Solution Manual For Introduction To Mathematical Statistics, 7th Edition

Preview (16 of 109 Pages)

100%

Purchase to unlock

Loading page image...

SOLUTIONSMANUALINTRODUCTION TOMATHEMATICALSTATISTICSSEVENTHEDITIONRobert HoggUniversity of IowaJoseph McKeanWestern Michigan UniversityAllen Craig

Loading page image...

Contents1Probability and Distributions12Multivariate Distributions113Some Special Distributions194Some Elementary Statistical Inferences315Consistency and Limiting Distributions496Maximum Likelihood Methods537Sufficiency658Optimal Tests of Hypotheses779Inferences about Normal Models8310 Nonparametric and Robust Statistics9311 Bayesian Statistics103iii

Loading page image...

Chapter 1Probability and Distributions1.2.1 Part (c):C1∩C2={(x, y) : 1< x <2,1< y <2}.1.2.3C1∩C2={mary,mray}.1.2.6Ck={x: 1/k≤x≤1−(1/k)}.1.2.7Ck={(x, y) : 0≤x≤1/k,0≤y≤1/k}.1.2.8 limk→∞Ck={x: 0< x <3}.Note: neither the number 0 nor the number 3is in any of the setsCk,k= 1,2,3, . . .1.2.9 Part (b): limk→∞Ck=φ, because no point is in all the setsCk,k= 1,2,3, . . .1.2.11 Becausef(x) = 0 when 1≤x <10,Q(C3) =∫100f(x)dx=∫106x(1−x)dx= 1.1.2.13 Part (c): Draw the regionCcarefully, noting thatx <2/3 because 3x/2<1.ThusQ(C) =∫2/30[∫3x/2x/2dy]dx=∫2/30x dx= 2/9.1.2.16 Note that25 =Q(C) =Q(C1) +Q(C2)−Q(C1∩C2) = 19 + 16−Q(C1∩C2).Hence,Q(C1∩C2) = 10.1.2.17 By studying a Venn diagram with 3 intersecting sets, it should be true that11≥8 + 6 + 5−3−2−1 = 13.It is not, and the accuracy of the report should be questioned.1

Loading page image...

2Probability and Distributions1.3.3P(C) = 12 + 14 + 18 +· · ·=1/21−(1/2) = 1.1.3.6P(C) =∫∞−∞e−|x|dx=∫0−∞exdx+∫∞0e−xdx= 26= 1.We must multiply by 1/2.1.3.8P(Cc1∪Cc2) =P[(C1∩C2)c] =P(C) = 1,becauseC1∩C2=φandφc=C.1.3.11 The probability that he does not win a prize is(9905)/(10005).1.3.13 Part (a):We must have 3 even or one even, 2 odd to have an even sum.Hence, the answer is(103)(100)(203)+(101)(102)(203).1.3.14 There are 5 mutual exclusive ways this can happen: two “ones”, two “twos”,two “threes”, two “reds”, two “blues.” The sum of the corresponding proba-bilities is(22)(60)+(22)(60)+(22)(60)+(52)(30)+(32)(50)(82).1.3.15(a)1−(485)(20)(505)(b)1−(48n)(20)(50n)≥12,Solve for n.1.3.20 Choose an integern0>max{a−1,(1−a)−1}. Then{a}=∩∞n=n0(a−1n, a+1n).Hence by (1.3.10),P({a}) =limn→∞P[(a−1n , a+ 1n)]= 2n= 0.1.4.2P[(C1∩C2∩C3)∩C4] =P[C4|C1∩C2∩C3]P(C1∩C2∩C3),and so forth. That is, write the last factor asP[(C1∩C2)∩C3] =P[C3|C1∩C2]P(C1∩C2).

Loading page image...

