Solution Manual For Introduction To Mathematical Statistics, 7th Edition

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SOLUTIONSMANUALINTRODUCTION TOMATHEMATICALSTATISTICSSEVENTHEDITIONRobert HoggUniversity of IowaJoseph McKeanWestern Michigan UniversityAllen Craig

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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

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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

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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).

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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.

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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.

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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.

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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).

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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.

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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.

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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.

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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

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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).

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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.

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