Some Discrete Distributions

CHAPTER 6Some discrete distributions6.1. Examples: Bernoulli, binomial, Poisson, geometric distributionsBernoulli distributionA random variablebe aXsuch thatBernoulli random variableP(X 1) pwith parameterVar X p p2 p(1 p).We denote such a random variable byBinomial distributionA random variable X has a k) nk pk (1 p)n k .nwith parametersnandpcumbersome calculations one can deriveX Y1 · · · Yn ,EX EY1 · · · EYn np.soifP(X X Binom (n, p).Bernoulli trials is a binomial random variable.is binomial, thentoX Bern (p).binomial distributionWe denote such a random variable byThe number of successes inP(X 0) 1 p is saidp. Note EX p and EX 2 p,andEX np. An easier wayYi are independentwhere theAfter someis to realize that ifXBernoulli variables, soWe have not de ned yet what it means for random variables to be independent, but here wemean that the events such as(Yi 1)are independent.Proposition 6.1X : Y1 · · · Yn , where {Yi }ni 1 are independent Bernoulli random variablesparameter p, thenEX np, Var X np(1 p).SupposewithProof. First we use the de nition of expectation to see that nnXXn in in ip (1 p) ip (1 p)n i .EX iiii 0i 1Then81

826. SOME DISCRETE DISTRIBUTIONSEX nXii 1 npn!pi (1 p)n ii!(n i)!nXi 1(n 1)!pi 1 (1 p)(n 1) (i 1)(i 1)!((n 1) (i 1))!n 1X(n 1)!pi (1 p)(n 1) ii!((n 1) i)!i 0n 1X n 1 nppi (1 p)(n 1) i np,ii 0 npwhere we used the Binomial Theorem (Theorem 1.1).To get the variance ofX,we rst observe thatEX 2nXi 1EYi2 XEYi Yj .i6 jNowEYi Yj 1 · P(Yi Yj 1) 0 · P(Yi Yj 0) P(Yi 1, Yj 1) P(Yi 1)P(Yj 1) p2n22using independence of random variables {Yi }i 1 . Expanding (Y1 · · · Yn ) yields n terms,22of which n are of the form Yk . So we have n n terms of the form Yi Yj with i 6 j . HenceVar X EX 2 (EX)2 np (n2 n)p2 (np)2 np(1 p). Later we will see that the variance of the sum of independent random variables is the sumof the variances, so we could quickly get Var X np(1 p). Alternatively, one can computeE(X 2 ) EX E(X(X 1)) using binomial coe cients and derive the variance of X fromthat.Poisson distributionA random variableXhas thePoisson distributionP(X i) e λWe denote such a random variable byX Pois(λ). Xi 0so the probabilities add up to one.with parameterλi.i!Note thatλi /i! eλ ,λif

6.1. EXAMPLES: BERNOULLI, BINOMIAL, POISSON, GEOMETRIC DISTRIBUTIONS83Proposition 6.2SupposeXis a Poisson random variable with parameterλ,thenEX λ,Var X λ.Proof. We start with the expectationEX Xiei λ λi 0i! λ e Xλi 1λ λ.(i 1)!i 1Similarly one can show thatE(X ) EX EX(X 1) 2 λ2 e λ Xi 22 λ ,soEX 2 E(X 2 X) EX λ2 λ,Example 6.1. Xi 0i(i 1)e λλii!i 2λ(i 2)!and hence Var X λ.Suppose on average there are 5 homicides per month in a given city. Whatis the probability there will be at most 1 in a certain month?Solution :IfXEX 5. Since the expectationP(X 0) P(X 1) e 5 5e 5 .is the number of homicides, we are given thatfor a Poisson isExample 6.2.λ,thenλ 5.ThereforeSuppose on average there is one large earthquake per year in California.What's the probability that next year there will be exactly 2 large earthquakes?Solution : λ EX 1, so P(X 2) e 1 ( 21 ).We have the following proposition connecting binomial and Poisson distributions.Proposition 6.3 (Binomial approximation of Poisson distribution)Xn is a binomial random variable with parameters n and pnP(Xn i) P(Y i), where Y is Poisson with parameter λ.Ifandnpn λ,then