31.4.5[(43)(4810)+(44)(489)]/(5213)[(42)(4811)+(43)(4810)+(44)(489)]/(5213).1.4.10P(C1|C) =(2/3)(3/10)(2/3)(3/10) + (1/3)(8/10) = 37<23 =P(C1).1.4.12 Part (c):P(C1∪Cc2)=1−P[(C1∪Cc2)c] = 1−P(C∗1∩C2)=1−(0.4)(0.3) = 0.88.1.4.14 Part (d):1−(0.3)2(0.1)(0.6).1.4.16 1−P(T T) = 1−(1/2)(1/2) = 3/4, assuming independence and thatHandTare equilikely.1.4.19 LetCbe the complement of the event; i.e.,Cequals at most 3 draws to getthe first spade.(a)P(C) =14+3414+(34)2 14.(b)P(C) =14+13513952+135038513952.1.4.22 The probability that A wins is∑∞n=0(5646)n16=38.1.4.27 LetYdenote the bulb is yellow and letT1andT2denote bags of the first andsecond types, respectively.(a)P(Y) =P(Y|T1)P(T1) +P(Y|T2)P(T2) = 2025.6 + 1025.4.(b)P(T1|Y) =P(Y|T1)P(T1)P(Y).1.4.30 Suppose without loss of generality that the prize is behind curtain 1.Con-dition on the event that the contestant switches.If the contestant choosescurtain 2 then she wins, (In this case Monte cannot choose curtain 1, so hemust choose curtain 3 and, hence, the contestant switches to curtain 1). Like-wise, in the case the contestant chooses curtain 3. If the contestant choosescurtain 1, she loses. Therefore the conditional probability that she wins is23.1.4.31(1) The probability is 1−(56)4.(2) The probability is 1−[(56)2+1036]24.

Loading page image...

4Probability and Distributions1.5.2 Part (a):c[(2/3) + (2/3)2+ (2/3)3+· · ·] =c(2/3)1−(2/3) = 2c= 1,soc= 1/2.1.5.5 Part (a):p(x) ={(13x)(395−x)(525)x= 0,1, . . . ,50elsewhere.1.5.9 Part (b):50∑x=1x/5050 =50(51)2(5050) =51202.1.5.10 For Part (c): LetCn={X≤n}. ThenCn⊂Cn+1and∪nCn=R. Hence,limn→∞F(n) = 1. Letǫ >0 be given. Choosen0such thatn≥n0implies1−F(n)< ǫ. Then ifx≥n0, 1−F(x)≤1−F(n0)< ǫ.1.6.2 Part (a):p(x) =(9x−1)(10x−1)111−x=110,x= 1,2, . . .10.1.6.3(a)p(x) =(56)x−1(16),x= 1,2,3, . . .(b)∞∑x=1(56)x−1(16)=1/61−(25/36) =611.1.6.8Dy={1,23,33, . . .}. The pmf ofYisp(y) =(12)y1/3,y∈ Dy.1.7.1 If√x <10 thenF(x) =P[X(c) =c2≤x] =P(c≤ √x) =∫√x0110dz=√x10.Thusf(x) =F′(x) ={120√x0< x <1000elsewhere.1.7.2C2⊂Cc1⇒P(C2)≤P(Cc1) = 1−(3/8) = 5/8.

Loading page image...

51.7.4 Among other characteristics,∫∞−∞1π(1 +x2)dx= 1πarctanx∣∣∣∣∣∣∞−∞= 1π[π2−(−π2)]= 1.1.7.6 Part (b):P(X2<9)=P(−3< X <3) =∫3−2x+ 219dx=118[x22 + 2x]3−2=118[212−(−2)]= 2536.1.7.8 Part (c):f′(x) = 12 2xe−x= 0;hence,x= 2 is the mode because it maximizesf(x).1.7.9 Part (b):∫m03x2dx= 12 ;hence,m3= 2−1andm= (1/2)1/3.1.7.10∫ξ0.204x3dx= 0.2 :hence,ξ40.2= 0.2 andξ0.2= 0.21/4.1.7.13x= 1 is the mode because for 0< x <∞becausef(x)=F′(x) =e−x−e−x+xe−x=xe−xf′(x)=−xe−x+e−x= 0,andf′(1) = 0.1.7.16 Since ∆>0X > z⇒Y=X+ ∆> z.Hence,P(X > z)≤P(Y > z).1.7.19 Sincef(x) is symmetric about 0,ξ.25<0. So we need to solve,∫ξ.25−2(−x4)dx=.25.The solution isξ.25=−√2.

Loading page image...

6Probability and Distributions1.7.20 For 0< y <27,x=y1/3,dxdy= 13y−2/3g(y) ==13y2/3y2/39=127.1.7.22f(x)=1π ,−π2< x < π2.x=arctany,dxdy=11 +y2,−∞< y <∞.g(y)=1π11 +y2,−∞< y <∞.1.7.23G(y)=P(−2 logX4≤y) =P(X≥e−y/8) =∫1e−y/84x3dx= 1−e−y/2,0< y <∞g(y)=G′(y) ={e−y/20< y <∞0elsewhere.1.7.24G(y)=P(X2≤y) =P(−√y≤X≤ √y)=∫√y−√y13dx=2√y30≤y <1∫√y−113dx=√y3+131≤y <4g(y)=13√y0≤y <116√y1≤y <40elsewhere.1.8.4E(1/X) =100∑x=511x150.The latter sum is bounded by the two integrals∫101511xdxand∫100501xdx.An appropriate approximation might be150∫101.550.51x dx=150 (log 100.5−log 50.5).

