On lower and upper bounds for probabilities of unions and the Borel Cantelli lemma

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1 arxiv: v [mathpr] 6 Aug 204 On lower and upper bounds for probabilities of unions and the Borel Cantelli lemma Andrei N Frolov Dept of Mathematics and Mechanics St Petersburg State University St Petersburg, Russia address: AndreiFrolov@poboxspburu January 29, 208 Abstract We obtain new lower and upper bounds for probabilities of unions of events These bounds are sharp They are stronger than earlier ones General bounds maybe applied in arbitrary measurable spaces We have improved the method that has been introduced in previous papers We derive new generalizations of the first and second part of the Borel Cantelli lemma Key words: the Borel Cantelli lemma, Bonferroni inequalities, the Chung Erdős inequality, bounds for probabilities of unions of events, measure of unions Introduction Let Ω,F,P) be a probability space and {A n } be a sequence of events Put U n = n A i and ξ n = I A +I A2 + +I An for n =,2,, where I B denotes the indicator of event B Bounds for PU n ) play an important role in probability and statistics Many of them are based on moments α k n) = Eξ k n, k =,2,, of random variable ξ n For example, Bonferroni type inequalities use binomial moments of ξ n, but one can write them in terms of α k n) as well Chung and Erdős 952) have derived the most simple and applicable lower bound for PU n ) of this type that is PU n ) α2 n) α 2 n)

2 It follows from the Cauchy Buniakowski inequality It is very convenient because of α n) = PA i ), α 2 n) = PA i A j ) Various generalizations of the Chung Erdős inequality were obtained in Dawson and Sankoff 967), Gallot 966), Kounias 968), Kwerel 975), Boros and Prékopa 989), Galambos and Simonelli 996), de Caen 997), Kuai, Alajaji and Takahara 2000), Prékopa 2009), Frolov 202) and references therein For example, in Kwerel 975) and Boros and Prékopa 989), one can find upper and lower bounds for PU n ) that are based on α k n) for k 3 and α k n) for k 4 The lower bounds are stronger than the Chung Erdős inequality They are simple enough in applications as well Indeed, for every k n, moments α k n) are sums of probabilities of intersections of k events from A,A 2,,A n Note that a precision of bounds increases when numbers of used moments enlarge From the other hand, making use of the Hölder inequality, one can conclude that ) Eξn ) p /p ) ) Eξn ) p q/p PU n ) lim, p + Eξ p n where p > and /p+/q = It follows that applications of L p -norms of ξ n /Eξ n with p > 2 do not give lower bounds stronger than the Chung Erdős inequality In particular, one can not derive better bounds, using α 2 n) and α 3 n) instead of α n) and α 2 n), for instance Stronger bounds have to involve moments of smaller orders and then we have to use moments of non-integer orders Of course, these bounds are more complicated in calculations One can find such lower bounds in Frolov 202) and upper bounds in Frolov 204) Note that in Frolov 202), one part of bounds is obtained by applications of the Cauchy Buniakowski and Hölder inequalities and another one is proved by a method which we improve in this paper One can find a discussion on relationship of these bounds in there In this paper, we improve the method from Frolov 202) and derive new upper and lower bound for PU n ) Some of them include quantities similar to α k n), and may be calculated relatively simple Every new bound forpu n ) may be used to obtain generalizations of the Borel-Cantelli lemma, the classical variant of which is as follows The Borel-Cantelli lemma ) If series PA n ) converges, then PA n io) = 0, where {A n io} = n= n= k=n A k 2) If {A n } are independent and series PA n ) diverges, then PA n io) = n= 2 j= Eξ p n

3 The first part of the Borel-Cantelli lemma works in many situations, but the assumption of independence in the second one is restrictive Therefore, main attention was paid to generalization of the second part The most important generalization of the second part of the Borel-Cantelli lemma has been obtained by Erdős and Rényi 959) They proved that PA n io) = provided L =, where L = liminf n α 2 n) α 2 n) This result implies in particular that the independence maybe relaxed to pairwise independence in the second part Of course, the proof is based on the Chung Erdős inequality Kochen and Stone 964) and Spitzer 964) have proved that PA n io) /L Further generalizations of the second part of the Borel Cantelli lemma have been obtained in Kounias 968), Móri and Székely 983), Andel and Dupas 989), Martikainen and Petrov 990), Petrov 2002), Xie 2008), Feng, Li and Shen 2009), Frolov 202) and references therein Note that Móri and Székely 983) obtained the lower bound for PA n io) in terms of non-integer moments of ξ n /Eξ n Generalizations of the first part of the Borel Cantelli lemma may be found in Frolov 204) and references therein One can find results under additional assumptions on dependence of events, conditional Borel Cantelli lemma and further references in Chandra 202) The rest of the paper is organized as follows We present our method and general inequalities in section 2 Section 3 contains some new bounds for probabilities of unions Note that these results may be also applied to measures of unions in arbitrary measurable spaces In section 4, new generalizations of the Borel Cantelli lemma are proved 2 Method and general results Our method is based on the following result which is a generalization of Theorem in Frolov 202) Theorem Let l and N be natural numbers such that 2 l N Let {r i, i N} and {f ki, k l, i N} be arrays of non-negative real numbers For k l, put s k = f ki r i ) Assume that there exist real numbers c,c 2,,c N and a,a 2,,a l such that c i )r i = l a i s i 2) 3

