SHARP BOUNDS FOR EXPECTATIONS OF SPACINGS FROM DECREASING DENSITY AND FAILURE RATE FAMILIES

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1 APPLICATIONES MATHEMATICAE 3,4 (24), pp Katarzyna Danielak (Warszawa) Tomasz Rychlik (Toruń) SHARP BOUNDS FOR EXPECTATIONS OF SPACINGS FROM DECREASING DENSITY AND FAILURE RATE FAMILIES Abstract. We apply the metho of projecting functions onto convex cones in Hilbert spaces to erive sharp upper bouns for the expectations of spacings from i.i.. samples coming from restricte families of istributions. Two families are consiere: istributions with ecreasing ensity an with ecreasing failure rate. We also characterize the istributions for which the bouns are attaine.. Introuction. Let X,..., X n be i.i.. ranom variables with common cumulative istribution function (cf) F, mean µ, finite variance 2 an quantile function given by F (u) = sup{x : F (x) u}, u <. We write X :n,..., X n:n for orer statistics an consier spacings, that is, ifferences of consecutive orer statistics, R j:n = X j+:n X j:n, j n. Spacings are wiely use in gooness-of-fit tests, quality control problems an characterizations of istributions. For a eeper iscussion of their properties an applications we refer the reaer to Pyke []. Moriguti [8] presente sharp upper bouns for spacings in the class of istributions with finite variance, expresse in units. López-Blázquez [6] erive bouns for the expectations of X j+k:n X j:n in j:n k = (Var X j:n k ) /2 units for general istributions with finite secon moments an for iscrete istributions of N points [7]. Danielak an Rychlik [3] obtaine bouns in the classes of istributions with ecreasing ensity on 2 Mathematics Subject Classification: Primary 6E5, 62G3; Seconary 62N5. Key wors an phrases: sharp boun, convex cone, ecreasing ensity, ecreasing failure rate, orer statistics, projection, spacings. The secon author was supporte by the KBN (Polish State Committee for Scientific Research) Grant No. 5 P3A 2 2. [369]

2 37 K. Danielak an T. Rychlik the average (DDA for brevity) an ecreasing failure rate on the average (DFRA). Bouns for arbitrary ifferences X k:n X j:n, j < k n, expresse in ifferent scale units generate by various central absolute moments of the parent istribution of a single observation are presente in Danielak [2]. In this paper we present sharp upper bouns for the expectations of spacings, in units, when the parent istribution F belongs to the class of istributions with ecreasing ensity (DD) or with ecreasing failure rate (DFR). Let U an V enote the istribution function of stanar uniform istribution an stanar exponential istribution, respectively. We say that F belongs to the class DD if F U = F is convex on (, ). Similarly F belongs to the class DFR if F V is convex in (, ). These two classes can be treate together as a family of istributions F such that F W is convex on the support of W, where W = U, V. We then say that F succees W in convex orer (F c W ), a notion introuce for continuous life istributions by van Zwet [4]. Denote the ensity function an the cf of the ith orer statistic from the stanar uniform sample of size n by n f i:n (x) = nb i,n (x), F i:n (x) = B k,n (x), respectively, where ( ) k B i,k (x) = x i ( x) k i, x, i =,..., k, k =,,..., i are Bernstein polynomials. Using the representation E F X i:n = an setting r j:n = f j+:n f j:n, we obtain (.) E F R j:n = Changing variables in (.) we get (.2) E F R j:n = k=i F (x)f i:n (x) x, [F (x) µ]r j:n (x) x. [F W (x) µ]r j:n W (x)w(x) x, where W is an absolutely continuous cf with ensity w, support [, ) = [, W ) an a finite variance. The last integral can be treate as the inner prouct in the real Hilbert space H = L 2 ([, ), w(x)x) of square integrable functions on [, ) with respect to the weight function w. Applying

3 the Schwarz inequality to (.2) an noting that we obtain the boun (.3) Sharp bouns for spacings 37 F W µ W =, E F R j:n r j:n W W, which is attaine iff the two factors of the integran in (.2) are proportional. If F is an arbitrary cf with finite variance an F c W, then the transformation F W µ efines a family of functions { } (.4) CW = g C W : g(x)w(x) x =, where (.5) C W = {g H : g is nonecreasing an convex}. In general, the functions r j:n W are neither nonecreasing nor convex. In orer to erive sharp bouns for (.3) we apply the projection metho presente in Gajek an Rychlik [4]. For a thorough justification an numerous applications we refer the reaer to Rychlik [2]. Below we only briefly sketch some basic ieas. Observe that (.4) is a convex cone in the Hilbert space H. We nee to replace a fixe function r j:n W by its projection onto (.4) enote by PW r j:nw. The norm of the projection is the optimal boun in units, which is achieve by F such that F W µ is proportional to PW r j:nw. Note that (.5) is a translation invariant convex cone: g C W implies that g + c C W for any real c. Due to the following lemma (cf. Rychlik []) we can replace the original projection problem by a simpler one of fining the projection P W r j:n W of the function r j:n W onto (.5). Lemma. Let H =L 2 ([, ), w(x)x) with w(x) x = an C be a translation invariant convex cone in H. If the projection P h of an arbitrary h H onto C exists, then (.6) P h(x)w(x) x = h(x)w(x) x. As r j:nw (x)w(x) x =, we have PW r j:nw = P W r j:n W, an finally the boun E F R j:n P W r j:n W W is sharp an is attaine by a unique F satisfying (.7) F W (x) µ = P W r j:n W (x) P W r j:n W W. In Section 2 we escribe the shape of the projection in terms of three parameters an etermine them. Section 3 contains the main results of the paper. The proofs (quite long) are given in Section 4.

