On the variance of the number of extreme points of a random convex hull

Statistics & Probability Letters 44 (1999) 123 – 130

On the variance of the number of extreme points of a random convex hull Bruno MassÃe UniversitÃe du Littoral, 1, Place de l’Yser, B 1022, 59375, Dunkerque Cedex 1, France Received July 1997; received in revised form October 1998

Abstract By an appropriate choice of the underlying law, the variance of the number of extreme points of the convex hull of c 1999 n i.i.d. random points in the plane can be inÿnitely often as close to n2 as we want (as n grows to inÿnity). Elsevier Science B.V. All rights reserved Keywords: Random convex hull; Variance; Extreme points

1. Introduction Let (Xn )n¿1 be a sequence of i.i.d. Rd -valued random vectors deÿned on a probability space ( ; A; Pr). And let P be their common distribution. For n¿1, we shall use the following notations: • Cn is the convex hull of {X1 ; : : : ; Xn }, • Nn is the number of vertices of Cn , • Mn is the P-measure of the complementary of Cn . Note that Nn and Mn are random variables. The properties of Cn ; Nn and Mn have been investigated in a good many papers in the last 30 years or so. Schneider (1988) gives a survey. Two remarks can be made about these papers. Firstly, the distribution and the asymptotic behaviour of Cn , Nn and Mn depend heavily on P. As far as we know, the only general results are in Efron (1965), Buchta (1990) and MassÃe (1993,1995). Secondly, many papers investigate the expectations E(Nn ) and E(Mn ), but very little is known about the variances. Moreover, what is known generally relies on quite restrictive hypotheses on P (e.g. Mannion, 1994; Hsing, 1994; Groeneboom, 1988). Two exceptions deserve to be mentioned. Efron (1965) gives an explicit and general formula linking V (Nn ) to V (Mn ), and MassÃe (1995) gives an explicit and distribution-free bound, namely n(d + 1), for E-mail address: [email protected] (B. MassÃe) c 1999 Elsevier Science B.V. All rights reserved 0167-7152/99/$ - see front matter PII: S 0 1 6 7 - 7 1 5 2 ( 9 8 ) 0 0 2 9 8 - 3

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Pk=n V (Nn − k=1 Mn; k ) where Mn; k is the P-measure of the complementary of the convex hull of {Xi | i6n; i 6= k} (k = 1; : : : ; n). It is natural to ask if we can get such a distribution-free bound for V (Nn ). With the help of a very simple example, the ÿrst result of the present paper shows that the answer is no. But we can do better and this is the object of the present paper’s main result which says that, by an adequate choice of P; V (Nn ) can be inÿnitely often (as n grows to inÿnity) as close to n2 as we want. This is, in a way, partially doing for V (Nn ) what Devroye (1991) did for E(Nn ). 2. Results The ÿrst result of the present paper states that V (Nn ) has no non-trivial distribution-free bound. Proposition 1. For each integer n¿4; there exists a probability measure Pn on R2 such that; if P = Pn ; V (Nn )¿n2 =4 − 2n − 4: Now here is the present paper’s main result. V (Nn ) can be inÿnitely often (as n grows to inÿnity) as close to n2 as we want since the sequence (w(n))n¿1 (see hereafter) grows to inÿnity as slowly as we want. Theorem 1. Let (w(n))n¿1 be a sequence of real numbers; strictly increasing; positive and such that w(n) → +∞ (as n → +∞) and w(n)6n (n¿1). There exists a probability measure Pw on R2 such that; if P = Pw V (Nn )¿n2 =w(n)

inÿnitely often

(n → +∞):

3. Proof of the proposition Let n¿4 and  ∈ R such that n + (1 − )n = 1=2: Let A and B be two arcs of circles in the plane such that every straight line tangent to A or B seperates A from B. Let Pn be a probability measure on the plane such that Pn (A) = ; Pn (B) = 1 − ; Pn ({z}) = 0

for every point z:

Note that • If X1 ; : : : ; Xn are all in A (resp. B); then each of them is a vertex of Cn . • If there is exactly one of them in A (resp. B); then Cn is a triangle. • In other cases; Cn has exactly four vertices. Hence E(Nn2 ) = n2 (n + (1 − )n ) + 9n(n−1 (1 − ) + (1 − )n−1 ) +16[1 − n − (1 − )n − n−1 (1 − ) − (1 − )n−1 ] ¿ n2 =2

