NORMALIZED INCOMPLETE BETA FUNCTION: LOG-CONCAVITY IN PARAMETERS AND OTHER PROPERTIES

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1 DOI 117/s z Journal of Mathematical Sciences, Vol 217, No 1, August, 216 NORMALIZED INCOMPLETE BETA FUNCTION: LOG-CONCAVITY IN PARAMETERS AND OTHER PROPERTIES D B Karp UDC The logarithmic concavity/convexity in parameters of the normalized incomplete beta function has been established by Finner and Roters in 1997 as a corollary of a rather difficult result, based on generalized reproductive property of certain distributions In the first part of this paper, a direct analytic proof of the logarithmic concavity/convexity mentioned above is presented These results are strengthened in the second part, where it is proved that the power series coefficients of the generalized Turán determinants formed by the parameter shifts of the normalized incomplete beta function have constant sign under some additional restrictions The method of proof suggested also leads to various other new facts, which may be of independent interest In particular, linearization formulas and two-sided bounds for the above-mentioned Turán determinants, and also two identities of combinatorial type, which we believe to be new, are established Bibliography: 16 titles 1 Motivation and introduction The beta distribution is perhaps the single most important continuous compactly supported probability distribution Its particular cases include the Wigner semicircle, the Marchenko Pastur and the arcsine laws; it is important in Bayesian analysis as conjugate prior to binomial and geometric distributions [8]; it plays a role in a large number of applications ranging from population genetics to project management The beta distribution is defined by the density x a 1 1 x) b 1 /Ba, b) supported on [, 1], where Ba, b) is Euler s beta function [3] The cumulative distribution function CDF) of the beta density is given by the normalized incomplete beta function x I x a, b) B xa, b) Ba, b) 1 t a 1 1 t) b 1 dt t a 1 1 t) b 1 dt In the recent paper [4], containing the mathematical analysis supporting a recent investigation related to lottery frauds in Florida, the authors solve the following optimization problem: m c i α i min under the constraint i1 fα 1,α 2,,α m ) m I pi w i,α i w i +1) ε i1 For efficient numerical solution of such a problem, it is desirable that the function f be quasiconcave, which implies that the feasible set of the problem is convex and the local minimizer is also global Since monotonic transformations do not alter quasi-concavity, it is sufficient to show that logf is a concave function This fact, in turn, is implied by the log-concavity of b I x a, b) on, ) The purpose of this paper is to investigate the log-concavity/convexity Far Eastern Federal University, Vladivostok, Russia; Institute of Applied Mathematics, Far Eastern Branch of the Russian Academy of Sciences, Vladivostok, Russia, dimkrp@gmailcom Published in Zapiski Nauchnykh Seminarov POMI, Vol 44, 215, pp Original article submitted September 21, /16/ Springer Science+Business Media New York 91

2 properties of the function I x a, b) viewed as a function of the parameters a and b Inparticular, we demonstrate that b I x a, b) is indeed log-concave on, ), whereas a I x a, b) is logconvex on, ) forb, 1) and log-concave for b>1 The log-concavity of b I x a, b) is equivalent to the positivity of the generalized Turán determinant I x a, b + α)i x a, b + β) I x a, b)i x a, b + α + β) for all α, β >, whereas the log-convexity log-concavity) of a I x a, b) isequivalenttothe negativity positivity) of I x a + α, b)i x a + β,b) I x a, b)i x a + α + β,b) We go one step further and study the signs of the power series coefficients in powers of x) of the above Turán determinants Under some additional restrictions, we demonstrate that these coefficients are of the same sign We further conjecture that such restrictions can be removed without altering the results Our method of proof also leads to various companion results, which may be of independent interest In particular, we establish linearization formulas and two-sided bounds for the above Turán determinants Further, we find two combinatorial-style identities for finite sums, which we believe to be new When this paper was nearly completed, we discovered that the log-concavity/convexity of the CDF of beta distribution was demonstrated in 1997 by Finner and Roters in their fundamental work [7] Partial results in this direction were given by the same authors in [6] and by Das Gupta and Sarkar in [9] However, their log-concavity proofs are very involved and indirect they appear as a by-product of generalized reproductive property of certain probability measures Finner and Roters note in their paper that we are now able to conclude a result for the Beta distribution, which seems to be very hard to obtain by usual analytic methods The proofs presented in this paper are precisely by usual analytic methods The power series coefficients of the Turán determinants were not considered by the above authors, so our results in this direction strengthen the achievements of [7] We conclude the introduction by presenting the following alternative expression for the normalized incomplete beta function in terms of the Gauss hypergeometric function [3]: 2F 1 α, β; γ; z) n α) n β) n z n, α) n γ) n n! Indeed, using Euler Pochhammer integral representation 2F 1 α, β; γ) Γα + n) Γα) Γγ) 1 u β 1 1 u) γ β 1 1 ux) α du, Γβ)Γγ β) we derive, by a change of variable and application of Euler s formula Ba, b) Γa)Γb)/Γa + b), Γa + b)xa I x a, b) Γ)Γb) 2 F 1 1 b, a; ;x) Further, on applying the other Euler formula 92 2F 1 α, β; γ) 1 x) γ α β 2F 1 γ α, γ β; γ),