4 372 K. Danielak an T. Rychlik 2. The projection problem. We present assumptions on the projecte functions h = r j:n W chosen so as to cover the cases W = U, V. (A) Let h be a boune, twice ifferentiable function on [, ) such that h() =, lim x h() an h(x)w(x) x =, where w is a positive weight function satisfying w(x) x =. Moreover, we assume that h is ecreasing on (, a), convex increasing on (a, b), concave increasing on (b, c) an ecreasing on (c, ) for some < a < b < c. The lemma below escribes the behavior of the functions r j:n W in [, ) for W = U, V. Lemma 2. (a) Let W = U. The function r :2 is linear increasing. If n 3, then r :n is first concave increasing, then ecreasing; for 2 j n 2 the function r j:n is ecreasing, convex increasing, concave increasing an ecreasing; an r n :n is first ecreasing, then convex increasing. Moreover, r j:n has a unique zero in (, ) at θ = j/n. (b) Let W = V. The function r :2 V is concave increasing. If n 3, then r :n V is first concave increasing, then ecreasing; for 2 j n 2 the function r j:n V is ecreasing, convex increasing, concave increasing an ecreasing; an r n ;n V is first ecreasing, then convex increasing an ultimately concave increasing. The function r j:n V has a unique zero in (, ) at θ = ln( j/n). It follows that r j:n W satisfies (A) for W = U, 2 j n 2 an W = V, 2 j n. From now on we assume that h satisfies (A). It follows that h has exactly one zero θ (a, c) an the sign of h at the inflection point b may be arbitrary. The following lemma escribes the shape of the projection of an arbitrary function h satisfying (A) onto the convex cone (.5). Lemma 3. Let C C W be the class of functions of the form h(α), x < α, (2.) g (x) = h(x), α x <, λ(x ) + h(), x <, for some a α < b an λ h (), or { γ, x <, (2.2) g (x) = λ(x ) + γ, x <. for λ an γ R. Then for any g C W there exists a function g C such that h g h g.

5 Then (2.3) (2.4) Lemma 4. Let h : (, ] R be given by Sharp bouns for spacings 373 h() = h(x)w(x) x. w(x) x (i) h() = an h() < for any (, ), (ii) there exists a unique α (a, θ) such that h(α) = h(α), the function h is ecreasing with h > h in (, α], an h is increasing with h < h in (α, ). We introuce the following notations: λ () = (x )[h(x) h()]w(x) x (x, )2 w(x) x Y () = λ () h (), Z() = [h(x) λ ()(x ) h()]w(x) x. Proposition. Assume that α satisfies (2.3). If the set Y ={ (α, b) : Y () an Z() = } is not empty, then h(α), x α, (2.5) P W h(x) = h(x), α < x, h() + λ(x ), < x <, for = = sup{ Y} an λ = λ ( ). Otherwise, [ ] (x )I[,) (x) (2.6) P W h(x) = h() (x )w(x) x for the greatest < α satisfying (2.7) h(x)w(x) x [ = (x ) 2 w(x) x w(x) x ( (x )w(x) x ) 2 ] (x )w(x) x (x )h(x)w(x) x.

6 374 K. Danielak an T. Rychlik 3. Main results. Sharp upper bouns for E F R j:n, 2 j n 2, an F belonging to the class DD are presente in the following Proposition 2. Let X,..., X n be i.i.. ranom variables with ecreasing ensity, cf F, finite E F X = µ an Var F X = 2. Put j 2 [ ] (n j + 3)! (3.) Y (x) = f m:n+2 (x) + + f j :n+2 (x) 3(n j)! m= (3.2) Z (x) = If (3.3) + [ + (n j + 2)(n j + ) ( (n j))] f j:n+2 (x) + [ + (n j + )(n j) ( (n j ))] f j+:n+2 (x), j m= f m:n+2 (x) + [ 6 (n j + 2)(n j + ) ] f j:n+2 (x) [ 6 (n j + )(n j 4) + ] f j+:n+2 (x). Y (α) >, Z (α) < < Z (y), where α = (j )/(n ) an y is the smallest positive zero of (3.), then (3.4) for (3.5) E F R j:n B = B(j, n) B 2 = α[f j+:n (α) f j:n (α)] 2 {( )( + (n!)2 2j 2n 2j 2 (2n )! j n j ( )( 2j 2n 2j 2 j n j ( )( 2j 2 2n 2j + j n j ) [F 2j+:2n () F 2j+:2n (α)] ) [F 2j:2n () F 2j:2n (α)] ) [F 2j :2n () F 2j :2n (α)] + ( ) { 3 λ2 ( ) 2 + λ( )[f j+:n () f j:n ()] + [f j+:n () f j:n ()] 2}, with being the smallest positive zero of (3.2) an λ = λ () = F j+:n+() n j+ 2(n+2) [(n j)f j+:n+2() (n j + 2)f j:n+2 ()] 3 (. )3 (n + ) Equality hols in (3.4) for }

7 (3.6) F (x), ( x µ rj:n = x µ λ, If (3.3) fails, then (3.7) ) B, B r j:n() λ E F R j:n Sharp bouns for spacings 375 +, x µ < r j:n(α) B, r j:n (α) B x µ < r j:n() B, r j:n () B x µ x µ < λ( ) + r j:n(), B λ( ) + r j:n(). B f j+:n+() + 3 (n + ) 3( ) for being the smallest positive solution to n j + (3.8) [4(j + )f j+2:n+3 (x) + (n j + 2)f j+:n+3 (x)] 6 Equality hols in (3.7) for, (3.9) F (x) = with, + 2 ( )2 ( + x µ a = = j+ m= mf m+:n+3 (x). x µ < a, ) + 3, a x µ 3( ) x µ 3( ) + 3, a 2 = + 3( ) + 3. a 2, < a 2, Distributions (3.6) an (3.9) are not absolutely continuous. The former has a jump of size α at the left en of its support, then is the inverse function of a nonecreasing polynomial, an has a right uniform tail. The cf (3.9) is a mixture of an atom an a cf of uniform istribution. However, it is easy to fin sequences of absolutely continuous F k c U, k, which attain the bouns asymptotically. For j =, n = 2 the boun erive by Plackett [9] is optimal an is attaine by a uniform istribution belonging to the class DD. If j =, n 3, then the projection is a linear function (cf. Danielak []) of the form P U r :n (x) = 2(2x )/(n + ) an E F R :n / 2 3/(n + ). The boun is attaine for the uniform istribution on [ µ 3, µ + 3 ]. If j = n, then the optimal boun coincies with that obtaine in the class of arbitrary

