Generating uniform random vectors over a simplex with implications to the volume of a certain polytope and to multivariate extremes

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1 DOI /s Generating uniform random vectors over a simplex with implications to the volume of a certain polytope and to multivariate extremes Shmuel Onn Ishay Weissman Springer Science+Business Media, LLC 2009 Abstract A uniform random vector over a simplex is generated. An explicit expression for the first moment of its largest spacing is derived. The result is used in a proposed diagnostic tool which examines the validity of random number generators. It is then shown that the first moment of the largest uniform spacing is related to the dependence measure of random vectors following any extreme value distribution. The main result is proved by a geometric proof as well as by a probabilistic one. Keywords Convex polytope Multivariate extreme value distribution Pickands dependence function Simplex Triangulation Uniform spacings 1 Introduction Let k := v = v 1,v 2,...,v k ) : v i 0, } v i = 1 be the standard unit-simplex in R k k 2). Recall that a simplex is an iterated pyramid, that is, a polytope whose number of vertices exceeds its dimension by one, such as a triangle and the standard pyramid. Consider the real function A 0 : k [k 1, 1] defined by A 0 v) := max 1 i k v i, v k. Dedicated to Reuven Rubinstein on his seventieth birthday. S. Onn I. Weissman ) Faculty of Industrial Engineering and Management, Technion Israel Institute of Technology, Haifa 32000, Israel ieriw01@ie.technion.ac.il

2 The main purpose of this paper is to prove that the volume under A 0 is given by vola 0 ) := A 0 v)dv = k k1/2 k! 1 k1/2 =: i k! ζ 1k). 1) Why should one be interested in vola 0 )? In Sect. 2 we show its relevance to the generation of uniform random variables and in particular to the generation of uniform random vectors over a simplex. A diagnostic tool for the validity of a random number generator is proposed. In Sect. 3 we show the connection between vola 0 ) and multivariate extreme value distributions. The proof of 1) is given in two versions a geometric one in Sect. 4 and a probabilistic proof in Sect. 5. Along the way, we discover another interesting connection between vola 0 ) and the volume of a certain convex polytope P k defined by 3)), namely vola 0 ) = k 1/2 k!volp k ). Remark 1 Volume computations in fact volume approximations) of convex bodies of dimension n are in general quite complex. Most of the published algorithms involve multiphase Monte-Carlo techniques, based on random walks in R n. In a recent survey, Vempala 2005) lists the improvements in complexity from n 23 in 1991 to n 4 in In the present paper we were able to compute the volume under A 0 and get an exact expression for all k 2. 2 Generating a uniform random vectors over k Suppose one wants to generate a random vector, uniformly distributed over k. Rubinstein and Kroese 2007), Algorithm 2.5.3, suggest the following: Algorithm 1 1. Generate k 1 independent random variables U 1,...,U k 1 from U0, 1). 2. Sort the U i } into the order statistics and let U 0) = 0 and U k) = Define the spacings U 1) U 2) U k 1) V i = U i) U i 1), and return the vector V = V 1,...,V k ). i = 1,...,k For the proof that V is indeed uniformly distributed over the simplex k see David 1981), pp A more efficient algorithm, which does not require sorting the major time-consuming step) is given by Rubinstein and Melamed 1998) as Algorithm Algorithm 2 1. Generate k independent unit-exponential random variables Y 1,...,Y k and compute T k = k 1 Y i. 2. Define E i = Y i /T k and return E = E 1,...,E k ).

