Common Families of Distributions

Chaper 3 Common Families of Disribuions 3 The pmf of X is f N N +, N, N +,, N Then EX N N N N + N + N N N N + N N N + N N + N N + Similarly, using he formula for N, we obain E N N +N + N N N N N + 6 VarX EX EX N N N N + 3 Le X number of defecive pars in he sample Then X hypergeomericn, M, K where M number of defecives in he lo and K sample size a If here are 6 or more defecives in he lo, hen he probabiliy ha he lo is acceped X is a mos 6 94 P X M, N 6, K K K K K 5 95 By rial and error we find P X 56 for K 3 and P X 98 for K 3 So he sample size mus be a leas 3 b Now P accep lo P X or, and, for 6 or more defecives, he probabiliy is a mos 6 94 6 94 P X or M, N 6, K K + K By rial and error we find P X or for K 5 and P X or 933 for K 5 So he sample size mus be a leas 5 33 In he seven seconds for he even, no car mus pass in he las hree seconds, an even wih probabiliy p 3 The only occurrence in he firs four seconds, for which he pedesrian does no wai he enire four seconds, is o have a car pass in he firs second and no oher car pass This has probabiliy p p 3 Thus he probabiliy of waiing eacly four seconds before saring o cross is p p 3 p 3 K K

3- Soluions Manual for Saisical Inference 35 Le X number of effecive cases If he new and old drugs are equally effecive, hen he probabiliy ha he new drug is effecive on a case is 8 If he cases are independen hen X binomial, 8, and P X 85 85 8 85 So, even if he new drug is no beer han he old, he chance of 85 or more effecive cases is no oo small Hence, we canno conclude he new drug is beer Noe ha using a normal approimaion o calculae his binomial probabiliy yields P X 85 P Z 5 33 37 Le X Poissonλ We wan P X 99, ha is, P X e λ + λe λ Solving e λ + λe λ by rial and error numerical bisecion mehod yields λ 66384 38 a We wan P X > N < where X binomial, / Since he cusomers choose randomly, we ake p / We hus require which implies ha P X > N N+ < < N+ This las inequaliy can be used o solve for N, ha is, N is he smalles ineger ha saisfies The soluion is N 537 < N+ b To use he normal approimaion we ake X n5, 5, where we used µ 5 and σ 5Then P X > N P X 5 5 > N 5 5 < hus, P Z > N 5 5 < where Z n, From he normal able we ge P Z > 33 99 < N 5 5 33 N 537 Therefore, each heaer should have a leas 537 seas, and he answer based on he approimaion equals he eac answer

Second Ediion 3-3 39 a We can hink of each one of he 6 children enering kindergaren as 6 independen Bernoulli rials wih probabiliy of success a win birh of approimaely 9 The probabiliy of having 5 or more successes approimaes he probabiliy of having 5 or more ses of wins enering kindergaren Then X binomial6, 9 and 3 a P X 5 4 6 6 6, 9 9 which is small and may be rare enough o be newsworhy b Le X be he number of elemenary schools in New York sae ha have 5 or more ses of wins enering kindergaren Then he probabiliy of ineres is P X where X binomial3,6 Therefore P X P X 698 c Le X be he number of Saes ha have 5 or more ses of wins enering kindergaren during any of he las en years Then he probabiliy of ineres is P X where X binomial5, 698 Therefore P X P X 39 4 M/N p,m,n K!!K! M N M K N K M/N p,m,n M!N M!N K! N!M!N M K! In he i, each of he facorial erms can be replaced by he approimaion from Sirling s formula because, for eample, M! M!/ πm M+/ e M πm M+/ e M and M!/ πm M+/ e M When his replacemen is made, all he π and eponenial erms cancel Thus, M N M M/N p,m,n K K N K M/N p,m,n M M+/ N M N M+/ N K N K+/ N N+/ M M +/ N M K+ N M K +/ We can evaluae he i by breaking he raio ino seven erms, each of which has a finie i we can evaluae In some is we use he fac ha M, N and M/N p imply N M The firs erm of he seven erms is M M M M M M M M M + M M e e Lemma 34 is used o ge he penulimae equaliy Similarly we ge wo more erms, N M N M e K N M N M K and N N N K e K N