8 376 K. Danielak an T. Rychlik istributions with finite variance (see Danielak [2]). In this case the boun (3.4) is sharp an becomes an equality for the cf (3.6) with =. We now turn to the case when F belongs to the class DFR. Assume that 2 j n. Proposition 3. Let X,..., X n be i.i.. ranom variables with ecreasing failure rate, cf F, finite E F X = µ an Var F X = 2. Put Y 2 (x) = j 2 [ f m:n+2 (x)+ n j n j + 2(n j+2)! ] (3.) f j :n+2 (x) (n j)! m= [ ] + + (n j + )[ 4(n j)] f j:n+2 (x) n j [ ] + + (n j)[2(n j) 3] f j+:n+2 (x), n j (3.) If (3.2) Z 2 (x) = j m= f m:n+ (x) + [(n j)(n j + ) ] f j:n+ (x) + [(n j)(2 n + j) ]f j+:n+ (x). Y 2 (α ) >, Z 2 (α ) < < Z 2 (y), where α = (j )/(n ) an y is the smallest positive zero of (3.), then (3.3) for E F R j,j+:n B = B(j, n) (3.4) B 2 = rj:n(α 2 ) + ( )[2λ 2 2λr j:n ( ) + rj:n( 2 )] {( )( ) + (n!)2 2j 2n 2j 2 [F 2j+:2n ( ) F 2j+:2n (α )] (2n )! j n j ( )( ) 2j 2n 2j 2 [F 2j:2n ( ) F 2j:2n (α )] j n j ( )( ) } 2j 2 2n 2j + [F 2j :2n ( ) F 2j :2n (α )], j n j where is the smallest positive zero of (3.) an { λ = λ (V j+ ( )) = f m:n+ ( ) 2( ) n j m= } (n j)f j+:n+ ( ) + (n j + )f j:n+ ( ).

9 The boun (3.3) is achieve for, ( x µ (3.5) F (x) = rj:n V Sharp bouns for spacings 377 ) B x µ, a x µ ( x µ λ B r j:n( ) + V ( ) λ with a = r j:n (α )/B, a 2 = r j:n ( )/B. If (3.2) oes not hol, then (3.6) E F R j,j+:n where ϱ is the smallest positive solution to j+ m= ), f j+:n+(ϱ) + ϱ ϱ(n + ) ϱ, x µ < a, <a 2, a 2, m n j f m+:n+2(x) = (n j + )f j+:n+2 (x) + 2(j + )f j+2:n+2 (x). The boun (3.6) is attaine for (3.7) F (x), = ( [ ] ) x µ + ϱ V ( ϱ) ϱ + + V (ϱ), x µ ϱ + ϱ, x µ ϱ > + ϱ. The cf (3.5) has a jump of size α = (j )/(n ) at the left en of its support, then is the inverse function of a nonecreasing polynomial, an finally has an exponential tail. The cf (3.7) is a mixture of an atom an an exponential istribution. If j =, n 2, then P V r :n V (x) = (x )/(n ) an E F R :n / /(n ). The boun is attaine for the exponential istribution with location an scale parameters equal to µ an, respectively. 4. Proofs. We shall frequently apply the following lemma: Lemma 5. The number of zeros of a linear combination of Bernstein polynomials m (4.) W (x) = a k B k,m (x), x (, ), k= oes not excee the number of sign changes of the sequence a,..., a m. The initial an final signs of (4.) in (, ) are ientical with the signs of the first an last nonzero elements of a,..., a m, respectively. The proof of the former statement, known as variation iminishing property of Bernstein polynomials, can be foun in Schoenberg [3], an of the latter was presente in Gajek an Rychlik [5].

10 378 K. Danielak an T. Rychlik We also use the formulae below (with the convention that B l,m (x) = for l > m or l < ): (4.2) xb l,m (x) = l + m + B l+,m+(x), ( x) s B l,m (x) = y (m l + s)!m! (m l)!(m + s)! B l,m+s(x), B l,m (x) = m[b l,m (x) B l,m (x)], B l,m (x) x = l B s,m+ (y). m + Proof of Lemma 2. (a) Let W = U. We have r :2 (x) = 4x 2. Assume that n 3. Using (4.2) we get s= r j:n(x) = n(n )[ B j 2,n 2 (x) + 2B j,n 2 (x) B j,n 2 (x)], r j:n(x) = n(n )(n 2) [ B j 3,n 3 (x) + 3B j 2,n 3 (x) 3B j,n 3 (x) + B j,n 3 (x)]. If 2 j n 2, then, by Lemma 5, r j:n is either first positive, then negative an ultimately positive (+ +, for brevity) or negative everywhere in [, ). The latter is impossible, because r j:n integrates to in (, ) an vanishes at an. Thus, r j:n has first a minimum, then a maximum, an it is convex an concave about the minimum an maximum, respectively. This combine with Lemma 5 implies that r j:n is + +, an our claim follows. Similar consierations apply to the remaining cases. (b) Assume that W = V. The function r :2 V (x) = 2( 2e x ) is concave increasing on [, ). Take n 3. Defining C j,m (x) = B j,m V (x) we get r j:n V (x) = n [ C j,n (x) + C j,n (x)] an (r j:n V ) (x) = n(n )e x [ C j 2,n 2 (x) + 2C j,n 2 (x) C j,n 2 (x)], (r j:n V ) n(n ) (x) = n 2 e x { (n j + )C j 3,n 2 (x) + (3n 3j + )C j 2,n 2 (x) (3n 3j )C j,n 2 (x) + (n j )C j,n 2 (x)}. Since each C l,m is a superposition of an increasing function V an a Bernstein polynomial, the statement of Lemma 5 hols for linear combinations of C l,m as well. Analyzing the signs of (r j:n V ) an (r j:n V ), analogously to the proof of part (a) we easily obtain the esire conclusions. Proof of Lemma 3. We show that for any g C W we can fin a function g C which is closer to h than g. Our proof starts with the observation

