6586Author(s): Charles S. Allen and Thomas W. StarbirdSource: The American Mathematical Monthly, Vol. 97, No. 10 (Dec., 1990), pp. 932-933Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2324346 .

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932 PROBLEMS AND SOLUTIONS [December

-Xi has a negative exponential distribution, with EXi =,u < 0, where ,-t lies in the interval specified below.

In particular, if Xi (or -Xi) has a negative exponential distribution with mean ,u, then Sk (or -Sk) has a Gamma distribution with parameters k and A. Hence, letting I. = (0, oo), (or (- oo, 0)),

skl1 ESke-Sk = fskes e-s/ ds

r(2k) X k

r(k) (l+ /)2)

Now,

r(2k) (2k - 1)! 1 (2k)! 1( -1/2 ( k _~~ ~ ~~~~ = = = _ -4 r(k)k! (k- l)!k! 2 k!k! 2 k

Therefore,

001 r(2k) bA \k 1 00 -1 2 -4/t \k

M=E- =

II 2 Mk=E r((k)) (1 + IL)2 )=2 kE ( k) + bt)2)

= +((i- ( 2 ) } 1{ - -1

where the series converges if (and only if) - t/(l + ,)2 < -. This inequality holds except when - 3 - C < , < - 3 + V8 (approximately - 5.83 < , < - .17) or when /u - 1. Thus

A + 3 + A- < A < 1 1 -+

1

00 A = 1

undefined - 3 - 8 < ,t < - 3 + V . 0. P. Lossers conjectures that this is always true whenever the original series converges.

Parts (a) and (b) were solved also, vi upcrossing probabilities, by Boris Pittel and by the proposer. Solutions via Laplace transforms and Lagrange's formula were given also by Terence R. Shore, Western Maryland College Problems Group, and a referee.

Rays in Normed Linear Spaces

6586 [1988, 963]. Proposed by Charles S. Allen, Drury College, SprinVield, MO.

In an infinite-dimensional normed linear space does there exist a convex set which does not contain a ray but whose closure does contain a ray?

Solution by Thomas W. Starbird, Jet Propulsion Laboratory, Pasadena, CA. We show the answer is yes by following Ann Griesel's suggestion to construct such a

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1990] PROBLEMS AND SOLUTIONS 933

set in an arbitrary infinite-dimensional topological linear space X. In fact, we shall construct in such an X a sequence xi1 (i > 1, i > 1) whose convex hull contains no ray, but whose closure (and hence the closure of its convex hull) contains a ray.

Let {ei}, i = 0, 1, 2,..., be a collection of linearly independent points in X. Let {ri}, i = 1, 2,. .. be an enumeration of the positive rationals. Set

Xij =rieO +ei/i, i)> 1 ,j 1.

Since xij - rieo as j -> cc, the closure of {xij} contains the ray from 0 in the direction eo. We now show that the convex hull of {xij} cannot contain a ray.

Fix two distinct points a and b in a ray R, and assume that they are in the convex hull of {xi}. Hence they are convex combinations of finitely many {xij}, and belong to the linear span L of {eo, el,... , e,,} for some integer n. Thus R itself is contained in L. If R is also in the convex hull of {xij1 then any z in R has the form

z = aEcx1ijxij, ij

where the aij are nonnegative and

E aij=1. ii}

But the aij are 0 for i > n since the ei are linearly independent-the formula for xij has only nonnegative coefficients so no cancellation can occur. Thus R is in the convex hull of the xij with i < n. But this is merely the convex hull of n line segments, namely the segments (rieo, rie0 + ei] for i = 1,2,..., n, and hence cannot contain a ray.

Editorial comment. Victor Klee showed in 1953 that in every infinite-dimen- sional real vector space X there is even a ubiquitous convex set containing no ray. A subset S of X is called ubiquitous if every point of X is an endpoint of some open segment contained in S. See pp. 107-108 of V. Klee, Convex sets in linear spaces III, Duke Math. J. 20(1953), 105-111. In the opposite direction S. K. Chung (Hong Kong) showed that a convex set in a finite-dimensional Hausdorff topologi- cal vector space contains a ray if its closure does.

Solved also by S. K. Chung (Hong Kong), Nir Cohen (Israel), A. A. Jagers (The Netherlands), Victor Klee, 0. P. Lossers (The Netherlands), Charles Riley, John Henry Steelman, Richard B. Tucker, and the proposer.

Conjugacy Classes of Mobius Transformations

6589 [1989, 65]. Proposed by Kevin Brown, Kent, WA.

Given a positive integer g, let n(g) denote the number of conjugacy classes of Mobius transformations each member of which generates a cyclic group of order g. Characterize those g greater than 6 for which n(g) 1 (mod 6). (A Mobius transformation is a linear fractional transformation of the extended complex plane. Cf. Chapter II of Caratheodory's Theory of Functions, Volume I, Chelsea, 1954.)

Solution by the proposer. The numbers g that occur are all numbers of the form p2d and 2p2d, where d is any positive integer and p is any prime congruent to 11 modulo 12 (the smallest such g is 121). For the proof we represent each M6bius

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