3 we obtain the following well-known expression: Γa + b)xa I x a, b) Γ)Γb) 1 x)b 2F 1 a + b, 1; ;x) x a 1 x) b Γa + b + n) Γb)Γ+n) xn n 2 Log-concavity of I x a, b) in a and b The proof of the next theorem has been inspired by the log-concavity proof for incomplete gamma function given in [1]; also see a related result in [2] Theorem 1 For every fixed a> and x, 1), the function b I x a, b) is strictly logconcave on, ) Proof We need to show that 2 b 2 logi xa, b)) < It is easy to compute 2 b 2 logba, b)) ψ b) ψ a + b) >, 2) where ψz) Γ z)/γz) is the logarithmic derivative of the gamma function, and the inequality follows from the fact that z ψ z) isdecreasingon, ) because Further, ψ z) B x a, b) 2 2 b 2 logi xa, b)) B x a, b) n t a 1 1 t) b 1 log1 t) dt 1 n + z) 2 t a 1 1 t) b 1 [log1 t)] 2 dt 2 B x a, b) 2 ψ b) ψ a + b)) : U a,b x) We have U a,b ) U a,b 1) The second equality can be seen by setting x 1 in logi x a, b)) and then differentiating with respect to b, which is legitimate by continuous double differentiability; alternatively, one can set x 1 in the above formula and compute the integrals We aim at demonstrating that U a,b x) is decreasing on,x 1 ) and increasing on x 1, 1) for a certain x 1, 1) This will prove that U a,b x) < forx, 1) for all a, b > We have U a,b V a,b x) : x) x a 1 1 x) b 1 t a 1 1 t) b 1 [log1 t)] 2 dt + B x a, b)[log1 x)] 2 2log1 x) t a 1 1 t) b 1 log1 t) dt 2B x a, b)ψ b) ψ a + b)) It is rather straightforward to see that V a,b ) and V a,b 1) + We aim at demonstrating that V a,b x) isdecreasingon,x 2 ) and increasing on x 2, 1) for a certain x 2, 1) This will 1) 93

4 imply that it changes sign on, 1) exactly once from minus to plus, which, in turn, implies the same conclusion for U a,b x) By taking the next derivative, after cancellations we obtain W a,b x) : 1 1 x)v a,b 2 x) B x a, b)log1 x)+ t a 1 1 t) b 1 log1 t) dt x a 1 1 x) b ψ b) ψ a+b)) It is clear that W a,b 1) + The value at zero depends on a: if<a<1, then W a,b ) ; ifa 1,thenW a,b ) ψ b) ψ a + b)) < ; if a>1, then W a,b ) Below, we will demonstrate that W a,b x) isincreasingon, 1) if <a 1andW a,b x) is decreasing on,x 3 ) and increasing on x 3, 1) for a certain x 3, 1) if a>1 This will imply that it changes sign on, 1) exactly once from minus to plus, which, in turn, implies the same conclusion for V a,b x) By taking the next derivative, we obtain Z a,b x) :1 x)w a,b x) B x a, b) ψ b) ψ a + b))x a 2 1 x) b a 1)1 x) bx) We have Z a,b 1) Ba, b) > At x, five cases reveal themselves: Case I: if <a<1, then Z a,b ) + In this case, Z a,b x) > on, 1) because a 1)1 x) bx) < andψ b) ψ a + b) > Case II: if a 1,thenZ a,b ) bψ b) ψ a+b)) > ; below, we will show that Z a,b x) on, 1), so that Z a,b x) bψ b) ψ a + b)) > for all x, 1) Case III: if 1 <a<2, then Z a,b ) Case IV: if a 2,thenZ a,b ) ψ b) ψ a + b)) < Case V: if a>2, then Z a,b ) By taking one more derivative, we obtain Q a,b x) :Z a,b x)x3 a 1 x) 1 b x 2 ψ b) ψ a + b))a + b 1) 2 x 2 a 1)2a +2b 3)x +a 1)a 2)) Straightforward calculation yields Q a,b ) ψ b) ψ a + b))a 1)a 2), Q a,b 1) 1 b 2 ψ b) ψ a + b)) We see that Q,b 1) 1 and Q 1,b 1) because ψ 1 + b) ψ b) 1/b 2 Sincea Q a,b 1) is decreasing on, ), these formulas lead us to the following conclusions Case I: if <a<1, then Q a,b ) < andq a,b 1) > for all b> Since Q a,b x) is quadratic, this implies that it has exactly one change of sign on, 1) Case II: if a 1,thenQ a,b x) on[, 1] for all b> because ψ 1 + b) ψ b) 1/b 2 Case III: if 1 <a<2, then Q a,b ) > andq a,b 1) < for all b> Since Q a,b x) is quadratic, this implies that it has exactly one change of sign on, 1) Case IV: if a 2,thenQ a,b ) and Q a,b 1) < for all b> It follows that Q a,b x) > on,α)andq a,b x) < onα, 1) for a certain α, 1) An apparent alternative Q a,b x) < on, 1) cannot hold because Z a,b x) would then be decreasing on, 1), whereas Z a,b ) < and Z a,b 1) > rules out this possibility Case V: if a > 2, then Q a,b ) < andq a,b 1) < for all b > This implies that Q a,b x) follows the sign pattern, +, ) on, 1) An apparent alternative Q a,b x) < on, 1) cannot hold because Z a,b x) would then be decreasing on, 1), whereas Z a,b ) and Z a,b 1) > rules out this possibility 94