11 Sharp bouns for spacings 379 that it suffices to consier functions g satisfying g() <. Monotonicity of g an the fact that g integrates to imply that either g() < < lim x g(x) or g(x) = for x [, ). We exclue the latter case since there exists a function that vanishes in [, θ], is linear increasing in (θ, ) an is a better approximation to h than the constant (see Gajek an Rychlik [5]). As max{g, h(a)} is nonecreasing convex an is closer to h than g, it suffices to restrict our attention to functions g satisfying > g() h(a). Since h g is continuous, the set {x [, ) : h(x) = g(x)} is close. It follows that there exist at most countably many close intervals (possibly egenerate) where h = g. Note that it suffices to consier those g for which the set {h = g} contains at most one nonegenerate interval. Inee, suppose that there are at least two such intervals. They must be subsets of [a, b], because g is nonecreasing convex. If h = g in some [α, ] [α 2, 2 ] with < α 2 an h g in (, α 2 ) then { h(x), x (, α 2 ), g(x) = g(x), x (, α 2 ), is nonecreasing convex an h g h g, a contraiction. Now we nee to consier two cases: (I) the set {g = h} contains a nonegenerate interval, (II) the set {g = h} oes not contain any interval. (I) Suppose that h = g on some [α, ] [a, b], α <. We are going to show that there exists a function g of the form (2.), closer to h than g. Take an arbitrary ξ [α, ] an enote by h the nonecreasing function closest to h [,ξ] taking value h(ξ) at ξ, an by h 2 the nonecreasing convex function closest to h [ξ,) such that h 2 (ξ) = h(ξ). We are now in a position to show that h is either constantly h(ξ), or for some a η < ξ, constantly h(η) on [, η] an equal to h on [η, ξ], an h 2 is continuous an equal to h on [ξ, ν] an increasing linear on [ν, ) for some ν [ξ, ]. Note that h is convex, an so it is the best approximation of h [,ξ] in the class of nonecreasing convex functions. Furthermore, the function { h (x), x [, ξ], g(x) = h 2 (x), x (ξ, ), is nonecreasing an convex, satisfies (2.) an is closer to h than g. Now, our goal is to fin the nonecreasing function P h closest to h [,ξ]. Applying the moification of the Moriguti metho of obtaining greatest convex minorants, presente in Rychlik [2, Example 3, pp. 4 6], we observe that either P h(x) = ζ an ζ > h(ξ), or P h is constantly h(η) on [, η] for some a η < ξ, an equal to h on [η, ξ]. Only in the latter case the projection has the require form. We procee to show that if the former hols, then the nonecreasing function closest to h [,ξ] taking value h(ξ) at ξ is

12 38 K. Danielak an T. Rychlik constantly h(ξ) on [, ξ]. Lemma yiels ξ h(x)w(x) x = ξ P h(x)w(x) x = ζ ξ w(x) x. Any nonecreasing function g such that g(ξ) = h(ξ) can be represente as where g is nonecreasing an ξ g(x) = g (x) + h(ξ) g (ξ), g (x)w(x) x = an g () ζ g (ξ). Therefore h g 2 = h g [h(ξ) g (ξ)] 2 ξ h(x)w(x) x = ζ ξ w(x) x, = h g 2 2[h(ξ) g (ξ)](h g, ) + [h(ξ) g (ξ)] 2 (, ). Combining P h h g h, (h g, ) =, an (h P h, ) = with h(ξ) ζ g (ξ) we euce that h g 2 h P h 2 + [h(ξ) P h(ξ)] 2 (, ) = h P h [h(ξ) P h(ξ)] 2 = h h(ξ) 2. It follows that the nonecreasing function closest to h [,ξ] taking value h(ξ) at ξ is either constantly h(ξ) on [, ξ] or constantly h(η) on [, η] for some η < ξ, an equal to h on [η, ξ]. It remains to observe that h 2 is of the form escribe above. This was prove in Lemma of Gajek an Rychlik [5]. (II) Assume now that the set {g = h} oes not contain any interval. We will prove that there exists a function g of the form (2.) or (2.2) such that h g h g. Suppose that there are some a α < b such that g(α) = h(α), g() = h() an g(x) < h(x) for x (α, ). Set { h(x), x (α, ), g (x) = g(x), x (α, ). Obviously, h g h g an g is nonecreasing. If g(α ) < h(α ), then g (α) = h (α). If g(α ) > h(α ), then g (α) < h (α). Similarly, the inequalities g(+) < h(+) an g(+) > h(+) imply g () = h () an g () > h (), respectively. Consequently, g has nonecreasing erivative. Thus g is convex. Note that the set {g = h} contains an interval, so we are in case (I) an can fin an approximation of the form (2.). Suppose now that g(a+) h(a+). Our claim is that there exists a function g of the form (2.) which is closer to h than g. As we have assume that g(a) g() h(a), it suffices to consier the case g(a) = h(a). If

13 Sharp bouns for spacings 38 g(a+) = h(a+), which means that g = h on some interval, then we procee as in case (I). Assume then that g(a+) < h(a+). If there exists (a, b] such that g() = h(), then g can be replace by a function equal to h on some interval, which leas to case (I). Otherwise, two subcases are possible: either g crosses h at δ (b, ), or g runs beneath h in the whole (a, ). But then < an we set δ =. Let l t enote the tangent line to h at t. The slope of l t continuously increases in [a, b) an so oes the value l t (δ) ranging from l a (δ) < h(δ) to l b (δ) > h(δ). It follows that there exists φ (a, b) such that l φ (δ) = h(δ). Set h(a), x < a, g(x) = h(x), a x < φ, l(x), φ x <. Obviously g is of the form (2.), it is nonecreasing an convex, an closer to h than g. Consier the case g(a+) > h(a+). Since g() <, it follows that there exists α [, a] such that h(α) = g(α). Clearly, there also exists α [a, b] such that h(α ) = h(α). Note that the constant h(α) is closer to h than g on [, α ]. Our next objective is to show that if g(a+) > h(a+) an the set of egenerate intervals {g = h} (a, b] contains at least two points then one of the following two cases hols: ) there exists a function g of the form (2.) closer to h than g; 2) there are only two such points < φ, an then g > h on (, φ) an g(φ+) < h(φ+). If h(x) = g(x) for x {, φ}, < φ an g < h on (, φ), then, as alreay shown, g can be replace by some function of the form (2.). If h(x) = g(x) for x {, φ}, h(x) g(x) for x (a, ) an g( ) > h( ), g > h on (, φ) an g(φ+) > h(φ+), then the function { h(x), x (, φ), g (x) = g(x), x (, φ), is closer to h than g, belongs to (.5) an can be replace by a function of the form (2.), because {g = h} contains an interval. Therefore the inequality g(a+) > h(a+), combine with the assumption that {g = h} oes not contain an interval, implies four cases: (i) {g = h} (a, b) =, i.e. g > h on (a, b), (ii) {g = h} (a, b) = {φ} an g(φ+) > h(φ+), (iii) {g = h} (a, b) = {φ} an g(φ+) < h(φ+), (iv) {g = h} (a, b) = {, φ} an g > h on (, φ) an g(φ+) < h(φ+).