4 If c i 0 for all i =,2,,N, then R = If c i 0 for all i =,2,,N, then r i l a i s i 3) R l a i s i 4) Inequalities 3) and 4) turn to equalities if, for some i < i 2 < < i l N, the numbers r i,r i2,,r il are solutions of the linear system l f kij r ij = s k, k l, 5) j= r i = 0 for all i i k and c ik = 0 for all k l In this case, R = r i +r i2 + +r il Proof By 2), we have R l a i s i = c i r i, that yields the assertion of Theorem Now we describe the method which gives sharp lower and upper bounds for R We first choose number l and array {f ki } The next step is to take c i To satisfy relation 2), the simplest choice of c i is c i = l a j f ji j= for all i N Since the bounds for R have to be sharp ie they have to turn to equalities for some set of numbers r,r 2,,r N ), we will take coefficients a j such that c ik = 0 for some i < i 2 < < i l N To this end, we need a way to determine i,i 2,,i l Note that if f ki = i k, k l, then putting c i = l k= i ) i k for all i N gives a simple way to find i,i 2,,i l such that c i 0 or c i 0 for every i Indeed, assume first that l = 2 If we put i = m and i 2 = m, where 2 m N, 4

5 then c i 0 for all i If we take i = and i 2 = N, then c i 0 for every i Here and in the sequel, natural number m is a parameter which will be specified in proofs below Further, suppose that l = 3 If we choose i = m, i 2 = m and i 3 = N with 2 m N, then c i 0 for all i If we take i =, i 2 = m and i 3 = m, where 3 m N, then c i 0 for every i For l = 4, putting i = m, i 2 = m, i 3 = N and i 4 = N yields that c i 0 for all i If l = 4 and i =, i 2 = m, i 3 = m and i 4 = N, 4 m N, then c i 0 for every i And so on We will only use this way in the sequel to choose i,i 2,,i l When we know i,i 2,,i l, coefficients a j may be found as solutions of the following linear system: l a j f jik =, k =,2,,l j= Taking into account that f ki may differ from i k, we have to make certain that the choice of i,i 2,,i l yields desired inequalities for c i If we construct a lower bound, then we have to check that c i 0 for all i If we deal with an upper bound, then we have to verify that c i 0 for all i By Theorem, we get either inequality 3), or inequality 4) Since indices i k depend on m, we make an optimization over m Note that in the case f ki = i k, we get l c i = i ) l = a k i k i k k= for all i It means that a j are coefficients in the decomposition of c i over degrees of i and all i,i 2,,i l are in this decomposition Unfortunately, we can not follow this pattern to find a j in general case Indeed, if f ki = i γ k, γk > 0, for example, we could put l ) ) γk γ i k c i = k= In this case we will not obtain a desired decomposition with all i γ,i γ 2,,i γ l A variant with γ k instead of γ k γ k yields the same problem as well The above reasons lead us to the following result Corollary Assume that for some i < i 2 < < i l N, coefficients a j are solutions of the following linear system: l a j f jik =, k =,2,,l 6) Put j= c i = i k k= l a j f ji 7) j= 5

6 for all i N Assume that the numbers r i,r i 2,,r i l are solutions of the linear system 5) and r i = 0 for all i i k, i N If c i 0 for all i =,2,,N, then R R = r i +r i 2 + +r i l If c i 0 for all i =,2,,N, then R R Proof By 7), we have 2) If c i 0 for all i =,2,,N, then inequality 3) holds By definition of r,r 2,,r N, we get N s k = f ki ri = s k, k =,2,,l The last equality follows from 5) It yields that R l a i s i By 6), we have c ik = 0 for all k =,2,,l Applying Theorem to r,r 2,,r N, we conclude that l R = a i s i R The case c i 0 for all i =,2,,N may be considered in the same way We choose {f ik } such that results will be more simple To this end, in the sequel, we put f ik = i a+k ) for all i N and k l, where a > 0 and > 0 Then relation ) turns to s k = i a+k ) r i, k l 8) For the case a = =, we will use a special notation s k = i k r i, k l 9) We start with the case l = 2 Our first result is a lower bound for R Theorem 2 Define s and s 2 by 8) Put δ = s 2 / s ) /, θ = δ [ δ] and θ = δ ρ δ θ) )/ δ + θ) δ θ) ) [0,), where [ ] denotes the integer part of the number in brackets Here and in the sequel, we suggest that 0/0 = 0 Then R s / θ s a+ )/ ) a θ) s / a+ )/ θ) s ) a 0) s / 2 θ s / 6