5 Looking carefully at each case, we see that for all b>the function Z a,b x) is positive on, 1) if <a 1and changes sign from minus to plus if a > 1 This implies that W a,b x) behaves in the same way, so that in all cases, W a,bx) changes sign exactly once from minus to plus This, in turn, implies that V a,b x) has the same behavior, whence V a,bx) isfirst decreasing and then increasing In view of the boundary values at x andx 1, this means that V a,b x) changes sign exactly once from minus to plus Finally, it follows that U a,b x) is negative on,x 1 )andpositiveonx 1, 1) for a certain x 1, 1), yielding U a,b x) < on, 1), as is claimed The definition of I x a, b) immediately leads us to the reflection formula I x a, b) 1 I 1 x b, a) This implies 2 a 2 logi xa, b)) [I x a, b)] 2 { I 1 x b, a) 2 a 2 I 1 xb, a) [ ] 2 } a I xb, a) [I x a, b)] a 2 I 1 xb, a) In accordance with Theorem 1, the first term is negative The second term, however, may change sign depending on the values of b and x, as is demonstrated by numerical evidence Hence we cannot draw any definitive conclusion about the log-concavity of a I x a, b) from the reflection formula Indeed, it turns out that the result depends on the value of b Nevertheless, a method similar to that used in the proof of Theorem 1 also works here when combined with the following lemma Lemma 1 For a fixed α>, define f α x) :, ), ) by f α x) ψx + α) ψx)) 2 + ψ x + α) ψ x) Then f α x) < if <α<1 and f α x) > if α>1 Proof Using the recurrence relations ψx +1)ψx)+1/x and ψx +1)ψx) 1/x 2,we compute fx +1) fx) ψx + α +1) ψx +1)) 2 ψx + α) ψx)) 2 + ψ x + α +1) ψ x + α)+ψ x) ψ x +1) ψx + α +1) ψx +1)+ψx + α) ψx))ψx + α +1) ψx +1) ψx + α)+ψx)) + 1 x 2 1 x + α) 2 2ψx + α)+ 1 x + α 2ψx) 1 ) 1 x x + α 1 ) + 1 x x 2 1 x + α) ) 2 α α α2x + α) 2ψx) 2ψx + α)+ + xx + α) xx + α) x 2 x + α) 2 2α ψx) ψx + α)+ 1 ) xx + α) x By writing g α x) for the function in the parentheses and using the integral representations ψx) γ + e t e xt 1 e t dt and 1 x e xt dt, 95

6 where γ stands for the Euler Mascheroni constant, we obtain g α x) e xt e αt e t ) 1 e t dt The latter formula makes it obvious that g α x) > for<α<1andg α x) < forα>1 Further, from the asymptotic formulas ψx) logx) 1 2x + Ox 2 ), ψ x) 1 x + 1 2x 2 + Ox 3 ), x, we conclude that lim fx) Altogether this implies that for <α<1, x and, for α>1, fx) <fx +1)<fx +2)< < lim fx + m) m fx) >fx +1)>fx +2)> > lim fx + m), m which proves the lemma Stronger results for a function similar to f α x) but still slightly different from it can be found in [16, Theorem 12] Theorem 2 Let x, 1) be fixed Then a) if <b<1, then the function a I x a, b) is strictly log-convex on, ); b) if b 1, then the function a I x a, b) is equal to x a, whence it is log-neutral; c) if b>1, then the function a I x a, b) is strictly log-concave on, ) Proof The case b 1 can be verified directly It remains to show that 2 a 2 logi xa, b)) > if <b<1 and 2 a 2 logi xa, b)) < if b>1 For completeness, however, we will include b 1 into the forgoing considerations By symmetry, from 2) we have Further, 2 a 2 logba, b)) ψ a) ψ a + b) > 96 B x a, b) 2 2 a 2 logi xa, b)) B x a, b) t a 1 1 t) b 1 logt) dt t a 1 1 t) b 1 [logt)] 2 dt 2 B x a, b) 2 ψ a) ψ a + b)) : U a,b x)

