Exact Computation of Minimum Sample size for Estimation of Poisson Parameters

Exact Computation of Minim um Sample size f or Estimation of P oisson P a ramete rs ∗ Xinjia Chen July 2007 Abstract In this pap er, we dev elop an approa ch f o r the exa ct determination of the minimu m sa mple size for the estimation of a Poisson par ameter with prescrib ed margin of error a nd confidence level. The exact computation is made p ossible by reducing infinite many ev aluations of cov er- age probability to fi nite man y ev aluations. Such reduction is based on o ur discov ery that the minim um of cov erag e probability with resp ect to a Poisson parameter bo unded in an in terv al is attained at a discr ete set of finite many v alues. 1 In tro du ction The estimation of a Poisson parameter fi n ds numerous ap p lications in v ario us fields of sciences and engineering [3 ]. The problem is formulat ed as follo ws. Let X b e a P oisson random v ariable defined in a probabilit y space (Ω , F , Pr) such th at Pr { X = k } = λ k e − λ k ! , k = 0 , 1 , · · · , wh ere λ > 0 is referred to as a Poisson parameter. It is a frequent problem to estimate λ based on n ident ical and indep enden t samples X 1 , · · · , X n of X . An estimate of λ is conv en tionally tak en as b λ n = P n i =1 X i n . The nice prop erty of such estimate is that it is of maxim um lik ely-ho o d and p ossesses minim um v ariance among all u nbia sed estimates. A crucial qu estion in the estimation is as follo ws: Given the know le d ge that λ b elongs to interval [ a, b ] , what is the minimum sample size n that guar ant e es the diffe r enc e b etwe en b λ n and λ b e b ounde d within so me pr escrib e d mar gin of e rr or with a c onfidenc e level higher than a pr escrib e d value? The main con tribu tion of this pap er is to pro vide exact ans w er to this imp ortant qu estion. T he pap er is organized as follo ws. In Section 2, the tec hn iques f or computing the min im um sample size is dev elop ed with the m argin of er r or taken as a b ound of absolute error. In Section 3, we ∗ The au thor had b een previously working with Lou isiana S tate Un ivers ity at Baton Rouge, LA 70803, USA , and is now with D epartment of Electrical Engineering, Sou thern Universit y and A&M College, Baton Rouge, LA 70813, USA ; Email: chenxinjia@gmail.co m 1 deriv e corresp ondin g samp le size metho d b y u sing relativ e error b ound as the margin of error. In Section 4, we dev elop tec hniques for computing minimum sample size with a mixed error