Chaer 3 Common Families of Disribuions Secion 31 - Inroducion Purose of his Chaer: Caalog many of common saisical disribuions (families of disribuions ha are indeed by one or more arameers) Wha we should know abou hese disribuions: Definiion Background Descriive Measures: mean, variance, momen generaing funcion Tyical alicaions Ineresing and useful inerrelaionshis Secion 32 Discree Disribuions Discree Uniform Disribuion (1, N) Definiion: A random variable has a discree uniform (1, N ) disribuion if P ( = N ) = 1/ N, = 1, 2,, N, where N is a secified osiive ineger 1
N + 1 ( N + 1)( N 1) Descriive Measures: E =, Var =, 2 12 n nn ( + 1) n 2 nn ( + 1)(2n+ 1) i = and i= 1 i = i= 1 2 6 1 N k M () = e using he formula k= 1 N Generalized Discree Uniform ( N0, N1): P ( = N 0, N1) = 1/( N1 N0 + 1), = N,, N 0 1 How o calculae E and Var Hyergeomeric Disribuion ( NMK,, ) Definiion: A discree random variable has he hyergeomeric disribuion wih arameers ( NMK,, ) if M N M K P( = N, M, K) =,ma(0, M ( N K)) min( M, K) N K Assumions of he hyergeomeric disribuion: 1 The oulaion or se o be samled consiss of N individuals, objecs, or elemens (or a finie oulaion) 2 Each individual can be characerized as a success ( S ) or failure ( F ), and here are M successes in he oulaion 3 A samle of K individuals is seleced wihou relacemen in such a way ha each subse of size n is equally likely o be chosen, ie, obain a random samle 2
4 Number of success is he random variable Hyergeomeric random variable: = he number of success among he K rials Descriive Measures: E M M ( N M)( N K) M M N K = K, Var = K ( ) = K (1 )( ) N N N( N 1) N N N 1 N K Noe: is known as he finie oulaion correcion facor which is always less han one Hence, he N 1 hyergeomeric disribuion has smaller variance han does he binomial random variable Noe: There is an easy way o calculae he mean of he hyergeomeric disribuion Le Y i = 1 if he i h rial is K M success and Y i = 0 if he i h rial is failure Then = Y i= 1 i and PY ( i = 1) = In he ne chaer (mulile N K M random variables), we will rove ha E = EY i= 1 i Then E = K In addiion, he calculaion of Var can N also be simlified by he similar way 3
Alicaion (Eamle 321 - Acceing samling) Define N =number of iem, M =number of defecive iems Then we can calculae he robabiliy ha a samle size K conains defecive iems For eamle, if N = 25, 6 25 6 0 10 0 M = 6, K = 10, = 0, hen P ( = 0) = = 0028 25 10 Alicaion (Coninue) Suose N = 100 The lo will acceed if M 6 We randomly check K ars and acce hem if and only if we find here is no defecive iem Wha is he minimum K ha can ensure ha we acce an unacceable lo wih a robabiliy less han 010? M 100 M 0 K Soluion: P ( = 0 MK, ) = 010for M > 5 However, we do no know M We need o 100 K calculae his robabiliy for 0 M 100 To find he minimum K, we only need o calculae M = 6 We can ge K = 32 P ( = 0 K= 31) = 01005 and P ( = 0 K= 32) = 00918 Bernoulli Disribuion ( ) Definiion: A random variable has a Bernoulli disribuion wih he arameer if P ( = 1 ) = and P ( = 0 ) = 1, where 0 1 4