7 Direct verification yields U a,b ) U a,b 1) Next, we have U a,b V a,b x) : x) x a 1 1 x) b 1 t a 1 1 t) b 1 [logt)] 2 dt + B x a, b)[logx)] 2 2logx) Repeatedly using L Hôpital s rule, we compute the limits Similarly, lim x B x a, b) lim [logx)] 2 x so that V a,b ) Further, 1 t a 1 1 t) b 1 logt)dt 2B x a, b)ψ a) ψ a + b)) x a 1 x) b 1 2log1/x)) lim log1/x)) 3 x x a 1 2 lim 3log1/x)) 2 x ax a lim logx) x 3 2 lim 2log1/x) x a 2 x a 3 lim t a 1 1 t) b 1 logt) dt, a 3 x a x 1 t a 1 1 t) b 1 [logt)] 2 dt Ba, b)[ψa) ψa + b)) 2 + ψ a) ψ a + b)], so that V a,b 1) Ba, b)[ψa) ψa + b)) 2 ψ a) ψ a + b))] From Lemma 1 it follows that a) V a,b 1) < for<b<1; b) V a,b 1) for b 1;c) V a,b 1) > forb>1 We are going to demonstrate that V a,b x) has exactly one change of sign in cases a) and c) and is identically zero in case b) By taking the next derivative, upon cancellations we obtain W a,b x) : x 2 V a,b x) B x a, b)logx) t a 1 1 t) b 1 logt)dt x a 1 x) b 1 ψ a) ψ a + b)) The boundary values are W a,b ) and, <b<1, W a,b 1), b 1, Ba, b)ψa + b) ψa)) >, b > 1 These values follow from the evaluation t a 1 1 t) b 1 logt) dt Ba, b)ψa) ψa + b)) 97

8 and the recurrence relations ψ) ψa) 1/a, ψ ) ψ a) 1/a 2 Bytakingthe next derivative, we obtain Z a,b x) :xw a,b x) B xa, b)+ψ a) ψ a + b))x a 1 x) b 2 b 1)x a1 x)) Clearly, Z a,b ) At x 1, the following five cases reveal themselves: Case I: if <b<1, then Z a,b 1) Case II: if b 1,thenZ a,b 1) Below, we will show that Z a,b x) on, 1), whence Z a,b x) for all x, 1) Case III: if 1 <b<2, then Z a,b 1) + Case IV: if b 2,thenZ a,b 1) Ba, b)+ψ a) ψ a + b) > Case V: if b>2, then Z a,b 1) Ba, b) > Finally, computing one more derivative and making some rearrangements, we arrive at Z a,b Q a,b x) x) x a 1 1 x) b 3 1 x) 2 ψ a) ψ a + b))a + b 1) 2 x 2 2)b 1) + 2a 2 )x + a 2 ) The boundary values are Q a,b ) 1 a 2 ψ a) ψ a + b)), Q a,b 1) ψ a) ψ a + b))b 1)b 2) We see that Q a, ) 1 and Q a,1 ) because ψ 1 + a) ψ a) 1/a 2 Sinceb Q a,b ) is decreasing on, ), these formulas lead us to the following conclusions: Case I: if <b<1, then Q a,b ) > andq a,b 1) < for all a> Since Q a,b x) is quadratic, this implies that it has exactly one change of sign on, 1) Case II: if b 1,thenQ a,b x) on[, 1] for all a>becauseψ 1 + a) ψ a) 1/a 2 Case III: if 1 <b<2, then Q a,b ) < andq a,b 1) > for all b> Since Q a,b x) is quadratic, this implies that it has exactly one change of sign on, 1) Case IV: if b 2,thenQ a,b ) < andq a,b 1) for all a> It follows that Q a,b x) < on,α)andq a,b x) > onα, 1) for a certain α, 1) An apparent alternative Q a,b x) < on, 1) cannot hold because Z a,b x) would then be decreasing on, 1), whereas Z a,b ) and Z a,b 1) > rules out such a possibility Case V: if b > 2, then Q a,b ) < andq a,b 1) < for all a > This implies that Q a,b x) follows the sign pattern, +, ) on, 1) An apparent alternative Q a,b x) < on, 1) cannot hold because Z a,b x) would then be decreasing on, 1), whereas Z a,b ) and Z a,b 1) > rules out such a possibility Tracing back the primitives Q a,b Z a,b W a,b V a,b U a,b,weseethatforalla>, the function Z a,b x) changes sign from plus to minus on, 1) if <b<1, is identically zero if b 1, and changes sign from minus to plus if b>1 This implies that W a,b x) andv a,b x) behaveinthesameway,sothatu a,b x) is first increasing and then decreasing if <b<1, is identically zero if b 1, and is first decreasing and then increasing if b>1 In view of the boundary values at x andx 1, this means that U a,b x) > on, 1) for <b<1and U a,b x) < forb>1, as is claimed 3 Coefficient-wise log-concavity of I x a, b) As was mentioned in Sec 1, the logarithmic concavity of b I x a, b) demonstrated in Theorem 1 is equivalent to the inequality φ a,b,α,β x) :I x a, b + α)i x a, b + β) I x a, b)i x a, b + α + β) >, 3) valid for all positive values of a, b, α, andβ and all x, 1) This inequality expresses the strict Wright log-concavity of b I x a, b) a property equivalent to log-concavity for 98

