in Lemma 1. 16
Let x and y be column vectors of dimension V , that is, x = fxi gVi=1 and y = fyi gVi=1 . We de ne the scalar product hx; yi by x1 y1 + + xV yV . Then, the next crucial theorem captures some important combinatorial properties of fC1 ; C` g that allow to estimate the generating function Hm;2 (u; v) for ` = 2, and nally Hm;` (u; v) for any ` (cf. Theorem 9). Now we are ready to establish a closed-form formula for the generating function Hm;2 (u; v). We construct a recurrence relationship between the distributions of fC1 (b); C2 (b)g and fC2 (b0 ); C3 (b0 )g where b0 is the sux of b of length m ? 1. In the above we write Ci (b) instead of Ci in order to show explicitly a dependency of Ci on the string b. With this in mind, we can proceed to the following key theorem.
Theorem 8 We have the identity Hm;2(u; v) = hx(u); Am?1 (u; v)y(v)i, where A(u; v) is a V V square matrix whose generic element aij (u; v) satis es the following aij (u; v) = pi (1 + pi (v ? 1) + pj (u ? 1)) when i 6= j , and
aii (u; v) = pi (1 + pi (uv ? 1)) for i = j . The row vectors x(u) and y(v) are de ned as x(u) = f1 + pi(u ? 1)gVi=1 and y(v) = fpi (1 + pi (v ? 1))gVi=1 .
Proof: Let us de ne a random variable ?i as the number of matches between string a and i (b) without counting the eventual rst matching at position i (recall that i (b) is the shifted version of b by i positions on the cycle). For example, ?1 = C1 if there is no matching at position 1, and ?1 = C1 ? 1 otherwise. De ne next the generating function Pi;m (u; v) as
Pi;m (u; v) =
X
r1 ;r2
Prf?1 = r1 & C2 = r2 & string b starts with symbol igur1 vr2 ;
and let Pm (u; v) denote the row vector fPi;m (u; v)gVi=1 . Note that P1 (u; v) = y(v) and Hm;2 (u; v) = Pii==1V (1 + pi(u ? 1))Pi;m (u; v), thus Hm;2 (u; v) = hx(u); Pm (u; v)i. The most interesting fact that we prove next is the following relationship Pm (u; v) = A(u; v)Pm?1 (u; v) when m > 1. A proof of this relies on building a recurrence relationship between fC1 (b); C2 (b)g and fC2 (b0 ); C3 (b0 )g, as explained above. Let i and j be the two rst symbols of string b and let k be the second symbol of string a. When i 6= j we have ?1 (b) = ?2 (b0 ) + 1 and C2 (b) = C3 (b0 ) if k = j , ?1 (b) = ?2 (b0 ) and C2 (b) = C3 (b0 ) + 1 if k = i, and ?1 (b) = ?2 (b0 ) and C2 (b) = C3 (b0 ) otherwise. When i = j , we have 17
a = 01100101110101011110101110010111 b = 01011001010111 b0 = 1011001010111 0
1
1
1
0
0
1
0
1
1
1
0
1
0
0
1
0
1
1
0
0
1
0
1
0
1
1
1
0
1
0
1
1
0
0
1
0
1
0
1
1
1
1
1
1
0
0
1
0
1
1
1
0
1
0
1
0
1
1
1
1
0
1
0
1
1
1
0
0
1
0
1
1
1
C1 (bfb) = 6, ?1 (b) = 5, C2 (b0 ) = 5, ?2 (b0 ) = 4 C2 (bfb) = 7, ?2 (b) = 7, C3 (b0 ) = 7, ?3 (b0 ) = 6 1
1
1
0
1
0
1
1
1
0
0
1
0
1
1
1
Figure 4: Illustration of the relationship between fC1 ; C2 g and f?2 ; ?3 g (boxes show matches). ?1 (b) = ?2 (b0 )+1 and C2 (b) = C3 (b0 )+1 if k = i, and ?1 (b) = ?2 (b0 ) and C2 (b) = C3 (b0 ) otherwise. This is illustrated in Figure 4. Since fC2 (b0 ); C3 (b0 )g has the same distribution as fC1 (b0 ); C2 (b0 )g, we obtain the following identity X Pi;m (u; v) = (1 + p (uv ? 1))P ( u; v ) + (1 + pi(v ? 1) + pj (u ? 1))Pj;m?1 (u; v) ; i i;m ? 1 pi j 6=i
which proves our theorem. The next theorem extends Theorem 8 and give a formula for the generating function Hm;` (u; v), which is of independent interest.
Theorem 9 For all q < m the following holds: Hm;1+q (u; v) = (Hh;2(u; v))q?` (Hh+1;2(u; v))` , where h = b mq c and ` = m ? hq. Proof. For i q de ne b(i) as a subsequence of string b obtained by selecting the ith symbol of b, then the i + qth, then i + 2qth, and so forth. For 1 i `, string b(i) is of length h + 1, whereas for ` + 1 i q its length is h. We can do the same with string a and obtain subsequences a(1) ; : : : ; a(q) . Let ha; i (b); j (b)i be a new notation for the two dimensional row vector [Ci ; Cj ], that is, it represents the number of matches between a and simultaneously i (b) and j (b). It is easy to see that [C1 ; C1+q ] = ha(1) ; 1 (b(1) ); 2 (b(1) )i + + ha(q) ; 1 (b(q) ); 2 (b(q) )i. Note that ha(i) ; 1 (b(i) ); 2 (b(i) )i's has the same distribution as [C1 ; C2 ] when b(i) is of length 18
h + 1 for i `, and the same distribution as [C1 ; C2 ] when b(i) is of length h for ` < i q. This nally proves the theorem.
Theorem 9 establishes a closed form formula for the generating function of the joint distribution PrfC1 = r1 ; C` = r2 g. Therefore, in principle we can recover the probabilities PrfC1 = r1 ; C` = r2 g from Hm;` (u; v) by Cauchy's formula, as we did in Lemma 1. The diculty is that the generating function Hm;`(u; v) is expressed in terms of matrix A(u; v), so we need some tools from linear algebra to apply the saddle point method. However, before we plunge into this, we should treat the symmetric case (i.e., pk = 1=V ) separately since, as the next lemma shows, it possesses a very special property.
Lemma 5 In the symmetric case, for all i 6= j , the random variables Ci and Cj are pairwise independent.
Proof: It suces to prove that C1 is independent of C1+q for all 1 q m. In the symmetric case the aij (u; v)'s are all identical and equal to V1 (1+ V1 (u?1+v ?1)) except when i = j where aii (u; v) = V1 (1+ V1 (uv ? 1)). Note that y(v) coincides with an eigenvector of the matrix A and A(u; v)y(v) = (1 + V1 (u ? 1))(1 + V1 (v ? 1))y(v), and therefore Hm;2 (u; v) = (1 + V1 (u ? 1))m (1 + V1 (v ? 1))m . This last formula shows that C1 and C2 are independent. Applying Theorem 9 one concludes that also Hm;1+q (u; v) = (1+ V1 (u ? 1))m (1+ V1 (v ? 1))m . Therefore C1 and C1+q are also independent.
Corollary 2 In the symmetric case we have Mm;n ? mP p2m(1=V ? 1=V 2) log n when m; n ! 1, given log n = o(m). Proof: We already know that the above is true when < 1 (cf. Theorem 2). Using
the result stated in Lemma 5 about pairwise independence of the Ci 's, we can rewrite inequality (9) as PrfC1 > rg)2 PrfMm;n > rg nPrfC > r(gn+ (27) (n2 ? n)(PrfC1 > rg)2 : 1 It is clear that the RHS of the above tends to 1 if and only if nPrfC1 > rg ! 1. Referring to Lemma 1, it is clear that this occurs for = 2(1 ? ")(P ? P 2 ) log n and any arbitrarily chosen " > 0.
Hereafter, we concentrate on the asymmetric case. Let (u; v) be the principal eigenvalue of the matrix A(u; v). Let (u; v) (resp. (u; v)) be the corresponding right (resp. left) eigenvector of A(u; v) such that h(u; v); (u; v)i = 1, that is, A(u; v) (u; v) = (u; v) (u; v) and AT (u; v)(u; v) = (u; v)(u; v) (cf. [25, 24]). We distinguish the following three cases: 19
1. When u = v = 1, (1; 1) = 1, (1; 1) = y(1) and (1; 1) = x(1), the other eigenvalues are null. 2. When v = 1, (u; 1) = 1 + P (u ? 1), (u; 1) = (1; 1) = y(1) and (u; 1) = 1=(u; 1) x(u), the other eigenvalues are null. 3. When u = 1, (1; v) = (v; 1) = 1 + P (v ? 1), (1; v) = y(v) and (1; v) = 1=(1; v) x(1), the other eigenvalues are null. It follows that the other eigenvalues are O((u ? 1)(v ? 1)), and therefore we immediately prove the following fact.
