math.PR 2012-02-01 0

The limiting distribution for the number of symbol comparisons used by QuickSort is nondegenerate (extended abstract)

In a continuous-time setting, Fill (2010) proved, for a large class of probabilistic sources, that the number of symbol comparisons used by QuickSort, when centered by subtracting the mean and scaled by dividing by time, has a limiting distribution, …

Authors: Patrick Bindjeme, James Allen Fill

THE LIMITING DISTRIBUTION F OR THE NUMBER OF SYMBOL COMP ARISONS USED BY QUICKSOR T IS NONDEGENER A TE (EXTENDED ABSTRA CT) P A TRICK BINDJEME JAMES ALLEN FILL Abstract In a con tinuous-time setting, Fill [2] prov ed, for a large class of probabilistic sources, that the num b er of symbol comp arisons used by Quic kSort , when centered b y subtracting the mean and scaled by d ividing by time, has a limiting d istribution, but prov ed little abou t that limiting random v a riable Y —not even that it is nondegenerate. W e establis h the nondegeneracy of Y . The pro of is p erhaps surprisingly difficult. 1. The number of symbol comp ariso ns used by Quic kSort : Brief review of a limiting-distribution resul t In this section we br iefly review the main theorem of [2]. An infinite sequence of independent and identically distributed keys is generated; each key is a rando m word ( w 1 , w 2 , . . . ) = w 1 w 2 · · · , that is, an infinite sequence, or “string”, of symbols w i drawn fr om a totally ordered finite alphab et Σ. The co mmon distribution µ of the keys (called a pr ob abilistic sour c e ) is allo wed to b e an y distribution over w ords, i.e., the distribution o f any sto chastic pro cess with time par ameter set { 1 , 2 , . . . } and state space Σ . W e know thanks to Kolmogor ov’s consistency criter ion (e.g., Theorem 3 . 3 . 6 in [1 ]) that the p ossible distributions µ are in one-to-o ne cor re- sp ondence with consisten t sp ecifica tions o f finite-dimensio nal ma rginals, i.e., of the fundamental pr ob abilities (1.1) p w := µ ( { w 1 w 2 · · · w k } × Σ ∞ ) with w = w 1 w 2 · · · w k ∈ Σ ∗ . This p w is the probabilit y that a w ord drawn from µ ha s w as its leng th- k prefix. F or each n , Hoa r e’s [6 ] Qui ckSort a lg orithm c an b e us ed to sor t the first n keys to be g enerated. W e may and do assume that the first k ey in the sequence is chosen as the piv ot, and tha t the same is true r ecursively (in the sense, for example, that the pivot used to s ort the k e ys smaller than the original pivot is the first key to b e generated that is smaller than the o r iginal pivot). A compar ison of tw o keys is done by scanning the tw o words from left to right, comparing the symbols of matching index one by one un til a difference is found. W e let S n denote the total n um b er of symbol comparisons needed when n k ey s are so rted by QuickSor t . Theorem 1.1 (Fill [2], Theo rem 3 .1) . Consider the c ontinuous-time sett ing in which keys ar e gener ate d fr om a pr ob abilistic sour c e at t he arrival t imes of an in- dep endent Poisson pr o c ess N with u n it r ate. L et S ( t ) = S N ( t ) denote the numb er Date : Jan uar y 27, 2012. Researc h s upported by the Acheso n J. Duncan F und for the Adv ancement of R esearc h in Statistics. 