9 all continuous functions [13, Chapter I4] and [14, Sec 11] On the other hand, formula 1) implies that the function of φ a,b,α,β x) defined in 3) has the power series expansion φ a,b,α,β x) x 2a 1 x) 2b+α+β k φ k x k 4) It is then natural to ask whether and when the coefficients φ k are positive In general, for a given formal power series fμ) f k μ)x k 5) k with nonnegative coefficients f k μ) thatdependcontinuouslyonaparameterμ from a real interval I, we will say that μ fμ) is coefficient-wise Wright log-concave on I for α A [, ), β B [, ) if the formal power series for the product difference fμ + α)fμ + β) fμ)fμ + α + β) has nonnegative coefficients at all powers of x for α A, β B and all μ I If the coefficients are strictly positive we say that the corresponding property holds strictly If the coefficients are nonpositive we talk about coefficient-wise Wright log-convexity These concepts, using a slightly different terminology, were introduced by the author jointly with S I Kalmykov) in [1] The claims of Theorem 2 can be restated in terms of the function ψ a,b,α,β x) :I x a + α, b)i x a + β,b) I x a, b)i x a + α + β,b) x 2a+α+β 1 x) 2b ψ k x k 6) k as follows: If a>and<x<1, then ψ a,b,α,β x) < for<b<1andψ a,b,α,β x) > for b>1 Just like with φ a,b,α,β x) it is then natural to consider the coefficient-wise log-convexity and log-concavity of a x a 1 x) b I x a, b), ie, study the sign of the coefficients ψ k The main purpose of this section is to give a partial solution to the following conjectures, which are substantial strengthenings of Theorems 1 and 2, respectively Conjecture 1 The coefficients φ k are positive for all a, b, α, β > and x, 1), so that b x a 1 x) b I x a, b) is strictly coefficient-wise Wright log-concave on, ) for these values of parameters Conjecture 2 The coefficients ψ k are negative for all a, α, β > and x, 1) if <b<1 and positive if b>1, so that a x a 1 x) b I x a, b) is strictly coefficient-wise Wright logconvex on, ) for α, β >, x, 1), and <b<1; anda x a 1 x) b I x a, b) is strictly coefficient-wise Wright log-concave on, ) for α, β >, x, 1), andb>1 Our approach also leads us to a number of related results, which may be of independent interest These results include alternative formulas for the coefficients φ k and ψ k, linearization identities for the product differences of hypergeometric functions, two-sided inequalities for the Turán determinants formed by normalized incomplete beta functions, and two presumably new combinatorial identities We will need the following lemma see [1, Lemma 3]) Lemma 2 Assume that μ fμ) is coefficient-wise Wright log-concave on [, ) for α 1, β, ie, fμ +1;x)fμ + β) fμ)fμ + β +1;x) 99

10 has nonnegative coefficients at all powers of x for all μ, β Thenμ fμ) is coefficientwise Wright log-concave on [, ) for α N, β α 1 Theorem 3 Let α N Then the following identity holds true: α 1 ψ a,b,α,β x) x 2a+α+β 1 x) 2b j Γa + b + j)γa + b + β + α j 1) [Γb)] 2 Γa + j +1)Γa + β + α j) { ) a + b + j 1,a+ b + j +1 a + j +1 2 F 1 a + j +2 a + b + β + α 1 j a + β + α j 2F 1 1,a+ b + β + α j a + β + α j +1 )} 7) Proof First we investigate the function ψ a,b,1,μ x) defined in 6) Using 1), we compute ψ a,b,1,μ x)[γb)] 2 x 2a+μ+1 1 x) 2b A a,b,1,µ {}}{ Γa + b)γa + b + μ) Γ)Γa + μ +1) { a + b 2 F 1 a + b + μ a + μ +1 2 F 1 A a,b,1,μ m 1,a+ b +1 )2F a ,a+ b )2F 1 x m m k 1,a+ b + μ a + μ +1 1,a+ b + μ +1 a + μ +2 ) )} { a + b)a + b +1) k a + b + μ) m k )a +2) k a + μ +1) m k a + b + μ)a + b) ka + b + μ +1) m k a + μ +1)) k a + μ +2) m k A a,b,1,μb 1) m x m a + b) k a + b + μ) m k m 2k + μ) )a + μ +1) a +2) m k a + μ +2) m k k { ) a + b 1,a+ b +1 A a,b,1,μ 2 F 1 a + b + μ 1,a+ b + μ +1 a +2 a + μ +1 2 F 1 a + μ +2 } )}, wherewehaveusedthesummationformula m k [ a + b) k a + b + μ) m k )a + μ +1) a + b)m+1 m 2k + μ) 1 a + b + μ) ] m+1, a +2) k a + μ +2) m k b 1) ) m+1 a + μ +1) m+1 demonstrated in [11, p 72] Next, we set ga) I x a, b)/x a 1 x) b and compute 1