7 Note that if s = 0, then r i = 0 for all i, s 2 = δ = θ = θ = 0 and 0) is trivial If s > 0, then δ Moreover, if δ =, then r i = 0 for all i 2, s 2 = s = r, θ = θ = 0 and 0) turns to R r Proof For every natural m, 2 m N, put i = m and i 2 = m By 6) and 7), we have c i = a i a a 2 i a+ for i =,2,,N, where a and a 2 satisfy the following linear system: It follows that m ) a a +m ) a+ a 2 =, m a a +m a+ a 2 = a = m a+ m ) a+ m a+ m ) a m a m ) a+, a m a m ) a 2 = m a+ m ) a m a m ) a+, and c i = m a+ m ) a+ m a m ) a m a+ m ) a m a m ) a+ ia + m a+ m ) a+ ia+ a m a m ) for all i =,2,,N Now we check that c i 0 for all i Consider function fx) = a x a a 2 x a+ for real x 0 We have f0) =, fm ) = fm) = 0, f+ ) = + and f x) = x a aa + a + )a 2 x ) It is clear that there exists a unique solution of equation f x) = 0 It follows that function fx) takes its minimum at x 0 m,m) and fx) 0 for x m,m) and fx) 0 otherwise Hence, c i = fi) 0 for all i =,2,,N Linear system 5) is Solving this system, we get r m = m ) a r m +m a r m = s, m ) a+ r m +m a+ r m = s 2, s m a+ s 2 m a m a+ m ) a m a m ) a+, r m = s 2m ) a s m ) a+ m a+ m ) a m a m ) a+ Inequalities r m 0 and r m 0 imply that s 2/ s ) / m + s 2 / s ) /, which coincides with δ m + δ This inequality defines m uniquely for non-integer δ If δ is integer then there are two variants Anyone of them maybe excluded Hence, without loss of generality, we assume in the sequel that δ < m + δ Then m = +[ δ] = + δ θ, 7

8 provided δ < N It follows in this case that θ = δ m ) m m ) [0,) Taking into account that s 2 N s, we see that δ N Hence, we finally put m = min{+[ δ],n} N Assume that δ < N Then m = +[ δ] and m ) a = s a/ s / 2 θ s / ) a It follows that Similarly, This implies that r m = s m δ ) m ) a m m ) ) m a = s a/ a+ )/ θ) s = ) a s / 2 θ s / ) a s / 2 + θ) s / r m = s δ m ) ) m a m ρ m ) ρ ) = s / θ s a+ )/ ) a 2 + θ) s / By Corollary, we have R r m + r m Substituting of r m and r m in the last inequality yields inequality 0) Assume that δ = N Then N 0 = s 2 N s = i N )i a r i, where i N )i a < 0 for i N The latter implies that r = r 2 = = r N = 0 Moreover, δ = N yields that θ = θ = 0 Hence inequality 0) turns to inequality R r N which holds obviously Theorem 2 yields more simple bounds as follows Corollary 2 Define s and s 2 by 8) If, then If <, then R sa+ )/ R ) s a/ 2 θ θ s a+ )/ 2) s a/ 2 8

9 Proof Put gθ, θ) = s / θ s a+ )/ ) a θ) s / a+ )/ θ) s ) a s / 2 θ s / We have gθ, θ) = gθ,θ)+ θ θ) s / s a+ )/ ) a 2 + θ) s / s / If θ θ, then gθ, θ) gθ,θ) It follows from inequality 0) that s a+ )/ 2 θ s / ) a R s / θ s a+ )/ θ) sa+ )/ ) a + ) a 2 + θ) s / s / 2 θ s / The right-hand side of the last inequality takes its minimum over θ for θ = 0 Hence, inequality 0) implies inequality ) provided θ θ Inequality θ θ is equivalent to δ θ) δ θ) + θ δ θ + ) The latter inequality holds when x is a convex function of x, ie for For <, x is a concave function of x and, therefore, θ > θ So, we have proved inequality ) To check inequality 2), we note that for θ > θ, } { θ θ θ gθ, θ) min, gθ,θ) = θ θ θ gθ,θ) Inequality 2) now follows from 0) Theorem 2 and Corollary 2 yield the following result in the case a = = Corollary 3 Define s and s 2 by 9) Put δ = s 2 /s and θ = δ [δ] The following inequality holds: R θs 2 s 2 + θ)s + θ)s2 s 2 θs s2 s 2 Note that the middle part of the last inequality takes its minimum over θ for θ = 0 Now we turn to upper bounds for l = 2 Our next result is as follows Theorem 3 Define s and s 2 by 8) The following inequality holds: R Na+ N a+ N a s Na N a+ N a s 2 3) 9

