实例介绍
【实例简介】
这是Introduction to Mathematical Statistics经典教材的答案，下载之后好好学习哦~
Contents 1 Probability and Distributions 2 Multivariate Distributions 11 3 Some Special Distributions 17 4 Unbiasedness, Consistency, and Limiting Distributions 29 5 Some Elementary Statistical Inferences 35 6 Maximum Likelihood methods 49 7 Sufficiency 59 8 Optimal Tests of Hypotheses 69 9 Inferences about Normal models 75 10 Nonparametric Statistics 85 11 Bayesian Statistics 93 12 Linear Models 99 Chapter 1 Probability and Distributions 12.1Part(c):C1∩C2={(x,y):1<x<2,1<y<2} 1.2.3 C1n C2=mary, mray) 12.6Ck={x:1/k≤x≤1-(1/k) 1.2.7 {(x,y):0≤x≤1/k,0≤y≤1/k} 1.2.8 limk-oo Ck=a:0<x<3. Note: neither the number 0 nor the number 3 is in any of the sets Ck, k= 1, 2, 3 1.2.9 limk_oo Ck=, because no point is in all the sets Ck, k= 1, 2, 3, 1.2. 11 Because f(a=0 when 1<x< 10 10 ∫(x)dx=/6(1-x) 1.2.13 Part(c): Draw the region C carefully, noting that a 2/3 because 3 c/2<1 Thus 2/33a/2 2/9 /2 1. 2. 16 Note that Q(C)=Q(C1)+Q(2)-Q(C1∩C2)=19+10-QC1nC2 Hence,Q(C1∩C2)=10 1.2. 17 By studying a venn diagram with 3 intersecting sets, it should be true that 11>8+6+5-3-2-1=13 It is not, and the accuracy of the report should be questioned Probability and Distributions 1.3.1 P(C) 1/2 1.3.6 P(C) edx+/e-adr=2≠1 We must multiply by 1/2 1.3.8 PCC2)=I[C1nC2)9=P(C)=1, because Cl∩C2= P and°=C 1.3. 11 Compute the probabilities of obtaining 2 red, 1 white, 1 blue and 1 red, 2 white, 1 blue and 1 red, 1 white, 2 blue and sum those probabilities 1.3.14 Part(a]: We must have 3 even or one even, 2 odd to have an even sum Hence, the answer is 10/10 10/10 3 3 1.3.15 There are5 mutual exclusive ways this can happen:two“ones”,two“twos”, two“ threes”,two“ reds”, two"blues.3 The sum of the corresponding proba- bilities is )+ +(。)()+ 1.3.16 ≥, Solve for n 1.3. 23 Choose an integer no>maxfa-l(1-a)-). Then fa)=nmno(a-i,a+1) Hence by(1.3.10) P(a=li 1.4.2 P[(C1∩C2∩C3)nCa=PC4C1nC2∩C3]P(C1∩c2nC) and so forth. That is write the last factor as P(C1∩C2)nC3]=PC3|Ci1∩C2]P(C1∩C2) 1.4.5 48 +(2) [(4)G8)(3)(16)+(4)()/(3) 1.4.9Part(b): (1/2)(6/10) (1/2)(3/10)+(1/2)(6/10)3 1.4.10 (2/3)3/10) 32 P(CIC) (2/3)(3/10)+(1/3)(8/10 )-73=P(Ch 1.4.12Part(c) P(C1UC2)=1-P[(1UC)9=1-P(C1∩C2) 1-(0.4)(0.3)=0.88 1.414Part(d): 1-(03)2(0.)(6) 1.4. 1-P(TT)=1-(1/ 2)(1/ 2)=3/4, assuming independence and that h an T' are equilikely 1.4.19 Let C be the complement of the event; i.e., C equals at most 3 draws to get the first spade (a)P(O)=+是+() (b)P(C)=+部+3 1.2 The probability that a wins is=0()”喜=善 1.4.27 Let y denote the bulb is yellow and let Ti and T2 denote bags of the first and second types, respectively P(Y)=P(Y)P(Ti)+P(YT2)P(2)=26+ P(T1Y) P(YT1P(T1 P(Y 1.4. 30 Suppose without loss of generality that the prize is behind curtain 1 Condition on the event that the contestant switches if the contestant chooses curtain 2 then she wins, (In this case Monte cannot choose cur- tain 1. so he must choose curtain 3 and hence. the contestant switches to curtain 1). Likewise, in the case the contestant chooses curtain 3. If the contestant chooses curtain 1. she loses Therefore the conditional probability that she wins is 4 Probability and Distributions 1.4.31(1)The probability is 1-(6) (2) The probability is 1- (6)2+30 15. 