14 382 K. Danielak an T. Rychlik If either (i) or (ii) hol, then there are two possibilities: (a) g h on the whole (b, ) (g an h may be tangent at some δ (b, c)), (b) there exist δ (b, c) an (δ, ) such that g > h on (b, δ) an (, ) an g < h on (δ, ). If (a) hols, then there exists a nonecreasing linear function l such that h l g on (b, c), where g is convex an h concave. Moreover, h l g on (c, ), an l g on (, b). In case (i) we have l() g() g(α) = h(α) = h(α ), so l crosses h = max {h(α), h} at some α. The function { h(α), x α, g 2 (x) = l(x), α < x <, belongs to (.5) an is of the form (2.) provie α > α. Otherwise it is of the form (2.2). The constantly h(α) function is closer to h than g on (, α ), an h l g on (α, ). If case (ii) hols, then g(φ) = h(φ). It follows that l crosses h at α φ. Hence g 2 is of the form (2.) an is closer to h than g. Assume now that (b) hols. Take the line l 2 passing through (δ, h(δ)) an (, h( )). Then l 2 (x) g(x) for x [, δ], g(x) l 2 (x) h(x) for x [δ, ] an h(x) l 2 (x) g(x) for x > an h(δ ) < l 2 (δ ). Applying the same arguments as in (a) we conclue that l 2 crosses h = max {h(α), h} at some α. Therefore, we improve the approximation if we replace g by { h(α), x α, g 3 (x) = l 2 (x), α < x <, which is of the form (2.) or (2.2). It remains to consier cases (iii) an (iv). Then g runs beneath h on [φ, b] where h is convex an either g < h on (b, ) or g crosses h at a unique (b, ). If the former hols, then we set = <. Let l 3 be the line passing through (φ, h(φ)) an (, h( )), with h( ) g( ). Then l 3 (x) g(x) for x [, φ], h(φ ) < l 3 (φ ) an g(x) l 3 (x) h(x) for x [φ, ] an h(x) l 3 (x) g(x) for x >. Since l 3 (x) g(x) for x < φ an g() < h(), we see that l 3 (x) = h(x) for some x > φ. Moreover l 3 crosses h = max{h(α), h} at some α (α, φ]. Let { h(α), x α, g 4 (x) = l 3 (x), α < x <. If (iv) hols, then g 4 is of the form (2.). If (iii) hols, then g 4 is either of the form (2.) or (2.2). Obviously, g 4 h g h. This ens the proof. Proof of Lemma 4. (i) From (A) we have h() =. Since h() for any θ, we see that h(x)w(x) x = an clearly h(x)w(x) x <,

15 Sharp bouns for spacings 383 which implies h() < for θ. Likewise, h() for > θ, which gives < h(x)w(x) x = h(x)w(x) x. Thus h() < for > θ. (ii) We have h () = h()w() w(x) x w() h(x)w(x) x [ w(x) x]2 = w()[h() h()] w(x) x. It suffices to show that h > h in [, α] an h < h in (α, ). Let W () = w(x) x. Then W is strictly increasing an W () =. With the notation for γ [, ] we have (4.3) H W (W ()) = H W (γ) = W () γ hw (x) x hw (x) x = h(x)w(x) x. Note that H W (γ) = hw (γ) an H W (W ()) = h(). Thus H W () =, H W ecreases in (, W (a)) an in (W (c), ), increases in (W (a), W (c)), an H W (W (θ)) =. Hence H W is concave ecreasing in (, W (a)), convex ecreasing in (W (a), W (θ)), convex increasing in (W (θ), W (c)), concave increasing in (W (c), ) an attains its local minimum at γ = W (θ). Moreover H W () = an H W () = hw (x) x = h(x)w(x) x =. By efinition of W an (4.3), we have h() = H W (W ())/W (). It follows that h() is the slope of the linear function l passing through (, ) an (W (), H W (W ())). By concavity of H W in [, W (a)], l lies beneath H W for W () (, W (a)), that is, for [, a]. Every line tangent to H W at W () (, W (a)) lies over H W in that interval. As H W (W ()) = h(), we see that h() > h() for [, a]. It follows from the convexity of H W in [W (a), W (θ)] that there is a unique α [a, θ] such that the ifference h h changes its sign from positive to negative. Precisely, α is the point such that the line through (, ) an (W (α), H W (W (α))) is tangent to H W at W (α). Moreover, for (θ, ) we have h() < < h(). Lemma 6. Assume that α satisfies (2.3). Then for any (, c] the function { h(), α, (4.4) g(x) = h(max{x, α}), α < c,

16 384 K. Danielak an T. Rychlik is the projection of h [,] onto the convex cone of nonecreasing functions in the Hilbert space L 2 ([, ), w(x)x). Proof. Rychlik ([2, Example 3, pp. 4 6]) presente the solution to the problem of projecting functions onto the convex cone of nonecreasing functions in spaces of the type L 2 with nonuniform weight function. In particular, the projection of h [,) L 2 ([, ), w(x)x) is (H W ) W, where H W is the greatest convex minorant of H W on [, ). From the properties of H W an H W W, given above, we conclue that (H W ) W (x) has the form (4.4), which completes the proof. Clearly, for any max{b, α} the function (4.4) is the projection of h [,] onto the convex cone of nonecreasing an convex functions in the space L 2 ([, ), w(x)x). Proof of Proposition. Suppose first that the assumptions of the first claim are satisfie. Take an arbitrary ξ (α, ) an g efine by (2.5). By Lemma 6, g [,ξ] is the projection of h [,ξ] onto the cone of nonecreasing functions in L 2 ([, ξ), w(x)x). Proposition in Danielak [] implies that g [ξ,] is the projection of h [ξ,] onto the convex cone of nonecreasing convex functions in L 2 ([ξ, ), w(x)x). Therefore, for any f C W (see (.5)) we have [f(x) h(x)] 2 w(x) x [g(x) h(x)] 2 w(x) x because the inequality hols for the integrals over both [, ξ] an [ξ, ]. It follows that g is the projection of h, because it is closer to h than any other function f from the cone. Suppose now that P W h is of the form (2.5), but the parameters α, o not satisfy the assumptions of the proposition. We will show that we can fin better approximations of h than g, which gives a contraiction. First, assume that α (a, ) oes not satisfy (2.3). Consier g of the form (2.5) restricte to [, ] with fixe an α [a, ]. Let Then D(α) = [g(x) h(x)] 2 w(x) x. D (α) = 2h (α)[h(α) h(α)] α w(x) x. If D (α) <, then D(α) ecreases if α increases. If D (α) >, then D(α) ecreases if α ecreases. In both cases approximation of h can be improve, a contraiction.