10 Proof Take i = and i 2 = N By 6) and 7), we have c i = a i a a 2 i a+ for all i =,2,,N, where a and a 2 are such that Hence and a +a 2 =, N a a +N a+ a 2 = a = Na+ N a+ N a, a 2 = Na N a+ N a, Na+ c i = + Na aia+ N a+ N aia N a+ N for all i =,2,,N Let us check that c i 0 for all i Consider again function fx) = a x a a 2 x a+ for real x 0 We have f0) =, f) = fn) = 0, f+ ) = + and f x) = x a aa + a + )a 2 x ) We see that there exists a unique solution x 0 of equation f x) = 0 It follows that function fx) takes its minimum at x 0,N), fx) 0 for x,n) and fx) 0 otherwise Hence, c i = fi) 0 for all i =,2,,N Therefore Theorem yields inequality 3) Theorem 3 implies the following result for a = = Corollary 4 Define s and s 2 by 9) The following inequality holds: R N + N s N s 2 Comparing Theorems 2 and 3, we see that lower bounds seem more interesting for l = 2 The situation will change in the case l = 3, to which we turn now We start again with a lower bound for R Theorem 4 Define s, s 2 and s 3 by 8) Put δ = N s s 2, δ 2 = N s 2 s 3, δ = δ 2 / δ ) /, θ = δ [ δ] and θ = δ δ θ) )/ δ+ θ) δ θ) ) [0,) The following inequality holds: R δ θ)n a δ θ) a ) N a δ θ) a N δ θ) ) + δ θn a δ θ+) a ) N a δ θ+) a N δ θ +) ) + s Na 4) Note that if δ = 0, then r = r 2 = = r N = 0, δ = δ 2 = θ = θ = 0 and 4) turns to R r N that holds obviously If δ > 0, then δ in view of N N δ = N i ρ )i a r i N i ρ )i a+ r i = δ 2 0

11 The latter also yields that δ = only if r 2 = = r N = 0 In the last case, δ = δ 2 = N )r, θ = θ = 0 and 4) turns to R r +r N It follows that we may assume that δ > in the sequel Proof Take natural m, 2 m N, and put i = m, i 2 = m, i 3 = N By 6) and 7), we get c i = a i a a 2 i a+ a 3 i a+2, where a, a 2 and a 3 are determined by the following system of linear equations Then we have a = m ) a a +m ) a+ a 2 +m ) a+2 a 3 =, m a a +m a+ a 2 +m a+2 a 3 =, N a a +N a+ a 2 +N a+2 a 3 = ma+ N a+ N m ) m ) a+ N a+ N m ) )+m a+ m ) a+ m m ) ), a 2 = ma N a N 2 m 2 ) m ) a N a N 2 m ) 2 )+m a m ) a m 2 m ) 2 ), a 3 = ma N a N m ) m ) a N a N m ) )+m a m ) a m m ) ), where = N a m a m ) a m m ) )N m ) )N m ) Considering function fx) = a x a a 2 x a+ a 3 x a+2, one can check that c i 0 for all i =,2,,N Solving linear system 5), that is we get m ) a r m +m a r m +N a r N = s, m ) a+ r m +m a+ r m +N a+ r N = s 2, m ) a+2 r m +m a+2 r m +N a+2 r N = s 3, r m = ma N a N m ) m N s N +m ) s 2 + s 3, r m = m ) a N a N m ) ) m ) N s N +m ) ) s 2 + s 3, r N = m )a m a m m ) ) m ) m s m +m ) ) s 2 + s 3 Making use of inequalities rm 0 and r m 0, we conclude that ) / ) / δ2 δ2 m + δ δ

12 The latter is equivalent to δ m + δ By the same reason as in the proof of Theorem 2, we assume that δ < m + δ Taking into account that δ 2 N ) δ, we obtain δ N Hence, we put m = min{+[ δ],n } It is not difficult to check that rm = δ θ) δ θ) a N δ θ) ), δ θ rm = δ θ+) a N δ θ+) ), rn = s N δ θ a N a N δ θ+) ) + θ N δ θ) ) Corollary implies that R rm +r m +r N Substituting the formulae for r m, r m and rn in the last inequality, we arrive at 4) Theorem 4 implies the next result Corollary 5 Under notations of Theorem 4, if a, then If a, then R δ θ)n a δ a ) N a δ θ) a N δ ) + δ θn a δ +) a ) N a δ θ+) a N δ +) ) + s Na 5) R δ θ)n a δ ) a ) N a δ θ) a N δ ) ) + δ θn a δ a ) N a δ θ+) a N δ ) + s Na 6) Proof We need the following technical result Lemma If either 0 < u < v <, or < u < v, then fx) = ux v x, x > 0, is a decreasing function We omit the proof of Lemma Put u = δ θ)/n and v = δ/n Since δ 2 N ) δ, we have v < If a <, then by Lemma, u a u v > v, a which is equivalent to N a δ θ) a N δ θ) N a δ > a N δ We will have an opposite inequality for a > It follows that N a δ θ) a N δ θ) 2 )