Corollary 3 Hm;2(u; v)=m?1 (u; v) = hx(u); (u; v)i h(u; v); y(v)i + O((u ? 1)m (v ? 1)m ). Proof: This is a classical property of the principal eigenvalue and follows from the PerronFrobenius theorem (the interested reader is referred to [3, 24, 25] for details).
As a consequence of Corollary 3 we have the following important expansion of the generating function Hm;` (u; v).
Corollary 4 Let Fm (u; v) = Pmi=2 Hm;i(u; v). We have Fm (u; v) = a(u; v) ? am (u; v) + O((u ? 1)(v ? 1)) ; ((u; v))m 1 ? a(u; v) with a(u; v) = hx(u); (u; v)i h(u; v); y(v)i=(u; v)
(28)
Proof: From Corollary 3 and Theorem 9 we have the estimate Hm;1+q (u; v) = ((u; v))m (a(u; v))q + O((u ? 1)h+1 (v ? 1)h+1 )) : Therefore
?1 m Fm (u; v) = mX q + X O((u ? 1)h+1 (v ? 1)h+1 ) : ( a ( u; v )) ((u; v))m q=1 h=2
The proof is easily completed by summing the geometric series in the last expression. The following two lemmas present more detailed Taylor's expansions of the principal eigenvalue of A(u; v) de ned in Corollary 4. These expansions are next used in the saddle p point method to obtain a sharp estimate of Fm;n (r) around r = mP + 2mP (1 ? P ) log n.
Lemma 6 The Taylor expansion of (u; v) to the second order is 1 + (u ? 1)P + (v ? 1)P + (u ? 1)(v ? 1)(2T ? P 2 ), with T = p31 + + p3V . 20
Proof. We know that (u; v) = 1 + (u ? 1)P + (v ? 1)P + O((u ? 1)(v ? 1)). We adopt
the following notation. If f (u; v) is a function of two variables u and v, then we denote by fu(u; v) (resp. fv (u; v)) the partial derivative of f (u; v) with respect to u (resp. to v). We have = h; A i, where the variables (u; v) have been dropped in the last expression to simplify the presentation. Thus, u = hu ; A i + h; Au i + h; Au i. Since A = , AT = , and since hu ; i + h; u i = 0 (because we assume h; i = 1), we get u = h; Au i. Substituting u = 1 in the last expression, we obtain the identity (1; v) = P (P + 2T (v ? 1) + ii==1V p4i (v ? 1)2 )=(1 + P (v ? 1)), and after some simple algebra the proof is completed.
Lemma 7 The Taylor expansion of a(u; v) to the second order is 1 ? (u ? 1)(v ? 1)(T ? P 2). Proof. Easy computations give a(u; 1) = a(1; v) = 1, therefore a(u; v)?1 is O((u?1)(v?1)).
Di erentiating twice a(u; v) and setting u = v = 1 leads to a formula beginning with huv ; i + h; uv i and ending with a linear combination of scalar products involving rst partial derivatives of , , x and y. These rst derivatives are already known since and are completely determined when u = 1 or v = 1. For huv ; i + h; uv i, we di erentiate twice both sides of h; i = 1 in order to get huv ; i + h; uv i + hu ; v i + hv ; u i = 0, which leads to a complete determination of auv (1; 1). Lemmas 6 and 7 are crucial to apply Cauchy's formula in order to estimate Fm;n (r) for r > mP . To do that, we can use the double Cauchy formula I I 1 Fm;n (r) = (2i)2 Fm (u; v) ur (u ?d1)udvrv(v ? 1) : This kind of integration is rather unusual. Since PrfCi > r; Cj > rg PrfCi + Cj > 2rg (r) = Pm PrfC1 + Ci > 2rg, which leads to a single integration we can estimate Fm;n i=2 I
1 (r) = 1 F ( u; u ) Fm;n m 2 r 2i u (u ? 1) du : Finally, we can prove the following asymptotics for the tail of Fm;n (r), which establishes Theorem 6.
Proof of Theorem 6. We parallel the proof of Lemma 1. By Cauchy and (28) we have I m a(u; u) ? am (u; u)) du ; 1 (29) Fm;n (r) 2i (u; uu)( 2r (1 ? a(u; u)) u?1 21
the integration path encircling the unit disk. Let 1 + h = u. Using Lemmas 18 and 19, we obtain the expansion log(m (u; u)=u2r ) = = = +
m log(1 + 2hP + h2 (2T ? P 2 ) + O(h3 )) ? 2r log(1 + h) ?2(r ? mP )h + (r + 2mT ? 3mP 2 )h2 + O((m + r)h3) ?(r ? mP )2(r ? m(3P 2 ? 2T ))?1 + (r ? 3mP 2 + 2mT )(h ? h0 )2 + O((m + r)h3 ) ;
p
p
with h0 = (r ? mP )=(r ? 3mP 2 +2mT ). Let r = mP + mx. Substituting h = h0 + it= m, and using 1 ? a(1 + h; 1 + h) = h2 (T ? P 2 ) + O(h3 ) we get the rst estimate 1 I m (u; u)a(u; u)du = 2i u2r (u ? 1)(1 ? a(u; u)) ! Z 2 2 2 p = exp ? P ? 3Px 2 + 2T 2(T ?mP 2 ) exp((?(P x? 3P ++it2)T3 )t ) dt(1 + O(1= m)) : P ?3P 2 +2T p Since x = O( log n), we obtain ! 1 I m (u; u)a(u; u)du m(P ? 3P 2 +p2T )5=2 exp ? x2 2i u2r (u ? 1)(1 ? a(u; u)) 2(T ? P 2 ) x3 P ? 3P 2 + 2T : (30) It remains to evaluate the second term in (29), that is, 1 I m (u; u)am (u; u)du : (31) 2i u2r (u ? 1)(1 ? a(u; u)) Using the estimates from Lemmas 18 and 19 we nd log(m (u; u)am (u; u)=u2r ) = ?(r ? mP )h + (r + mT ? 2mP 2 )h2 + O((m + r)h3 ) : Hence (31) becomes ! 1 I m (u; u)am (u; u)du m(P ? 2pP 2 + T )5=2 exp ? x2 2i u2r (u ? 1)(1 ? a(u; u)) 2 x3 P ? 2P 2 + T : (32) Since P ? 2P 2 + T < P ? 3P 2 + 2T in the asymmetric case, the exponent in (32) is larger (r). This than in (30), so (30) is the leading term in the asymptotic expansion of Fm;n concludes the proof of Theorem 6.
APPENDIX
Proof of Lemma 2: We rst prove the existence of a sequence m such that (16) holds.
Since for all p and r we have B (0; p; r) = 0 and limx!1 B (x; p; r) = 1, we can nd for the Bernoulli distribution a sequence m such that
p B (m ? m m; p; rm ) > exp(? 2(p m?p2 ) ) ; 22
(33)
and
p B (m ? m m ? 1; p; rm ) exp(? 2(p m?p2 ) ) < 1 :
(34)
Our aim is to prove that limm!1 m = 1. Let us assume contrary that lim inf m!1 m < 1, thus there exists and a subsequence of m bounded from the above by the constant
. For simplicity of the presentation, we assume that m < for all m. Then, due p to monotonicity of B (m; p; r) with respect to m, we have B (m ? qm m ? 1; p; rm ) > B (m ? pm ? 1; p; rm ). Furthermore, for rm = (m ? pm ? 1)p + (m ? pm ? 1) m with p + p )2 p m p (35) m = ( 1 ?
= m = m + O( m ) ; we obtain
p ) p + O ( 1 m m m B (m ? m ? 1; p; rm ) p exp ? 2(p ? p2 ) exp ? 2(p ? p2 ) ; 2(p ? p2 ) m
(36) where the last inequality holds for large m due to > 1 and m ! 1. Clearly, the above contradicts (34), hence, m ! 1, as needed. The proof for the second inequality is similar and is only sketched. We set m such that and
p B (m + m m; p; rm ) < exp(? 2(p m? p2) )
(37)
p B (m + m m + 1; p; rm ) exp(? 2(p m? p2 ) ) :
(38)
If one assumes m < < 1, then
p
p
exp(? 2(p ?m p2 ) ) ; 2 2(p ? p ) m (39) p p with m = m + O( m ), which contradicts B (m + m m +1; p; rm ) exp(? m =2(p ? p2 )).
B (m + m m + 1; p; rm ) < B (m + m + 1; p; rm ) p
1
ACKNOWLEDGEMENT The authors sincerely thank a referee who pointed out an error in an earlier version of the paper.