17 Sharp bouns for spacings 385 Suppose now that g is of the form (2.5), α satisfies conition (2.3), but λ λ (). Set D(, λ) = = α [g(x) h(x)] 2 w(x) x [h() + λ(x ) h(x)] 2 w(x) x. We fix an look for λ h () for which D(, λ) is minimize. The function D(, λ) is quaratic in λ an attains its minimum for λ () efine by (2.4). Hence, we minimize D(, λ) by taking λ () = max{h (), λ ()}. Thus we can exclue all λ λ (). Furthermore, if λ () = h (), then we improve the approximation by ecreasing (cf. Gajek an Rychlik [5, pp. 7 7]). So, we can also exclue λ = λ() λ (). Suppose then that g is of the form (2.5), α satisfies (2.3) an λ = λ (). If Y () <, then g is not convex. If Z(), then the necessary conition (.6) fails an g is not a projection. Therefore, conitions (2.3), Y () an Z() = are necessary for g of the form (2.5) with λ = λ () to be the projection of h onto C W. If there exist α < < 2 satisfying these conitions, then (2.5) with 2 an λ ( 2 ) approximates h better on [α, ) than (2.5) with an λ ( ) (cf. Danielak [, proof of Proposition ]). Therefore we take the greatest satisfying the above conitions. Summing up, we have prove that the assumptions of the first part of the proposition are necessary an sufficient. If they are not satisfie, then by Lemma 3, the projection is of the form (2.2). We now show that the function satisfying the assumptions of the secon claim is the best approximation of h of this form. Consier the function D(γ, λ, ) = g h 2 = [h(x) γ] 2 w(x) x + [h(x) λ(x ) γ] 2 w(x) x. Defining g λ, (x) = h(x) λ(x )I [,) (x), we can write D(γ, λ, ) = [ g λ, (x) γ] 2 w(x) x. The function D(γ, λ, ), with λ an fixe, is quaratic convex an attains

18 386 K. Danielak an T. Rychlik its minimum at (4.5) γ = γ (λ, ) = Set D(λ, ) = D(γ, λ, ) an Then the quantity g λ, (x)w(x) x = λ g (x) = (x )I [,) (x) D(λ, ) = for any fixe (, ) is minimize at where We have λ = λ () = L() = M() = (x )w(x) x. (x )w(x) x. [h(x) λ g (x)] 2 w(x) x h(x) g (x)w(x) x g (x) 2 w(x) x = L() M(), (x )h(x)w(x) x, (x ) 2 w(x) x L () = M () = 2 [ h(x)w(x) x, (x )w(x) x (x )w(x) x] 2. w(x) x. Note that L() = an L () <, an the same hols for M. It follows that L(), M() an λ () are positive for any [, ). Therefore, for arbitrary fixe the function of the form (2.2) with optimal parameters γ an λ is nonecreasing an convex. Let D() stan for D(λ (), ). Then D() = h 2 (x)w(x) x 2 L() h(x) g (x)w(x) x M() + ( ) L() 2 M() g 2 (x)w(x) x = h 2 (x)w(x) x L2 () M()

19 an Sharp bouns for spacings 387 D () = 2 L() [ λ () M ] () L (). M() 2 Since D () is continuous an D() oes not attain its minimum at = or =, we see that the conition D () = is necessary for (, ) to be optimal. Thus we nee (4.6) λ () = 2 L () M () = h() (x )w(x) x. Substituting (4.6) into (4.5), we conclue that has to satisfy γ h(x)w(x) x () = w(x) x = h(). Plugging the optimal γ () an λ () into (2.2) we obtain (2.6). The function g is the best constant approximation of h in [, ] because g(x)w(x) x = h(x)w(x) x. It is also the best linear approximation of h in [, ] since it is of the form λ ()(x ) + γ. It remains to fin (, ) satisfying D () =, for which the function of the form (2.6) is the projection of h onto (.5). We have state that the function (2.6) is the projection if the following conitions are satisfie: ) a constant approximation in [, ] is optimal in this interval, 2) the point (, γ()) lies on the curve (, h()), 3) the increasing linear part of (2.6), say l, is the optimal linear approximation of h on [, ). Assume that D () = for some > α. Then either g runs beneath h on (, c), or g an h have at least one common point over (, c). In both cases we can fin a better approximation of the form (2.5). If conitions ) 3) hol for = α, then the constant an linear parts are the optimal nonecreasing approximations of h on [, α] an (α, ), respectively, an they together efine the esire projection of h on [, ). Assume that l(α) < h(α) = h(α) for = α, where l(α) enotes the value of the optimal linear approximation of h on [α, ) at α. Then g can be replace by a function of the form (2.5), passing through (α, h(α)). So we have a contraiction. If the function (2.6) with = α is not the projection, then the best linear approximation of h on (α, ) is the best nonecreasing an convex approximation on this interval an l(α) > h(α). Now we ecrease starting from = α. Linear functions are still the best nonecreasing

20 388 K. Danielak an T. Rychlik convex approximations of h on (, ), the values l() change continuously until they reach the level h(), earlier than h(), because h() > h() for < α. Then the function that is constantly h() on [, ] an equal to l on (, ) is the projection of h onto (.5), because h() an l are the projections of h on [, ) an (, ), respectively. The following lemma is a simplifie version of Lemma 4 in Gajek an Rychlik [5]. Lemma 7. If {y (, b] : Y (y) > } = (, v) an Z has a finite number of zeros, then Z is either positive or negative or changes its sign once from to + in (, v). Proof of Proposition 2. We begin by fining α efine by (2.3). Since h(x) = s j:n (x) = n[b j,n (x) B j,n (x)], α is the unique solution to α n[b j,n (x) B j,n (x)] x = nα[b j,n (α) B j,n (α)], which is equivalent to (4.7) B j,n (α) = (j + )B j+,n (α) jb j,n (α), an finally α = (j )/(n ). Note that, by Lemma 4(ii), α > a. The next step is to evaluate λ () Since = n (x )[B j,n (x) B j,n (x) B j,n () + B j,n ()] x (x. )2 x n x[b j,n (x) B j,n (x)] x n j m= = B m,n+() + (j + )B j+,n+ (), n + [B j,n (x) B j,n (x)] x = j + n + B j+,n+(), n[b j,n () B j,n ()] (x ) x = (n j)(n j + )B j,n+() (n j + )(n j + 2)B j,n+ (), 2(n + ) we finally obtain