3-4 Soluions Manual for Saisical Inference Noe, he produc of hese hree is is one Three oher erms are and The only erm lef is N M M M N M N M K N / M / / N K N M N M K K M/N p,m,n N K K M/N p,m,n p p K M N K N M K N K K b If in a we in addiion have K, p, MK/N pk λ, by he Poisson approimaion o he binomial, we heurisically ge M N M K K N p K p K e λ λ! c Using Sirling s formula as in a, we ge M N M K N K N,M,K, M N, KM N λ N,M,K, M N, KM N! N,M,K, M N, KM N e λ! λ! λ N,M,K, M N, KM N e λ λ! K e M e N M K e K N K e K K KM N M 3 Consider a sequence of Bernoulli rials wih success probabiliy p Define X number of successes in firs n rials and Y number of failures before he rh success Then X and Y have he specified binomial and hypergeomeric disribuions, respecively And we have F r P X r λ N MK N K N K P rh success on n + s or laer rial P a leas n + r failures before he rh success P Y n r + P Y n r F Y n r

Second Ediion 3-5 33 For any X wih suppor,,, we have he mean and variance of he runcaed X T are given by EX T P X T P X > P X In a similar way we ge EX T EX P X> Thus, VarX T P X P X > EX P X > P X > P X EX P X > a For Poissonλ, P X > P X e λ λ! e λ, herefore P X T e λ λ! e λ EX T λ/ e λ,, VarX T λ + λ/ e λ λ/ e λ EX P X > b For negaive binomialr, p, P X > P X r p r p p r Then r+ p r p P X T p r,,, 34 a p log p log p b EX T r p p p r VarX T r p + r p p p r r p p p r p, since he sum is he Taylor series for log p EX p p log p log p log p p log p Since he geomeric series converges uniformly, EX p p d p log p log p dp p d p p d p p log p dp log p dp p p log p Thus VarX p p log p + p log p Alernaively, he mgf can be calculaed, M pe log+pe e log p log p and can be differeniaed o obain he momens p p

3-6 Soluions Manual for Saisical Inference 35 The momen generaing funcion for he negaive binomial is r p M pe + r pe r r pe, he erm r pe pe λe λe as r, p and rp λ Thus by Lemma 34, he negaive binomial momen generaing funcion converges o e λe, he Poisson momen generaing funcion 36 a Using inegraion by pars wih, u α and dv e d, we obain Γα + α+ e d α e b Making he change of variable z, ie, z /, we obain α α e d + αγα αγα 37 Γ/ / e d z e z where he penulimae equaliy uses 334 EX ν ν Γαβ α α e /β d Γν+αβν+α Γαβ α βν Γν+α Γα / zdz Γαβ α e z / dz π π ν+α e /β d Noe, his formula is valid for all ν > α The epecaion does no eis for ν α r p 38 If Y negaive binomialr, p, is momen generaing funcion is M Y pe, and, r p from Theorem 35, M py pe Now use L Hôpial s rule o calculae p p p pe p p p e p +e p, so he momen generaing funcion converges o r, he momen generaing funcion of a gammar, 39 Repeaedly apply he inegraion-by-pars formula Γn z n z z dz n e n! + Γn z n z z dz, unil he eponen on he second inegral is zero This will esablish he formula If X gammaα, and Y Poisson The probabilisic relaionship is P X P Y α 3 The momen generaing funcion would be defined by π e + d > hus he momen generaing funcion does no eis + d, e + d On,, e >, hence

Second Ediion 3-7 3 a b EXX EXX e λ λ! e λ λ e λ λ y λ! λ y y! EX λ + EX λ + λ le y e λ λ e λ λ VarX EX EX λ + λ λ λ r + pr p r + rr + pr p r + + y p rr + p y y pr + p y p rr p, where in he second equaliy we subsiued y, and in he hird equaliy we use he fac ha we are summing over a negaive binomialr +, p pmf Thus, c EX VarX EXX + EX EX p r p rr + p + r p p p r p p Γαβ α α e /β d Γαβ α Γαβ α Γα + βα+ αα + β VarX EX EX αα + β α β αβ α+ e /β d d Use 338 EX Γα+Γα+β Γα+β+Γα EX Γα+Γα+β Γα+β+Γα VarX EX EX αγαγα+β α α+βγα+βγα α+β α+αγαγα+β α+β+α+βγα+βγα αα+ α+βα+β+ α α+β αα+ α+βα+β+ αβ α+β α+β+