References [1] K. Abrahamson, Generalized String Matching, SIAM J. Comput., 16, 1039-1051, 1987. 23
[2] Abramowitz, M. and Stegun, I., Handbook of Mathematical Functions, Dover, New York (1964). [3] Aldous, D., Probability Approximations via the Poisson Clumping Heuristic, Springer Verlag, New York 1989. [4] Apostolico, A., Atallah, M., Larmore, L., and McFadin, S., Ecient Parallel Algorithms for String Editing and Related Problems, SIAM J. Computing, 19, 968-988, 1990. [5] Arratia, R., Gordon, L., and Waterman, M., An Extreme Value Theory for Sequence Matching, Annals of Statistics, 14, 971-993, 1986. [6] Arratia, R., and Waterman, M., The Erdos-Renyi Strong Law for pattern matching with a Given Proportion of Mismatches, Annals of Probability, 17, 1152-1169, 1989. [7] Arratia, R., Gordon, L., and Waterman, M., The Erdos-Renyi Law in Distribution, for Coin Tossing and Sequence Matching, Annals of Statistics, 18, 539-570, 1990. [8] Atallah, M., Jacquet, P., and Szpankowski, W., Pattern Matching with Mismatches: A Simple Randomized Algorithm and Its Analysis, Proc. Combinatorial Pattern Matching, Tuscon 1992. [9] Chang, W.I., and Lawler, E.L., Approximate String Matching in Sublinear Expected Time, Proc. 31st Ann. IEEE Symp. on Foundations of Comp. Sci., 116-124, 1990. [10] Chung, K.L. and Erdos, P., On the Application of the Borel-Cantelli Lemma, Trans. of the American Math. Soc., 72, 179-186, 1952. [11] DeLisi, C., The Human Genome Project, American Scientist, 76, 488-493, 1988. [12] Feller, W., An Introduction to Probability Theory and its Applications, Vol. II, John Wiley & Sons, New York (1971). [13] Flajolet, P., Analysis of Algorithms, in Trends in Theoretical Computer Science (ed. E. Borger), Computer Science Press, 1988. [14] Galambos, J., The Asymptotic Theory of Extreme Order Statistics, John Wiley & Sons, New York (1978). [15] Z. Galil, Open Problems in Stringology, Combinatorial Algorithms on Words (Eds. A. Apostolico and Z. Galil), 1-8 (1984). [16] L. Guibas and A. Odlyzko, Periods in Strings Journal of Combinatorial Theory, Series A, 30, 19-43 (1981). [17] L. Guibas and A. W. Odlyzko, String Overlaps, Pattern Matching, and Nontransitive Games, Journal of Combinatorial Theory, Series A, 30, 183-208 (1981). [18] Henrici, P., Applied and Computational Complex Analysis, vol. I., John Wiley& Sons, New York 1974. 24
[19] Jacquet, P. and Szpankowski, W., Autocorrelation on Words and Its Applications. Analysis of Sux Trees by String-Ruler Approach, INRIA Technical report No. 1106, October 1989; submitted to a journal. [20] Karlin, S. and Ost, F., Counts of Long Aligned Matches Among Random Letter Sequences, Adv. Appl. Probab., 19, 293-351, 1987. [21] Knuth, D.E., J. Morris and V. Pratt, Fast Pattern Matching in Strings, SIAM J. Computing, 6, 323-350, 1977. [22] Lander, E., Langridge, R., and D. Saccocio, Mapping and Interpreting Biological Information, Comm. of the ACM, 34, 33-39, 1991. [23] Louchard, G., and Szpankowski, W., String Matching: A Preliminary Probabilistic Results, Universite de Bruxelles, TR-217, 1991. [24] Noble, B. and Daniel, J., Applied Linear Algebra, Prentice-Hall, New Jersey 1988 [25] Seneta, E., Non-Negative Matrices and Markov Chains, Springer-Verlag, New York 1981. [26] Szpankowski, W., On the Height of Digital Trees and Related Problems, Algorithmica, 6, 256-277 (1992). [27] M. Zuker, Computer Prediction of RNA Structure, Methods in Enzymology, 180, 262288, 1989.
25
Philippe Jacquety INRIA Rocquencourt 78153 Le Chesnay Cedex France
Wojciech Szpankowskiz Dept. of Computer Science Purdue University W. Lafayette, IN 47907 U.S.A.
Abstract The study and comparison of strings of symbols from a nite or an in nite alphabet is relevant to various areas of science, notably molecular biology, speech recognition, and computer science. In particular, the problem of nding the minimum "distance" between two strings (in general, two blocks of data) is of great importance. In this paper we investigate the (string) pattern matching problem in a probabilistic framework, namely, it is assumed that both strings form an independent sequences of i.i.d. symbols. Given a text string a of length n and a pattern string b of length m, let Mm;n be the maximum number of matches between p b and all m-substrings of a. Our main probabilistic result shows that Mm;n ? mP = ( 2m log n) in probability (pr.) provided m; n ! 1 such that log n = o(m), where P is the probability of a match between any two symbols of these strings. If the alphabet is symmetric (i.e., all symbols occur with the same probability), p a stronger result holds, namely (Mm;n ? mP ) 2m(1=V ? 1=V 2 ) log n (pr.), where V is the size of the alphabet. In either case, symmetric alphabet or not, we also prove that Mm;n =m ! P almost surely (a.s).
This author's research was supported by the Oce of Naval Research under Grants N0014-84-K-0502
and N0014-86-K-0689, and in part by AFOSR Grant 90-0107, and the NSF under Grant DCR-8451393, and in part by Grant R01 LM05118 from the National Library of Medicine. y This research was primary supported by NATO Collaborative Grant 0057/89. z This author's research was supported by AFOSR Grant 90-0107 and NATO Collaborative Grant 0057/89, and, in part by the NSF Grant CCR-8900305, and by Grant R01 LM05118 from the National Library of Medicine.
1
1. INTRODUCTION Pattern matching is one of the most fundamental problems in computer science. It is relevant to various areas of science, notably molecular biology, speech recognition, theory of languages, coding, and so forth. For example, in molecular biology often one compares two, say DNA, sequences to detect an unusual homology (similarity) between these strings. Comparisons of blocks of data representing, say time series of physical measurements, prices, etc., also lead to string comparisons, but with possibly in nite alphabets. In general, the problem of nding the minimum "distance" between two strings is of great importance. The version of the string matching problem we investigate here is the following one. Consider two strings, a text string a = a1 a2 :::an and a pattern string b = b1 b2 :::bm of lengths n and m respectively, such that symbols ai and bj belong to a V -ary alphabet = f1; 2; :::; V g. The alphabet may be nite or not. Let Ci be the number of positions at which the substring ai ai+1 :::ai+m?1 agrees with the pattern b (the index j that is out of P range is understood to stand for 1+(j mod n)). That is, Ci = mj=1 equal(ai+j ?1 ; bj ) where equal(x; y) is one if x = y, and zero otherwise. In other words, the "distance" between a and b is the Hamming distance. We are interested in the quantity Mm;n = max1in fCi g which represents the best matching between b and any m-substring of a. We analyze Mm;n under the following probabilistic assumption: symbols from the alphabet are generated independently, and symbol i 2 occurs with probability pi . This probabilistic model is known as the Bernoulli model [26]. An estimate of Mm;n can be used in various statistical tests, for example in the usual hypothesis-testing context in which one needs to detect a "signi cant" matching. In that context, a matching is considered signi cant when it is much larger than the one obtained for random strings, that is, than Mm;n ; indeed, if even random independent strings give rise to a match of Mm;n , it is reasonable not to attach particular signi cance to this amount of \similarity" between two strings. Furthermore, an estimate of Mm;n can be used to control some parameters of algorithms designed for pattern matching, as we did in Atallah et al. [8]. Our main probabilistic result concerning Mm;n can be summarized as follows. We prove p that with high probability Mm;n ?mP = ( 2m log n) for n; m ! 1 such that log n = o(m) where P is the probability that two symbols match. The restriction on n and m is important. We explicitly give the constants that are implicit in the () notation. Our simulation experp iments indicate that the second term in Mm;n , that is, ( 2m log n), spreads in probability between these two bounds, that is, we conjecture that the second term is rather a random variable than a constant. However, for the symmetric alphabet (i.e., all symbols occur with 2
p
the same probability) stronger result holds, namely (Mm;n ?mP ) 2m(1=V ? 1=V 2 ) log n (pr.). In either case { symmetric alphabet or not { we demonstrate that almost surely (a.s.) Mm;n =m ! P as long as log n = o(m). A probabilistic analysis of any pattern matching is a "very complicated\ problem as asserted by Arratia, Gordon and Waterman [7]. In fact, the literature on probabilistic analysis of pattern matching is very scanty. To the best of our knowledge, it is restricted to three papers of Arratia, Gordon and Waterman [5, 6, 7], however, only [7] is relevant to our study. Papers [5, 6] investigate the Erdos-Renyi law for the longest contiguous run of matches. Only in [7] is the number of matches in a segment of length m investigated. This last problem resembles the one discussed in this paper, however the authors of [7] restrict themselves to log n = (m) while we analyze the case when log n = o(m). Finally, it should be mentioned that our formulation of the problem is an attempt to study a more important problem of deletion, insertion and substitution developed during the evolution of nucleotide sequences (cf. Apostolico et al. [4], Louchard and Szpankowski [23]). Algorithmic issues of pattern matching (with possible mismatches) are discussed in several papers. The problem of pattern matching with mismatches was posed by Galil [15] for the case of more than one mismatch. Of course, it has a linear time solution for the case of zero mismatch (i.e., k = m). For the case of a single mismatch (i.e., k = m ? 1) a linear time solution is also known (attributed to Vishkin in [15]). The best known time p bound for the general case of arbitrary k is O(n mpolylog(m)) and is due to Abrahamson [1]. The elegant probabilistic method of Chung and Lawler [9] led to an algorithm of O((n=m)k log m) time complexity on the average, assuming that k < m= log m + O(1) and that the alphabet size is O(1). The paper is organized as follows. The next section makes a more precise statement of our results. In particular, Theorem 2 contains our main probabilistic result. All proofs are delayed until Section 3, which is also of independent interest. It discusses a fairly general approach that can be used to analyze pattern matching in strings. In that section we apply extensively the saddle point method [18] to evaluate necessary asymptotic approximations.