21 Sharp bouns for spacings 389 (4.8) λ () j m= = B m,n+() (n j+)! 2(n j )! B j,n+() + (n j+2)! 2(n j)! B j,n+() 3 (n + )(. )3 We next evaluate Y () = λ () h (), where h () = n(n )[ B j 2,n 2 () + 2B j,n 2 () B j,n 2 ()]. Multiplying h () by the enominator of (4.8) we get h ()(n + )( ) 3 (n j + 3)! = B j 2,n+ () 3 3(n j)! 2(n j + 2)! + 3(n j )! B (n j + )! j,n+() 3(n j 2)! B j,n+(). Therefore (4.9) where j m= Y () = a mb m,n+ () 3 (n + )(, )3 a m =, m =,..., j 3, (n j + 3)! a j 2 = +, 3(n j)! a j = + (n j + 2)(n j + ) [ (n j)], a j = + (n j)(n j + ) [ 3 (n j ) 2]. Since the enominator in (4.9) is positive, Y has the same sign as the polynomial (3.) in the numerator. The coefficients a m are positive for m =,..., j 2, an a j is negative, because such is the expression in square brackets. Furthermore, a j = for j = n 2, an a j > for j n 3. Thus, by Lemma 5, (3.) is positive near an negative near, provie j = n 2. If j n 3, then (3.) is either + + or positive in (, ). Since the line tangent to h at b lies over the graph of h for > b, we have h (b) > λ (b), which implies Y (b) <. It follows that (3.) is positive near an has exactly one zero y (, b). The conition y > α, which is equivalent to Y (α) >, is necessary for the existence of a projection of the form (2.5). If y α, then the projection is of the form (2.6). Next, we fin the exact form of the polynomial Z() = n [B j,n (x) B j,n (x)] x n( )[B j,n () B j,n ()] 2 λ ()( ) 2.

22 39 K. Danielak an T. Rychlik Since n [B j,n (x) B j,n (x)] x = B j,n (), n( )[B j,n () B j,n ()] = (n j)b j,n () (n j + )B j,n (), λ () = 2 j m= B m,n+() (n j+)! (n j )! B j,n+() + (n j+2)! (n j)! B j,n+ () 2 3 (n + )(, )3 with the notation Z () = 2 3 (n + )( )Z(), we obtain Z () = 2 3 (n j )(n j + )B j,n+() + j B m,n+ ()+ (n j+)! 2(n j )! B j,n+() m= 2(n j + 2)! 3(n j)! (n j + 2)! 2(n j)! B j,n+ () B j,n+ (). The polynomial Z () can be represente as Z () = jm= b mb m,n+ (), where b m =, m =,..., j 2, b j = 6 (n j + )(n j + 2), b j = 6 (n j + )(n j 4) (cf. (3.2)). The coefficient b j is positive, because j < n. If j = n 2, then b j =, an b j < otherwise. It follows that for j = n 2 the polynomial Z (an so Z) changes its sign once from to + at some z (, ). If j < n 2, then (3.2) is either negative in the whole (, ) an then the projection is of the form (2.5), or it is +. By Lemma 7, (3.2) changes its sign in (, y) at most once an only from to +. Therefore, for 2 j n 2 the polynomial (3.2) is either negative on (, y], or there exists a unique z (, y] such that (3.2) changes its sign at z from to +. If the former hols, then the necessary conition (.6) fails an the projection is not of the form (2.5). In the latter case, the projection is of the form (2.5) if z > α. Summing up, (3.3) are necessary an sufficient conitions for (2.5) to be the projection, with α = α = (j )/(n ) an the smallest positive zero of the polynomial (3.2). Then P U h 2 U = α h 2 (α ) x + α h 2 (x) x + [h( ) + λ(x )] 2 x, where λ = λ ( ), an we finally get (3.5). Using (.7), we fin the cf (3.6) for which the boun (3.4) is achieve.

23 Sharp bouns for spacings 39 If (3.3) fails, then P W h is of the form (2.6). Conition (2.7) takes on the form B j,n () [ 3 ( )3 4 ( )4] ( )2 j = B m,n+ (), 2(n + ) which can be rewritten as m= 2 3 (j + )(n j + )B (n j + 2)! j+,n+2() + B j,n+2 () = 6(n j)! This is equivalent to where K() = j+ m= c m B m,n+2 () =, c m = m >, m =,..., j, c j = j 6 (n j + )(n j + 2), c j+ = (j + ) [ 2 3 (n j + )] j+ m= mb m,n+2 (). (cf. (3.8)). Since n j 2, we have c j+ <. This combine with Lemma 5 implies that K is positive near, negative near an has a unique zero (, ). It uniquely etermines P U h of the form (2.6). Its norm gives the boun in (3.7). Applying (.7), we obtain the cf (3.9) attaining the boun (3.7). Proof of Proposition 3. We procee analogously to the proof of Proposition 2. Set h(x) = r j:n V (x) = f j+:n V (x) f j:n V (x). Suppose first that P V h is of the form (2.5). The task is now to fin α satisfying α [B j,n ( e x ) B j,n ( e x )]e x x (cf. (2.3)) or equivalently = ( e α )[B j,n ( e α ) B j,n ( e α )] B j,n ( e α ) = (j + )B j+,n ( e α ) jb j,n ( e α ). Writing α = e α, we obtain equation (4.7) an so α = (j )/(n ). Next we etermine (4.) λ () = (x )[s j:nv (x) s j:n V ()]e x x (x )2 e x. x