1 2 P A TRICK BINDJEM E JAMES ALLE N FI LL of symb ol c omp arisons r e qu ir e d by Quic kSort t o sort the keys gener ate d thr ough ep o ch t , and let (1.2) Y ( t ) := S ( t ) − E S ( t ) t , 0 < t < ∞ . Assume that (1.3) ∞ X k =0  X w ∈ Σ k p 2 w  1 / 2 < ∞ with p w as in (1.1 ) . Then ther e exists a r andom variable Y such t hat Y ( t ) → Y in L 2 as t → ∞ . In p articular, Y ( t ) L → Y and [ b e c ause E Y ( t ) → E Y ] we have E Y = 0 , and V ar Y ( t ) → V ar Y . In the full-length pa p er (in prepa ration) corre s po nding to [2 ], this theorem will be extended by replacing the power 1 / 2 in (1.3) by 1 / p for any given p ∈ [2 , ∞ ) and concluding that Y ( t ) → Y in L p . F rom Theorem 1.1 we know that V ar S ( t ) = O ( t 2 ) as t → ∞ , but we don’t know that V ar S ( t ) = Θ( t 2 ) b eca use the theorem do e s not contain the imp ortant information that the limiting random v ariable Y is nondeg enerate (i.e., do es not almost surely v anish). The purp ose of the presen t extended abstract is to show that Y is nondeg enerate ; this is stated as o ur main Theor e m 2.1 below. The pro o f turns out to b e surprisingly difficult; we do not know the v alue of V ar Y , a nd the pro of of Theorem 2.1 do es not pr ovide it. The consequence V ar S ( t ) = Θ( t 2 ) of our Theorem 2.1 settles a que s tion that ha s been op en since the work of Fill and Janson [4] even in the sp ecia l case o f the standar d bina r y sour ce with Σ = { 0 , 1 } and the fundamen tal probabilities of (1.1) equal to 2 − k . 2. Main resul ts The following is the main theorem of this extended abstract. Theorem 2.1. The limit distribution in The or em 1.1 is nonde gener ate. Throughout this extended abstr act, we work in the setting o f Theor em 1.1. Theorem 2.1 follows immediately from Pro po sitions 2.3 – 2.4 in this section. Definition 2.2 . F o r an in teg er k and a prefix w ∈ Σ k we define (with little pos sibility of nota tio nal co nfusion), for compar is ons among keys that hav e ar rived by e p o ch t , the coun ts S k ( t ) := num b er of comparisons of ( k + 1 )s t symbols , S w ( t ) := num b er of comparisons of ( k + 1 )s t symbols betw een k eys with prefix w . The following tw o pro po sitions combine to establish Theo rem 2 .1. W e wr ite Σ ∗ := ∪ 0 ≤ k< ∞ Σ k for the s e t of all prefixes. Prop ositio n 2.3. If t he r andom vari ables S w ( t ) , w ∈ Σ ∗ , ar e nonne gatively c orr e- late d for e ach fix e d t , then the limit distribut ion in The or em 1.1 is nonde gener ate. A proo f o f Prop osition 2 .3 ca n b e found in Section 3 (see Subsection 3 .2). Prop ositio n 2.4 . F or e ach fixe d t , the ra ndom variables S w ( t ) , w ∈ Σ ∗ , ar e non- ne gatively c orr elate d. F ollowing six lemmas in Sec tion 4, a pro o f of Pro p os ition 2.4 can b e