A Recursive Formula for Moments of a Binomial Distribution


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1 A Recursive Formula for Momets of a Biomial Distributio Árpád Béyi Uiversity of Massachusetts, Amherst, MA ad Saverio M Maago Naval Postgraduate School, Moterey, CA While teachig a course i probability ad statistics, oe of the authors came across a apparetly simple questio about the computatio of higher order momets of a radom variable The topic of momets of higher order is rarely emphasized whe teachig a statistics course The textboos we came across i our classes, for example, treat this subject rather scarcely; see [3, pp ], [4, pp ], also [2, p 206] Most of the examples give i these boos stop at the secod momet, which of course suffices if oe is oly iterested i fidig, say, the dispersio or variace of a radom variable X, D 2 X = M 2 X MX 2 Nevertheless, momets of order higher tha 2 are relevat i may classical statistical tests whe oe assumes coditios of ormality These assumptios may be checed by examiig the sewess or urtosis of a probability distributio fuctio The sewess, or the first shape parameter, correspods to the the third momet about the mea It describes the symmetry of the tails of a probability distributio The urtosis, also ow as the secod shape parameter, correspods to the fourth momet about the mea ad measures the relative peaedess or flatess of a distributio Sigificat sewess or urtosis idicates that the data is ot ormal However, we arrived at higher order momets uitetioally The discussio that follows is the outcome of a questio raised by our studets without owig much about the motivatio for studyig momets of higher order It has to do with discrete radom variables x1 x X 2 x m p 1 p 2 p m ad their pmomets M p X = MX p = m i=1 x p i p i 1 Here, p deotes a positive iteger; o the first lie of the table of X we have the values of the radom variable i icreasig order, that is x 1 < x 2 < < x m, while o the secod lie we read the probabilities with which X taes the values x i, amely p i = P {X = x i }, 1 i m We ote also that for 1 i m, p i [0, 1] ad m i=1 p i = 1 Let 1 be a iteger Give ow the discrete radom variable X B, 1 such that for all = 1,,, P {X = } = 1 2 2, the studets were ased to compute the first momet MX = 1 2 = c THE MATHEMATICAL ASSOCIATION OF AMERICA
2 The th colum of the variable X refers to the evet of drawig white balls out of a bowl cotaiig a total of equally umbered white ad blac balls I this drawig scheme a particular case of the biomial or Beroulli scheme we require that a extracted ball is reitroduced i the bowl after each drawig Goig bac to the computatio of the first momet 2, it is easy to see that MX = 1 2 =1 1 1 = = 2 3 The trasitio from the first to the secod equality follows from the wellow Newto biomial idetity 2 q = q = q q r r=0, q N After explaiig the calculatio of 2, as a couterquestio from the studets, it was ased whether the same id of argumet would give a simplified aswer to M 2 X or eve, say, M 2003 X? Ad how about a aswer i closed form for M p X, p 1, i geeral? I what follows, we pla to aswer satisfactorily, we hope the former questio by displayig a recurrece relatio for the geeral pmomets The reader should ote that the recursive formula is useful for calculatios usig pecil ad paper as log as p is i a relatively small rage Observe also that, eve for the particular case of X i discussio, the recursio does ot fall ito a very ice shape Most of the studets would rarely thi of usig pecil ad paper for such computatios cosiderig the sigificat use of software such as STS whe teachig a statistics class However, our decisio of writig this short ote was made by a eed of showig our studets what we as mathematicias do i certai situatios: we try to prove rigorously ad i most geerality the observatios we mae by looig at several particular cases displayed o a computer scree Our mai goal is to show the followig result: Propositio Let p 1 The a M p X = M p 1 X 1 2 M p 1X 1 ; 4 b M p X = 2 Q p 1, where Q p p 1 is the moic polyomial of degree p 1 i the variable which solves the recursio Q p = 2Q p 1 1Q p 1 1 Proof Let us redeote S, p = 2 M p X = p =0 Sice p C = p 1 = p 1 1 1, VOL 36, NO 1, JANUARY 2005 THE COLLEGE MATHEMATICS JOURNAL 69