criterion. Section 5 is the conclusion. The pro ofs are giv en in Ap p end ices. Throughout th is pap er, w e sh all use th e follo wing notations. The set of in tegers is denoted b y Z . Th e ceiling fun ction and flo or function are d enoted resp ectiv ely b y ⌈ . ⌉ and ⌊ . ⌋ (i.e., ⌈ x ⌉ represent s the smallest int eger n o less than x ; ⌊ x ⌋ represents the largest int eger no greater than x ). The m ultiv ariate function S ( n, k , l, λ ) means S ( n, k, l , λ ) = P l i = k λ i e − λ i ! . The left limit as η tends to 0 is denoted as lim η ↓ 0 . Th e other notations will b e made clear as w e pro ceed. 2 Con trol of Absolute Error Let ε ∈ (0 , 1) b e the margin of absolute err or and δ ∈ (0 , 1) b e the confidence parameter. In many applications, it is desirable to fi n d the minimum sample size n suc h th at Pr n | b λ n − λ | < ε o > 1 − δ for an y λ ∈ [ a, b ]. He re Pr n | b λ n − λ | < ε o is referr ed to as the co v erage probabilit y . The interv al [ a, b ] is introdu ced to tak e in to accoun t the kno wledge of λ . Th e exact determination of m inim um sample size is r eadily tractable with mo dern compu tational p o we r by taking adv an tage of the b ehavio r of the co verag e p robabilit y c haracterized by Theorem 1 as follo ws. Theorem 1 L e t 0 < ε < 1 and 0 ≤ a < b . L et X 1 , · · · , X n b e identic al and indep endent Poisson r andom variables with me an λ ∈ [ a, b ] . L et b λ n = P n i =1 X i n . Then, the minimum of Pr {| b λ n − λ | < ε } with r esp e ct to λ ∈ [ a, b ] is achieve d at the finite set { a, b } ∪ { ℓ n + ε ∈ ( a, b ) : ℓ ∈ Z } ∪ { ℓ n − ε ∈ ( a, b ) : ℓ ∈ Z } , which has less than 2 n ( b − a ) + 4 elements. See App endix A f or a pr o of. The ap p lication of Theorem 1 in the computation of minimum sample size is ob vious. F or a fixed sample size n , since the minim um of cov erage probabilit y with λ ∈ [ a, b ] is attained at a finite set, it can determined b y a computer wh ether the sample size n is large en ou gh to ens ure Pr n | b λ n − λ | < ε o > 1 − δ for an y λ ∈ [ a, b ]. Starting from n = 2, one can find the min imum sample size by gradually increment ing n and chec king wh ether n is large enough. 