2. MAIN RESULTS This section presents our main results derived under the Bernoulli model discussed P above. Note that, in such a model, P = Vi=1 p2i represents the probability of a match between a given position of the text string a and a given one in the pattern string b. All our asymptotic results are valid in the so called right domain asymptotic, that is, for 3
log n = o(m). In fact, the most interesting range of values for m and n is such that m = (n ) for some 0 1. However, to cover the whole right domain asymptotics, we assume that may also slowly converge to zero, that is, 1 ?1 o(m= log m) (for simplicity of notation we write instead of n ). Finally, for simplicity of the presentation, it helps to imagine that a is written on a cycle of size n 2m. Then, we call i (b) the version of b written on the same cycle, and cyclically shifted by i positions relative to a. Note that Ci is the number of matches between a and b when b is shifted by i positions. In our cyclic representation, Ci can alternatively be thought of as the number of places on the cycle in which a and i (b) agree. It is easy to see that the distribution of Ci is binomial, hence for any i ! m PrfCi = `g = ` P ` (1 ? P )m?` : (1) Naturally, the average number of matches EC1 is equal to mP . Furthermore, Ci tends almost surely to its mean mP (by the Strong Law of Large Numbers [12]). Now we can present our rst result regarding Mm;n = max1in fCi g which can be interpreted as the amount of similarity between a and b.
Theorem 1 If m and n are in the right domain asymptotic, that is , log n = o(m), then Mm;n = P lim m!1 m
(a:s:)
(2)
Proof. A lower bound on Mm;n follows from the fact that, by its de nition, Mm;n must be greater than C1 which tends { by the Strong Law of Large Numbers { almost surely to mP . So, now we concentrate on an upper bound. We rst establish the bound in probability (pr.) sense. From Boole's inequality we have PrfMm;n > rg = PrfC1 > r or C2 > r or ::: Cn > rg nPrfC1 > rg :
(3)
It suces to show that for r mP the above probability becomes o(1), that is, nPrfC1 > (1 + )mP g = o(1). For this we need an estimate of the tail for the binomial distribution (1). Such an estimate is computed in Section 2 by the saddle point method. A simpler (and more general) approach, however, is necessary for the purpose of this proof. We note that C1 can be represented as a sum of m independent Bernoulli distributed random variables Xi , where Xi is equal to one when there is a match at the ith position, and zero p otherwise. From the Central Limit Theorem we know that (C1 ? EC1 )=( mP (1 ? P )) ! 4
N (0; 1), where N (0; 1) is the standard normal distribution. Let Gm(x) be the distribution of p P ( mi=1 Xi ? mEX1 )= m varX1 , and let (x) be the standard normal distribution. Then, from Feller [12]
?x2 =2 p e p o( m) ; Gm(x) = (x) + 2 2 p where (x) ex?px2=2 . In (3), set r = mP + (1 + ") m2P (1 ? P ) log n. Then q PrfMm;n > mP + (1 + ") 2mP (1 ? P ) log ng n1" :
(4)
The above rate of convergence does not yet warrant the application of the Borel-Cantelli lemma to prove (2), but clearly (4) proves Mm;n =m ! P (pr.) in the right domain asymptotic. To establish the stronger (a.s.) convergence, we proceed as follows. De ne a subsep f = (M quence nr = s2r for some integers r and s. Let M n m;n ? mP )= 2m. Then, by (4) p p f = log n P ? P 2 (a.s.). To nish the proof we need to extend this to all n. Note M nr r that for a xed s we can always nd an r such that s2r n (s + 1)2r . Furthermore, since f is a stochastically increasing sequence, we have M n p f f M r log( s + 1)2r = pP ? P 2 (a:s:) : M ( s +1)2 n p p p lim sup lim sup r n!1 log n r!1 log s2r log(s + 1)2 This proves (2). Theorem 1 does not provide much useful information, and an estimate of Mm;n based on it would be a very poor one. From the proof of Theorem 1 we learn, however, that p Mm;n ? EC1 = O( m log n), hence the next term in the asymptotics of Mm;n plays a very signi cant role, and de nitely cannot be omitted in any reasonable computation. The next theorem { our main result { provides an extension of Theorem 1, and shows how much the maximum Mm;n di ers from the average EC1 . The proof of the above theorem will be given in Section 3.
Theorem 2 Under the hypothesis of Theorem 1 we also have for every " > 0 Mpm;n ? mP < (1 + ")g = 1 ; lim Pr f (1 ? " ) n!1 2m log n
where
q
q
= maxf (1 ? )(P ? T ); minf(P ? T ); (1 ? )1 gg p
= minf P ? P 2 ;
V X
j =1
q
pj 1 ? pj g : 5
(5) (6) (7)
In the above, P
where T = Vi=1 p3i .
)(P ? 3P + 2T ) ; 1 = (P ? T6( T ? P 2) 2
Remark 1. The lower bound (6) and the upper bound (7) are composed of two di erent
expressions. None of the expressions is better than the other for the entire range of the probabilities 0 < pi < 1. With respect to the lower bound, we can show that there exist values of the probabilities fpi gVi=1 such that either 2 = (1 ? )(P ? T ) or 2 = minf(P ? T ); (1 ? )1 g. The latter case clearly can occur. Surprisingly enough, the former case may happen too. Indeed, let one of the probability pi dominates all the others, e.g., p1 = 1 ? ". Then, we have either 2 = (1 ? )" + O("2 ) or 2 = minf2=3 (1 ? )"; "g + O("2 ). For close to one the former bound is tighter. With respect to the upper bound, we have a p P similar situation. In most cases, 2 = P ? P 2 . However, sometimes = Vi=1 pi 1 ? pi . Indeed, this occurs for example for a binary alphabet with p1 < 0:1. 2 Note that, when all the pi ! 1=V (the symmetric case) we have 1 ! 1. Thus, for all p < 1 we have = = 1=V ? 1=V 2 . Thus, we obtain the following corollary.
Corollary 1 If the alphabet is symmetric and < 1, then for every " > 0 the following holds
Mm;n ? mP p lim Pr f 1 ? " < < 1 + "g = 1 ; n!1 2m(1=V ? 1=V 2 ) log n p that is, Mm;n ? mP 2m(1=V ? 1=V 2 ) log n (pr.).
(8)
In the last subsection we will extend the condition of the corollary to the limiting case = 1. In order to verify our bounds for Mm;n we have performed some simulation experiments, which are presented in Figures 1, 2, and 3. In these gures we plot the experimental distrip bution function Prf(Mm;n ? mP )= mP (1 ? P ) xg versus x. Our simulation experiments presented in Figure 1 con rm our theoretical result of Corollary 1. Namely, for the symp metric alphabet, (Mm;n ? mP )= m log n) converges to a constant. But { surprisingly { this is not true for an asymmetric alphabet, as indicated by Figures 2 and 3. In fact, based on p these simulations we conjecture, that (Mm;n ? mP )= m log n) converges in probability to a random variable, say Z , which is not a degenerate one, except for the symmetric alphabet. A study of the limiting distribution of Z seems to be very dicult. Our Theorem 2 provides some information regarding the behavior of the tail of the distribution of Z . In particular, 6
1.0
m;n ?mP < xg Prf pMmP (1?P ) 0.8
0.6
0.4
0.2
0 0.0
x
Lower=Upper bound 2
4
6
8
10
Figure 1: Distribution function of (Mm;n ? mP )(mP (1 ? P ))?1=2 for V = 2 with p1 = 0:5 (symmetric case), via simulations. we know that Z degenerates for the symmetric alphabet, that is, possibly the variance of Z becomes asymptotically zero in this case.