found in Section 5 (sp ecifically: in Subsection 5.2). LIMIT DISTN. FOR QUICKSOR T SYMBOL COMP ARISONS IS NONDE GENERA TE 3 3. Proof of Proposition 2.3 3.1. A lo w er b ound for the v ariance of K ( t ) . Definition 3.1 . If K n is the num b er o f key comparisons needed to sort the firs t n keys to arr ive using Quicks ort , and N is the Poisson pro cess in Theorem 1 .1 (independent of the gener ation o f the keys), we define K ( t ) := K N ( t ) . In order to prove P rop osition 2.3, we first establish the follo wing lemma. Lemma 3.2. We have V ar K ( t ) ≥ (1 + o (1)) σ 2 t 2 as t → ∞ wher e σ 2 := 7 − 2 3 π 2 . Pr o of. By the law o f total v ar iance (namely , v ar iance equals the sum of exp ectation of conditional v a riance and v a riance of co nditional expectation) we hav e (3.1) V ar K ( t ) ≥ E V ar [ K ( t ) | N ( t )] = e − t ∞ X n =0 t n n ! V ar K n . F rom (for e x ample) (1 . 2) in [3] w e ha ve V ar K n = 7 n 2 − 4 ( n + 1) 2 H (2) n − 2 ( n + 1) H n + 1 3 n, so lim n →∞ V ar K n n 2 = σ 2 . It follo ws tha t, giv en α > 0, there exists n α such that V ar K n ≥ (1 − α ) σ 2 n 2 for all n ≥ n α . W e therefore have from (3.1) that V ar K ( t ) ≥ (1 − α ) σ 2 e − t ∞ X n = n α t n n ! n 2 = (1 + o (1))(1 − α ) σ 2 e − t ∞ X n =0 t n n ! n 2 as t → ∞ = (1 + o (1))(1 − α ) σ 2 ( t 2 + t ) = (1 + o (1))(1 − α ) σ 2 t 2 . Since α > 0 is arbitrar y , the lemma follows.  3.2. Pro o f of Prop osi tion 2.3. Definition 3 .3 . F or any nonnega tiv e in teger k , with S k ( t ) as in Definition 2.2 we define Y k ( t ) := S k ( t ) − E S k ( t ) t . Pr o of of Pr op osition 2. 3 . With Y ( t ) a s in Theorem 1.1, w e hav e Y ( t ) = ∞ X k =0 Y k ( t ) and, from Theor em 1 .1, V ar Y ( t ) → V ar Y as t → ∞ . (3.2) 4 P A TRICK BINDJEM E JAMES ALLE N FI LL Knowing that E Y k ( t ) = 0 for an y nonnegative integer k and t ∈ (0 , ∞ ) , that E Y ( t ) = 0 for t ∈ (0 , ∞ ) , and finally that the random v ariables Y k ( t ) satisfy the hypo theses of the elementary probabilistic Lemma 2.8 of [2] for p 0 = 2, w e hav e for any t ∈ (0 , ∞ ) that (3.3) V ar n X k =0 Y k ( t ) ! → V ar Y ( t ) as n → ∞ . Now, from the fa c t that S k ( t ) = X w ∈ Σ k S w ( t ) for any k, we hav e V ar n X k =0 Y k ( t ) ! = n X k =0 V ar Y k ( t ) + 2 X 0 ≤ i 0 .  LIMIT DISTN. FOR QUICKSOR T SYMBOL COMP ARISONS IS NONDE GENERA TE 5 4. Six lemmas In the next section, we will need the following six lemmas . W e write κ n := E K n for the exp ected num b er o f key compariso ns required to so r t the first n keys to arrive. Lemma 4.1. If ∆ 2 ( n, a, b ) := 2 b X j =1 1 n + 2 − j − a ( b − 1 + κ j − 1 + κ b − j − κ b ) for any nonne gative inte gers a , b , and n with a + b ≤ n , then we have ∆ 2 ( n, a, b ) ≥ 0 . Pr o of. It is well known, and ca n easily b e chec ked, using the ex plicit for mu la κ n = 2( n + 1) H n − 4 n , tha t b − 1 + κ j − 1 + κ b − j − κ b is symmetric ab out j = ( b + 1) / 2 , (4.1) and decreasing in j = 1 , . . . , ⌊ ( b + 1) / 2 ⌋ . Thu s ∆ 2 ( n, a, b ) = 2 ⌊ b/ 2 ⌋ X