Noes: 1 A Bernoulli rial (named afer James Bernoulli) is an eerimen wih wo, and only wo, ossible oucomes 2 Bernoulli random variable = 1 if success occurs and = 0 if failure occurs where he robabiliy of a success is Descriive Measures: E =, Var = (1 ), M ( ) = e + (1 ) Binomial Disribuion ( n, ) Definiion: A random variable has a binomial disribuion wih arameers ( n, ) if Noes: n P ( = n, ) = (1 ) n, where = 0,1,, n, n > 0, and 0 1 1 A Binomial eerimen consiss of n indeenden idenical Bernoulli rials, ie, he eerimen consiss of a sequence of n rials, where n is fied in advance of he eerimen 2 The rials are idenical, and each rial can resul in one of he same wo ossible oucomes, which we denoe by success (S) and failure (F) 3 The rials are indeenden, so ha he oucome on any aricular rial does no influence he oucome on any oher rial 4 The robabiliy of success is consan from rial o rial; we denoe his robabiliy by 5
5 n = Y i= 1 i, where Y,, 1 Yn are n idenical, indeenden Bernoulli random variables Hence, he sum of idenical, indeenden Bernoulli random variables has a Binomial disribuion ( n, ) n Descriive Measures: E n = n, Var = n(1 ), M ( ) = ( e + (1 )) Alicaion (Eamle 323 - Dice robabiliies) Suose we are ineresed in finding he robabiliy of obaining a leas on 6 in four rolls of a fair die This eerimen can be modeled as a sequence of four Bernoulli rails wih success robabiliy = 1/ 6 Le =number of 6 in four rolls, hen ~binomial(4,1/ 6) Thus P 4 1 5 0 6 6 0 4 ( 1) = 1 P ( = 0) = 1 ( ) ( ) = 0518 Poisson Disribuion (λ ) Definiion: A random variable has a Poisson disribuion wih he arameer ( λ > 0) if e λ λ P ( = λ) =, = 0,1,! Noes: 1 A Poisson disribuion is yically used o model he robabiliy disribuion of he number of occurrences (wih λ being he inensiy rae) er uni ime or er uni area 2 A basic assumion: for small ime inervals, he robabiliy of an even occurring is roorional o he lengh of waiing ime 6
3 I was shown in Secion 23 ha Binomial mf aroimaes Poisson mf Poisson mf is also a limiing disribuion of a negaive binomial disribuion 4 A useful resul: By Taylor series eansion, we have Descriive Measures: E = λ, Var = λ, M () = e λ ( e 1) e λ λ = = 0! Alicaion (Eamle 324 Waiing Time) If here are five calls in 3 minues in average, wha is he robabiliy ha here will be no calls in he ne minue? Soluion: Le = number of calls in a minue, hen has a Poisson disribuion wih E = λ = 5/3, hus 5/3 0 e (5/ 3) P= ( 0) = = 0189, P ( 2) = 1 P ( = 0) P ( = 1) = 0496 0! Alicaion (Eamle 325 Poisson aroimaion) A yeseer, on he average, makes one error in every 500 words yese A yical age conains 300 words Wha is he robabiliy ha here will be no more han wo errors in five ages Soluion: Le = number of errors in five ages, hen ~ binomial ( n, ), where n = 1500 and = 1/500 Thus 2 1500 1 1 1500 P ( 2) = ( ) (1 ) 04230 = 0 = 500 500 3 2 If we use he Poisson aroimaion, we have λ = n = 3, and P ( 2) e (1 + 3 + 3 / 2) = 04232 Negaive Binomial Disribuion (, r ) 7