Loading page image...

71.8.6E[X(1−X)] =∫10x(1−x)3x2dx.1.8.8 When 1< y <∞G(y)=P(1/X≤y) =P(X≥1/y) =∫11/y2x dx= 1−1y2g(y)=2y3E(Y)=∫∞1y2y3dy= 2,which equals∫10(1/x)2x dx.1.8.10 The expectation ofXdoes not exist becauseE(|X|) = 2π∫∞0x1 +x2dx= 1π∫∞11u du=∞,where the change of variableu= 1 +x2was used.1.9.2M(t) =∞∑x=1(et2)x=et/21−(et/2),et/2<1.FindE(X) =M′(0) and Var(X) =M′′(0)−[M′(0)]2.1.9.40≤var(X) =E(X2)−[E(X)]2.1.9.6E[(X−μσ)2]=1σ2σ2= 1.1.9.8K(b)=E[(X−b)2] =E(X2)−2bE(X) +b2K′(b)=−2E(X) + 2b= 0⇒b=E(X).1.9.11 For a continuous type random variable,K(t)=∫∞−∞txf(x)dx.K′(t)=∫∞−∞xtx−1f(x)dx⇒K′(1) =E(X).K′′(t)=∫∞−∞x(x−1)tx−2f(x)dx⇒K′′(1) =E[X(X1)];and so forth.

Loading page image...

8Probability and Distributions1.9.123=E(X−7)⇒E(X) = 10 =μ.11=E[(X−7)2] =E(X2)−14E(X) + 49 =E(X2)−91⇒E(X2) = 102 and var(X) = 102−100 = 2.15=E[(X−7)3].Expand (X−7)3and continue.1.9.16E(X)=0⇒var(X) =E(X2) = 2p.E(X4)=2p⇒kurtosis = 2p/4p2= 1/2p.1.9.17ψ′(t)=M′(t)/M(t)⇒ψ′(0) =M′(0)/M(0) =E(X).ψ′′(t)=M(t)M′′(t)−M′(t)M′(t)[M(t)2]⇒ψ′′(0) =M(0)M′′(0)−M′(0)M′(0)[M(0)2]=M′′(0)−[M′(0)]2= var(X).1.9.19M(t) = (1−t)−3= 1 + 3t+ 3·4t22! + 3·4·5t33! +· · ·Considering the coefficient oftr/r!, we haveE(Xr) = 3·4·5· · ·(r+ 2),r= 1,2,3. . . .1.9.20 Integrating the parts withu= 1−F(x),dv=dx, we get{[1−F(x)]x}b0−∫b0x[−f(x)]dx=∫b0xf(x)dx=E(X).1.9.23E(X)=∫10x14dx+ 0·14 + 1·12 = 58.E(X2)=∫10x214dx+ 0·14 + 1·12 =712.var(X)=712−(58)2=37192.1.9.24E(X) =∫∞−∞x[c1f1(x) +· · ·+ckfk(x)]dx=k∑i=1ciμi=μ.

Loading page image...

9Because∫∞−∞(x−μ)2fi(x)dx=σ2i+ (μi−μ)2, we haveE[(X−μ)2] =k∑i=1ci[σ2i+ (μi−μ)2].1.10.2μ=∫∞0xf(x)dx≥∫∞2μ2μf(x)dx= 2μP(X >2μ).Thus12≥P(X >2μ).1.10.4 If, in Theorem 1.10.2, we takeu(X) = exp{tX}andc= exp{ta}, we haveP(exp{tX} ≥exp{ta}]≤M(t) exp{−ta}.Ift >0, the events exp{tX} ≥exp{ta}andX≥aare equivalent. Ift <0,the events exp{tX} ≥exp{ta}andX≤aare equivalent.1.10.5 We haveP(X≥1)≤[1−exp{−2t}]/2tfor all 0< t <∞, andP(X≤ −1)≤[exp{2t} −1]/2tfor all−∞< t <0. Each of these bounds has the limit 0 ast→ ∞andt→ −∞, respectively.