13 take its minimum over θ for θ = 0, if a, and for θ =, if a Making use of Lemma, one can check that the same holds true for N a δ θ+) a N δ θ+) Now Corollary 5 follows from the latter and Theorem 4 For, inequalities 5) and 6) imply more simple bounds Corollary 6 Assume that Then θ θ and one can replace θ by θ in 5) and 6) Moreover, if a in addition, then If a in addition, then R δ N a δ a ) N a δ a N δ ) + s N a R δ N a δ ) a ) N a δ a N δ ) ) + s N a Proof One can check that θ θ and one can put θ = θ = 0 in 5) and 6) in the same way as in the proof of Corollary 2 The next result is an upper bound for R Theorem 5 Define s, s 2 and s 3 by 8) Put ˆδ = s 2 s, ˆδ 2 = s 3 s 2, ˆδ = ˆδ 2 /ˆδ ) /, θ = ˆδ [ˆδ] and ˆθ = ˆδ ˆδ θ) )/ˆδ + θ) ˆδ θ) ) [0,) The following inequality holds: R s ˆδ ˆθ)ˆδ θ) a ) ˆδ θ) a ˆδ θ) ) ˆδˆθˆδ θ +) a ) ˆδ θ+) a ˆδ θ+) ) 7) Note that if ˆδ = 0, then then r i = 0 for all i 2, ˆδ 2 = ˆδ = θ = ˆθ = 0, s = r and 7) holds If ˆδ > 0, then taking into account that ˆδ 2 = s 3 s 2 = i a+ i )r i 2 i=2 i a i )r i = 2 s 2 s ) = 2 ˆδ, i=2 we arrive at ˆδ 2 Proof Take natural m, 3 m N, and put i =, i 2 = m, i 3 = m By 6) and 7), c i = a i a a 2 i a+ a 3 i a+2, where a, a 2 and a 3 are solutions of linear system a +a 2 +a 3 =, m ) a a +m ) a+ a 2 +m ) a+2 a 3 =, m a a +m a+ a 2 +m a+2 a 3 = 3

14 Then we have a = ma+2 )m ) a+ ) m a+ )m ) a+2 ), a 2 = ma+2 )m ) a ) m a )m ) a+2 ), a 3 = ma+ )m ) a ) m a )m ) a+ ), where = m a m ) a m m ) )m ) )m ) Considering function fx) = a x a a 2 x a+ a 3 x a+2, one can check that c i 0 for all i =,2,,N Linear system 5) is as follows We have r +m ) a r m +m a r m = s, r +m ) a+ r m +m a+ r m = s 2, r +m ) a+2 r m +m a+2 r m = s 3, r = ma m ) a m m ) ) m m ) s m +m ) ) s 2 + s 3, r m = ma m ) m s m +) s 2 + s 3, r m = m ) a m ) ) m ) s m ) +) s 2 + s 3 Again making use of rm 0 and r m 0, we get ) / ˆδ2 ˆδ2 m + ) / ˆδ ˆδ This inequality is equivalent to ˆδ m ˆδ + By the same reason as in the proof of Theorem 2, we may assume that ˆδ < m ˆδ + Remember that ˆδ 2 It follows that we can put m = min{+[ˆδ],n} It is not difficult to check that r = s ˆδ ˆθ ) ˆδ θ) + ˆθ, ˆδ + θ) ˆδ rm = ˆθ) ˆδ θ) a ˆδ θ) ), r m = ˆδˆθ ˆδ + θ) a ˆδ + θ) ) 4