846. SOME DISCRETE DISTRIBUTIONS6.1 (Approximation of Poisson by binomials)Note that by settingpn : λ/nforn λwe can approximate the Poisson distribution with parametertions with parametersnandλby binomial distribu-pn .This proposition shows that the Poisson distribution models binomials when the probabilityof a success is small. The number of misprints on a page, the number of automobile accidents,the number of people entering a store, etc. can all be modeled by a Poisson distribution.Proof. For simplicity, let us suppose thatcan useλn npn λ.n λ npnforn λ.In the general case weWe writeP(Xn i) n!pi (1 pn )n ii!(n i)! n i n iλn(n 1) · · · (n i 1) λ1 i!nnnin(n 1) · · · (n i 1) λ (1 λ/n) .nii! (1 λ/n)iObserve that the following three limits existn(n 1) · · · (n i 1) 1,n nii(1 λ/n) 1,n (1 λ/n) e λ ,nn which completes the proof.In Section 2.2.3 we consideredk 1, 2, . . . , n.discrete uniform distributionsP(X k) n1die (with n 6),withThis is the distribution of the number showing on aforforexample.Geometric distributionA random variableXhas the geometric distribution with parameterP(X i) (1 p)i 1pforp, 0 p 1,ifi 1, 2, . . . .Using a geometric series sum formula we see that Xi 1 XP(X i) (1 p)i 1 p In Bernoulli trials, if we leti 1X1p 1.1 (1 p)get a heads, thenXX will be a geometricX is the rst time webe the rst time we have a success, thenrandom variable. For example, if we toss a coin over and over andwill have a geometric distribution. To see this, to have the rst success

6.1. EXAMPLES: BERNOULLI, BINOMIAL, POISSON, GEOMETRIC DISTRIBUTIONSk thoccur on thek 1(1 p)k 1 p.trial, we have to havesuccess. The probability of that isProposition 6.4failures in the rstk 1p, 0 p 1,If X is a geometric random variable with parameter85trials and then athen1EX ,p1 p,p2Var X FX (k) P (X 6 k) 1 (1 p)k .Proof. We will use X1 nrn 1(1 r)2n 0which we can show by di erentiating the formula for geometric seriesThenEX Xi 1Then the variancei · P(X i) Xi 11/(1 r) P 112 ·p .p(1 (1 p))i · (1 p)i 1 p 2 X 2 11Var X E (X EX) E X i · P(X i)ppi 12To nd the variance we will use another sum. First Xrnrn , (1 r)2n 0which we can di erentiate to see that X1 r n2 rn 1 .3(1 r)n 1ThenEX 2 Xi 1i · P(X i) 2 Xi 1i2 · (1 p)i 1 p Thus2 pVar X EX (EX) p2222 p(1 (1 p)).3 ·p p2(1 (1 p)) 211 p .pp2n 0rn .

866. SOME DISCRETE DISTRIBUTIONSThe cumulative distribution function (CDF) can be found by using the geometric series sumformula1 FX (k) P (X k) XP(X i) i k 1 Xi 1(1 p)i k 1(1 p)kp (1 p)k .p 1 (1 p) Negative binomial distributionA random variableXhasnegative binomial distributionP(X n) with parameters n 1 rp (1 p)n r , n r, r 1, . . . .r 1randpifA negative binomial represents the number of trials until r successes. To get the aboveththsuccess in the ntrial, we must exactly have r 1 successes in theformula, to have the rthrst n 1 trials and then a success in the ntrial.Hypergeometric distributionA random variableXhas hypergeometric distribution with parameters mN min i .P(X i) Nnm, nandNifN balls, of which m are oneN m are another, and we choose n balls at random without replacement,the probability of having i balls of the rst color.This comes up in sampling without replacement: if there arecolor and the otherthenXrepresentsAnother model where the hypergeometric distribution comes up is the probability of a successchanges on each draw, since each draw decreases the population, in other words, when weconsider sampling without replacement from a nite population). Thensize,mis the number of success states in the population,quantity drawn in each trial,inNis the populationis the number of draws, that is,is the number of observed successes.