3. ANALYSIS In this section, we prove our main result, namely Theorem 2. In the course of deriving it, we establish some interesting combinatorial properties of pattern matching that have some similarities with the work of Guibas and Odlyzko [16, 17] (see also [19]). The proof itself consists of two parts: the upper bound (cf. Section 3.2) and the more dicult lower bound (cf. Section 3.3). Section 3.1 provides necessary mathematical background, and presents some preliminary results. Before we plunge into technicalities, we present an overview of our approach. An upper bound on Mm;n is obtained from (3), or a modi cation of it, and the real challenge is in establishing a tight lower bound. The idea was to apply the second moment method, in the form of Chung and Erdos [10] which states that for events fCi > rg, the following holds P n [ > rg)2 ( P PrfMm;n > rg = Prf (Ci > r)g PrfC > rg + iPPrfCi Pr : (9) i (i6=j ) fCi > r & Cj > rg i i=1 7
1.0
m;n ?mP < xg Prf pMmP (1?P ) 0.8
0.6
0.4
0.2
0 0.0
Lower bound 2
x
Upper bound 4
6
8
10
Figure 2: Distribution function of (Mm;n ? mP )(mP (1 ? P ))?1=2 for V = 2 when n = 50; 000 and m = 600, when n = 50; 000 and m = 600, with p1 = 0:2, via simulations. Thus, one needs to estimate the joint distribution PrfCi > r & Cj > rg. Ideally, we would p expect that the right-hand side of (9) goes to one when r = mP +(1 ? ") m2(P ? P 2 ) log n for any " > 0, which would match the upper bound. Provided the above works, one needs an estimate of PrfCi > r & Cj > rg. This is a dicult task since C1 and Cj are strictly positively correlated random variables for all values of 1 < j n. In fact, there are two types of dependency. For j > m, the patterns aligned at positions 1 and j do not overlap, but nevertheless C1 and Cj are correlated since the same pattern is aligned. Although this correlation is not dicult to estimate, it is strong enough to "spoil" the second moment method. In addition, an even stronger correlation between C1 and Cj occurs for j < m since in addition overlapping takes place. We treat this type of dependency in Section 3.4. To resurrect the second moment method, we introduce a conditional second moment method. The idea is to x a string of length m built over the same alphabet as the pattern b, and to estimate C1 and Cj under the condition b = . Then, C1 and Cj are conditionally independent for all j > m, and the dependency occurs only for j m, that is, when C1 and Cj overlap. This idea we shall develop below. 8
1.0
m;n ?mP < xg Prf pMmP (1?P ) 0.8
0.6
0.4
0.2
0 0.0
Lower bound
x
Upper bound
2
4
6
8
10
Figure 3: Distribution function of (Mm;n ? mP )(mP (1 ? P ))?1=2 for V = 2 when n = 100; 000 and m = 1000, with p1 = 0:1, via simulations. We introduce some notation. For a given , we de ne
G(r; ) = PrfC1 > rjb = g ; m X F (r; ) = PrfC1 > r & Ci > r jb = g i=2
p() = Prfb = g :
(10) (11) (12)
=V p j where the denotes the number of Considering our Bernoulli model, p() = jj =1 j j occurrence of the symbol j in the string . When varies, we may consider the conditional probabilities G(r; ) and F (r; ) as random variables with respect to . We adopt this point of view. Q
3.1 Preliminary Results In this subsection, we put together several results concerning the tail of a binomially distributed random variable, that are used in the proof of our main results. If X is a binomially distributed random variable with parameters m and p, then we write B (m; p; r) = PrfX > rg for the tail of X . We need the following result concerning the tail of C1 . 9
Lemma 1 When m and both tend to in nity such that = O(log m), then p p ? 1 exp 2(P ? P 2 ) : (13) B (m; P; mP + m ) = PrfC1 mP + m g p 2(P ? P 2 ) Proof: According to (1), C1 is binomially distributed, that is, PrfC1 = rg = ?mrP r (1 ? P )m?r . Introducing the generating function C (u) = Pmr=1 PrfC1 = rgur for u complex, we easily get the formula C (u) = (1 + P (u ? 1))m . Then, by Cauchy's celebrated formula [18] I 1 (14) PrfC1 rg = 2i (1 + P (u ? 1))m ur (u1? 1) du ; where the integration is along a path around the unit disk for u complex. The problem is how to evaluate this integral for large m. In this case the best suited method seems to be the saddle point method [18, 13]. This method applies to integrals of the following form
I (m) =
Z
P
(x)e?mh(x) dx ;
(15)
where P is a closed curve, and (x) and h(x) are analytical functions inside P . We evaluate I (m) for large m. The value of the integral does not depend on the shape of the curve P . The idea is to run the path of integration through the saddle point, which is de ned to be a place where the derivative of h(x) is zero. To apply this idea to our integral (14) we represent it in the form of (15) and nd the minimum of the exponent. De ne u = 1 + h, and then the exponent in our integral can be expanded as log((1 + P (u ? 1))m =ur ) = m log(1 + hP ) ? r log(1 + h) = (Pm ? r)h ? 1=2(P 2 m ? r)h2 + O(m + r)h3 ) 2 = ? 12 (r ? mP )2 (r ? mP 2 )?1 + (r ? mP 2 ) (h?2h0 ) + O(m + r)h3 )
p
with h0 = (r ? mP )=(r ? mP 2 ). Let r = mP + mx, with x > 0. Changing the scale of p the variable under the integrand: h = h0 + it= m(P ? P 2 ), we obtain !
Z log m 2 p exp(?t2 =2) dt ?1 + O(1=pm) : 1 x p PrfC1 mP + mxg = exp 2(P ? P 2 + x= m) 2 ? log m pP x?P 2 + it p Therefore, when x ! 1 (e.g., as log m) we get p p Z log m exp(?t2 =2) dt P ? P 2 Z 1 exp(?t2 =2)dt = 2(P ? P 2 )
? log m pP x?P 2 + it
x
?1
x
where the last integral can be computed from the error function [2]. This completes the proof of our lemma. 10
To estimate the conditional probability G(r; ), we need some additional asymptotics for the binomial distribution, which are summarized in the lemma below.
Lemma 2 Let rm = mp + pm m where m = O(log m). Then for all > 1 there exists a sequence m ! +1 such that the following hold pm; p; r ) exp( m ) 1 (16) lim inf B ( m ?
m m m!1 2(p ? p2 ) and
p
lim sup B (m + m m; p; rm ) exp( 2(p m? p2 ) ) 1 : m!1
(17)
Proof: These are computational consequences of Lemma 1. Details are deferred to the Appendix.
Now, we use Lemma 2 to estimate a typical behavior of the probability G(r; ) (recall that we treat G(r; ) as a random variable with respect to ). For a string 2 m , we de ne j as the number of occurrences of the symbol j 2 in . If varies, then j can be treated as a random variable, too. Let also C1 (j ) be the number of matches that involve P symbol of j 2 between b = and a1 a2 am . Clearly, C1 = Vj=1 C1 (j ), and C1 (j ) is binomially distributed with parameters j and pj . We have the following simple estimate of G(r; ).
Lemma 3 For all rj such that PVj=1 rj = r, the following holds V Y j =1
B ( j ; pj ; rj ) G(r; )
V X j =1
B ( j ; pj ; rj ) :
(18)
Proof: This follows immediately from the next two trivial bounds PrfC1 > rg Prf8j : C1 (j ) > rj g ; and
PrfC1 > rg Prf9j : C1 (j ) > rj g :
This completes the proof. Our main result of this subsection is contained in the next theorem. It provides (a.s.) bounds for the probability G(r; ). 11
Theorem 3 Let rm = mP + pm m with m = O(log m), and P and T as de ned in Theorem 2. For all > 1, the following estimates hold: (i) m mlim !1 Prf : G(rm ; ) exp ? 2(P ? T ) g = 1 ; (ii) m mlim !1 Prf : G(rm ; ) exp ? 2S 2 g = 1 ; p P where S = Vj=1 pj 1 ? pj .