3 Con trol of Rela tiv e Error Let ε ∈ (0 , 1) b e the margin of relativ e error and δ ∈ (0 , 1) b e the confiden ce parameter. It is in teresting to determine the minim um sample size n so th at Pr (      b λ n − λ λ      < ε ) > 1 − δ 2 for any λ ∈ [ a, b ]. As has b een p oin ted out in Section 2, an essential machinery is to r educe in fi nite man y ev aluations of the co v erage p robabilit y Pr {| b λ n − λ | < ελ } to finite m any ev aluations. Such reduction can b e accomplished by making u se of Theorem 2 as follo ws. Theorem 2 L e t 0 < ε < 1 and 0 < a < b . L et X 1 , · · · , X n b e identic al and indep endent Poisson r andom variables with me an λ ∈ [ a, b ] . L et b λ n = P n i =1 X i n . Then, the minimum of Pr n | b λ n − λ | λ < ε o with r esp e ct to λ ∈ [ a, b ] is achieve d at the finite set { a, b } ∪ { ℓ n (1+ ε ) ∈ ( a, b ) : ℓ ∈ Z } ∪ { ℓ n (1 − ε ) ∈ ( a, b ) : ℓ ∈ Z } , which has less than 2 n ( b − a ) + 4 elements. See App endix B f or a pro of. 4 Con trol of Absolute Error or Relativ e Error Let ε a ∈ (0 , 1) and ε r ∈ (0 , 1) b e resp ectiv ely th e margins of absolute error and relativ e error. Let δ ∈ (0 , 1) b e the confid ence parameter. In many situ ations, it is desirable to fi nd the smallest sample size n suc h that Pr ( | b λ n − λ | < ε a or      b λ n − λ λ      < ε r ) > 1 − δ (1) for any λ ∈ [ a, b ]. T o make it p ossible to compu te exactly th e minim u m s ample size asso ciated with (1), we h a v e Theorem 3 as follo w s. Theorem 3 L e t 0 < ε a < 1 , 0 < ε r < 1 and 0 ≤ a < ε a ε r < b . L et X 1 , · · · , X n b e identic al and indep endent Poisson r andom variables with me an λ ∈ [ a, b ] . L et b λ n = P n i =1 X i n . Then, the minimum of Pr n | b λ n − λ | < ε a or    b λ n − λ λ    < ε r o with r esp e ct to λ ∈ [ a, b ] is achieve d at the finite set { a, b, ε a ε r } ∪ { ℓ n + ε a ∈ ( a, ε a ε r ) : ℓ ∈ Z } ∪ { ℓ n − ε a ∈ ( ε a ε r , b ) : ℓ ∈ Z } ∪ { ℓ n (1+ ε r ) ∈ ( a, ε a ε r ) : ℓ ∈ Z } ∪ { ℓ n (1 − ε r ) ∈ ( ε a ε r , b ) : ℓ ∈ Z } , which has less than 2 n ( b − a ) + 7 elements. Theorem 3 can b e sho wn b y applyin g Theorem 1 and Th eorem 2 with the observ ation th at Pr ( | b λ n − λ | < ε a or      b λ n − λ λ      < ε r ) =    Pr n | b λ n − λ | < ε a o for λ ∈ h a, ε a ε r i , Pr n    b λ n − λ λ    < ε r o for λ ∈  ε a ε r , b i . By virtue of Chernoff b oun ds, it can b e sh own th at, for an y ε ∈ (0 , 1), Pr { b λ n ≤ (1 − ε ) λ } <  e − ε (1 − ε ) 1 − ε  nλ < exp  − λnε 2 2  , Pr { b λ n ≥ (1 + ε ) λ } <  e ε (1 + ε ) 1+ ε  nλ < exp  − (2 ln 2 − 1) λnε 2  . 3 As a result, Pr {| b λ n − λ | > ελ } < δ if λ > ln 2 δ (2 ln 2 − 1) nε 2 . Therefore, to c heck whether (1) is satisfied f or an y λ ∈ [ a, b ], it suffices to c hec k (1) for a ≤ λ ≤ min ( b, ln 2 δ (2 ln 2 − 1) nε 2 r ) . Finally , we would like to p oin t out that similar c haracteristics of the co v erage probability can b e sho wn f or the problem of estimating binomial p arameter or th e prop ortion of fin ite p opulation, whic h allo ws for the exact compu tation of minim um sample size. F or details, see our recen t pap ers [1, 2]. 