3-8 Soluions Manual for Saisical Inference e The double eponenialµ, σ pdf is symmeric abou µ Thus, by Eercise 6, EX µ VarX µ σ e µ /σ d σz e z σdz σ z e z dz σ Γ3 σ 33 a α β d β β α βα β, hus f inegraes o b EX n βαn n β, herefore EX EX αβ β αβ β VarX αβ β αβ β c If β < he inegral of he second momen is infinie 34 a f β e /β, > For Y X /γ, f Y y γ /β β y γ, y > Using he ransformaion z y γ /β, we calculae e yγ EY n γ n y γ+n e yγ /β dy β n/γ z n/γ e z dz β n/γ Γ β γ + Thus EY β /γ Γ γ Γ + and VarY β/γ γ + Γ γ + b f β e /β, > For Y X/β /, f Y y ye y /, y > We now noice ha EY y e y / dy since π y e y /, he variance of a sandard normal, and he inegrand is symmeric Use inegraion-by-pars o calculae he second momen EY y 3 e y / dy π ye y / dy, where we ake u y, dv ye y / Thus VarY π/4 c The gammaa, b densiy is f X Γab a a e /b Make he ransformaion y / wih d dy/y o ge f Y y f X /y /y Γab a a+ e /by y

Second Ediion 3-9 The firs wo momens are EY EY Γab a Γa ba Γab a a e /by y a a b, Γa ba Γab a a b and so VarY a a b d f 3/ e /β, > For Y X/β /, f Γ3/β 3/ Y y Γ3/ y e y, y > To calculae he momens we use inegraion-by-pars wih u y, dv ye y o obain EY Γ3/ y 3 e y dy Γ3/ ye y dy Γ3/ and wih u y 3, dv ye y o obain EY Γ3/ y 4 e y dy 3 Γ3/ y e y dy 3 π Γ3/ Using he fac ha π y e y, since i is he variance of a n,, symmery yields y e y dy π Thus, VarY 6 4/π, using Γ3/ π α y α y e e γ γ e f e, > For Y α γ log X, f Y y e γ of EY and EY canno be done in closed form If we define I log e d, I, < y < Calculaion log e d, hen EY Eα γ log α γi, and EY Eα γ log α αγi + γ I The consan I 57757 is called Euler s consan 35 Noe ha if T is coninuous hen, P T +δ T P T +δ, T P T P T +δ P T F T +δ F T F T Therefore from he definiion of derivaive, h T F T δ F T + δ F T δ F T F T f T F T Also, d d log F T F T f T h T 36 a f T β e /β and F T h T β e /β d e /β e /β Thus, f T F T /βe /β e /β β

3- Soluions Manual for Saisical Inference b f T γ β γ e γ /β, and F T γ β γ e γ /β d γ/β e u du e u γ/β e γ /β, where u γ/β Thus, h T γ/βγ e γ /β e γ /β γ β γ c F T +e µ/β and f T e µ/β +e µ/β Thus, h T β e µ/β+e µ/β e µ/β β F T +e µ/β 37 a The uniform pdf saisfies he inequaliies of Eercise 7, hence is unimodal d b For he gammaα, β pdf f, ignoring consans, d f α e /β β βα, which only has one sign change Hence he pdf is unimodal wih mode βα c For he nµ, σ d µ pdf f, ignoring consans, df σ e /β /σ, which only has one sign change Hence he pdf is unimodal wih mode µ d For he beaα, β pdf f, ignoring consans, d d f α β α α+β, which only has one sign change Hence he pdf is unimodal wih mode 38 a i µ known, f σ ep µ, πσ σ α α+β h, cσ πσ I, σ, w σ σ, µ ii σ known, h ep b i α known, h α Γα ii β known, f µ ep πσ σ σ, cµ πσ ep f β µ σ ep µ σ ep µ σ,, w µ µ, σ Γαβ α α e β,, >, cβ β α, w β β, f α e /β epα log, Γαβα h e /β, >, cα Γαβ α w α α, log iii α, β unknown, f α, β Γαβ α epα log β, h I {>}, cα, β Γαβ α, w α α, log, w α, β /β, c i α known, h α I,, cβ Bα,β, w β β, log ii β known, h β I,, cα Bα,β, w α, log