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125

and E(Nn ) 6 n(n + (1 − )n ) + 4(1 − n − (1 − )n ) 6 4 + (n − 4) (n + (1 − ))n 6 n=2 + 2: Hence V (Nn ) = E(Nn2 ) − (E(Nn ))2 ¿ n2 =4 − 2n − 4: 4. Proof of theorem Let w(x) be a one-to-one map from [o; +∞[ into itself, strictly increasing and such that 06w(x)6x (x ∈ [0; +∞[). The method used in the following construction of Pw has been partially inspired by Devroye (1991). 4.1. Construction of Pw Let  ∈ ]0; 1=2[ and  = [(1 − 2) (1 − e−=2 )]−1=2 . For every x in [0; +∞[, put u(x) =  the sequences (nm )m¿1 , (qm )m¿0 and (rm )m¿1 , by q1 ∈ ]0; 1=2[, −1 ) nm − 1is the integer part of u−1 (qm 2 ; rm−1 ) rm = Min(qm

p w(x). Now deÿne

(m¿1);

(m¿1);

qm+1 = Min(rm2 ; m−1 n−2 m )

(m¿1);

and +∞ X

q0 = 1 −

(qm + rm ):

m=1

P+∞ P+∞ P+∞ l (note that m=1 (qm + rm )6 l=0 2−2 ¡ l=1 2−l ): Now we give the support of Pw (see Fig. 1). Let A0 be a point and (Am )m¿1 and (Bm )m¿1 two sequences of arcs of circles such that: every tangent to Am or Bm separates A0

k=m−1 [

(Ak ∪ Bk ) ∪ Am

k=1

from Bm

+∞ [

(Ak ∪ Bk ):

k=m+1

Finally, let Pw be a probability measure on the plane such that: Pw ({z}) = 0

for every point z 6= A0 ;

Pw (Am ) = qm

(m¿0)

and Pw (Bm ) = rm

(m¿1):

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Fig. 1. Support of Pw .

4.2. Auxiliary results For clarity’s sake we collect below some auxiliary results. Lemma 1 gives some properties of the support of Pw . Lemma 2 gives some properties of the sequences (nm )m¿1 , (qm )m¿0 and (rm )m¿1 . S+∞ Lemma 1. Let I be a subset of A0 m=1 (Am ∪ Bm ) and denote by J its convex hull. Let m¿1. S+∞ (1) If there is no point of I in Bm k=m+1 (Ak ∪ Bk ); then every point of I which is in Am is a vertex of J. then Bm contains at most two vertices of J. (2) If there is at least one point of I in A0 ; S +∞ (3) If there is at least one point of I in Bm k=m+1 (Ak ∪ Bk ); then Am contains at most two vertices of J. Proof. (1) Let z ∈ I ∩ Am . The tangent to Am at z is the boundary of a halfspace which contains I . So z is a vertex of J . (2) If there are k points S+∞ of I in Bm (k¿2); then the convex hull of A0 and these points is a triangle. (3) Let z0 ∈ I ∩ (Bm k=m+1 (Ak ∪Bk )); and let {z1 ; : : : ; zk } ⊂ I ∩ Am (k¿2). Then the convex hull of {z0 ; z1 ; : : : ; zk } is a triangle. Lemma 2. For every m¿1; put pm = p0 + nm ) = 0; (1) lim nm (1 − pm m→+∞

(2) (pm−1 + qm )nm 6e− (m¿2); (3) lim nm (pm−2 + qm−1 )nm = 0; m→+∞

Pk=m

k=1 (qk

+ rk ). Then

B. MassÃe / Statistics & Probability Letters 44 (1999) 123 – 130

(4)

−1 lim mn−1 m qm = 0;

m→+∞

(5) (pm−1 + qm )nm ¿1 − 2

(m¿2):

Proof. (1) Simple arguments give " i=n !# " # +∞ +∞ [m [ [ nm (Ak ∪ Bk ) 6nm Pr X1 ∈ (Ak ∪ Bk ) Xi ∈ 1 − pm = Pr i=1