(19) (20)
Proof: We rst deal with part (i). For convenience of presentation we drop the subscript m from rm and m. Let rj = mp2j + j pm with PVj=1 j = 1. By (16) of Lemma 2, for any > 1 there exists a sequence m ! 1 such that the following holds for m large enough V V Y Y 2 p B (mpj ? m m; pj ; rj ) exp(? 2p2 (1j ? p ) ) : j j j =1 j =1 P Taking j = p2j (1 ? pj )=( p2i (1 ? pi )), after some algebra involving an optimization of the right-hand side of the above, we obtain (? =2(P ? T )) in the exponent of the right-hand p Q side (RHS) of the above. But, by Lemma 3, G(r; ) Vj=1 B (mpj ? m m; pj ; rj ) where is such that for all j : j mpj ? m pm. Thus, p Prf : G(r; ) exp(? 2(P ? T ) )g Prf8j : j mpj ? m mg :
By the Central Limit Theorem, every random variable
j tends to the Gaussian distribution q with mean mpj and with the standard deviation m(pj ? p2j ). So, it is clear that mlim !1 Prf8j :
p
j mpj ? m mg = 1
as m ! 1. The proof of part (ii) goes along the same lines. For completeness, we provide the details p P below. Let rj = mp2j + j m with Vj=1 j = 1. By Lemma 2, for every > 1 we can nd a sequence m ! 1 such that the following holds for m large enough: V X
p
B (mpj + m m; pj ; rj ) V ?1
V X
2
exp(? 2p2 (1j ? p ) )
j j j =1 j =1 p Taking j = S ?1 pj 1 ? pj , we obtain (? =2S 2 ) in the exponent of the RHS of the above. p P But, by Lemma 3, G(r; ) Vj=1 B (mpj + m m; pj ; rj ) where is such that for all j : p
j mpj + m m. Thus,
p
Prf : G(r; ) exp ? 2 S 2 g Prf8j : j mpj + m mg : 12
By the Central Limit Theorem, every
j tends to the Gaussian random variable with mean q mpj and with standard deviation m(pj ? p2j ). Hence, mlim !1 Prf8j :
p
j mpj + m mg = 1
for m ! 1.
Remark 2. If we denote by Gm the set of 2 m such that G(r; ) exp ? 2(P ?T ) , then (19) asserts that PrfGm g ! 1. As above, if we denote by Gm0 a set of 2 m such that G(r; ) exp(? 2 S 2 ) ), then (20) implies that PrfGm0 g ! 1. 3.2 Upper Bounds We establish here the two upper bounds claimed in Theorem 2. We prove the following theorem.
Theorem 4 Let " > 0 be an arbitrary non-negative real number. Then (i) If = 2(1 + ")(P ? P 2 ) log n, then the following holds limn;m!1 PrfMm;n < rg = 1 p with r = mP + m .
p P (ii) If = 2(1 + ")S 2 log n, with S = Vi=1 pi 1 ? pi, then limn;m!1 PrfMm;n < rg = 1 p with r = mP + m .
Proof: Part (i) was in fact already proved in Theorem 1, and it really follows directly from PrfMm;n < rg nB (m; P; r). We concentrate on proving part (ii). We have PrfMm;n > rg
X
2Gm0
n p()G(r; ) + Prf 62 Gm0 g
max fnG(r; )g + Prf 62 Gm0 g ; 2Gm0 p where the set Gm0 is de ned in Theorem 3 (ii) with = 1 + ". Note that Prf 2= Gm0 g ! 0, by Theorem 3. Then by (20) G(r; ) exp(? =(2S 2 )) for 2 Gm0 , thus p PrfMm;n > rg n1? 1+" + Prf 62 Gm g :
But both terms of the above tend to zero, and this establishes the second upper bound in Theorem 2.
3.3 Lower Bounds 13
The lower bounds are much more intricate to prove. We start with a simple one that does not take into account overlapping. Then, we extend this bound to include the overlapping. This extension is of interest to us since it leads to the exact constant in the symmetric case. Mm;n where In the rst lower bound we ignore overlapping by considering Mm;n = max1ibn=mc fC1+im g, that is, the maximum is taken over all nonoverlapping Mm;n positions of a. Note also that C1 and C1+m are conditionally independent (i.e., under b = ). Our rst lower bound is contained in the theorem below.
Theorem 5 Let 0 < " < 1. If = 2(1 ? ")(P ? T )(1 ? ) log n and r = mP + pm , then limn;m!1 PrfMm;n < rg ! 0. Proof. Due to conditional independence we can write < rg = PrfMm;n
X
2m
p()(1 ? G(r; ))n=m
max f(1 ? G(r; ))n=m g + Prf 2= Gm g ; 2Gm where Gm is de ned in remark 2, that is, for = (1?")?1=2 we have G(r; ) > exp(? =(2(P ? T ))) for all 2 Gm . Note also that Prf 2= Gmg ! 0. We concentrate now on the rst term of the above. Using log(1 ? x) ?x, we obtain (1 ? G(r; ))n=m e?(n=m)G(r;) : Now, it suces to show that (n=m)G(r; ) ! 1 for 2 Gm . By Theorem 3, we have for
2 Gm
n G(r; ) n exp ? (n=m)1?p1?" ! 1 ; m m 2(P ? T ) where the convergence is a consequence of our restrictions log n = o(m). This completes the proof of the "easier" lower bound in Theorem 2.
The second lower bound is more elaborate. It requires an estimate of PrfC1 > r & Ci > = rg also for i < m. Let Fm;n = Pmi=2 PrfC1 > r & Ci > rg. Note that Fm;n Fm;n Pm i=2 PrfC1 + Ci > 2rg. In the next subsection we prove the following theorem.
Theorem 6 For m = O(log m) ! 1 we have 2 2T )5=2 (mP + pm m ) m(P ? 3P + p exp(? P ? 3P m2 + 2T ) : Fm;n 2 3 2(T ? P ) m for m ! 1. 14
(21)
Assuming that Theorem 6 is available, we proceed as follows. Using the second moment method we have X PrfMm;n > rg p()S (r; ) ; (22) 2m
where following (9)
(r; ))2 S (r; ) = nG(r; ) + nF (r; )(nG + (n2 ? n(2m + 1))(G(r; ))2 : P In the above, the probability F (r; ) is de ned in (12) (to recall, F (r; ) = mi=2 PrfC1 > r & Ci > r jb = g). We rewrite S (r; ) as ?1 S (r; ) = nG(1r; ) + n(FG((r;r; )))2 + 1 ? 2mn+ 1 :
(23)
We also have the following identity X
2m
p()F (r; ) = Fm;n (r) ;
(r) is estimated in Theorem 6. The almost sure where the probability Fm;n (r) Fm;n behavior of F (r; ) { as a random function of { is discussed in the next lemma.
Lemma 4 If ! 1, then for any constant we have
mlim !1 Prf : F (r; ) m exp ? P ? 3P 2 + 2T g = 1 :
(24)
Proof: It is a simple consequence of Markov's inequality: Since P2m p()F (r; ) = Fm (r), it is clear that PrfF (r; ) > m Fm (r)g 1= m . Remark 3. We denote by Gm00 the set of 2 m such that F (r; ) m exp(? P ?3P 2+2T ) holds. The lemma shows that limm!1 Prf 2 Gm00 g = 1. 2 Now we are ready to prove the main result of this subsection that provides the most elaborate lower bound. Let 1 be de ned as in Theorem 2, that is, )(P ? 3P + 2T ) : 1 = (P ? T3( T ? P 2) 2
Let also 2 = minf(1 ? )1 ; 2(P ? T )g. 15
Theorem 7 Let 0 < " < 1. If = (1 ? ")2 log n, then limm;n!1 PrfMm;n > rg = 1 for r = mP + pm . Proof: Note that it suces to prove the theorem for any positive and arbitrarily small ".
By (22) and (23) we need to show that
nG(r; ) ! 1 (25) F (r; ) ! 0 : (26) nG2(r; ) The rst identity is easy to prove. For 2 Gm by Theorem 3 with (1 ? ")?1=2 we have p nG(r; ) n1?1=2 1?"2 =(P ?T ) ! 1
since 2 =(2(P ? T )) 1. Now, we deal with (26). Note that 1 > 0 since P 2 < T . For 2 Gm \ Gm00 where Gm and Gm00 are de ned as in (respectively) Theorem 3 and Lemma 4 (with = 1), we have F (r; ) m exp( ? 2 nG (r; ) n P ? T P ? 3P 2 + 2P ) 1? ?(11?")2 =1 exp(( ? 1) P ? T )
n
We know that 1 ? ? 2 =1 (1 ? ") "2 =1 . Choosing ? 1 = O("2 ) in the above, we nally obtain 1 F (r; ) !0; 2 1 ? ? = 2 1 (1?")+O ("2 ) nG (r; ) n since " can be arbitrary small. Putting everything together, we have just proved that S (r; ) ! 1 for all 2 Gm \ Gm00 . But, by (22), Theorem 3 and Lemma 4 PrfMm;n > rg
X
2m
p()S (r; )
Prf 2 Gm \ Gm00 g 2Gmin fS (r; )g ! 1 ; m \G 00 m
which completes the proof.