3 The vectors E and V are identically distributed. The reason is that the Y i can be thought of as the times between successive events of a homogeneous Poisson process. It is well known see for instance Karlin and Taylor 1975, p. 126 or Epstein and Weissman 2008, p. 17) that the conditional joint distribution of Y 1,Y 1 + Y 2,...,Y 1 + +Y k 1 ), given T k, is the same as the joint distribution of k 1 order statistics from U0,T k ). Obviously, there is a direct connection between vola 0 ) and the random vector V. Ifwe let V k) = max V i = A 0 V) be the largest spacing, then By 1), EV k) = k A 0 v)dv k dv = k 1/2 k 1)!volA 0 ). EV k) = 1 1 k i = 1 k ζ 1k). 2) This and V k) are useful when one wants to check the validity of a certain random number generator. Suppose we generate N = m k 1) uniform on [0, 1] random variables. For each block sample) of size k 1, we compute V k).havingm independent and identically distributed i.i.d.) replications of V k), one can compare their sample-mean with EV k) and test the significance of the difference. A more detailed diagnostic would be to plot the V k) } against their block number from 1 to m). As in control charts, we also plot the target value EV k) and an upper and a lower control limits. If the random number generator is valid, the m points should scatter around the target value, within the control limits. We take the control limits from Devroye 1982), who proved that lim infkv k) log k + log 3 k) = log 2 a.s., and lim supkv k) log k)/2log 2 k) = 1 Here, log j is the j times iterated logarithm. a.s. Example 1 As an example, we generated N = = uniform random numbers in S-Plus. Thus we have 100 independent replications of V 101), which are plotted in Fig. 1 as described above. Although the lower and upper limits are based on the asymptotic formulas to improve the approximation for finite k = 101, we replaced log k by ζ 1 k)), there is only one outlier out of 100 samples. Using 5), it turns out that the upper and lower limits correspond, respectively, to the upper and lower second percentiles more precisely, to 1.988% and 2.054%). Hence, even the most trustful random number generator may produce, on the average, two upper and two lower outliers. The sample-mean in this example is.05091, while EV 101) = With standard deviation of.01164, the difference is negligible. Example 2 Here we generated N = random variables from the autoregressive model. That is, U 1 is uniform on 0, 1) and for i = 2, 3,...,10000, U i = 0.1U i E i,

4 Fig. 1 Control Chart for V k) k = 101), i.i.d., S-Plus generated Fig. 2 Control Chart for V k) k = 101), autoregressive model, S-Plus generated where the E i } are i.i.d. uniform on 0, 1), generated by S-Plus. Note, the autocorrelation of lag1is0.1 and we expect a different behavior of the V k). We repeated the same procedure as before and the results are shown in Fig. 2. Compared with the previous case, we observe a general upward shift, 6 upper outliers and sample-mean of , which is significantly larger than expected if the model was i.i.d. uniform. The authors have experimented with a variety of examples. It seems that this diagnostic tool is quite sensitive to deviations from the i.i.d. uniform model, but it would be worthwhile to do more research. In particular, given a sample of size N, what is the most effective partition into m sub-samples of size k each?

5 3 Multivariate extreme value distributions Another interesting connection is between vola 0 ) and multivariate extremes. Suppose that X = X 1,X 2,...,X k ) is a random vector in R k,havingamultivariate extreme value distribution G with margins G i x) = P X i x}.ifwelet Bx) := log Gx) k 1 log G ix i ), then obviously Gx) = exp Bx) 1 } log G i x i ). Pickands 1981) showed that Bx) = Av)v k ),wherev i = log G i x i )/ k 1 log G j x j ). Further, A is convex and satisfies 1 k A 0v) Av) 1, v k. For a detailed account of the subject and the properties of the Pickands dependence function A, see Beirlant et al. 2004). Since we are only interested in the dependence among the X i }, without any loss of generality we can choose a convenient set of margins G i }. It is customary in the extreme value literature to choose the Fréchet distribution function G i x) = exp 1/x)x > 0) for all i = 1, 2,...,k. Thus, our G has the form Gx) = exp Av) 1 ) x 1 i, x i > 0, v k, where v i = x 1 i / k 1 x 1 j. If the X i } are independent, then A 1 and if they are completely dependent, then A A 0. Thus, a natural coefficient of dependence for the random vector X is τ = k 1 Av))dv k 1 A 0 v))dv = k1/2 /k 1)! vola) k 1/2 /k 1)! vola 0 ). Clearly 0 τ 1, τ = 0, 1 correspond to independence and complete dependence, respectively. Now, the numerator of τ is case-specific, while the denominator depends on k only. Thus, it will be useful to have an explicit expression for the latter. Weissman 2008) gives several examples with their τ -values, including the logistic model. In the latter case, ) α Av) = v 1/α i, v k, 1 <α 1. Clearly, α = 1 corresponds to total independence and α 0 corresponds to complete dependence. The case k = 2 is shown in Fig. 3. For each α = 0,.25,.50,.75 and 1, the function Av, 1 v) is plotted as a function of v [0, 1]. For each α, thevalueofτ is 4 times the volume area) between A and 1.