6 nm

+∞ X

k=m+1

k=m+1

(qk + rk )6 nm

k=m+1

+∞ X

2l qm+1 6 nm

l=0

+∞ X

l qm+1 = nm qm+1 (1 − qm+1 )−1 :

l=1

Hence nm )6n2m qm+1 (1 − qm+1 )−1 6m−1 (1 − qm+1 )−1 : nm (1 − pm

(2) Let m¿2 2

(pm−1 + qm )nm 6(1 − rm )nm 6e−nm rm 6Min(e−nm qm ; e− ): (3) Let us recall that −1 ¿(m − 1)n2m1 qm

(m¿2):

Hence, we obtain successively nm ¿u−1 (m − 1)n2m−1 u(nm ) (m − 1)−1 ¿n2m−1

(because u−1 is non-decreasing); (because u is non-decreasing);

and 1=2 ¿n2m−1 nm

(because w(x)6x):

So part 2 of the present lemma implies nm (pm−2 + qm−1 )nm 6nm (pm−2 + qm−1 )nm−1 

−1=2 3=4 nm

6nm e−

−1=2 3=4 nm

and we can conclude by using limm→+∞ nm = +∞. (4) Because of deÿnition of nm −1 −1 −1=2 mn−1 m qm 6 mnm u(nm )6mnm

(because w(x)6x ) m

6 2 m(u(nm ))−1 62 mqm 62 mq1(2

−1)

and it remains only to remember that q1 ∈ ]0; 1[. (5) Since 0 ¡ rm 6q1 ¡ 12 (pm−1 + qm )nm =

1−

rm +

+∞ X

!!nm

(qk + rk )

k=m+1

¿ 1−

+∞ X l=0

!nm l rm2

¿ 1−

+∞ X

!nm rml

l=1

¿ (1 − rm (1 − rm )−1 )nm ¿1 − nm rm (1 − rm )−1 which gives the announced result because of the deÿnition of rm .

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We collect now in Lemma 3 two well-known results of combinatorial calculus which do not deserve any proof. Lemma 3. For any positive integer n; any a ∈ R and any b ∈ R; k=n P

(1)

k=1 k=n P

(2)

k=1

n k

k2 k

n k







ak bn−k = n(n − 1)a2 (a + b)n−2 + na(a + b)n−1 ;

ak bn−k = na(a + b)n−1 :

4.3. Proof of the theorem Sl=k For any k¿1, let Ck = A0 l=1 (Al ∩ Bl ). Now let m be a ÿxed positive integer. We are going to prove that V (Nnm )¿n2m =w(nm ) for large enough m. For any plane’s subset D, let #D be the number of points X1 ; : : : ; Xnm which are in D. We need to consider the following events: G1 = (#Cm ¡ nm ) ∪ (#Ao = 0) ∪ (#(Cm−2 ∩ Am−1 ) = nm ); G2 = (#Cm ¡ nm ) ∩ (#Ao ¿ 0) ∩ (#Bm ¿ 0); G3 = (#Cm−1 ¡ nm ) ∩ (#Ao ¿ 0) ∩ (#Bm−1 ¿ 0); Hk = (#Am = k) ∩ (#Cm−1 = nm − k) (k = 1; : : : ; nm ); Hk0 = Hk ∩ (#A0 ¿ 0) ∩ (#Bm−1 ¿ 0) (k = 1; : : : ; nm ): Let us remember that pm = p0 +

k=m X

(qk + rk ):

k=1

Since G1 ; G2 ; G3 ; H10 ; : : : ; Hn0m form a partition of , E(Nnm ) = E(Nnm | G1 )Pr(G1 ) + E(Nnm |G2 ∪ G3 )Pr(G2 ∪ G3 ) +

k=n Xm

E(Nnm | Hk0 )Pr(Hk0 ):

k=1

Now nm ) + (1 − q0 )nm + (pm−2 + qm−1 )nm ; Pr(G1 )6(1 − pm

E(Nnm | G2 ∪ G3 )62m + 1 E(Nnm | Hk0 )6k + 2m − 1

(see Lemma 1); (see Lemma 1)

and Pr(Hk0 )6Pr(Hk )

(k = 1; : : : ; nm ):