5 Conclusion W e hav e dev elop ed an exact metho d for the computation of min im um sample size for th e estima- tion of P oisson parameters, whic h only requires finite man y ev aluations of the co v erage p robabilit y . Our sample size metho d p ermits rigorous con trol of s tatistica l sampling error. A Pro of of Theorem 1 Define K = P n i =1 X i and C ( λ ) = Pr      K n − λ     < ε  = Pr { g ( λ ) ≤ K ≤ h ( λ ) } where g ( λ ) = max (0 , ⌊ n ( λ − ε ) ⌋ + 1) , h ( λ ) = ⌈ n ( λ + ε ) ⌉ − 1 . It should b e noted that C ( λ ) , g ( λ ) and h ( λ ) are actually m ultiv ariate functions of λ, ε and n . F or simp licit y of n otations, we dr op th e argument s n and ε thr oughout the pro of of Theorem 1. W e need some preliminary results. Lemma 1 L et λ ℓ = ℓ n − ε wher e ℓ ∈ Z . Then, h ( λ ) = h ( λ ℓ +1 ) = ℓ for any λ ∈ ( λ ℓ , λ ℓ +1 ) . Pro of . F or λ ∈ ( λ ℓ , λ ℓ +1 ), we ha ve 0 < n ( λ − λ ℓ ) < 1 and h ( λ ) = ⌈ n ( λ + ε ) ⌉ − 1 = ⌈ n ( λ ℓ + ε + λ − λ ℓ ) ⌉ − 1 =  n  ℓ n − ε + ε + λ − λ ℓ  − 1 = ℓ − 1 + ⌈ n ( λ − λ ℓ ) ⌉ = ℓ =  n  ℓ + 1 n − ε + ε  − 1 = h ( λ ℓ +1 ) . 4 ✷ Lemma 2 L et λ ℓ = ℓ n + ε wher e ℓ ∈ Z . Then, g ( λ ) = g ( λ ℓ ) = max { 0 , ℓ + 1 } for any λ ∈ ( λ ℓ , λ ℓ +1 ) . Pro of . F or λ ∈ ( λ ℓ , λ ℓ +1 ), we ha ve − 1 < n ( λ − λ ℓ +1 ) < 0 and g ( λ ) = max (0 , ⌊ n ( λ − ε ) ⌋ + 1) = max(0 , ⌊ n ( λ ℓ +1 − ε + λ − λ ℓ +1 ) ⌋ + 1) = max  0 ,  n  ℓ + 1 n + ε − ε  + ⌊ n ( λ − λ ℓ +1 ) ⌋ + 1  = max  0 ,  n  ℓ + 1 n + ε − ε  − 1 + 1  = max { 0 , ℓ + 1 } = max  0 ,  n  ℓ n + ε − ε  + 1  = g ( λ ℓ ) . ✷ Lemma 3 L et α < β b e two c onse cutive elements of the asc e nding arr angement of al l distinct elements of { a, b } ∪ { ℓ n + ε ∈ ( a, b ) : ℓ ∈ Z } ∪ { ℓ n − ε ∈ ( a, b ) : ℓ ∈ Z } . Then, b oth g ( λ ) and h ( λ ) ar e c onstants for any λ ∈ ( α, β ) . Pro of . Since α and β are t wo consecutiv e elemen ts of the ascending arr angemen t of all distinct elemen ts of the set, it m u st b e true that ther e is no in teger ℓ suc h that α < ℓ n + ε < β or α < ℓ n − ε < β . It follo ws that there exist t w o int egers ℓ and ℓ ′ suc h that ( α, β ) ⊆  ℓ n + ε, ℓ +1 n + ε  and ( α, β ) ⊆  ℓ ′ n − ε, ℓ ′ +1 n − ε  . Applying L emm a 1 and Lemma 2, we hav e g ( λ ) = g  ℓ n + ε  and h ( λ ) = h  ℓ ′ +1 n − ε  for an y λ ∈ ( α, β ). ✷ Lemma 4 F or any λ ∈ (0 , 1) , lim η ↓ 0 C ( λ + η ) ≥ C ( λ ) and lim η ↓ 0 C ( λ − η ) ≥ C ( λ ) . Pro of . Observing that h ( λ + η ) ≥ h ( λ ) for any η > 0 and th at g ( λ + η ) = max(0 , ⌊ n ( λ + η − ε ) ⌋ + 1) = max(0 , ⌊ n ( λ − ε ) ⌋ + 1 + ⌊ n ( λ − ε ) − ⌊ n ( λ − ε ) ⌋ + nη ⌋ ) = max(0 , ⌊ n ( λ − ε ) ⌋ + 1) = g ( λ ) for 0 < η < 1+ ⌊ n ( λ − ε ) ⌋− n ( λ − ε ) n , w e h a v e S ( n, g ( λ + η ) , h ( λ + η ) , λ + η ) ≥ S ( n, g ( λ ) , h ( λ ) , λ + η ) (2) 5 for 0 < η < 1+ ⌊ n ( λ − ε ) ⌋− n ( λ − ε ) n . Since h ( λ + η ) = ⌈ n ( λ + η + ε ) ⌉ − 1 = ⌈ n ( λ + ε ) ⌉ − 1 + ⌈ n ( λ + ε ) − ⌈ n ( λ + ε ) ⌉ + nη ⌉ , w e h a v e h ( λ + η ) =    ⌈ n ( λ + ε ) ⌉ for n ( λ + ε ) = ⌈ n ( λ + ε ) ⌉ and 0 < η < 1 n , ⌈ n ( λ + ε ) ⌉ − 1 for n ( λ + ε ) 6 = ⌈ n ( λ + ε ) ⌉ and 0 < η < ⌈ n ( λ + ε ) ⌉− n ( λ + ε ) n . It follo ws that b oth g ( λ + η ) and h ( λ + η ) are ind ep endent of η if η > 0 is small enough. Since S ( n, g , h, λ + η ) is con tinuous with r esp ect to η for fi xed g and h , w e h a v e that lim η ↓ 0 S ( n, g ( λ + η ) , h ( λ + η ) , λ + η ) exists. As a result, lim η ↓ 0 C ( λ + η ) = lim η ↓ 0 S ( n, g ( λ + η ) , h ( λ + η ) , λ + η ) ≥ lim η ↓ 0 S ( n, g ( λ ) , h ( λ ) , λ + η ) = S ( n, g ( λ ) , h ( λ ) , λ ) = C ( λ ) , where the inequalit y follo ws from (2). Observing that g ( λ − η ) ≤ g ( λ ) for any η > 0 and that h ( λ − η ) = ⌈ n ( λ − η + ε ) ⌉ − 1 = ⌈ n ( λ + ε ) ⌉ − 1 + ⌈ n ( λ + ε ) − ⌈ n ( λ + ε ) ⌉ − nη ⌉ = ⌈ n ( λ + ε ) ⌉ − 1 = h ( λ ) for 0 < η < 1+ n ( λ + ε ) −⌈ n ( λ + ε ) ⌉ n , w e h a v e S ( n, g ( λ − η ) , h ( λ − η ) , λ − η ) ≥ S ( n, g ( λ ) , h ( λ ) , λ − η ) (3) for 0 < η < min n λ, 1+ n ( λ + ε ) −⌈ n ( λ + ε ) ⌉ n o . Since g ( λ − η ) = max(0 , ⌊ n ( λ − η − ε ) ⌋ + 1) = max(0 , ⌊ n ( λ − ε ) ⌋ + 1 + ⌊ n ( λ − ε ) − ⌊ n ( λ − ε ) ⌋ − nη ⌋ ) , w e h a v e g ( λ − η ) =    max(0 , ⌊ n ( λ − ε ) ⌋ ) for n ( λ − ε ) = ⌊ n ( λ − ε ) ⌋ and 0 < η < 1 n , max(0 , ⌊ n ( λ − ε ) ⌋ + 1 ) for n ( λ − ε ) 6 = ⌊ n ( λ − ε ) ⌋ and 0 < η < n ( λ − ε ) −⌊ n ( λ − ε ) ⌋ n . It follo w s that b oth g ( λ − η ) and h ( λ − η ) are indep enden t of η if η > 0 is sm all enough. Since S ( n, g , h, λ − η ) is con tinuous with r esp ect to η for fi xed g and h , w e h a v e that lim η ↓ 0 S ( n, g ( λ − η ) , h ( λ − η ) , λ − η ) exists. Hence, lim η ↓ 0 C ( λ − η ) = lim η ↓ 0 S ( n, g ( λ − η ) , h ( λ − η ) , λ − η ) ≥ lim η ↓ 0 S ( n, g ( λ ) , h ( λ ) , λ − η ) = S ( n, g ( λ ) , h ( λ ) , λ ) = C ( λ ) , where the inequalit y follo ws from (3). ✷ 