11 ψ a,b,α,β x 2a+α+β ga + α)ga + β) ga)ga + α + β) 1 x) 2b [ga + α)ga + β) ga + α 1)ga + β +1)] +[ga + α 1)ga + β +1) ga + α 2)ga + β +2)]+ +[ga +2)ga + β + α 2) g)ga + β + α 1)] +[g)ga + β + α 1) ga)ga + β + α)] 1 α 1 { a + b + j 1,a+ b + j +1 [Γb)] 2 A a+j,b,1,α+β 2j 1 a + j +1 2 F 1 a + j +2 j ) a + b + β + α 1 j 1,a+ b + β + α j 2F 1 a + β + α j a + β + α j +1 )} Substituting the value of A a+j,b,1,α+β 2j 1 into this formula yields 7) The next corollary refines Theorem 2 and gives a partial solution to Conjecture 2 Corollary 1 Let α N Then the coefficients ψ k defined in 6) can be computed by ψ k α 1 j Γa + b + j)γa + b + β + α j 1) [Γb)] 2 Γa + j +1)Γa + β + α j) { a + b + j)k+1 a + j +1) k+1 a + b + β + α 1 j) k+1 a + β + α j) k+1 } 8) Furthermore, if <b<1, thenψ k < for all α, β > and if b>1, thenψ k > for β α 1, β>, andα N Proof Formula 8) follows immediately from 7) The case <b<1 falls under conditions of [1, Theorem 3], and the claim follows from this theorem Further, in accordance with Lemma 2, for β α 1 the sign of ψ k ψ k α, β) is determined by the sign of ψ k 1,β) In view of a) k Γa + k)/γa), for α 1wehave [ ] Γa + b)γa + b + μ) Γa + k + b)γ) Γa + μ + k + b)γa + μ +1) ψ k 1 1,β) Γ)Γa + μ +1) Γa + b)γa + k +1) Γa + μ + b)γa + μ + k +1) Define the function Γa + μ + k + x)γa + x) fx) Γa + k + x)γa + μ + x) This function is decreasing according to [5, Theorem 6] It is straightforward that ψ k 1 1,β) < isequivalenttof1) <fb), whereas ψ k 1 1,β) > isequivalenttof1) >fb), so that the claim follows from the decrease of fx) This corollary can also be derived from our previous results in [1, Theorem 3] and [11, Theorem 2]; however, the above derivation has the advantage of presenting explicit formulas for the coefficients ψ k, which do not follow from these references Remark Substituting 1) for I x into the definition of ψ a,b,α,β given in 6) and changing notation, we can write 7) as the following linearization identity for the quadratic form in 11

12 Gauss hypergeometric functions: ) ) a) n+1 a + n +1, 1 a + β,1 2F 1 c) n+1 c + n +1 2F 1 c + β a + β) ) ) n+1 a, 1 a + β + n +1, 1 2F 1 c + β) n+1 c 2F 1 c + β + n +1 n { ) a) j a + β) n j a + j a + j +1, 1 c) j j c + β) n j c + j 2 F 1 c + j +1 a + β + n j a + β + n +1 j, 1 c + β + n j 2 F 1 c + β + n +1 j )} 9) Here, n α 1 is an arbitrary nonnegative integer, whereas the other parameters, as well as x, can be arbitrary complex numbers by analytic continuation In particular, for n the above identity reduces after some rearrangement) to ) ] a 1,a+1 1,a+ β c 2 F 1 )[2F c c + β ) ] a + β 1,a+ β +1 1,a c + β 2 F 1 )[2F c + β c Formula 9) leads to a curious combinatorial identity, which is presumably new Corollary 2 For arbitrary nonnegative integers m, n, the following identity holds: m k a) k a + β) m k b) k b + β) m k { a + k)n+1 a + β + m k) n+1 b + k) n+1 b + β + m k) n+1 n j a) j a + β) n j b) j b + β) n j } { a + j)m+1 b + j) m+1 a + β + n j) m+1 b + β + n j) m+1 Proof The identity claimed is obtained by equating the coefficients at x m on both sides of 9) and using the formula μ) n+1 μ + n +1) k μ) k μ + k) n+1 The next corollary presents two-sided bounds for the Turán determinant formed by shifts of I x a, b) in the first parameter Corollary 3 Let b>1, ν N, anda>ν 1 Then for all x, 1) Here, mx a 1 x) b I x a, b) 2 I x a ν, b)i x a + ν, b) MI x a, b) 2 1) m [Γa + b)] 2 [Γb)] 2 [Γ)] 2 Γa + b ν)γa + b + ν) [Γb)] 2 Γa ν +1)Γa + ν +1), M ) νa + b ν) ν ν) ν a + b) ν ) ν a + b ν) ν Proof The left-hand-side inequality in 1) is just a rewriting of the inequality 12 ψ a ν,b,ν,ν x 2a 1 x) 2b ψ, }