Hence nm ) + (1 − q0 )nm + (pm−2 + qm−1 )nm ] + 2m + 1 + E(Nnm )6nm [(1 − pm

k=n Xm k=1

(k + 2m − 1)

n  m

k

k nm −k qm pm−1 :

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129

This and part 2 of Lemma 3 imply nm ) + (1 − q0 )nm + (pm−2 + qm−1 )nm ] E(Nnm ) 6 nm [(1 − pm

+nm qm (pm−1 + qm )nm −1 + 4m: Thus parts 1 and 3 of Lemma 2, implies, if m is large enough, E(Nnm ) 6 nm qm (pm−1 + qm )nm−1 + 4m + 1 −1 −nm +1 + 1]: 6 nm qm (pm−1 + qm )nm−1 [(4m + 1)n−1 m qm (pm−1 + qm)

(4.1)

Now parts 4 and 5 of Lemma 2 give −1 −nm +1 = 0: lim (4m + 1)n−1 m qm (pm−1 + qm )

m→+∞

So, if m is large enough to provide (4:1) and −1 −nm +1 6e=4 − 1; (4m + 1)n−1 m qm (pm−1 + qm )

we obtain E(Nnm )6nm qm (pm−1 + qm )nm −1 e=4 :

(4.2)

Let us deal now with the minorization of E(Nn2m ). Since H1 ; : : : ; Hnm have empty intersection E(Nn2m ) ¿

k=n Xm

E(Nn2m =Hk )Pr(Hk )

k=1

¿

k=n Xm

k 2 Pr(Hk )

(part 1 of Lemma 1):

k=1

Thus, using part 1 of Lemma 3, we get 2 (pm−1 + qm )nm −2 + nm qm (pm−1 + qm )nm −1 E(Nn2m ) ¿ nm (nm−1 )qm 2 ¿ n2m qm (pm−1 + qm )nm −2 :

So, using the deÿnition of nm and parts 2 and 5 of Lemma 2, if m is large enough to provide (4.2) 2 (pm−1 + qm )nm −2 (1 − (pm−1 + qm )nm e=2 ) V (Nnm ) ¿ n2m qm 2 (1 − 2) (1 − e−=2 ) ¿ n2m qm

¿ n2m (w(nm ))−1 : 5. Concluding remarks Intuitively, if P = Pw ; V (Nn ) seems to be very small inÿnitely often too (as Devroye (1991) obtains for E(Nn ) and for quite similar probability distribution). But until now this has not been proved. It would be interesting to try to get results like the present paper’s theorem for less pathological distributions. But it seems to be very dicult.

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The present paper is part of a larger study of the dispersion of Nn for distributions like Pw . A full length paper will be submitted later. It will contain the proof that, for a quite similar distribution, (E(Nn ))n¿1 has an inÿnite limit (as n grows to inÿnity) but (Nn (E(Nn ))−1 )n¿1 does not converge to 1 in probability. Which contradicts a quite famous conjecture. References Buchta, C., 1990. Distribution-independent properties of the convex hull of random points. J. Theor. Probab. 3, 387–393. Devroye, L., 1991. On the oscillation of the expected number of extreme points of a random set. Statist. Probab. Lett. 11, 281–286. Efron, B., 1965. The convex hull of a random set of points. Biometrika 52, 331–343. Groeneboom, P., 1988. Limit theorems for convex hulls. Probab. Theory Related Fields 79, 327–368. Hsing, T., 1994. On the asymptotic distribution of the area outside a random convex hull in a disk. Ann. Appl. Probab. 4, 478–493. Mannion, D., 1994. The volume of a tetrahedron whose vertices are choosen at random in the interior of a parent tetrahedron. Adv. Appl. Probab. 2, 577–596. MassÃe, B., 1993. Principes d’invariance pour la probabilitÃe d’un dilatÃe de l’enveloppe convexe d’un eà chantillon. Ann. Inst. H. PoincarÃe Probab. Statist. 29, 37–55. MassÃe, B., 1995. Invariance principle for the deviation between the probability content and the interior point proportion of a random convex hull. J. Appl. Probab. 32, 1041–1047. Schneider, R., 1988. Random approximation of convex sets. J. Microsc. 151, 211–227.