6 Lemma 5 L et α < β b e two c onse cutive elements of the asc ending arr angement of al l distinct ele- ments of { a, b } ∪ { ℓ n + ε ∈ ( a, b ) : ℓ ∈ Z } ∪ { ℓ n − ε ∈ ( a, b ) : ℓ ∈ Z } . Then, C ( λ ) ≥ min { C ( α ) , C ( β ) } for any λ ∈ ( α, β ) . Pro of . By Lemma 3, b oth g ( λ ) and h ( λ ) are constan ts for any λ ∈ ( α, β ). Hence, w e can drop the argument and wr ite g ( λ ) = g , h ( λ ) = h and C ( λ ) = S ( n, g , h, λ ). F or λ ∈ ( α, β ), define in terv al [ α + η , β − η ] w ith 0 < η < min  λ − α, β − λ, β − α 2  . Then, C ( λ ) ≥ min µ ∈ [ α + η,β − η ] C ( µ ). Note th at ∂ S ( n, 0 ,l ,λ ) ∂ λ = − λ l e − λ l ! and thus, for g > 0, ∂ S ( n, g , h, λ ) ∂ λ = ∂ S ( n, 0 , h, λ ) ∂ λ − ∂ S ( n, 0 , g − 1 , λ ) ∂ λ = λ g − 1 e − λ ( g − 1)! − λ h e − λ h ! =  h ! ( g − 1)! − λ h − g +1  λ g − 1 e − λ h ! > 0 if λ < h h ! ( g − 1)! i 1 h − g +1 . F rom such inv estigatio n of the deriv ativ e of S ( n, g , h, λ ) with resp ective to λ , we can see that, f or 0 < η < min  λ − α, β − λ, β − α 2  , one of the follo wing thr ee cases must b e true: (1) C ( µ ) decreases monotonically for µ ∈ [ α + η , β − η ]; (2) C ( µ ) in creases monotonically for µ ∈ [ α + η , β − η ]; (3) there exists a n umb er θ ∈ ( α + η , β − η ) s uc h that C ( µ ) increases monotonically for µ ∈ [ α + η , θ ] and d ecreases monotonically for µ ∈ ( θ , β − η ]. It follo ws that C ( λ ) ≥ min µ ∈ [ α + η,β − η ] C ( µ ) = min { C ( α + η ) , C ( β − η ) } for 0 < η < min  λ − α, β − λ, β − α 2  . By Lemma 4, b oth lim η ↓ 0 C ( α + η ) and lim η ↓ 0 C ( β − η ) exist and C ( λ ) ≥ lim η ↓ 0 min { C ( α + η ) , C ( β − η ) } = min  lim η ↓ 0 C ( α + η ) , lim η ↓ 0 C ( β − η )  ≥ min { C ( α ) , C ( β ) } for an y λ ∈ ( α, β ). ✷ Finally , to sho w Theorem 1, note that the statemen t ab out the co verag e p robabilit y follo ws immediately from Lemma 5. The num b er of elemen ts of th e finite set can b e calculated by using the pr op ert y of the ceiling and flo or fu nctions. B Pro of of Theorem 2 Define C ( λ ) = Pr      K n − λ     < ελ  = Pr { g ( λ ) ≤ K ≤ h ( λ ) } 7 where g ( λ ) = ⌊ nλ (1 − ε ) ⌋ + 1 , h ( λ ) = ⌈ nλ (1 + ε ) ⌉ − 1 . It should b e noted that C ( λ ) , g ( λ ) and h ( λ ) are actually m ultiv ariate functions of λ, ε and n . F or simp licit y of n otations, we dr op th e argument s n and ε thr oughout the pro of of Theorem 2. W e need some preliminary results. Lemma 6 L et λ ℓ = ℓ n (1+ ε ) wher e ℓ ∈ Z . Then, h ( λ ) = h ( λ ℓ +1 ) = ℓ for any λ ∈ ( λ ℓ , λ ℓ +1 ) . Pro of . F