6 Fig. 3 Pickands dependence function for the logistic model α = 0, 0.25, 0.50, 0.75, 1) Fig. 4 The Lebesgue measures of k vs. k 4 A geometric proof of 1) In what follows it will be convenient to use the full-dimensional simplex k := v = v 1,v 2,...,v k 1 ) : v i 0, k 1 } v i 1, that is, the projection of k into R k 1. In this case, v k = 1 k 1 v i. Note that while L k ), the Lebesgue measure of k, is equal to 1/k 1)!, one has L k ) = k 1/2 L k ) = k 1/2 /k 1)!. The reason is this: As we project k onto k,thevertex0,...,0, 1) goes to 0,...,0), all other vertices are unaffected. The altitude of k i.e., the distance between 1/k 1),...,1/k 1), 0) and 0,...,0, 1)) isk/k 1)) 1/2,whichisk 1/2 times the altitude of k. This point is illustrated in Fig. 4. Goodman and O Rourke 2004, p. 374), give the volume and surface area of some standard polytopes. We now show how to evaluate the integral vola 0 ) = k A 0 v)dv. Some simple reductions The standard simplex k is the union k = σ σ over all k! permutations σ :1,...,k} 1,...,k}, where the simplex corresponding to σ is σ := v 1,...,v k ) : 0 v σk) v σk 1) v σ2) v σ1), } v i = 1.

7 Clearly, the desired integral is the sum of the integrals over these simplices, vola 0 ) = σ σ A 0 v)dv. By symmetry, it is enough to evaluate one of these integrals, say the one corresponding to the identity permutation σ = e. But over e we have A 0 v) = max i v i ) = v 1, and therefore, we obtain vola 0 ) = k! A 0 v)dv = k! v 1 dv. e e Now, e = v 1,v 2,...,v k ) : 0 v k v k 1 v 2 v 1, } v i = 1 so, writing v k = 1 k 1 v i, we can evaluate e v 1 dv by integrating v 1 over k 1 } e = v 1,v 2,...,v k 1 ) : 0 1 v i v k 1 v 2 v 1 and multiplying by k 1/2 ). This reduces to computing the volume of the following convex polytope in R k : k 1 P k = v 0,v 1,...,v k 1 ) : 0 1 v i v k 1 v 1, 0 v 0 v 1 }, 3) since v 1 dv = e e v1 0 dv 0 dv = volp k ). Triangulating P k and computing its volume In order to compute the volume volp k ),we determine a triangulation of the polytope P k, compute the volumes of the simplices in that triangulation, and add them up. For this, we need first to determine the vertices of P k. Lemma 1 The polytope P k has 2k vertices a i, b i,i = 1,...,k, as follows: 0, j = 0 aj i = 1 1 j i i, 0, i<j<k 1 j = 0 bj i i, = 1 1 j i 1 i k,0 j k 1) i, 0, i<j<k Proof The vertices are obtained as the elements of P k satisfying with equality k linearly independent inequalities from the system of k + 2 inequalities defining P k. It is easy to see that in each such choice, precisely one of the inequalities 0 v 0 or v 0 v 1 must hold with equality. For each, there are k choices of forcing equality on k out of the k + 1 remaining inequalities, and each choice gives exactly one of the vectors appearing in the above claimed list.