3.4 Proof of Theorem 6 In this subsection, we prove Theorem 6 concerning the asymptotic behavior of the probability PrfC1 > r; Ci > rg for i < m. In fact, we evaluate the generating function Hm;` (u; v) = Pr1 ;r2 PrfC1 = r1 & C` = r2 gur1 vr2 , and then compute the probability PrfC1 = r1 & C` = r2 g through the Cauchy integral as it was done in Lemma 1. 16
Let x and y be column vectors of dimension V , that is, x = fxi gVi=1 and y = fyi gVi=1 . We de ne the scalar product hx; yi by x1 y1 + + xV yV . Then, the next crucial theorem captures some important combinatorial properties of fC1 ; C` g that allow to estimate the generating function Hm;2 (u; v) for ` = 2, and nally Hm;` (u; v) for any ` (cf. Theorem 9). Now we are ready to establish a closed-form formula for the generating function Hm;2 (u; v). We construct a recurrence relationship between the distributions of fC1 (b); C2 (b)g and fC2 (b0 ); C3 (b0 )g where b0 is the sux of b of length m ? 1. In the above we write Ci (b) instead of Ci in order to show explicitly a dependency of Ci on the string b. With this in mind, we can proceed to the following key theorem.
Theorem 8 We have the identity Hm;2(u; v) = hx(u); Am?1 (u; v)y(v)i, where A(u; v) is a V V square matrix whose generic element aij (u; v) satis es the following aij (u; v) = pi (1 + pi (v ? 1) + pj (u ? 1)) when i 6= j , and
aii (u; v) = pi (1 + pi (uv ? 1)) for i = j . The row vectors x(u) and y(v) are de ned as x(u) = f1 + pi(u ? 1)gVi=1 and y(v) = fpi (1 + pi (v ? 1))gVi=1 .
Proof: Let us de ne a random variable ?i as the number of matches between string a and i (b) without counting the eventual rst matching at position i (recall that i (b) is the shifted version of b by i positions on the cycle). For example, ?1 = C1 if there is no matching at position 1, and ?1 = C1 ? 1 otherwise. De ne next the generating function Pi;m (u; v) as
Pi;m (u; v) =
X
r1 ;r2
Prf?1 = r1 & C2 = r2 & string b starts with symbol igur1 vr2 ;
and let Pm (u; v) denote the row vector fPi;m (u; v)gVi=1 . Note that P1 (u; v) = y(v) and Hm;2 (u; v) = Pii==1V (1 + pi(u ? 1))Pi;m (u; v), thus Hm;2 (u; v) = hx(u); Pm (u; v)i. The most interesting fact that we prove next is the following relationship Pm (u; v) = A(u; v)Pm?1 (u; v) when m > 1. A proof of this relies on building a recurrence relationship between fC1 (b); C2 (b)g and fC2 (b0 ); C3 (b0 )g, as explained above. Let i and j be the two rst symbols of string b and let k be the second symbol of string a. When i 6= j we have ?1 (b) = ?2 (b0 ) + 1 and C2 (b) = C3 (b0 ) if k = j , ?1 (b) = ?2 (b0 ) and C2 (b) = C3 (b0 ) + 1 if k = i, and ?1 (b) = ?2 (b0 ) and C2 (b) = C3 (b0 ) otherwise. When i = j , we have 17
a = 01100101110101011110101110010111 b = 01011001010111 b0 = 1011001010111 0
1
1
1
0
0
1
0
1
1
1
0
1
0
0
1
0
1
1
0
0
1
0
1
0
1
1
1
0
1
0
1
1
0
0
1
0
1
0
1
1
1
1
1
1
0
0
1
0
1
1
1
0
1
0
1
0
1
1
1
1
0
1
0
1
1
1
0
0
1
0
1
1
1
C1 (bfb) = 6, ?1 (b) = 5, C2 (b0 ) = 5, ?2 (b0 ) = 4 C2 (bfb) = 7, ?2 (b) = 7, C3 (b0 ) = 7, ?3 (b0 ) = 6 1
1
1
0
1
0
1
1
1
0
0
1
0
1
1
1
Figure 4: Illustration of the relationship between fC1 ; C2 g and f?2 ; ?3 g (boxes show matches). ?1 (b) = ?2 (b0 )+1 and C2 (b) = C3 (b0 )+1 if k = i, and ?1 (b) = ?2 (b0 ) and C2 (b) = C3 (b0 ) otherwise. This is illustrated in Figure 4. Since fC2 (b0 ); C3 (b0 )g has the same distribution as fC1 (b0 ); C2 (b0 )g, we obtain the following identity X Pi;m (u; v) = (1 + p (uv ? 1))P ( u; v ) + (1 + pi(v ? 1) + pj (u ? 1))Pj;m?1 (u; v) ; i i;m ? 1 pi j 6=i
which proves our theorem. The next theorem extends Theorem 8 and give a formula for the generating function Hm;` (u; v), which is of independent interest.
Theorem 9 For all q < m the following holds: Hm;1+q (u; v) = (Hh;2(u; v))q?` (Hh+1;2(u; v))` , where h = b mq c and ` = m ? hq. Proof. For i q de ne b(i) as a subsequence of string b obtained by selecting the ith symbol of b, then the i + qth, then i + 2qth, and so forth. For 1 i `, string b(i) is of length h + 1, whereas for ` + 1 i q its length is h. We can do the same with string a and obtain subsequences a(1) ; : : : ; a(q) . Let ha; i (b); j (b)i be a new notation for the two dimensional row vector [Ci ; Cj ], that is, it represents the number of matches between a and simultaneously i (b) and j (b). It is easy to see that [C1 ; C1+q ] = ha(1) ; 1 (b(1) ); 2 (b(1) )i + + ha(q) ; 1 (b(q) ); 2 (b(q) )i. Note that ha(i) ; 1 (b(i) ); 2 (b(i) )i's has the same distribution as [C1 ; C2 ] when b(i) is of length 18
h + 1 for i `, and the same distribution as [C1 ; C2 ] when b(i) is of length h for ` < i q. This nally proves the theorem.
Theorem 9 establishes a closed form formula for the generating function of the joint distribution PrfC1 = r1 ; C` = r2 g. Therefore, in principle we can recover the probabilities PrfC1 = r1 ; C` = r2 g from Hm;` (u; v) by Cauchy's formula, as we did in Lemma 1. The diculty is that the generating function Hm;`(u; v) is expressed in terms of matrix A(u; v), so we need some tools from linear algebra to apply the saddle point method. However, before we plunge into this, we should treat the symmetric case (i.e., pk = 1=V ) separately since, as the next lemma shows, it possesses a very special property.
Lemma 5 In the symmetric case, for all i 6= j , the random variables Ci and Cj are pairwise independent.
Proof: It suces to prove that C1 is independent of C1+q for all 1 q m. In the symmetric case the aij (u; v)'s are all identical and equal to V1 (1+ V1 (u?1+v ?1)) except when i = j where aii (u; v) = V1 (1+ V1 (uv ? 1)). Note that y(v) coincides with an eigenvector of the matrix A and A(u; v)y(v) = (1 + V1 (u ? 1))(1 + V1 (v ? 1))y(v), and therefore Hm;2 (u; v) = (1 + V1 (u ? 1))m (1 + V1 (v ? 1))m . This last formula shows that C1 and C2 are independent. Applying Theorem 9 one concludes that also Hm;1+q (u; v) = (1+ V1 (u ? 1))m (1+ V1 (v ? 1))m . Therefore C1 and C1+q are also independent.
Corollary 2 In the symmetric case we have Mm;n ? mP p2m(1=V ? 1=V 2) log n when m; n ! 1, given log n = o(m). Proof: We already know that the above is true when < 1 (cf. Theorem 2). Using
the result stated in Lemma 5 about pairwise independence of the Ci 's, we can rewrite inequality (9) as PrfC1 > rg)2 PrfMm;n > rg nPrfC > r(gn+ (27) (n2 ? n)(PrfC1 > rg)2 : 1 It is clear that the RHS of the above tends to 1 if and only if nPrfC1 > rg ! 1. Referring to Lemma 1, it is clear that this occurs for = 2(1 ? ")(P ? P 2 ) log n and any arbitrarily chosen " > 0.
Hereafter, we concentrate on the asymmetric case. Let (u; v) be the principal eigenvalue of the matrix A(u; v). Let (u; v) (resp. (u; v)) be the corresponding right (resp. left) eigenvector of A(u; v) such that h(u; v); (u; v)i = 1, that is, A(u; v) (u; v) = (u; v) (u; v) and AT (u; v)(u; v) = (u; v)(u; v) (cf. [25, 24]). We distinguish the following three cases: 19
1. When u = v = 1, (1; 1) = 1, (1; 1) = y(1) and (1; 1) = x(1), the other eigenvalues are null. 2. When v = 1, (u; 1) = 1 + P (u ? 1), (u; 1) = (1; 1) = y(1) and (u; 1) = 1=(u; 1) x(u), the other eigenvalues are null. 3. When u = 1, (1; v) = (v; 1) = 1 + P (v ? 1), (1; v) = y(v) and (1; v) = 1=(1; v) x(1), the other eigenvalues are null. It follows that the other eigenvalues are O((u ? 1)(v ? 1)), and therefore we immediately prove the following fact.