Second Ediion 3- iii α, β unknown, h I,, cα, β Bα,β, w α α, log, w β β, log d h! I {,,,}, cθ e θ, w θ log θ, r e h r I {r,r+,}, cp p p, w p log p, 39 a For he nµ, σ f e µ /σ e /σ +µ/σ, π σ so he naural parameer is η, η /σ, µ/σ wih naural parameer space {η,η :η <, < η < } b For he gammaα, β, f Γαβ α e α log /β, so he naural parameer is η, η α, /β wih naural parameer space {η,η :η >,η < } c For he beaα, β, Γα+β f e α log +β log, ΓαΓβ so he naural parameer is η, η α, β and he naural parameer space is {η,η :η >,η > } d For he Poisson e θ f e logθ! so he naural parameer is η log θ and he naural parameer space is {η: < η < } e For he negaive binomialr, p, r known, r+ P X p r e log p, 33 a so he naural parameer is η log p wih naural parameer space {η:η < } k hcθ ep w i θ i d θ i k hc θ ep w i θ i d + k Therefore E i i k k w i θ hcθ ep w i θ i i d θ i i j k k h logcθ cθ ep w i θ i d + E i i k logcθ + E w iθ i i w i θ i logcθ w i θ i

3- Soluions Manual for Saisical Inference b θ + + + + + k hcθ ep w i θ i d i k hc θ ep w i θ i d i k k hc θ ep w i θ i i i k k hc θ ep w i θ i i i k k hcθ ep w i θ i i i w i θ i d w i θ i d w i θ i d k k w i θ hcθ ep w i θ i i h + +E θ j i i θ j k logcθ cθ ep w i θ i d c k θ h cθ ep cθ i k w i θ logcθ E k i i w i θ i + E θj logcθ + logcθ k k w i θ E i E +E k Therefore Var i i k i w i θ i k θj logcθ + Var i i k w i θ i i i d w i θ θj i w i θ i θ i j k w i θ + E i i w i θ i θ j + E w k iθ i logcθ E θj i k i d w i θ θj i w iθ θj i 333 a i h e I { << }, cθ πθ ep θ θ >, w θ θ, ii The nonnegaive real line b i h I { << }, cθ πaθ w θ aθ, w θ aθ,, ii A parabola ep a < θ <, a >,

Second Ediion 3-3 c i h I {<< }, cα αα Γα α >, w α α, w α α, log, ii A line d i h C ep 4 I { << }, cθ epθ 4 < θ <, w θ θ, w θ θ, w 3 θ θ 3, 4 3, 6, 3 4 ii The curve is a spiral in 3-space iii A good picure can be generaed wih he Mahemaica saemen ParamericPlo3D{, ^, ^3}, {,, }, ViewPoin -> {, -, 5} 335 a In Eercise 334a w λ λ and for a neθ, e θ, w θ e θ b EX µ αβ, hen β µ α Therefore h I {<< }, α cα α, α >, w Γα µ α α, w α α α α µ, log, c From b hen α,, α n, β,, β n α,, α n, α µ,, αn µ 337 The pdf µ σ f σ is symmeric abou µ because, for any ɛ >, σ f µ+ɛ µ ɛ σ σ f σ Thus, by Eercise 6b, µ is he median σ f ɛ µ ɛ µ σ σ f σ 338 P X > α P σz + µ > σz α + µ P Z > z α by Theorem 356 339 Firs ake µ and σ a The pdf is symmeric abou, so mus be he median Verifying his, wrie P Z π +z dz π an z π π b P Z π an z π π π 4 4 By symmery his is also equal o P Z Wriing z µ/σ esablishes P X µ and P X µ + σ 4 34 Le X f have mean µ and variance σ Le Z X µ σ Then EZ EX µ σ and X µ VarZ Var σ σ VarX µ σ VarX σ σ Then compue he pdf of Z, f Z z f σz +µ σ σf σz +µ and use f Z z as he sandard pdf 34 a This is a special case of Eercise 34a b This is a special case of Eercise 34b 34 a Le θ > θ Le X f θ and X f θ Le F z be he cdf corresponding o fz and le Z fzthen F θ P X P Z + θ P Z θ F θ F θ P Z θ P Z + θ P X F θ