15 It follows from Corollary that R r +r m +r m Substituting of r, r m and r m in the latter inequality yields 7) Theorem 5 gives simpler bounds as well Corollary 7 Under notations of Theorem 5, if a, then If a, then R s ˆδ ˆθ)ˆδ a ) ˆδ θ) a ˆδ ) ˆδˆθˆδ +) a ) ˆδ θ+) a ˆδ +) ) R s ˆδ ˆθ)ˆδ ) a ) ˆδ θ) a ˆδ ) ) ˆδˆθˆδa ) ˆδ θ+) a ˆδ ) For, Corollary 7 implies the next result Corollary 8 Assume that If a, then If a, then R s ˆδ ˆδ a ) ˆδ a ˆδ ) R s ˆδ ˆδ ) a ) ˆδ a ˆδ ) ) Proofs of Corollaries 7 and 8 follow the same pattern as those of Corollaries 5 and 6 We omit details Remark The right-hand side of 7) may be obtained from the right-hand side of 4) by a formal replacement N =, δ = ˆδ, δ 2 = ˆδ 2, δ = ˆδ and θ = ˆθ Remark 2 All inequalities of Theorems 2 5 are sharp For each of these inequalities, there exists a set of numbers r,r 2,,r N such that the inequality turns to equality 3 Bounds for probabilities of unions In this section, we discuss bounds for probabilities of unions of events which follow from the results of section 2 Note that these bounds maybe applied to measures of unions of sets in arbitrary measurable spaces Let Ω,F,P) be a probability space For events A,A 2,,A N, put U = N A i Denote B i = {ω Ω : ω belongs to exactly i events of A,A 2,,A N } and p i = PB i ), i = 0,,,N Then PU) = p i 5

16 The simplest application of the above method is to put r i = p i for i =,2,,N Then the general results of the previous sections yield Theorems 2 5 and Corollaries and 2 in Frolov 202) and Theorem 3 and 4 and Corollaries 5 in Frolov 204) that are generalizations of earlier results Note that one may also consider more general events that maybe represented by sums p i +p i2 + +p im where i < i 2 < < i M, M N For example, sum p t +p t+ + +p N equals to probability that at least t events from A,A 2,,A N occur This requires a modification of the above method and will be done elsewhere We now turn to another representations of PU) which is a start point of our method as well By Lemma in Kuai, Alajaji and Takahara 2000), we have PU) = k= i p ik, where p ik = PB i A k ) We also give a simple proof of the last equality Putting ξ N = I A +I A2 + +I AN, we get ξ N I Bi = ii Bi and k= i p N ik = E k= For every fixed k, putting r ik = p ik /i and ) N ) i I N ) B i I Ak = E i I B i ξ N = E I Bi = PU) R k = r ik, we can take bounds for R k from our general results Denote s j k) = s j k) = i a+j ) r ik, 8) i j r ik, 9) for j l and k =,2,,N An application of Theorem 2 yields the next result Theorem 6 Define s k) and s 2 k) by 8), k =,2,,N Put δ k = s 2 k)/ s k)) /, θ k = δ k [ δ k ] and θ k = δ k δ k θ k ) )/ δ k + θ k ) δ k θ k ) ) [0,), where k =,2,,N 6

17 Then PU) k= θ k s a+ )/ k) s / 2 k)+ θ k ) s / k) ) a + θ k ) s a+ )/ k) s / 2 k) θ k s / k) ) a 20) For a = =, Theorem 6 implies Theorem in Kuai, Alajaji and Takahara 2000) By Corollary 3, we may put θ k = θ k = 0 in 20) and obtain a result in de Caen 997) It is clear that one can use all results from section 2 to derive upper and lower bounds similar to that of Theorem 6 Theorem 4 implies the following result Theorem 7 Define s k), s 2 k) and s 3 k) by 8), k =,2,,N Put δ k = N s k) s 2 k), δ 2k = N s 2 k) s 3 k), δ k = δ 2k / δ k ) /, θ k = δ k [ δ k ] and θ k = δ k δ k θ k ) )/ δ k + θ k ) δ k θ k ) ) [0,), k =,2,,N The following inequality holds: { δk θ k )N a δ k θ k ) a ) PU) N a δ k θ k ) a N δ k θ k ) ) + δk θk N a δ k θ k +) a ) N a δ k θ k +) a N δ k θ k +) ) + s } k) N a k= For a = =, we obtain the next result from Corollary 6 Corollary 9 Define s k), s 2 k) and s 3 k) by 9) and put δ k = Ns k) s 2 k), δ 2k = Ns 2 k) s 3 k) for k =,2,,N The following inequality holds: PU) } { δ2 k +s k) 2) N δ 2k Note that for all k =,2,,N, we have N ) s k) = p ik = E I Bi I Ak = EI U I Ak ) = PA k ), s 2 k) = s 3 k) = k= N ) ip ik = E ii Bi I Ak = Eξ N I Ak ) = PA i A k ), N ) i 2 p ik = E i 2 I Bi I Ak = E ξn 2 I ) N A k = It follows that δ k = PA i A k )=EN ξ N )I Ak ), δ 2k = PA i A j A k ) j= PA i A j A k )=Eξ N N ξ N )I Ak ), 22) for all k =,2,,N Now we turn to upper bounds The next result follows from Theorem 5 7 j=