6.2. FURTHER EXAMPLES AND APPLICATIONS876.2. Further examples and applications6.2.1. Bernoulli and binomial random variables.Example 6.3.A company prices its hurricane insurance using the following assumptions:(i) In any calendar year, there can be at most one hurricane.(ii) In any calendar year, the probability of a hurricane is 0.05.(iii) The numbers of hurricanes in di erent calendar years are mutually independent.Using the company's assumptions, nd the probability that there are fewer than 3 hurricanesin a 20-year period.Solution :X the number of hurricanesX Binom (20, 0.05), thereforedenote bywe see thatin a 20-year period. From the assumptionsP (X 3) P (X 6 2) 202020119020 (0.05) (0.95) (0.05) (0.95) (0.05)2 (0.95)18012 0.9245.Example 6.4.Phan has a0.6 probability of making a free throw.Suppose each free throwis independent of the other. If he attempts 10 free throws, what is the probability that hemakes at least 2 of them?Solution :IfX Binom (10, 0.6),thenP (X 2) 1 P (X 0) P (X 1) 1010010 1 (0.6) (0.4) (0.6)1 (0.4)901 0.998.6.2.2. The Poisson distribution.Recall that a Poisson distribution models well eventsthat have a low probability and the number of trials is high. For example, the probability ofa misprint is small and the number of words in a page is usually a relatively large numbercompared to the number of misprints.(1) The number of misprints on a random page of a book.(2) The number of people in community that survive to age100.(3) The number of telephone numbers that are dialed in an average day.(4) The number of customers entering post o ce on an average day.Example 6.5.Levi receives an average of two texts every3minutes. If we assume thatthe number of texts is Poisson distributed, what is the probability that he receives ve ormore texts in a 9-minuteperiod?Copyright 2017 Phanuel Mariano, Patricia Alonso Ruiz, Copyright 2020 Masha Gordina.

886. SOME DISCRETE DISTRIBUTIONSSolution :LetXbe the number of texts in a9 minuteperiod. ThenP (X 5) 1 P (X 6 4) 1 λ 3·2 6and4X6n e 6n 0n! 1 0.285 0.715.Example 6.6.tationλ.Solution :LetX1 ,.,Xkbe independent Poisson random variables, each with expec-What is the distribution of the random variableThe distribution ofYis Poisson with the expectationuse Proposition 6.3 and (6.1) to choosepn kλ1 /n λ1 /m λ/nY : X1 . Xk ?n mkλ kλ.To show this, weBernoulli random variables with parameterto approximation the Poisson random variables.them all together, the limit asn If we sumgives us a Poisson distribution with expectationlim npn λ. However, we can re-arrange the same n mk Bernoulli random variablesn in k groups, each group having m Bernoulli random variables. Then the limit gives us thedistribution ofX1 . Xk .This argument can be made rigorous, but this is beyond thescope of this course. Note that we do not show that the we have convergence in distribution.Example 6.7.Letλ1 , . . . , λk ,X1 . Xk ?pectationSolution :X1 , . . . , X kbe independent Poisson random variables, each with ex-respectively.Y λ λ1 . λk . To showthis, we again use Proposition 6.3 and (6.1) with parameter pn λ/n. If n is large, we canseparate these n Bernoulli random variables in k groups, each having ni λi n/λ Bernoullirandom variables. The result follows if lim ni /n λi for each i 1, ., k .The distribution ofYWhat is the distribution of the random variableis Poisson with expectationn This entire set-up, which is quite common, involves what is calleddistributed Bernoulli random variablesExample 6.8.independent identically(i.i.d. Bernoulli r.v.).Can we use binomial approximation to nd the mean and the variance ofa Poisson random variable?Solution :Yes, and this is really simple. Recall again from Proposition 6.3 and (6.1) that wecan approximate Poissonwherepn λ/n.Ywith parameterλby a binomial random variableEach such a binomial random variable is a sum onrandom variables with parameterpn .n n independent BernoulliThereforeEY lim npn lim nn λ λ,nλVar(Y ) lim npn (1 pn ) lim nn n nBinom (n, pn ), λ1 λ.n

6.2. FURTHER EXAMPLES AND APPLICATIONS6.2.3. Table of distributions.The following table summarizes the discrete distribu-tions we have seen in this chapter.N0 N {0}HereNstands for the set of positive integers, andis the set of nonnegative integers.PMF (kNameNotationParametersBernoulliBern(p)p [0, 1]1kBinomialBinom(n, p)n Np [0, 1]nkPoissonPois(λ)λ 0GeometricGeo(p)p (0, 1)NegativeNBin(r, p)r Np (0, 1)binomialHypergeometric89Hyp(N, m, n) N N0n, m N0 N0 )E[X] Var(X)pk (1 p)1 kpp(1 p)pk (1 p)n knpnp(1 p) ke λ λk!λ((1 p)k 1 p, for k 1,1p0,else.( k 1 rp (1 p)k r , if k r, rr 1p0,else. m(mk)(Nn k)N(n)nmNλ1 pp2r(1 p)p2nm(N n)m(1 N)N (N 1)