Nice Bounds for the Generalized Ballot Problem

Nice Bounds for the Generalized Bal lot Problem Delong Meng Massac h usetts Institute of T ec hnology Email: delong13@mit.edu Abstract This pap er gives tw o sharp b ounds for the generalized b allot p roblem with candidate A receiving at least µ times as candidate B for an arbitrary real num b er µ . In tro duction Suppo se in an election candidate A received a votes and ca n- didate B rec e ived b votes. W e coun t the votes o ne at a time in a ny of the  a + b a  po ssible sequences. Let a r and b r denote the n umber of votes A a nd B have after counting the r th vote where 1 ≤ r ≤ a + b (notice that a r + b r = r ). Let µ b e a ny p o sitive rea l num b e r. W e call a sequence desirable if a > µb and a r > µb r for a ll r . W e call a sequence cute if a ≥ µb and a r ≥ µb r for all r . Let P denote the pr obability tha t a sequence is desir able and P ∗ denote the probability that a seque nce is cute. Several authors star ted several articles quo ting se veral well-kno wn results of the Ballot Problem. F or br evity , Andre [7] a nd Barbier[8 ] dis c ov ered that if µ ∈ N , then P = a − µb a + b . Aeppli[9] show ed that if µ ∈ N , then P ∗ = a − µb + 1 a + 1 . Finally in 1 962 T ak acs[4] to o k a gia nt lea p a nd brav ely pr ov ed that for an arbitrar y µ ∈ R , P = a a + b b X j =0 C j  b j   a + b − 1 j  1 where C 0 = 1 and C j satisfies the following r ecurrence form ula: k X j =0 C j  k j   ⌊ k µ ⌋ + k − 1 j  = 0 for all p ositive in teger s k . This for mula gives th e exact v alue for P . How ever w e can hardly imag ine how big this num b er really is. Therefor e this article prov es the follo wing bo unds: Theorem 1. a − ⌊ µb ⌋ a + b ≤ P ≤ a − ⌊ µ ⌋ b a + b (1) Theorem 2. ⌊ a − µb + 1 ⌋ a + b ≤ P ∗ ≤ a + 1 − µb a + 1 . (2) W e prove the tw o upp er b ounds with the Pseudo-Re fle c tio n Principle and the t wo lower b o unds with Penetrating Analysis . Pseudo-Reflectio n Principle Let’s start with Theor em 1 . W e lo ok at the relationship betw een the undesira ble sequence a nd the sequence with a v otes for A and b − 1 votes for B . W e call this the P seudo-Reflection Pr inciple b ecause the case µ = 1 is es sentially the r eflection principle. 1 Both Goulden[5] a nd Renault[6] proved the e q uality ca se (when µ ∈ N ). They bo th considered the smallest r such that a r ≤ µb r . Ho wev er this appr oach do es not g eneralize to the case when µ ∈ R . Instead we consider the la rgest r such that a r ≤ µb r . When the r th vote is counted, we must have a r = ⌊ µb r ⌋ ≤ ⌊ µ ⌋ b r . There are  a r + b r a r  such undesira ble sequences. Now consider the num b er of seq uences with a r votes for A but b r − 1 votes for B . F or ea ch r there are  a r + b r − 1 b r − 1  such sequences. Consider the op eration of replac ing the first r votes in an undesira ble s e quence with these  a r + b r − 1 b r − 1  sequences. This o p er ation yields all sequences with a votes for A but b − 1 votes for B b ecause for any se q uence with a votes for A but b − 1 votes for B , there must ex ist a n r such tha t a r − 1 ≤ µ ( b r + 1) < a r . Since  a r + b r − 1 b r − 1  = b r a r + b r  a r + b r a r  ≥ 1 ⌊ µ ⌋ + 1  a r + b r a r  (3) we deduce that the num be r of undesirable s equence is a t mos t ⌊ µ ⌋ + 1 times the sequences with a r votes for A but b r − 1 votes for B . Ther e fore P ·  a + b a  ≤  a + b a  − ( ⌊ µ ⌋ + 1)  a + b − 1 b − 1  = a − ⌊ µ ⌋ b a + b  a + b a  . 1 How ev er we are not finding any bi jections here. W e are only counting the num b er of undesirable sequences. 2 R emark The conditions for equality to ho ld ar e not trivial. Dv o retzky [11 ] prov ed tha t that equalit y holds if and only if µ is sufficien tly close to a b or µ is sufficiently close to an integer. See Dvoretzky [11] for the precise definitions of sufficiently close. Now we move on to Theorem 2. Notice that the upp er bo und for Theorem 2 is a tr ivial cons equence of theorem 1 when µ is an integer. (W e can simply a dd one vote for A in the beg inning of the s e quence and then a r > b r .) But such a correla tion do es not g ive the sharp b o und in Theorem 2 for µ ∈ R . Using the Pseudo -Reflection technique w e similarly cons ide r the larg est r such that a r < µb r . W e hav e