or λ ∈ ( λ ℓ , λ ℓ +1 ), we ha ve 0 < n (1 + ε ) ( λ − λ ℓ ) < 1 and h ( λ ) = ⌈ nλ (1 + ε ) ⌉ − 1 = ⌈ nλ ℓ (1 + ε ) + (1 + ε )( λ − λ ℓ ) ⌉ − 1 =  n  ℓ n + (1 + ε )( λ − λ ℓ )  − 1 = ℓ − 1 + ⌈ n (1 + ε ) ( λ − λ ℓ ) ⌉ = ℓ =  n  ℓ + 1 n (1 + ε ) × (1 + ε )  − 1 = h ( λ ℓ +1 ) . ✷ Lemma 7 L et λ ℓ = ℓ n (1 − ε ) wher e ℓ ∈ Z . Then, g ( λ ) = g ( λ ℓ ) = ℓ + 1 for any λ ∈ ( λ ℓ , λ ℓ +1 ) . Pro of . F or λ ∈ ( λ ℓ , λ ℓ +1 ), we ha ve − 1 < n (1 − ε ) ( λ − λ ℓ +1 ) < 0 and g ( λ ) = ⌊ nλ (1 − ε ) ⌋ + 1 = ⌊ n [ λ ℓ +1 (1 − ε ) + (1 − ε )( λ − λ ℓ +1 )] ⌋ + 1 =  n × ℓ + 1 n (1 − ε ) × (1 − ε )  + ⌊ n (1 − ε )( λ − λ ℓ +1 ) ⌋ + 1 =  n × ℓ + 1 n (1 − ε ) × (1 − ε )  − 1 + 1 = ℓ + 1 =  n × ℓ n (1 − ε ) × (1 − ε )  + 1 = g ( λ ℓ ) . ✷ Lemma 8 L et α < β b e two c onse cutive elements of the asc e nding arr angement of al l distinct elements of { a, b } ∪ { ℓ n (1 − ε ) ∈ ( a, b ) : ℓ ∈ Z } ∪ { ℓ n (1+ ε ) ∈ ( a, b ) : ℓ ∈ Z } . Then, b oth g ( λ ) and h ( λ ) ar e c onstants for any λ ∈ ( α, β ) . 8 Pro of . Since α and β are t wo consecutiv e elemen ts of the ascending arr angemen t of all distinct elemen ts of the set, it m ust b e true that there is no int eger ℓ suc h that α < ℓ n (1 − ε ) < β or α < ℓ n (1+ ε ) < β . It follo ws that there exist t wo inte gers ℓ and ℓ ′ suc h that ( α, β ) ⊆  ℓ n (1 − ε ) , ℓ +1 n (1 − ε )  and ( α, β ) ⊆  ℓ ′ n (1+ ε ) , ℓ ′ +1 n (1+ ε )  . Applying L emm a 6 and Lemma 7 , we ha ve g ( λ ) = g  ℓ n (1 − ε )  and h ( λ ) = h  ℓ ′ +1 n (1+ ε )  for an y λ ∈ ( α, β ). ✷ Lemma 9 F or any λ ∈ (0 , 1) , lim η ↓ 0 C ( λ + η ) ≥ C ( λ ) and lim η ↓ 0 C ( λ − η ) ≥ C ( λ ) . Pro of . Observing that h ( λ + η ) ≥ h ( λ ) for any η > 0 and th at g ( λ + η ) = ⌊ n ( λ + η )(1 − ε ) ⌋ + 1 = ⌊ nλ (1 − ε ) ⌋ + 1 + ⌊ nλ (1 − ε ) − ⌊ nλ (1 − ε ) ⌋ + n η (1 − ε ) ⌋ = ⌊ nλ (1 − ε ) ⌋ + 1 = g ( λ ) for 0 < η < 1+ ⌊ nλ (1 − ε ) ⌋− nλ (1 − ε ) n (1 − ε ) , we hav e S ( n, g ( λ + η ) , h ( λ + η ) , λ + η ) ≥ S ( n, g ( λ ) , h ( λ ) , λ + η ) (4) for 0 < η < 1+ ⌊ nλ (1 − ε ) ⌋− nλ (1 − ε ) n (1 − ε ) . Sin ce h ( λ + η ) = ⌈ n ( λ + η )(1 + ε ) ⌉ − 1 = ⌈ nλ (1 + ε ) ⌉ − 1 + ⌈ nλ (1 + ε ) − ⌈ nλ (1 + ε ) ⌉ + n η (1 + ε ) ⌉ , w e h a v e h ( λ + η ) =    ⌈ nλ (1 + ε ) ⌉ for nλ (1 + ε ) = ⌈ nλ (1 + ε ) ⌉ and 0 < η < 1 n (1+ ε ) , ⌈ nλ (1 + ε ) ⌉ − 1 for nλ (1 + ε ) 6 = ⌈ nλ (1 + ε ) ⌉ and 0 < η < ⌈ nλ (1+ ε ) ⌉− nλ (1+ ε ) n (1+ ε ) . It follo w s that b oth g ( λ + η ) and h ( λ + η ) are indep enden t of η if η > 0 is sm all enough. Since S ( n, g , h, λ + η ) is con tinuous with r esp