j =1  1 n + 2 − a − j + 1 n + 2 − a − ( b + 1 − j )  (4.2) × ( b − 1 + κ j − 1 + κ b − j − κ b ) + 1 ( b is o dd) 2 n + 2 − a − ( b + 1) / 2 ( b − 1 + 2 κ ( b − 1) / 2 − κ b ) . The first o f the tw o terms in (4.2) equals 2[2( n + 2 − a ) − ( b + 1)] (4.3) × ⌊ b/ 2 ⌋ X j =1 1 ( n + 2 − a − j )( n + 2 − a − ( b + 1 − j )) ( b − 1 + κ j − 1 + κ b − j − κ b ) . Each of the t wo fa c tors in the sum in (4.3), namely 1 ( n +2 − a − j )( n +2 − a − ( b +1 − j )) and b − 1 + κ j − 1 + κ b − j − κ b , decreases in j over the range of summatio n, so by “Cheby- shev’s other ine q uality” [5] w e ha ve 1 ⌊ b/ 2 ⌋ ⌊ b/ 2 ⌋ X j =1 1 ( n + 2 − a − j )( n + 2 − a − ( b + 1 − j )) ( b − 1 + κ j − 1 + κ b − j − κ b ) ≥   1 ⌊ b/ 2 ⌋ ⌊ b/ 2 ⌋ X j =1 1 ( n + 2 − a − j )( n + 2 − a − ( b + 1 − j ))   ×   1 ⌊ b/ 2 ⌋ ⌊ b/ 2 ⌋ X j =1 ( b − 1 + κ j − 1 + κ b − j − κ b )   . 6 P A TRICK BINDJEM E JAMES ALLE N FI LL Thu s (4.3) is gr eater than or equal to 2[2( n + 2 − a ) − ( b + 1)] ⌊ b/ 2 ⌋ ×   1 ⌊ b/ 2 ⌋ ⌊ b/ 2 ⌋ X j =1 1 ( n + 2 − a − j )( n + 2 − a − ( b + 1 − j ))   ×   1 ⌊ b/ 2 ⌋ ⌊ b/ 2 ⌋ X j =1 ( b − 1 + κ j − 1 + κ b − j − κ b )   . If b is even, then by sy mmetr y w e hav e b/ 2 X j =1 ( b − 1 + κ j − 1 + κ b − j − κ b ) = 1 2 b X j =1 ( b − 1 + κ j − 1 + κ b − j − κ b ) = 0 , and ∆ 2 ( n, a, b ) ≥ 0, as desired. If b is o dd, then again by symmetry we ha ve ( b − 1) / 2 X j =1 ( b − 1 + κ j − 1 + κ b − j − κ b ) = 1 2   b X j =1 ( b − 1 + κ j − 1 + κ b − j − κ b ) − ( b − 1 + 2 κ ( b − 1) / 2 − κ b )   = − 1 2 ( b − 1 + 2 κ ( b − 1) / 2 − κ b ) . Hence ∆ 2 ( n, a, b ) ≥ − [2 ( n + 2 − a ) − ( b + 1 )] 1 ( b − 1) / 2 ( b − 1) / 2 X j =1 1 ( n + 2 − a − j )( n + 2 − a − ( b + 1 − j )) × ( b − 1 + 2 κ ( b − 1) / 2 − κ b ) + 2 n + 2 − a − b +1 2 ( b − 1 + 2 κ b − 1 2 − κ b ) = [ κ b − 2 κ b − 1 2 − ( b − 1)] × ( 2( n + 2 − a ) − ( b + 1) ( b − 1) / 2 ( b − 1) / 2 X j =1 1 ( n + 2 − a − j )( n + 2 − a − ( b + 1 − j )) − 4 2( n + 2 − a ) − ( b + 1) ) LIMIT DISTN. FOR QUICKSOR T SYMBOL COMP ARISONS IS NONDE GENERA TE 7 = [ κ b − 2 κ ( b − 1) / 2 − ( b − 1)][2 ( n + 2 − a ) − ( b + 1 )] × ( 1 ( b − 1) / 2 ( b − 1) / 2 X j =1 1 ( n + 2 − a − j )( n + 2 − a − ( b + 1 − j )) − 4 [2( n + 2 − a ) − ( b + 1)] 2 ) ≥ [ κ b − 2 κ ( b − 1) / 2 − ( b − 1)][2 ( n + 2 − a ) − ( b + 1 )] ×  1 ( n + 2 − a − ( b − 1) / 2)( n + 2 − a − ( b + 3) / 2) − 4 [2( n + 2 − a ) − ( b + 1)] 2  = 4[ κ b − 2 κ ( b − 1) / 2 − ( b − 1)][2 ( n + 2 − a ) − ( b + 1 )] ×  1 [2( n + 2 − a ) − ( b − 1)][2( n + 2 − a ) − ( b + 3)] − 1 [2( n + 2 − a ) − ( b + 1)] 2  = 16[ κ b − 2 κ ( b − 1) / 2 − ( b − 1)][2( n + 2 − a ) − ( b + 1)] [2( n + 2 − a ) − ( b − 1)][2( n + 2 − a ) − ( b + 3)][2( n + 2 − a ) − ( b + 1)] 2 ≥ 0 , the las t t wo inequalities following from the facts that 2 ( n + 2 − a ) − ( b + 3) and [ κ b − 2 κ ( b − 1) / 2 − ( b − 1 )] are nonneg ative—the former due to the fact that n ≥ a + b , and the latter to (4.1) and the identit y (4.4) b X j =1 ( b − 1 + κ j − 1 + κ b − j − κ b ) = 0 .  