a r = ⌈ µb r ⌉ − 1 . This time, how ever, we compare the ugly (non-cute) sequence to the s equence with a + 1 votes for A and b − 1 votes for B . W e similarly r eplace the first r votes with sequences of a r + 1 votes for A and b r − 1 votes for B . This o p er ation yields all p ossible s equences with a r + 1 votes for A and b r − 1 votes for B . Now we have  a r + b r b r − 1  = b r a r + 1  a r + b r a r  ≤ 1 µ  a r + b r a r  (4) which implies that the num b er of ugly sequenc e is at least µ times the num ber of se quences of a r + 1 votes for A and b r − 1 votes for B . Ther e fore P ∗ ·  a + b a  ≤  a + b a  − ( µ )  a + b b − 1  = a − µb + 1 a + 1  a + b a  . R emark W e can trans la te these sequences into lattice paths fr o m the or igin the p oint ( b , a ). A desira ble path never touches the line y = µx , and a cute path nev er go below the line. The inequality (3) shows that the n umber of un- desirable pa ths is a t lea s t ⌊ µ ⌋ times the n umber of paths fr o m the p oint (1 , 0) to ( b, a ). The inequalit y (4) shows that the num b er of ugly paths is at least µ times the num ber of pa th fro m the p oint (1 , − 1) to ( b, a ). But intuition do es not help muc h in this pr oblem since it inv o lves ca lculations and o ne to ⌊ µ ⌋ co r- resp ondence rather than a pure bijection. P enetrating Analysis W e fir st prove the low er b ound for Theorem 2. W e claim that at least ⌊ a − µb ⌋ + 1 of the a + b cyclic p ermutations of a ny given s e- quence of votes ar e cute. This metho d is ca lled p enetra ting a nalysis in Moha nt y [1]. F or any given sequence, define the weight ed partia l sum as S r = a r − µb r . Note that the sequence is cute if and only if S r ≥ 0 for all r . Supp ose S i is the 3 minim um (if ther e are multiple i s then we can take any of them). W e cyclic a lly per mute the fir s t i terms of the sequence to the end of the sequence. In other words we erase the fir st i ter ms and attach them to the end of the sequence. Now let S ′ be the weight ed partial sum for this new sequence. W e finish the pro of with three lemmas. L emma 1. This n ew se quenc e is cute. Pr o of: If r ≤ a + b − i , then S ′ r = S a + b − i − S i ≥ 0. If r > a + b − i , then S ′ r = S r − ( a + b − i ) + S a + b − S i ≥ 0 becaus e S a + b ≥ 0. Therefor e S ′ r ≥ 0 for all r . L emma 2. A cyclic p ermutation that b e gins with t he r th term of this se quenc e is cute if S ′ r ≤ S ′ t for al l r + 1 ≤ t ≤ a + b. F or c onvenienc e, we c al l such an r and also the r th vote cute. Pr o of: Let S ′′ denote the weigh ted partial sum for the c y clic p ermuta- tion that beg ins with the r th term of this sequence . If j ≤ a + b − r , then S ′′ j = S ′ a + b − j − S ′ r ≥ 0. If j > a + b − r , then S ′′ j = S ′ j − ( a + b − r ) + S ′ a + b − S ′ r ≥ 0 bec ause S ′ r ≤ S ′ a + b and S ′ j − ( a + b − r ) ≥ 0. Therefor e S ′′ r ≥ 0 for all r . L emma 3. Ther e ex ist at le ast ⌊ a − µb ⌋ + 1 cute votes. Pr o of: Let r 1 < r 2 < . . . < r k = a + b denote a ll the cute votes. W e hav e S ′ r k = a − µb and S ′ r 1 ≤ 1. Since S ′ r +1 ≤ S ′ r + 1, we must have S ′ r i +1 ≤ S ′ r i +1 ≤ S ′ r i + 1 (5) (If S ′ r i +1 > S ′ r i +1 , then there mu s t exist another cute vote b etw een r i + 1 and r i +1 , which contradicts the definition of r i .) Therefore k − 1 ≥ S ′ r k − S ′ r 1 ≥ a − µb − 1. Because k is an in tege r , we have tw o cases: 1. If a − µb − 1 is not a n integer, then k ≥ ⌈ a − µb ⌉ = ⌊ a − µb + 1 ⌋ . 2. If a − µb − 1 is an integer, then consider all r such that a r − µb r < 0. Since there are finitely many such negative v alues , there ex ist an ǫ such that a r − ( µ − ǫ ) b r ≥ 0 implies a r − µb r ≥ 0. Replacing µ with µ − ǫ would not affect the num b er of cute sequences . Ther efore k ≥ ⌈ a − ( µ − ǫ ) b ⌉ = a − µb + 1 . Now w e hav e shown that that at least ⌊ a − µb + 1 ⌋ of the a + b cyclic p er m uta tio ns of a ny g iven sequence are cute. Therefore P ∗ ≥ ⌊ a − µb + 1 ⌋ a + b . 