18 Theorem 8 Define s k), s 2 k) and s 3 k) by 8), k =,2,,N Put ˆδ k = s 2 k) s k), ˆδ 2k = s 3 k) s 2 k), ˆδ k = ˆδ 2k /ˆδ k ) /, θ k = ˆδ k [ˆδ k ] and ˆθ k = ˆδ k ˆδ k θ k ) )/ˆδ k + θ k ) ˆδ k θ k ) ) [0,), k =,2,,N The following inequality holds: { PU) s k) ˆδ k ˆθ } k )ˆδ k θ k ) a ) ˆδ k θ k ) a ˆδ k θ k ) ) ˆδkˆθ k ˆδ k θ k +) a ) ˆδ k θ k +) a ˆδ k θ k +) ) k= For a = =, we obtain the next result from Corollary 8 Corollary 0 Define s k), s 2 k) and s 3 k) by 9) and put ˆδ k = s 2 k) s k), ˆδ 2k = s 3 k) s 2 k) for k =,2,,N The following inequality holds: { } Note that ˆδ k = ˆδ 2k = PU) k= s k) ˆδ 2 k ˆδ 2k 23) PA i A k ) PA k ) = Eξ N )I Ak, 24) PA i A j A k ) j= PA i A k ) = Eξ N ξ N )I Ak 25) We finally mention that Theorems 6 8 and Corollaries 9 and 0 are new result 4 Borel Cantelli lemmas Let Ω,F,P) be a probability space and {A n } be a sequence of events Denote For m m, put U mn = n k=m {A n io} = limsupa n = A k Since n= k=n A k PA n io) = lim m lim n PU mn), every new upper or lower bound allows us to derive new variant of first or second part of the Borel Cantelli Lemma Our results of the previous section imply that Qm,n) PU mn ) Q m,n), 8

19 where Qm,n) and Q m,n) are the right-hand sides of the applied lower and upper bound, correspondingly It is clear that lim sup m lim supqm,n) PA n io) liminf liminf n m n Q m,n) It may happen that we cannot find these double limits But, if for every fixed m the inequality lim supqm,n) limsupq, n) n n holds, then we have PA n io) limsupq,n) n Similarly, if for every fixed m the inequality lim inf n Q m,n) liminf n Q,n) holds, then we get PA n io) liminf n Q,n) The most applicable variants of the Borel Cantelli lemma are proved by this way In the proof of the first part of the classical Borel Cantelli lemma, the upper bound for PU mn ) by s is used In the Erdős Rényi generalization of the second part of the Borel Cantelli lemma, the inequality with s and s 2 is applied Frolov 202) has applied the lower bound for PU mn ), based on s, s 2 and s 3 This yielded a generalization of the second part of the Borel Cantelli lemma in Theorem 9 of the last paper Note that mentioned here bounds are constructed for probabilities r i = p i In this section, we present new variants of the Borel Cantelli lemma based on inequalities 2) and 23) Note that the last inequalities are constructed from bounds for numbers r ik = p ik /i We start with the second part of the Borel Cantelli lemma Theorem 9 Denote ξ n = I A +I A2 + +I An and η n = n ξ n for all natural n Assume that Eη n I Ak 0 as n n Eη n ξ n I Ak Then It follows from 22) that k= PA n io) limsup n Eη n I Ak = n {PA k )+ Eη } ni Ak ) 2 Eη n ξ n I Ak k= P ) A i A k, Eηn ξ n I Ak = 9 P ) A i A j A k j=

20 Proof Inequality 2) and relation 22) yield that n ) P A k {PA k )+T k m,n)}, n m+ where We have k=m T k m,n) = k=m Eη n η m )I Ak ) 2 Eη n η m )ξ n ξ m )I Ak T k m,n) Eη n η m )I Ak ) 2 Eη n ξ n I Ak = Eη ni Ak ) 2 2Eη n η m I Ak +Eη m I Ak ) 2 Eη n ξ n I Ak Eη ni Ak ) 2 2m )Eη n I Ak Eη n ξ n I Ak By 2), the inequality n T k,n) = n k= holds for all natural n It implies that Hence m k= k= Eη n I Ak ) 2 Eη n ξ n I Ak m Eη n I Ak ) 2 Eη n ξ n I Ak n ) P A k {PA k )+ Eη ni Ak ) 2 2m )Eη } ni Ak n Eη n ξ n I Ak Eη n ξ n I Ak k=m k=m {PA k )+ Eη } ni Ak ) 2 2m n Eη n ξ n I Ak n 2m ) Eη n I Ak n Eη n ξ n I Ak k= This yields that for every fixed m, ) P A k limsup n n k=m k= k= {PA k )+ Eη } ni Ak ) 2 Eη n ξ n I Ak The last inequality implies the desired assertion Theorem 9 in Frolov 202) contains a lower bound for PA n io) constructed from p i There is an example in Frolov 202) which shows that this lower bound is better than previous ones One can check that for this example, the lower bounds of Theorem 9 in Frolov 202) and Theorem 9 of this section coincide Now we turn to the first part of the Borel Cantelli lemma 20