Lemma 4.2. If ∆ 1 ( n, a, b ) := 2 b X j =1 H n +1 − j − a ( b − 1 + κ j − 1 + κ b − j − κ b ) , for any nonne gative inte gers a , b , and n with a + b ≤ n , then we have ∆ 1 ( n, a, b ) ≤ 0 . Pr o of. W e hav e ∆ 1 ( n + 1 , a, b ) − ∆ 1 ( n, a, b ) = 2 b X j =1 ( H n +2 − j − a − H n +1 − j − a )( b − 1 + κ j − 1 + κ b − j − κ b ) = 2 b X j =1 1 n + 2 − j − a ( b − 1 + κ j − 1 + κ b − j − κ b ) = ∆ 2 ( n, a, b ) . 8 P A TRICK BINDJEM E JAMES ALLE N FI LL It follows from Lemma 4.1 that ∆ 1 ( n, a, b ) is no ndec r easing in n ≥ a + b . F rom the ident ity (4.4 ) w e ha ve ∆ 1 ( n, a, b ) = 2 b X j =1 ( H n +1 − j − a − H n )( b − 1 + κ j − 1 + κ b − j − κ b ) . As a result, lim n →∞ ∆ 1 ( n, a, b ) = 0 follows from the fact that lim n →∞ ( H n +1 − j − a − H n ) = 0 for an y 1 ≤ j ≤ b . Thus ∆ 1 ( n, a, b ) ≤ 0 for any n ≥ a + b , which finishes the pro of of the lemma.  Lemma 4.3. F or any nonne gative inte ger b , we have lim m →∞ b X j =1 κ m − j ( b − 1 + κ j − 1 + κ b − j − κ b ) = 0 . Pr o of. Since κ m = 2( m + 1) H m − 4 m = 2( m + 1)  ln m + γ + 1 2 m + O  1 m 2  − 4 m = 2 m ln m − (4 − 2 γ ) m + 2 ln m + (2 γ + 1) + O  1 m  , we hav e for e a ch fixed j that κ m − j = 2( m − j ) ln( m − j ) − (4 − 2 γ )( m − j ) + 2 ln( m − j ) + (2 γ + 1) + O  1 m − j  = 2( m − j )  ln m + ln  1 − j m  − (4 − 2 γ )( m − j ) + 2  ln m + ln  1 − j m  + (2 γ + 1 ) + O  1 m − j  = 2( m − j ) ln m − 2 j − (4 − 2 γ )( m − j ) + 2 ln m + (2 γ + 1 ) + O  1 m  = 2 m ln m − (4 − 2 γ ) m − 2 ( j − 1) ln m + 2(1 − γ ) j + (2 γ + 1 ) + O  1 m  . So b X j =1 κ m − j ( b − 1 + κ j − 1 + κ b − j − κ b ) = b X j =1  2 m ln m − (4 − 2 γ ) m − 2( j − 1) ln m + 2(1 − γ ) j + (2 γ + 1 ) + O  1 m  × ( b − 1 + κ j − 1 + κ b − j − κ b ) = − 2 [ln m − (1 − γ )] b X j =1 j ( b − 1 + κ j − 1 + κ b − j − κ b ) + O  1 m  , thanks once mor e to the iden tity (4.4). LIMIT DISTN. FOR QUICKSOR T SYMBOL COMP ARISONS IS NONDE GENERA TE 9 Now b X j =1 j ( b − 1 + κ j − 1 + κ b − j − κ b ) = 1 2 b X j =1 [ j + ( b + 1 − j )]( b − 1 + κ j − 1 + κ b − j − κ b ) = 1 2 ( b + 1) b X j =1 ( b − 1 + κ j − 1 + κ b − j − κ b ) = 0 by symmetry and (4.4 ), whic h finishes the pro of of the lemma.  Lemma 4.4. If Λ 1 ( a, b ) := 2 b X j =1 H j + a ( b − 1 + κ j − 1 + κ b − j − κ b ) for any nonne gative inte gers a and b , then Λ 1 ( a, b ) ≤ 0 . Pr o of. W e hav e Λ 1 ( a, b ) = 2 b X j =1 H b − j + a +1 ( b − 1 + κ j − 1 + κ b − j − κ b ) = ∆ 1 ( b + 2 a, a, b ) ≤ 0 , where the ineq uality follows from Lemma 4.2.  Lemma 4.5. If Λ( a, b ) := b X j =1 κ j + a − 1 ( b − 1 + κ j − 1 + κ b − j − κ b ) = b X j =1 κ a + b − j ( b − 1 + κ j − 1 + κ b − j − κ b ) for any nonne gative inte gers a and b , then Λ( a, b ) ≥ 0 . Pr o of. W e hav e Λ( a + 1 , b ) − Λ( a, b ) = 2 b X j =1 H j + a ( b − 1 + κ j − 1 + κ b − j − κ b ) = Λ 1 ( a, b ); so, in light of Lemma 4.4, Λ( a, b ) is nonincrea s ing in a . W e also ha ve from Lemma 4.3 that lim a →∞ Λ( a, b ) = 0 , which finishes the proo f.  10 P A TRICK BINDJEM E JAMES ALLE N FI LL Lemma 4.6. If Σ( n, a, b ) := a + b X j = a +1 ( κ j − 1 + κ n − j )( b − 1 + κ j − 1 − a + κ b + a − j − κ b ) = b X j =1 ( κ j + a − 1 + κ n − j − a )( b − 1 + κ j − 1 + κ b − j − κ b ) for any nonne gative inte gers a , b , and n with a + b ≤ n , then Σ( n, a, b ) ≥ 0 . Pr o of. W e hav e tha t Σ( n + 1 , a, b ) − Σ( n, a, b ) = b X j =1 ( κ n +1 − j − a − κ n − j − a )( b − 1 + κ j − 1 + κ b − j − κ b ) = 2 b X j =1 H n +1 − j − a ( b − 1 + κ j − 1 + κ b − j − κ b ) = ∆ 1 ( n, a, b ) , which implies from Lemma 4 .2 that Σ( n, a, b ) is nonincr e asing in n ≥ a + b . Reca ll that Σ( n, a, b ) = b X j =1 κ j + a − 1 ( b − 1 + κ j − 1 + κ b − j − κ b ) + b X j =1 κ n − j − a ( b − 1 + κ j − 1 + κ b − j − κ b ) = Λ( a, b ) + b X j =1 κ n − j − a ( b − 1 + κ j − 1 + κ b − j − κ b ); so, from Lemma 4.3 w e ha ve lim n →∞ Σ( n, a, b ) = Λ( a, b ) . The result fo llows fr om Lemma 4 .5.  5. The random v ariables S w ( t ) , w ∈ Σ ∗ , are nonnega tivel y co rrela ted In this s ection, w e first prov e the following (in Subsection 5.1) and then complete the proo f o f P rop osition 2.4 in Subsection 5.2. Prop ositio n 5.1. L et w ∈ Σ ∗ . Then the r andom variables S ∅ ( t ) and S w ( t ) ar e nonne gatively c orr elate d. 5.1. The random v ariables S ∅ ( t ) and S w ( t ) for an y w ∈ Σ ∗ are no nnega- tiv e ly correlated. In this Subsectio n 5.1 w e pr ov e Pr o p o sition 5.1, which states that (5.1) Cov ( S ∅ ( t ) , S w ( t )) ≥ 0 for an y w ∈ Σ ∗ , with the under standing that S ∅ ( t ) = K ( t ) = K N ( t ) . LIMIT DISTN. FOR QUICKSOR T SYMBOL COMP ARISONS IS NONDE GENERA TE 11 Pr o of of Pr op osition 5. 1 . W e hav e (5.2) Co v ( S ∅ ( t ) , S w ( t )) = Cov ( K ( t ) , S w ( t )) = T w ( t ) + V w ( t ) where (5.3) T w ( t ) := Cov ( E [ K ( t ) | N ( t )] , E [ S w ( t ) | N ( t )]) and (5.4) V w ( t ) := E Cov ( K ( t ) , S w ( t ) | N ( t )) . But Prop os itio ns 5 .2 – 5.3 will demonstrate that the expr essions T w ( t ) and V w ( t ) are each no nnegative.  5.1.1. Nonne gativity of T w ( t ) . Here w e pro ve the follo wing result. Prop ositio n 5.2 . The expr ession T w ( t ) define d in (5.3) is n onn e gative. Pr o of. W e hav e E [ K ( t ) | N ( t ) = n ] = κ n := E K n , which is increasing with n ; and E [ S w ( t ) | N ( t ) = n ] = n X j =0  n j  p j w (1 − p w ) n − j κ j is also increasing, following from the fact that the Binomia l( n, p w ) distributions increase stochastically with n . By “Chebyshev’s other inequality” [5 ], w e can conclude that Co v ( E [ K ( t ) | N ( t )] , E [ S w ( t ) | N ( t )]) ≥ 0 , which finishes the proo f o f the propo sition.  5.1.2. Nonne gativity of V w ( t ) . In this subsection we prov e the following prop os ition, thereby completing the proo f of Pro po sition 5.1. Prop ositio n 5.3 . The expr ession V w ( t ) define d in (5.4) is n onn e gative. This will b e a ccomplished using the next t wo pro po sitions, P rop ositions 5.4 a nd 5.7. Prop ositio n 5.4 . If ψ ( n, a, b ) := n − 1 X a

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