3 we ca write or S, p = =1 p = p 1 =0 = =0 p 1 =1 p S, p = S, p 1 S 1, p 1 ; 5 i the secod sum above we used the covetio 1 0 It is clear that 5 traslates to the pmomets exactly as the recursio a We ow prove part b Newto s biomial formula gives S, 0 = 2 Fromhere ad 5, we get S, 1 = S, 0 S 1, 0 = = 2 1 Note that, i particular, we recover the computatio of the first momet 3 Similarly, oe gets S, 2 = , S, 3 = , etc The formulas obtaied for the first four sums S, p, 0 p 3, suggest that S, p = 2 p Q p 1, 6 where Q p 1 is some polyomial of degree p 1ithevariable with uitary leadig coefficiet Ideed, oe ca prove this usig 5 ad a iductio argumet o p Assumig that 6 holds true, we have S, p + 1 = 2 p Q p p Q p 1 1 where we deoted = 2 p 1 2Q p 1 1Q p 1 1 = 2 p+1 Q p, Q p = 2Q p 1 1Q p 1 1 By our iductive assumptio, Q p 1 = p 1 + lower order terms To complete the iductio step, we oly eed to show that Q p is a polyomial of degree p i the variable with uitary leadig coefficiet To see this, ote that the leadig term of Q p comes from the differece 2 p p 1 = p + p 1 p = p + lower order terms, which proves our claim Observe that 6 ca be rewritte as which is equivalet to the statemet b 2 M p X = 2 p Q p 1, We remar that fidig explicitly the polyomials Q p for a geeral positive iteger p turs out to be a otrivial tas The iterested reader should chec for example that already for M 4 X, the polyomial Q 3 does ot factor completely over 70 c THE MATHEMATICAL ASSOCIATION OF AMERICA
4 the itegers as its predecessors I fact, Q 3 = Nevertheless, the recursio b ca be used to ifer certai properties of the polyomials Q p which defie the momets of X For example, usig agai a iductio argumet o p, oe ca show that the polyomial Q p has 0 as a root if p is eve, ad 1 as a root if p is odd I particular, we ca ifer further iformatio about the shape of the momets, amely that for all p 1, + 1 divides 2 2p M 2p X ad 2 divides 2 2p+1 M 2p+1 X It is also worth poitig out that there exists a alterative approach to the computatio of the sums S, p ad thus of the momets M p X If s 1, s 2,,s p 1 deote the symmetric sums associated to the umbers 1, 2,,p 1, that is, s 1 = p 1, s 2 = p 2 p 1,, s p 1 = 1 2 p 1, the the followig recursio holds: M p X s 1 M p 1 X p s p 1 M 1 X = 1 p + 12 p 7 Start by otig that for p we have 1 p + 1 p = 1 p + 1 p Hece, 1 p + 1 =0 = 1 p + 1 =p = 1 p + 12 p, p p where we used yet agai Newto s biomial idetity to obtai the last equality But 1 p + 1 = p s 1 p p s p 1, where s 1, s 2,,s p 1 are the symmetric sums defied above If we ow recall the defiitio of the sums S, p, we obtai S, p s 1 S, p 1 + s 2 S, p p s p 1 S, 1 = 1 p + 12 p, which is equivalet to the recurrece relatio 7 For more o the uses of recurrece relatios ad differece equatios, the iterested reader is referred to [1] To coclude, aswerig the questio raised by the studets about M 2003 X i a closed form usig oly pecil ad paper is ot coveiet The recursive formulas 4 ad 7 however are useful from a computatioal poit of view While 7 loos more elegat, formula 4 seems a bit easier to wor with sice it does ot require the a priori calculatio of symmetric sums I a way, pecil, paper ad computers all come together to solve i a satisfactory maer our problem of, ad for, the momet Acowledgmet The authors tha the referee for some useful suggestios Refereces 1 G E Box, G M Jeis, ad G C Reisel, Time Series Aalysis, Forecastig ad Cotrol, 3rd ed, Pretice Hall, 1994 VOL 36, NO 1, JANUARY 2005 THE COLLEGE MATHEMATICS JOURNAL 71
5 2 A T Craig ad R V Hogg, Itroductio to Mathematical Statistics, 5th ed, Macmilla, J L Devore, Probability ad Statistics: for Egieerig ad the Scieces, 4th ed, Broos/Cole, R J Larso ad M L Marx, A Itroductio to Mathematical Statistics ad Its Applicatios, 3rd ed, Pretice Hall, 2001 Proof Without Words From Richard Hammac of Robert Madiso Uiversity ad David Lyos of Lebao Valley College: Theorem A alteratig series a 1 a 2 + a 3 a 4 + a 5 a 6 + a 7 a 8 + coverges to a sum S if a 1 a 2 a 3 a 4 0 ad a 0 Moreover, if s = a 1 a 2 + a 3 ±a is the th partial sum the s 2 < S < s 2+1 Proof S a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 2 1 a 2 a 2+1 a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 2 1 a 2 s 2 s 2+1 A differet versio of this result with words is to appear i the Oxford Joural of Teachig Mathematics ad Its Applicatio 72 c THE MATHEMATICAL ASSOCIATION OF AMERICA
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