Corollary 3 Hm;2(u; v)=m?1 (u; v) = hx(u); (u; v)i h(u; v); y(v)i + O((u ? 1)m (v ? 1)m ). Proof: This is a classical property of the principal eigenvalue and follows from the PerronFrobenius theorem (the interested reader is referred to [3, 24, 25] for details).
As a consequence of Corollary 3 we have the following important expansion of the generating function Hm;` (u; v).
Corollary 4 Let Fm (u; v) = Pmi=2 Hm;i(u; v). We have Fm (u; v) = a(u; v) ? am (u; v) + O((u ? 1)(v ? 1)) ; ((u; v))m 1 ? a(u; v) with a(u; v) = hx(u); (u; v)i h(u; v); y(v)i=(u; v)
(28)
Proof: From Corollary 3 and Theorem 9 we have the estimate Hm;1+q (u; v) = ((u; v))m (a(u; v))q + O((u ? 1)h+1 (v ? 1)h+1 )) : Therefore
?1 m Fm (u; v) = mX q + X O((u ? 1)h+1 (v ? 1)h+1 ) : ( a ( u; v )) ((u; v))m q=1 h=2
The proof is easily completed by summing the geometric series in the last expression. The following two lemmas present more detailed Taylor's expansions of the principal eigenvalue of A(u; v) de ned in Corollary 4. These expansions are next used in the saddle p point method to obtain a sharp estimate of Fm;n (r) around r = mP + 2mP (1 ? P ) log n.
Lemma 6 The Taylor expansion of (u; v) to the second order is 1 + (u ? 1)P + (v ? 1)P + (u ? 1)(v ? 1)(2T ? P 2 ), with T = p31 + + p3V . 20
Proof. We know that (u; v) = 1 + (u ? 1)P + (v ? 1)P + O((u ? 1)(v ? 1)). We adopt
the following notation. If f (u; v) is a function of two variables u and v, then we denote by fu(u; v) (resp. fv (u; v)) the partial derivative of f (u; v) with respect to u (resp. to v). We have = h; A i, where the variables (u; v) have been dropped in the last expression to simplify the presentation. Thus, u = hu ; A i + h; Au i + h; Au i. Since A = , AT = , and since hu ; i + h; u i = 0 (because we assume h; i = 1), we get u = h; Au i. Substituting u = 1 in the last expression, we obtain the identity (1; v) = P (P + 2T (v ? 1) + ii==1V p4i (v ? 1)2 )=(1 + P (v ? 1)), and after some simple algebra the proof is completed.
Lemma 7 The Taylor expansion of a(u; v) to the second order is 1 ? (u ? 1)(v ? 1)(T ? P 2). Proof. Easy computations give a(u; 1) = a(1; v) = 1, therefore a(u; v)?1 is O((u?1)(v?1)).
Di erentiating twice a(u; v) and setting u = v = 1 leads to a formula beginning with huv ; i + h; uv i and ending with a linear combination of scalar products involving rst partial derivatives of , , x and y. These rst derivatives are already known since and are completely determined when u = 1 or v = 1. For huv ; i + h; uv i, we di erentiate twice both sides of h; i = 1 in order to get huv ; i + h; uv i + hu ; v i + hv ; u i = 0, which leads to a complete determination of auv (1; 1). Lemmas 6 and 7 are crucial to apply Cauchy's formula in order to estimate Fm;n (r) for r > mP . To do that, we can use the double Cauchy formula I I 1 Fm;n (r) = (2i)2 Fm (u; v) ur (u ?d1)udvrv(v ? 1) : This kind of integration is rather unusual. Since PrfCi > r; Cj > rg PrfCi + Cj > 2rg (r) = Pm PrfC1 + Ci > 2rg, which leads to a single integration we can estimate Fm;n i=2 I
1 (r) = 1 F ( u; u ) Fm;n m 2 r 2i u (u ? 1) du : Finally, we can prove the following asymptotics for the tail of Fm;n (r), which establishes Theorem 6.
Proof of Theorem 6. We parallel the proof of Lemma 1. By Cauchy and (28) we have I m a(u; u) ? am (u; u)) du ; 1 (29) Fm;n (r) 2i (u; uu)( 2r (1 ? a(u; u)) u?1 21
the integration path encircling the unit disk. Let 1 + h = u. Using Lemmas 18 and 19, we obtain the expansion log(m (u; u)=u2r ) = = = +
m log(1 + 2hP + h2 (2T ? P 2 ) + O(h3 )) ? 2r log(1 + h) ?2(r ? mP )h + (r + 2mT ? 3mP 2 )h2 + O((m + r)h3) ?(r ? mP )2(r ? m(3P 2 ? 2T ))?1 + (r ? 3mP 2 + 2mT )(h ? h0 )2 + O((m + r)h3 ) ;
p
p
with h0 = (r ? mP )=(r ? 3mP 2 +2mT ). Let r = mP + mx. Substituting h = h0 + it= m, and using 1 ? a(1 + h; 1 + h) = h2 (T ? P 2 ) + O(h3 ) we get the rst estimate 1 I m (u; u)a(u; u)du = 2i u2r (u ? 1)(1 ? a(u; u)) ! Z 2 2 2 p = exp ? P ? 3Px 2 + 2T 2(T ?mP 2 ) exp((?(P x? 3P ++it2)T3 )t ) dt(1 + O(1= m)) : P ?3P 2 +2T p Since x = O( log n), we obtain ! 1 I m (u; u)a(u; u)du m(P ? 3P 2 +p2T )5=2 exp ? x2 2i u2r (u ? 1)(1 ? a(u; u)) 2(T ? P 2 ) x3 P ? 3P 2 + 2T : (30) It remains to evaluate the second term in (29), that is, 1 I m (u; u)am (u; u)du : (31) 2i u2r (u ? 1)(1 ? a(u; u)) Using the estimates from Lemmas 18 and 19 we nd log(m (u; u)am (u; u)=u2r ) = ?(r ? mP )h + (r + mT ? 2mP 2 )h2 + O((m + r)h3 ) : Hence (31) becomes ! 1 I m (u; u)am (u; u)du m(P ? 2pP 2 + T )5=2 exp ? x2 2i u2r (u ? 1)(1 ? a(u; u)) 2 x3 P ? 2P 2 + T : (32) Since P ? 2P 2 + T < P ? 3P 2 + 2T in the asymmetric case, the exponent in (32) is larger (r). This than in (30), so (30) is the leading term in the asymptotic expansion of Fm;n concludes the proof of Theorem 6.
APPENDIX
Proof of Lemma 2: We rst prove the existence of a sequence m such that (16) holds.
Since for all p and r we have B (0; p; r) = 0 and limx!1 B (x; p; r) = 1, we can nd for the Bernoulli distribution a sequence m such that
p B (m ? m m; p; rm ) > exp(? 2(p m?p2 ) ) ; 22
(33)
and
p B (m ? m m ? 1; p; rm ) exp(? 2(p m?p2 ) ) < 1 :
(34)
Our aim is to prove that limm!1 m = 1. Let us assume contrary that lim inf m!1 m < 1, thus there exists and a subsequence of m bounded from the above by the constant
. For simplicity of the presentation, we assume that m < for all m. Then, due p to monotonicity of B (m; p; r) with respect to m, we have B (m ? qm m ? 1; p; rm ) > B (m ? pm ? 1; p; rm ). Furthermore, for rm = (m ? pm ? 1)p + (m ? pm ? 1) m with p + p )2 p m p (35) m = ( 1 ?
= m = m + O( m ) ; we obtain
p ) p + O ( 1 m m m B (m ? m ? 1; p; rm ) p exp ? 2(p ? p2 ) exp ? 2(p ? p2 ) ; 2(p ? p2 ) m
(36) where the last inequality holds for large m due to > 1 and m ! 1. Clearly, the above contradicts (34), hence, m ! 1, as needed. The proof for the second inequality is similar and is only sketched. We set m such that and
p B (m + m m; p; rm ) < exp(? 2(p m? p2) )
(37)
p B (m + m m + 1; p; rm ) exp(? 2(p m? p2 ) ) :
(38)
If one assumes m < < 1, then
p
p
exp(? 2(p ?m p2 ) ) ; 2 2(p ? p ) m (39) p p with m = m + O( m ), which contradicts B (m + m m +1; p; rm ) exp(? m =2(p ? p2 )).
B (m + m m + 1; p; rm ) < B (m + m + 1; p; rm ) p
1
ACKNOWLEDGEMENT The authors sincerely thank a referee who pointed out an error in an earlier version of the paper.
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