13 which follows from the nonnegativity of the coefficients ψ k for all k, 1, In order to prove the upper bound, consider the function a Ga) I xa, b)γ) x a 1 x) b Γa + b) 1 Γb) n a + b) n ) n x n According to [12, Theorem 2], the above function is log-convex for b>1, whence for a>ν 1 we have Ga ν)ga + ν) [Ga)] 2 This inequality can be written as I x a, b ν)i x a, b + ν) [Γ)]2 Γa + b + ν)γa + b ν) [Γa + b)] 2 Γa + ν +1)Γa ν +1) [I xa, b)] 2 Simple rearrangement of the above inequality gives the upper bound in 1) In order to investigate the power series coefficients of the function φ a,b,α,β x) defined in 3), we need the following lemma Lemma 3 The following identity holds: m k Proof Denote a + b) k a + b + β) m k ) k ) m k b2k m)+βa + k)) aa + b + β) m+1 a + b) m+1 ) ) m u k a + b) ka + b + β) m k ) k ) m k b2k m)+βa + k)), k, 1,,m An application of Gosper s algorithm yields the anti-differences u k v k+1 v k, v k a2 a + b) k a + b + β) m k+1 a) k a) m k+1, k, 1,,m+1 Straightforward substitution provides verification of this formula Hence we obtain m u k v m+1 v aa + b + β) m+1 a + b) m+1 ) ) m k Theorem 4 Let α N Then the following identity holds true: α 1 φ a,b,α,β x) x 2a 1 x) 2b+α+β aγa + b + j)γa + b + β + α j 1) [Γ)] 2 Γb + j +1)Γb + β + α j) j ) 1,a+ b + β + α j {a + b + β + α j 1) 2 F 1 1,a+ b + j +1 a + b + j) 2 F 1 )} 11) 13

14 Proof First we investigate the function φ a,b,1,β x) definedin3)wehave φ a,b,1,β x) x 2a 1 x) 2b+1+β m { x m Γa + b +1+k)Γa + b + β + m k) Γb +1)Γ+k)Γb + β)γ+m k) m k } Γa + b + k)γa + b + β +1+m k) Γb)Γ+k)Γb +1+β)Γ+m k) m x m Γa + b + k)γa + b + β + m k) Γb)Γb + β)γ+k)γ+m k) m k { a + b + k a + b + β + m k } b b + β Γa + b)γa + b + β) Γb +1)Γb + β +1)Γ) 2 m x m a + b) k a + b + β) m k b2k m)+βa + k)) ) k ) m k m Application of Lemma 3 then gives k φ a,b,1,β x 2a 1 x) 2b+1+β aγa + b)γa + b + β) Γb +1)Γb + β +1)Γ) 2 a + b + β) m+1 a + b) m+1 x m ) m m aγa + b)γa + b + β) Γb +1)Γb + β +1)Γ) 2 Next, denote fb) I x a, b)/x a 1 x) b Forα N, wehave { a + b + β) 2 F 1 1,a+ b + β +1 a + b) 2 F 1 1,a+ b +1 ) )} 12) φ a,b,α,β x 2a fb + α)fb + β) fb)fb + α + β) [fb + α)fb + β) 1 x) 2b+α+β fb + α 1)fb + β +1)]+[fb + α 1)fb + β +1) fb + α 2)fb + β +2)] + +[fb +2)fb + β + α 2) fb +1)fb + β + α 1)] +[fb +1)fb + β + α 1) fb)fb + β + α)] Every expression in brackets has the form x 2a 1 x) 2b α β φ a,b+α j,1,β α+2j 1, j 1, 2,,α Hence we can apply formula 12) to each of them, which yields 11) The next corollary refines Theorem 1 and gives a partial solution to Conjecture 1 14