Loading page image...

Chapter 2Multivariate Distributions2.1.2P(A5) = 78−48−38 + 28 = 28.2.1.5∫∞0∫∞0[2g(√x21+x22)/π√x21+x22]dx1dx2=∫∞0∫π/20[2g(ρ)/πρ]ρ dθdρ=∫∞0g(ρ)dρ= 1.2.1.6G(z)=P(X+Y≤z) =∫z0∫z−x0e−x−ydydx=∫z0[1−e−(z−x)]e−xdx= 1−e−z−ze−z.g(z)=G′(z) ={ze−z0< z <∞0elsewhere.2.1.7G(z)=P(XY≤z) = 1−∫1z∫1z/xdydx=1−∫1z(1−zx)dx=z−zlogzg(z)=G′(z) ={−logz0< z <10elsewhere.Why is−logz >0?11

Loading page image...

12Multivariate Distributions2.1.8f(x, y) =(13x)(13y)(2613−x−y)(5213)x≥0, y≥0, x+y≤13, xandyintegers0elsewhere.2.1.10P(X1+X2≤1) = 15∫1/20x21[∫1−x1x1x2dx2]dx1.2.1.14E[et1X1+t2X2]=∞∑i=1∞∑j=1et1i+t2j(12)i+j=∞∑i=1(et112)i∞∑j=1(et112)j=[11−2−1et1−1] [11−2−1et2−1],providedti<log 2,i= 1,2.2.2.1p(y1, y2) ={ (23)y2(13)2−y2(y1, y2) = (0,0),(−1,1),(1,1),(0,2)0elsewhere.2.2.2p(y1, y2) ={y1/36y1=y2,2y2,3y2;y2= 1,2,30elsewhere.y1123469p(y1)1/364/366/364/3612/369/362.2.4 The inverse transformation is given byx1=y1y2andx2=y2with JacobianJ=y2. By noting what the boundaries of the spaceS(X1, X2) map into, itfollows that the spaceT(Y1, Y2) ={(y1, y2) : 0< yi<1, i= 1,2}. The pdf of(Y1, Y2) isfY1,Y2(y1, y2) = 8y1y32.2.2.5 The inverse transformation isx1=y1−y2andx2=y2with JacobianJ= 1.The space of (Y1, Y2) isT={(y1, y2) :−∞< yi<∞, i= 1,2}.Thus thejoint pdf of (Y1, Y2) isfY1,Y2(y1, y2) =fX1,X2(y1−y2, y2),which leads to formula (2.2.1).

Loading page image...

132.3.2(a)c1∫x20x1/x22dx1=c12 = 1⇒c1= 2 andc2= 5.(b)10x1x22,0< x1< x2<1; zero elsewhere(c)∫1/21/42x1/(5/8)2dx= 6425(14−116)= 1225.(d)∫1/21/4∫1x110x1x22dx2dx1=∫1/21/4103x1(1−x31)dx1= 135512.2.3.3f2(x2)=∫x2021x21x32dx1= 7x62,0< x2<1.f1|2(x1|x2)=21x21x32/7x62= 3x21/x32,0< x1< x2.E(X1|x2)=∫x20x1(3x21/x32)dx1= 34x2.G(y)=P(34X2≤y)=∫4y/307x62dx2=(4y3)7,0< y <34g(y)={7(43)7y60< y <340elsewhere.E(Y)=7834 = 2132.Var(Y)=71024.E(X1)=2132.Var(X1)=55315360>71024.2.3.8 The marginal pdf ofXisfX(x) = 2∫∞xe−xe−ydy= 2e−2x,0< x <∞.Hence, the conditional pdf ofYgivenX=xisfY|X(y|x) = 2e−xe−y2e−2x=e−(y−x),0< x < y <∞,with conditional meanE(Y|X=x) =∫∞xye−(y−x)dy=x+ 1,x >0.

Preview Mode

This document has 109 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

Solution Manual for Introduction to Mathematical Statistics, 8th Edition

The Statistics of Inheritance

Estimation and Hypothesis Testing

Normal Distribution - Amount of Sleep

Hypothesis Testing � Comparing Two Groups

Inferential Statistics Week 2 Solution

STAT 250-004 Data Analysis Assignment 4

Correlation and Confidence Intervals

Two-Sample Hypothesis Tests

Probability and Statistics Assignment

Company

Explore

Study Tools