3-4 Soluions Manual for Saisical Inference The inequaliy is because θ > θ, and F is nondecreasing To ge sric inequaliy for some, le a, b be an inerval of lengh θ θ wih P a < Z b F b F a > Le a + θ Then F θ F θ F a + θ θ F a < F b F a + θ θ F θ F θ b Le σ > σ Le X f/σ and X f/σ Le F z be he cdf corresponding o fz and le Z fz Then, for >, F σ P X P σ Z P Z /σ F /σ F /σ P Z /σ P σ Z P X F σ The inequaliy is because /σ > /σ because > and σ > σ >, and F is nondecreasing For, F σ P X P X F σ To ge sric inequaliy for some, le a, b be an inerval such ha a >, b/a σ /σ and P a < Z b F b F a > Le aσ Then F σ F /σ F aσ /σ F a < F b F aσ /σ F /σ F σ 343 a F Y y θ F X y θ y >, by Theorem 3 For θ > θ, F Y y θ F X y θ F X y θ F Y y θ for all y, since F X θ is sochasically increasing and if θ > θ, F X θ F X θ for all Similarly, F Y y θ F X y θ < F X y θ F Y y θ for some y, since if θ > θ, F X θ < F X θ for some Thus F Y y θ is sochasically decreasing in θ b F X θ is sochasically increasing in θ If θ > θ and θ, θ > hen θ > θ Therefore F X θ F X θ for all and F X θ < F X θ for some Thus F X θ is sochasically decreasing in θ 344 The funcion g is a nonnegaive funcion So by Chebychev s Inequaliy, P X b E X /b Also, P X b P X b Since g is also nonnegaive, again by Chebychev s Inequaliy we have P X b P X b EX /b For X eponenial, E X EX and EX VarX + EX For b 3, E X /b /3 > /9 EX /b Thus EX /b is a beer bound Bu for b, Thus E X /b is a beer bound E X /b / < EX /b

Second Ediion 3-5 345 a M X e f X d a e f X d e a f X d e a P X a, a b where we use he fac ha e is increasing in for > M X e f X d a e f X d a e a f X d e a P X a, where we use he fac ha e is decreasing in for < c h, mus be nonnegaive 346 For X uniform,, µ and σ, hus P X µ > kσ P For X eponenialλ, µ λ and σ λ, hus P X µ > kσ P λ kλ X λ + kλ k X + k { k k < 3, k 3, { + e k+ e k k e k+ k > From Eample 36, Chebychev s Inequaliy gives he bound P X µ > kσ /k 347 Comparison of probabiliies k u, epλ Chebychev eac eac 94 96 5 7 67 4 43 35 5 34 8 44 3 65 33 498 5 4 674 65 67 So we see ha Chebychev s Inequaliy is quie conservaive P Z > P Z > e / d π + π + e / d π + e / d+ + e / d

3-6 Soluions Manual for Saisical Inference To evaluae he second erm, le u obain + e / d +, dv e / d, v e /, du +, o + e / + e / + + e + e / d / d Therefore, P Z π π π + e / + π + e / + π + e / + + + + e / d e / d 348 For he negaive binomial r + + P X + + 349 a For he hypergeomeric b p r p + r + pp X + M k +++ P X if < k, < M, M N k P X + + M k N M if M N k N k oherwise EgXX αβ g αβ Γα βα α e /β d Le u g, du g, dv αβ α e /β, v β α e /β Then EgXX αβ Γαβ α gβ α e /β + β g α e /β d Assuming g o be differeniable, E Xg X < and g α e /β, he firs erm is zero, and he second erm is βexg X E gx β α X Γα+β ΓαΓβ g β α α β d Le u g and dv β α α β The epecaion is Γα + β g α β ΓαΓβ + g α β d E Xg X, assuming he firs erm is zero and he inegral eiss

Second Ediion 3-7 35 The proof is similar o ha of par a of Theorem 368 For X negaive binomialr, p, EgX r + g p r p r + y gy p r p y se y + y y y r + y gy p r p y r + y y y y gy r + y p r p y he summand is zero a y r + y p y y X gx E, r + X p where in he hird equaliy we use he fac ha r+y y y r+y r+y y