8 Fig. 5 The convex polytope P 3 Note that for each i, the vectors a i and b i agree on all coordinates except for the 0th coordinate. For instance, for k = 3 these vectors are: a 1 = 0, 1, 0), a 2 = 0, 1 2, 1 ), a 3 = 0, 1 2 3, 1 ), 3 1 b 1 = 1, 1, 0), b 2 = 2, 1 2, 1 ), b 3 = 1 2 3, 1 3, 1 3 ). For k = 4 these vectors are: a 1 = 0, 1, 0, 0), a 2 = 0, 12, 12, 0 ), a 3 = 0, 1 3, 1 3, 1 ), a 4 = 0, 1 3 4, 1 4, 1 ), 4 1 b 1 = 1, 1, 0, 0), b 2 = 2, 1 2, 1 ) 1 2, 0, b 3 = 3, 1 3, 1 3, 1 ) 1, b 4 = 3 4, 1 4, 1 4, 1 ). 4 The case k = 3 is illustrated in Fig. 5. Lemma 2 For s = 1,...,k, the following polytope is a k-dimensional simplex whose volume is Here, conv stands for the convex hull. s := conva 1,...,a s, b s,...,b k }, vol s ) = 1 k!) 2 s. Proof The convex hull := convv 0, v 1,...,v k } of k + 1 points in R k is a simplex if and only if the determinant δ := detv 1 v 0,...,v k v 0 ) is nonzero, in which case its volume is

9 given by vol ) = 1 δ. So,fors = 1,...,k, we shall compute the following determinant, k! δ s := deta 1 a s,...,a s 1 a s, b s a s, b s+1 a s,...,b k a s ). For any vector v = v 0,v 1,...,v k 1 ) in R k,let v := v 1,...,v k 1 ) be its projection to R k 1 obtained by erasing the 0th coordinate. Then, for all i, wehave b i = ā i. Thus, expanding the determinant δ s on the column b s a s = 1 s, 0,...,0),wefindthat δ s = 1 s μ,where μ := detā 1 ā s, ā 2 ā s,...,ā s 1 ā s, ā s+1 ā s,...,ā k 1 ā s, ā k ā s ). Subtracting the first column from every other column, flipping its sign, and permuting, we get μ = 1) s 1 detā 2 ā 1, ā 3 ā 1,...,ā k 1 ā 1, ā k ā 1 ). Using the multilinearity of the determinant, μ can be expanded as a sum of 2 k 1 determinantal terms. Among these, each term in which ā 1 occurs more than once vanishes, and so does each term in which both ā k 1 and ā k occur, being a scalar multiple of one another see definition of the a i in Lemma 1). It is easy to see, then, that only two terms remain, giving μ = 1) s detā 2, ā 3,...,ā k 2, ā k 1, ā 1 ) + detā 2, ā 3,...,ā k 2, ā 1, ā k ) ). These two remaining terms are easy to compute, since by permuting a 1 to be the first column in each, the corresponding matrices become upper triangular. Since ā i i = 1, we finally obtain i μ =detā 1, ā 2,...,ā k 2, ā k 1 ) detā 1, ā 2,...,ā k 2, ā k ) 1 = 1 2 k 2) k 1) k 2) k = 1 k!. Summing up, we get, as claimed, for s = 1,...,k, vol s ) = 1 k! δ s = 1 1 k! s μ = 1 k!) 2 s. Lemma 3 The k simplices 1,..., k form a triangulation of P k. For example, when k = 3, 1 = conva 1, b 1, b 2, b 3 } 2 = conva 1, a 2, b 2, b 3 } 3 = conva 1, a 2, a 3, b 3 }. Proof We show by induction on s that 1,..., s form a triangulation of the polytope s ) Q s := conv i = conva 1,...,a s, b 1,...,b k }. For s = 1 this is surely true since Q 1 = 1. Suppose now the claim holds for Q s. Consider the next point a s+1 to be added so as to form Q s+1 = convq s a s+1 }). We claim that F s := conva 1,...,a s, b s+1,...,b k } is a facet of Q s, and that it is the only facet of Q s for