15 Corollary 4 Let α N Then the coefficients φ k defined in 4) can be computed by φ k α 1 j aγa + b + j)γa + b + β + α j 1) [Γ)] 2 Γb + j +1)Γb + β + α j)) k {a + b + β + α j 1) k+1 a + b + j) k+1 } 13) If, furthermore, β α 1 and β>, thenφ k > Proof Formula 13) follows immediately from 11) By virtue of Lemma 2, for β α 1the sign of φ k φ k α, β) is determined by the sign of φ k 1,β) In view of 12), for α 1wehave aγa + b)γa + b + β) φ k 1,β) Γb +1)Γb + β +1)Γ) 2 a + b + β) k+1 a + b) k+1 ), ) k which is clearly positive for β> Remark Substituting 1) for I x into the definition of φ a,b,α,β given in 3) and changing notation, we can write 11) as the following linearization identity for the quadratic form in Gauss hypergeometric functions: ) ) a + b) n+1 a + b + n +1, 1 a + b + β,1 2F 1 b) n+1 2F 1 a + b + β) ) ) n+1 a + b, 1 a + b + β + n +1, 1 2F 1 b + β) n+1 2F 1 a n a + b) j a + b + β) n j bb + β) b +1) j j b + β +1) n j ) a + b + β + n +1 j, 1 {a + b + β + n j) 2 F 1 )} a + b + j +1, 1 a + b + j) 2 F 1 14) Here, n α 1 is an arbitrary nonnegative integer, whereas the other parameters, as well as x, can be arbitrary complex numbers by analytic continuation In particular, for n the above identity after some rearrangement reduces to )[ ) a 1,a+1 1,b a c +1 2 F 1 c 2F 1 + c 1 ] c b c +1 )[ ) b 1,b+1 1,a b c +1 2 F 1 c 2F 1 + c 1 ] c a c +1 Formula 14) leads us to a curious combinatorial identity, which is presumably new Corollary 5 For arbitrary nonnegative integers m, n, the following identity holds: a n a + b) j a + b + β) n j {a + b + β + n j) m+1 a + b + j) m+1 } bb + β) b +1) j b +1+β) n j j ) m m k a + b) k a + b + β) m k ) k ) m k { a + b + k)n+1 a + b + β + m k) } n+1 b) n+1 b + β) n+1 15

16 Proof The identity claimed is obtained by equating the coefficients at x m on both sides of 14) and using the formula μ + n +1) k μ) n+1 μ) k μ + k) n+1 The next corollary presents two-sided bounds for the Turán determinant formed by shifts of I x a, b) in the second parameter Corollary 6 Let ν be a positive integer, b>ν, a> Then mx 2a 1 x) 2b [I x a, b)] 2 I x a, b + ν)i x a, b ν) M[I x a, b)] 2 15) for all x, 1) Here, [Γa + b)] 2 Γa + b ν)γa + b + ν) m [Γ)] 2 [Γb)] 2 [Γ)] 2 Γb ν)γb + ν), M b) νa + b ν) ν a + b) ν b ν) ν b) ν a + b ν) ν Proof The lower bound is just a rewriting of the inequality φ a,b ν,ν,ν x 2a 1 x) 2b φ, which follows from the nonnegativity of the coefficients φ k for all k, 1, In order to prove the upper bound, consider the function b F b) I x a, b)γb) x a 1 x) b Γa + b) a + b) n Γa + n +1) xn n n a + b) n n! x n Γa + n +1) n! It is easy to check that the coefficients {n!/γa + n +1)} n form a log-convex sequence Then, according to [12, Theorem 1], the above function is log-convex, so that for b>νwe have F b ν)f b + ν) [F b)] 2 This inequality can be written as I x a, b ν)i x a, b + ν) [Γb)]2 Γa + b + ν)γa + b ν) [Γa + b)] 2 [I x a, b)] 2 Γb + ν)γb ν) Simple rearrangement of the above inequality gives the upper bound in 15) 4 Acknowledgments I am indebted to Professor Skip Garibaldi Institute for Pure and Applied Mathematics at UCLA) for sharing the log-concavity problem with me and for explaining the motivation This research was supported by the Ministry of Education of the Russian Federation project No ) and by the Russian Foundation for Basic Research project No ) REFERENCES 1 H Alzer, Inequalities for the chi square distribution function, J Math Anal Appl, 223, ) 2 H Alzer and Á Baricz, Functional inequalities for the incomplete gamma function, J Math Anal Appl, 385, ) 3 G E Andrews, R Askey, and R Roy, Special Functions, Cambridge University Press 1999) 4 R Arratia, S Garibaldi, L Mower, and P B Stark, Some people have all the luck, Math Mag, 88, ) 5 J Bustoz and M E H Ismail, On gamma function inequalities, Math Comp, 47, ) 6 H Finner and M Roters, Distribution functions and log-concavity, Commun Statist Theory Meth, 22, ) 16

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