21 Theorem 0 Denote ξ n = I A +I A2 + +I An for all natural n and ξ 0 = 0 If k=m for all sufficiently large m, then PA n io) limsup m Eξ n ξ m )I Ak Eξ n ξ m ) 2 I Ak 0 as n 26) lim sup n {PA k ) Eξ } n ξ m )I Ak ) 2 Eξ n ξ m ) 2 I Ak k=m If condition 26) holds for m =, then PA n io) limsup m It follows from 24) and 25) that lim sup n {PA k ) Eξ } ni Ak ) 2 Eξn 2I A k k=m Eξ n I Ak = PA i A k ), EξnI 2 Ak = PA i A j A k ) j= where Proof Inequality 23) and assertions 24) and 25) imply that n ) P A k {PA k ) T k m,n)}, We have k=m k=m T k m,n) = Eξ n ξ m )I Ak ) 2 Eξ n ξ m )ξ n ξ m )I Ak T k m,n) Eξ n ξ m )I Ak ) 2 Eξ n ξ m ) 2 I Ak Eξ n ξ m )I Ak ) 2 2Eξ n ξ m )I Ak Eξ n ξ m ) 2 I Ak Eξ ni Ak ) 2 2Eξ n ξ m I Ak 2Eξ n I Ak Eξ 2 ni Ak Eξ ni Ak ) 2 2mEξ n I Ak Eξ 2 ni Ak It yields that n ) P A k {PA k ) Eξ } n ξ m )I Ak ) 2 +2 Eξ n ξ m ) 2 I Ak k=m k=m {PA k ) Eξ } ni Ak ) 2 Eξ n I Ak +2m EξnI 2 Ak EξnI 2 Ak k=m k= 2 k=m Eξ n ξ m )I Ak Eξ n ξ m ) 2 I Ak

22 Theorem 0 follows from the latter Theorem 0 generalizes the first part of the classic Borel Cantelli lemma If {A n } are independent and series PA n ) diverges, then by Theorem 0, the upper bound is n= So, the bound is sharp in this case References Andel J, Dupas V, 989 An extension of the Borel lemma Comment Math Univ Carolin 30, Boros E, Prékopa A, 989 Closed form two-sided bounds for probabilities that at least r and exactly r out of n events occurs Math Oper Research 4, de Caen D, 997 A lower bound on the probability of a union Discrete Math 69, Chandra TK, 202 The Borel-Cantelli lemma Springer, Heidelberg Chung KL, Erdős P, 952 On the application of the Borel-Cantelli lemma Trans Amer Math Soc 72, Dawson DA, Sankoff D, 967 An inequality for probabilities Proc Amer Math Soc 8, Erdős P, Rényi A, 959 On Cantor s series with convergent /q, Ann Univ Sci Budapest Sect Math 2, Feng C, Li L, Shen J, 2009 On the Borel Cantelli lemma and its generalization Comptes Rendus Math 347, Frolov AN, 202 Bounds for probabilities of unions of events and the Borel Cantelli lemma Statist Probab Lett 82, Frolov AN, 204 On inequalities for probabilities of unions of events and the Borel Cantelli lemma Vestnik Sankt-Peterburgskogo Universiteta, Seriya Matematika, Mekhanika, Astronomiya, N 2, 2 30 In Russian) English translation: Vestnik StPetersburg University, Mathematics, 204, N 2, Allerton Press, Inc Galambos J, Simonelli I, 996 Bonferroni-type inequalities with applications Springer-Verlag NY Gallot S 966 A bound for the maximum of a number of random variables J Appl Probab 3, Kochen S, Stone C, 964 A note on the Borel-Cantelli lemma Illinois J Math 8, Kounias EG, 968 Bounds for the probability of a union, with applications Ann Math Statist 39, Kwerel SM, 975 Bounds on the probability of the union and intersection of m events Adv Appl Probab 7, Kuai H, Alajaji F, Takahara G, 2000 A lower bound on the probability of a finite union of events Discrete Math 25, Móri TF, Székely GJ, 983 On the Erdős Rényi generalization of the Borel Cantelli lemma Studia Sci Math Hungar 8, Martikainen AI, Petrov VV, 990 On the Borel-Cantelli lemma Zapiski Nauch Semin LOMI 84, in Russian) English translation in: J Math Sci 994, 63, Petrov VV, 2002 A note on the Borel Cantelli lemma, Statist Probab Lett 58, Prékopa A, 2009 Inequalities for discrete higher order convex functions J Math Inequalities 4, Spitzer F, 964 Principles of random walk Van Nostrand, Princeton Xie YQ, 2008 A bilateral inequality for the Borel Cantelli lemma Statist Probab Lett 78,

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