24 392 K. Danielak an T. Rychlik The enominator of (4.) is equal to 2e, an its numerator can be rewritten as A() = n (x )[C j,n (x) C j,n (x)]e x x s j:n V ()e. Gajek an Rychlik [5, pp ]) showe that (x )C i,m (x)e x x = i S(i + k, m + k)c k,m+ (), m + where Therefore S(i, n) = E V (X i:n ) = j k= i m= n + m, i n. A() = [S(j + m, n m) S(j m, n m)]c m,n () m= + S(, n j)c j,n () s j:n V ()e. Since S(j + m, n m) S(j m, n m) = S(, n j) = /(n j) an it follows that A() = s j:n V ()e = (n j)c j,n () (n j + )C j,n (), j m= C m,n () n j an finally we obtain { j λ () = e C m,n () 2 n j m= Our next goal is to etermine (n j)c j,n () + (n j + )C j,n () } (n j)c j,n () + (n j + )C j,n (). Y () = λ () h () = e 2 [A() 2e h ()]. As we have h () (n j + 2)! e = C j 2,n () (n j)! 2(n j + )! + (n j )! C (n j)! j,n() (n j 2)! C j,n(), Y () = e 2 j a m C m,n (), m=

25 where Sharp bouns for spacings 393 a m =, m =,..., j 3, n j a j 2 = n j (n j + 2)! + 2, (n j)! a j = + (n j + )[ 4(n j)], n j a j = + (n j)[2(n j) 3]. n j We easily check that a m > for m =,..., j 2 an a j <. Moreover a j = for j = n an a j > for j < n. Hence Y is + for j = n, an it is either + or + + for the remaining values of j. By the argument use in the proof of the previous proposition, Y (b) <. Therefore, for any 2 j n, there exists y (, b) such that Y (y) = an Y () > for (, y). Setting Y 2 V () = 2(n+2)e Y (), we get (3.) for x = e. Next we analyze the behavior of Z() = where s j:n V (x)e x x s j:n V () e x x λ () (x )e x x, s j:n V (x)e x x = n [B j,n (y) B j,n (y)] y = C j,n (), e s j:n V () e x x = (n j)c j,n () (n j + )C j,n (). Finally, we obtain Z() = jm= b mc m,n (), where b m =, m =,..., j 2, 2(n j) b j = n j + 2 2(n j), b j = 2(n j) n j. 2 We see that b m < for m =,..., j 2, b j >, b j < for j < n an b j = for j = n. Therefore, for j = n the function Z has exactly one zero. If j < n, then Z is either +, or negative everywhere in (, ). Analysis similar to that in the proof of Proposition 2 shows that the latter is impossible. Set Z 2 V () = 2(n+)(n j)z(). As in the previous proof

26 394 K. Danielak an T. Rychlik we euce that relations (3.2) are necessary an sufficient for the existence of the projection of the form (2.5). Then α = V (α ) an = V ( ), where is the smallest positive zero of (3.). Determining the projection of the form (2.5) with λ = λ 2 ( ) enables us to calculate the boun (3.4) an the istribution function (3.5) attaining the boun. Suppose now that (3.2) fails. Then the projection is of the form (2.6). The aim is to fin satisfying (2.7). The conition can we rewritten as or equivalently C j,n ()(2e e 2 ) = ( e )e n j 2(j + )C j+,n+ () + (n j + )C j,n+ () = n j Further calculations lea to with K() = j+ m= m C m,n+ () = j C m,n (), m= m = m, m =,..., j, n j j = j (n j + ), n j ( ) j+ = (j + ) n j 2. j+ m= mc m,n+ (). Since m > for m =,..., j an j+ <, inepenently of the sign of j, the function K() has exactly one zero, say z. Thus, the projection is of the form P V h(x) = h(z)[e z (x z)i [z, ) (x) ], where Then h(z) = z [f j+:nv (x) f j:n V (x)]e x x z = f j+:n+( e z ) e x x ( e z )(n + ). P V h 2 V = [h(z)] 2 (2e z ) = f 2 j+:n+ ( e z ) ( e z ) 2 (n + ) 2 (2ez ) an substituting ϱ = e z, we obtain the square of the boun (3.6). Applying (.7) we etermine the cf (3.7), for which the boun (3.6) is achieve.

27 Sharp bouns for spacings 395 References [] K. Danielak, Sharp upper mean-variance bouns on trimme means from restricte families, Statistics 37 (23), [2], Sharp upper bouns for expectations of ifferences of orer statistics in various scale units, Comm. Statist. Theory Methos 33 (24), [3] K. Danielak an T. Rychlik, Exact bouns for expectations of spacing from DDA an DFRA families, Statist. Probab. Lett. 65 (23), [4] L. Gajek an T. Rychlik, Projection metho for moment bouns on orer statistics from restricte families. I. Depenent case, J. Multivariate Anal. 57 (996), [5],, Projection metho for moment bouns on orer statistics from restricte families. II. Inepenent case, ibi. 64 (998), [6] F. López-Blázquez, Cotas para el valor esperao e espaciamientos e estaísticos orenaos y recors, in: Actas XV Jornaas Luso-Espanholas e Matematícas, Universiae e Evora, Vol. IV (99), [7], Bouns for the expecte value of spacings from iscrete istributions, J. Statist. Plann. Inference 84 (2), 9. [8] S. Moriguti, A moification of Schwarz s inequality with applications to istributions, Ann. Math. Statist. 24 (953), 7 3. [9] R. L. Plackett, Limits of the ratio of mean range to stanar eviation, Biometrica 34 (947), [] R. Pyke, Spacings, J. Roy. Statist. Soc. Ser. B 27 (965), [] T. Rychlik, Mean-variance bouns for orer statistics from epenent DFR, IFR, DFRA an IFRA samples, J. Statist. Plann. Inference 92 (2), [2], Projecting Statistical Functionals, Lecture Notes in Statist. 6, Springer, New York, 2. [3] I. J. Schoenberg, On variation iminishing approximation methos, in: On Numerical Approximation (Maison, 958), R. E. Langer (e.), Univ. Wisconsin Press, Maison, WI, 959, [4] W. R. van Zwet, Convex Transformations of Ranom Variables, Math. Centre Tracts 7 (964), Math. Centrum, Amsteram, 964. Institute of Mathematics Polish Acaemy of Sciences P.O. Box 2 Śniaeckich Warszawa, Polan anielak@impan.gov.pl Institute of Mathematics Polish Acaemy of Sciences Chopina Toruń, Polan T.Rychlik@impan.gov.pl Receive on (744)