10 which a s+1 is beyond the hyperplane H s spanned by F s,thatis,h s separates a s+1 from Q s. For more details on the so-called beneath-beyond method for the inductive construction of polytopes see Sect. 5.2 of Grübaum 2003). To prove the claim, it suffices to show, for each vertex v of Q s which does not lie in F s, that the open line segment between a s+1 and v intersects F s. Now, the vertices of Q s which are not in F s are b 1,...,b s, so consider any of the b i with i s. By the definition of the a j and b j see Lemma 1), we have ib i a i ) = s + 1)b s+1 a s+1 ), which by suitable manipulation gives s + 1 s i as+1 + i s i bi = s + 1 s i bs+1 + i s i ai. The left hand side of this equation is in the open segment between a s+1 and b i whereas the right hand side is in F s since so are a i and b s+1, proving the claim. Since F s is a simplicial) facet of Q s and the only one for which a s+1 is beyond the hyperplane H s spanned by F s, a triangulation for Q s+1 = convq s a s+1 }) is obtained from the triangulation 1,..., s of Q s by adding the single new simplex convf s a s+1 }) = s+1. This completes the induction. Since P k = Q k, it follows that 1,..., k is a triangulation of P k and the lemma follows. Combining the above statements implies k! v 1 dv = k! volp k ) = k! e = k! s=1 vol s ) s=1 1 sk!) 2 = 1 k! k ) = 1 k! ζ 1k). This completes the geometric proof. 5 A probabilistic proof of 1) Let Y 1,Y 2,...,Y k be independent unit-exponential random variables and let T = k 1 Y j be their sum. Then, by Sect. 1, Y1 T, Y 2 T,..., Y ) k d = V 1,V 2,...,V k ), T both random vectors are uniform over k. For the probabilistic proof we need two Lemmas. Lemma 4 The random vector Y 1 T, Y 2 T,..., Y k T ) and T are independent. Proof Let E i = Y i /T and consider the transformation from Y 1,Y 2,...,Y k ) to E 1,E 2,..., E k 1,T). The inverse transformation is given by Y 1 = TE 1 Y 2 = TE 2

11 . Y k 1 = TE k 1 Y k = T1 E 1 E k 1 ). It is straight forward to show that the Jacobian of the transformation is equal to T k 1.It follows that the joint density function is given by f E1,...,E k 1,T e 1,...,e k 1,t)= 1e k } exp t)t k 1 1t >0}, where e = e 1,e 2,...,e k 1 ). Lemma 5 For any two random variables U and W with finite first moments and EW 0), E U W = EU EW COV W, U ) = 0. W Proof COV W, U ) = 0 EU = EW E U W W E U W = EU EW. Combining the last two Lemmas, EV k) = E max Y j T = E max Y j ET = k k. 4) In view of 2), the probabilistic proof of 1) is complete. Remark 2 Equation 4) can be derived by integrating on [0, 1]) P V k) >x}= ) k 1) j 1 1 jx) k 1 + j. 5) j=1 According to Darling1953), this expression dates back to W.A. Whitworth in However, we have not found any reference to 4) in the entire statistical and probabilistic literature. Remark 3 The argument which led us to 4) can be used to compute other moments of V k). For instance, if ζ 2 k) = k 1 j 2,then EVk) 2 = Emax Y j ) 2 = Varmax Y j ) + ζ1 2k) = ζ 2k) + ζ1 2k). ET 2 VarT ) + k 2 2k + k 2 Acknowledgements The research of Shmuel Onn was supported in part by a grant from ISF the Israel Science Foundation and by a VPR grant at the Technion. Both authors acknowledge the support of the Fund for Promotion of Research at the Technion.

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