4 F or Theorem 1, we c a n similar ly show that at least a − ⌊ µb ⌋ o f the a + b cyclic per mutations of any given sequence ar e desira ble. Notice that ⌊ a − µb + 1 ⌋ = a − ⌊ µb ⌋ if a − µb − 1 is not a n integer, and ⌊ a − µb + 1 ⌋ = a − ⌊ µb ⌋ + 1 otherwise . W e can imitate the pro of for Theor e m 2 until the last step. If a − µb − 1 is an in teg e r , there do es not ǫ such that a r − ( µ − ǫ ) b r > 0 can guara nt ee that a r − µb r > 0 b eca use a r − µb r can b e equal to 0 . Ther efore we can only conclude that P ≥ a − ⌊ µb ⌋ a + b . R emark 1 W e can also prov e the low er b ound by induction on b . Again let’s prov e the result o f Theorem 2, and we ca n follow the s a me pro cedur e for Theo- rem 1. Bas e case b = 1 is trivial. Suppo se that the bo und is v alid for all p ositive int eg ers less than b . W e fir st apply lemma 1 to p ermute any g iven sequence into a c ute sequence. Then w e consider t wo situations: 1. If there exist a cute r such that 0 < r < b , then we can cut the sequence int o t wo cute sequence s . By the inductive hypothesis, the num ber of cute votes no less than ⌊ a r − µb r + 1 ⌋ + ⌊ a − a r − µ ( b − b r ) + 1 ⌋ ≥ ⌊ a − µb + 1 ⌋ . 2. If there do es not exist a cute r such that 0 < r < b , then for all cute r , we m us t hav e b r = 0 or b r = b . The rest follo ws trivially from (5) . R emark 2 Notice that any cute sequence must start with a vote for A . W e can th us trea t all the v otes for B in betw een tw o votes for A as a sing le blo ck. Therefore the argument is still v alid even if B receives a n y n umber of weighted votes as lo ng as the w eig ht s a dd up to b . W e can reformulate the pro blem as follow. Suppo se in an election A rec e ived a votes all weigh ted 1 . Howev er B received b ′ weigh ted votes whos e sum is b . Both a and b are integers, and µ is a r eal num b er. W e count the a + b ′ votes in a random order. Let a r and b r denote the sum of the weighted votes A and B have after co un ting the r th vote where 1 ≤ r ≤ a + b ′ . Define P as the probability that a r > µb r for all 1 ≤ r ≤ a + b ′ (desirable sequences). Then a − ⌊ µb ⌋ a + b ′ ≤ P ≤ a a + b ′ The upp er b ound is b ecause each desirable seq uence must start with a vote for a . This b ound, although a chiev able, is extr emely weak compare to (1) and (2). Go ulden [5] discussed the e q uality ca se for the low er b ound when µ = 1 and all the weights a re integers. In fact equality holds for the low er b ound if 5 µ and all the weigh ts a re integers. Therefore if all the weigh ts are integers, then P ≤ a − ⌊ µ ⌋ b a + b ′ , but w e still cannot find a sharp upper b ound for arbitra r y weigh ts. F urther Though ts The inspiration of this pap er is to sear ch for a closed formula for any µ ∈ R . Intuitiv ely such a formula probably do esn’t exist. Even if it do es we must use more adv anced techniques. W e still hav e many unans wered questions in this pap er. F or example, can we find a shar p er b ound for Theorem 1 a nd Theorem 2? Can w e find an upper b ound if given sp ecific weigh ts in the last pro blem? Can we derive similar inequalities for a multi-candidate election? References 1. Mohant y , Gopal, Lattice Path Counting a nd Applications. Acedamic Pr ess, New Y ork, 1980. 2. Mar c Renault, F our P ro ofs of the Bollot The o rem, Mathematics Magazine 80, December 20 07, 3 45-35 2. 3. L. T a k ac s, Ballot Pr oblems. Z. W ahr s cheinlic hkeistheorie, 1 (196 2), 154-1 58. 4. L. T ak acs, A generaliza tion o f the ballo t problem and its application to the theory of queues. J. Amer. Statist. Asso c., 57 (1962), 327-337 5. I.P . Goulden and Luis G. Ser rano, Ma intaining the Spirit of the Reflection Principle when the Boundary has Arbitra ry In teger Slope, J. Combinatorial Theory (A) 1 0 4 (20 03) 31 7-326 . 6. Mar c Rena ult, Lost (and F o und) in T ranslatio n: Andre’s Actual Metho d and its Application to the Gener alized Ba llot Problem, Americ an Mathematic al Monthly 11 5 (200 8) 35 8 -363 . 7. D. Andre, Solution directe du pro ble me reso lution par M. Bertr and, Comptes R endus de l’A c ademie des S cienc es , Paris, 105 (1887 ) 436C437 . 8. E . Ba rbier, Genera lisation du pro bleme reso lution pa r M. J. B ertrand, Comptes R endus de l’A c ademie des Scienc es, Paris , 105 (1887) p. 407. 9. A. Aeppli, Z ur Theorie V erketteter W ahr scheinlic hkeiten. These, Zur ich (1924). 10. J. Sa r an a nd K. Sen, Some Distr ibution Results on Gener alized Ballo t P rob- lems. Sv azek 30 (19 85) 157-165 . 11. A. Dvoretzky and Th. Motzkin, A Pro blem of Arrangements, Duke Math. J. 14 (19 47) 30 5-313 . 6