Definiion: A random variable has a negaive binomial disribuion wih arameers ( r, ) if r+ 1 r P ( = r, ) = (1 ), = 0,1, Assumions and Noes: 1 The eerimen consiss of a sequence of indeenden rials 2 Each rial can resul in eiher a success (S) or failure (F) 3 The robabiliy of success,, is consan from rial o rial 4 The eerimen coninues (rials are erformed) unil a oal of r successes have been observed, where r is a secified osiive ineger 5 In conras o he binomial eerimen where he number of rials is fied and he number of successes is random, he negaive binomial eerimen has he number of successes fied and he number of rials random 6 Define =oal number of failures unil he r h success Then Y = + r is he oal number of rials unil he r h success and has a negaive binomial disribuion Then, y 1 r y r PY ( = y) = P( = y r) = (1 ) ( y r) r 1 r+ 1 r r r y 7 We have P ( = r, ) = (1 ) ( 1) (1 ), 0,1, hence he name negaive = = binomial 8
8 I is no easy o verify ha 1 r r (1 ) 1 = r = r 1 Descriive Measures: E r(1 ) r(1 ) =, Var =, M () 2 = 1 (1 ) e, hen r(1 ) EY = + r, r(1 ) r VarY =, M () 2 Y = e 1 (1 ) e How abou EY, VarYM, Y ( )? r Soluion: To calculae all he measures, we firs recall he Taylor series eansion for g ( ) = (1 ) r According k d o he Taylor series eansion g ( ) = g ( ) k 0 k = 0 = d k! d r k r k We have (1 ) = ( 1) ( r)( r 1) ( r k + 1)(1 ) k d Therefore, ( r)( r 1) ( r k + 1) r+ k 1 g ( ) ( 1) = k k k k= 0 k= 0 k! = k r Then we have (1) (2) (3) = 0 r+ 1 r r r (1 ) = (1 (1 )) = 1 r+ 1 r+ 1 r(1 ) E = = r = 1 r 1 r+ 1 1 (1 ) (1 ) (1 ) = 0 = 1 r+ 1 r+ 1 M Ee e e (1 (1 ) e ) r r r () = = (1 ) ((1 ) ) = 0 = = = 0 r 9
(4) If lim r r(1 ) = λ and 1, hen r(1 ) r(1 ) E = λ, Var = λ In addiion, 2 1+ (1 ) e ( e 1)(1 ) M e 1 (1 e ) 1 (1 e ) 1 (1 e ) r r r λ( e 1) () = [ ] = (1 + ) = (1 + ) Alicaion (Eamle 326 Inverse binomial samling) A echnique known as inverse binomial samling is useful in samling biological oulaions If he roorion of individuals wih cerain characerisics is and we samle unil we see r such individuals, hen he number of individuals samled is a negaive binomial random variable Geomeric Disribuion () Definiion: A random variable has a geomeric disribuion wih arameers ( ) if P 1 ( = ) = (1 ), = 1,2, Assumions and Noes: 1 The eerimen consiss of a sequence of indeenden rials 2 Each rial can resul in eiher a success (S) or failure (F) 3 The robabiliy of success,, is consan from rial o rial 4 The eerimen coninues (rials are erformed) unil he firs success 5 The geomeric disribuion is a secial case of he negaive binomial wih r = 1 10
Descriive Measures: E 1 1 1 e = + 1 =, Var =, M () 2 = 1 (1 ) e Memoryless roery of he Geomeric Disribuion P ( > s > ) = P ( > s ) for s > s s Proof: P ( > s > ) = P ( > s) / P ( > ) = (1 ) /(1 ) = (1 ) Alicaion (Eamle 327 Failure imes) The geomeric disribuion is someimes used o model lifeimes or ime unil failure of comonens 11
N= 30, M= 4, K= 16 N= 30, M= 16, K= 16 00 01 02 03 04 N= 30, M= 10, K= 16 00 01 02 03 04 00 01 02 03 04 N= 30, M= 22, K= 16 00 01 02 03 04 Figure 1 The mf of Hyergeomeric disribuion wih differen arameers ( NMK,, ) 12
n= 20, = 01 n= 20, = 07 00 02 04 0 5 10 15 20 0 5 10 15 20 n= 20, = 03 00 02 04 0 5 10 15 20 n= 20, = 05 00 02 04 00 02 04 n= 20, = 09 00 02 04 0 5 10 15 20 n= 20, = 095 00 02 04 0 5 10 15 20 0 5 10 15 20 Figure 2 The mf of Binomial disribuion wih differen arameers ( n, ) 13
mean= 05 mean= 2 00 01 02 03 04 mean= 1 00 01 02 03 04 00 01 02 03 04 mean= 4 00 01 02 03 04 Figure 3 The mf of Poisson disribuion wih differen arameers λ 14
= 01 = 05 00 02 04 06 00 02 04 06 00 02 04 06 = 025 = 06 00 02 04 06 Figure 4 The mf of Geomeric disribuion wih differen arameers 15