ect to η for fi xed g and h , w e h a v e that lim η ↓ 0 S ( n, g ( λ + η ) , h ( λ + η ) , λ + η ) exists. As a result, lim η ↓ 0 C ( λ + η ) = lim η ↓ 0 S ( n, g ( λ + η ) , h ( λ + η ) , λ + η ) ≥ lim η ↓ 0 S ( n, g ( λ ) , h ( λ ) , λ + η ) = S ( n, g ( λ ) , h ( λ ) , λ ) = C ( λ ) , where the inequalit y follo ws from (4). Observing that g ( λ − η ) ≤ g ( λ ) for any η > 0 and that h ( λ − η ) = ⌈ n ( λ − η )(1 + ε ) ⌉ − 1 = ⌈ nλ (1 + ε ) ⌉ − 1 + ⌈ nλ (1 + ε ) − ⌈ nλ (1 + ε ) ⌉ − nη (1 + ε ) ⌉ = ⌈ nλ (1 + ε ) ⌉ − 1 = h ( λ ) 9 for 0 < η < 1+ nλ (1+ ε ) −⌈ nλ (1+ ε ) ⌉ n (1+ ε ) , we hav e S ( n, g ( λ − η ) , h ( λ − η ) , λ − η ) ≥ S ( n, g ( λ ) , h ( λ ) , λ − η ) (5) for 0 < η < min n λ, 1+ nλ (1+ ε ) −⌈ nλ (1+ ε ) ⌉ n (1+ ε ) o . Since g ( λ − η ) = ⌊ n ( λ − η )(1 − ε ) ⌋ + 1 = ⌊ nλ (1 − ε ) ⌋ + 1 + ⌊ nλ (1 − ε ) − ⌊ nλ (1 − ε ) ⌋ − nη (1 − ε ) ⌋ , w e h a v e g ( λ − η ) =    ⌊ nλ (1 − ε ) ⌋ for nλ (1 − ε ) = ⌊ nλ (1 − ε ) ⌋ and 0 < η < 1 n (1 − ε ) , ⌊ nλ (1 − ε ) ⌋ + 1 for nλ (1 − ε ) 6 = ⌊ nλ (1 − ε ) ⌋ and 0 < η < nλ (1 − ε ) −⌊ nλ (1 − ε ) ⌋ n (1 − ε ) . It follo w s that b oth g ( λ − η ) and h ( λ − η ) are indep enden t of η if η > 0 is sm all enough. Since S ( n, g , h, λ − η ) is con tinuous with r esp ect to η for fi xed g and h , w e h a v e that lim η ↓ 0 S ( n, g ( λ − η ) , h ( λ − η ) , λ − η ) exists. Hence, lim η ↓ 0 C ( λ − η ) = lim η ↓ 0 S ( n, g ( λ − η ) , h ( λ − η ) , λ − η ) ≥ lim η ↓ 0 S ( n, g ( λ ) , h ( λ ) , λ − η ) = S ( n, g ( λ ) , h ( λ ) , λ ) = C ( λ ) , where the inequalit y follo ws from (5). ✷ By a similar argum ent as that of Lemm a 5, we hav e Lemma 10 L e t α < β b e two c onse cutive elements of th e asc e nding arr angement of al l dis- tinct elements of { a, b } ∪ { ℓ n (1 − ε ) ∈ ( a, b ) : ℓ ∈ Z } ∪ { ℓ n (1+ ε ) ∈ ( a, b ) : ℓ ∈ Z } . Then, C ( λ ) ≥ min { C ( α ) , C ( β ) } for any λ ∈ ( α, β ) . Finally , to sho w Theorem 2, note that the statemen t ab out the co verag e p robabilit y follo ws immediately from Lemma 10. The n umb er of elemen ts of the finite set can b e calculate d by using the pr op ert y of the ceiling and flo or fu nctions. References [1] X. Chen, “Exact computation of minim um sample size for estimation of binomial p arame- ters,” arXiv:0707.2 113 v1 [math.ST], Ju ly 2007. [2] X. C hen, “Exact computation of minimum sample size for estimating p r op ortion of finite p opulation,” arXiv:0707.211 5 v1 [math.ST], July 2007. [3] M. M. Desu an d D. Ragha v arao, Sample Size Metho dolo gy , Academic Press, 1990. 10