2 Part c[(2/3)+(2/3)2+(2/3)3+… c(2/3) 2c=1 1-(2/3) 1/2 1.5.5 Part(a) P(a) 5 0 where 1.5.9 Part(a) ∑ 50(51)51 /5050 T=1 2(5050)202 1.5. 11 For Part(c): Let Cn=X Sn. Then Cn C Cn+1 and Un Cn =R. Hence, F(n Let e>0 be given. Choose no such that n no implies 1-F(n)<E. Then if x 270, 1-F(r)<1-F(no)<e 1.6.2Part(a): 10 =1,2,10. 1.63 m) 1,2,3. (b)∑ 1/6 (25/36)11 16.8D={1,23,3,……}. The pmf of y is y p(3y) ∈ 171If√<10then F(x)=P[X()=c2≤x=P(c≤√) 102=10 T 0<x<100 elsewhere 5 1.7.2 1→P(C2)≤P(Ci=1-(3/8)=5/8 1.7.4 Among other characteristics 1「丌 丌(142a=- arctan 2 2 1.7.6 Part(b) P(X2<9 P(-3<X<3) d 1g 182 +2 1.7. 8 Part(c) 0: hence, x= 2 is the mode because it maximizes f(r) 1.7.9 Part(b) 32- d hence,m3=2-1andm=(1/2)1/3 l.7.10 4a3dx=0.2 hence,52=02and5o2=0.21/4 1.7.13 =l is the mode because for 0< oo because +re=re ∫() ce-x+e-=0, and∫(1)=0. 1.7.16 Since△>0 X>z→Y=X+△ Hence, P(X>2)< P(Y>a) 1.7. 19 Since f(ar)is symmetric about 0, 5.25 <0. So we need to solve, 52 The solution is $.25 Probability and Distributions 1.720For0<y<27, 1/3 dx 1 y 2/3 g(y)= 7 1.7.22 丌 ny dy 1 +y4 <y< q(3 ∞<y< T 1+y 2 1.7.23 (y)=P(-2logX4≤y)=P(X≥e-8) 4 xo dx= 1 /2 y g(y)= G(y) e-y/20<y<∞o 0 elsewhere 1.724 (X2≤y)=P(√≤X≤√) 0≤y<1 /￥=+31≤y<4 0≤y< 9(y) 6√y 1<y<4 elsewhere 1.8.5 B(1/X)=∑ 50 x=51 The latter sum is bounded by the two integrals 101 dx and 1001 An appropriate approximation might be r1015 (log100.5-log50.5) 50.5 【实例截图】
【核心代码】
这是Introduction to Mathematical Statistics经典教材的答案，下载之后好好学习哦~
Contents 1 Probability and Distributions 2 Multivariate Distributions 11 3 Some Special Distributions 17 4 Unbiasedness, Consistency, and Limiting Distributions 29 5 Some Elementary Statistical Inferences 35 6 Maximum Likelihood methods 49 7 Sufficiency 59 8 Optimal Tests of Hypotheses 69 9 Inferences about Normal models 75 10 Nonparametric Statistics 85 11 Bayesian Statistics 93 12 Linear Models 99 Chapter 1 Probability and Distributions 12.1Part(c):C1∩C2={(x,y):1<x<2,1<y<2} 1.2.3 C1n C2=mary, mray) 12.6Ck={x:1/k≤x≤1-(1/k) 1.2.7 {(x,y):0≤x≤1/k,0≤y≤1/k} 1.2.8 limk-oo Ck=a:0<x<3. Note: neither the number 0 nor the number 3 is in any of the sets Ck, k= 1, 2, 3 1.2.9 limk_oo Ck=, because no point is in all the sets Ck, k= 1, 2, 3, 1.2. 11 Because f(a=0 when 1<x< 10 10 ∫(x)dx=/6(1-x) 1.2.13 Part(c): Draw the region C carefully, noting that a 2/3 because 3 c/2<1 Thus 2/33a/2 2/9 /2 1. 2. 16 Note that Q(C)=Q(C1)+Q(2)-Q(C1∩C2)=19+10-QC1nC2 Hence,Q(C1∩C2)=10 1.2. 17 By studying a venn diagram with 3 intersecting sets, it should be true that 11>8+6+5-3-2-1=13 It is not, and the accuracy of the report should be questioned Probability and Distributions 1.3.1 P(C) 1/2 1.3.6 P(C) edx+/e-adr=2≠1 We must multiply by 1/2 1.3.8 PCC2)=I[C1nC2)9=P(C)=1, because Cl∩C2= P and°=C 1.3. 11 Compute the probabilities of obtaining 2 red, 1 white, 1 blue and 1 red, 2 white, 1 blue and 1 red, 1 white, 2 blue and sum those probabilities 1.3.14 Part(a]: We must have 3 even or one even, 2 odd to have an even sum Hence, the answer is 10/10 10/10 3 3 1.3.15 There are5 mutual exclusive ways this can happen:two“ones”,two“twos”, two“ threes”,two“ reds”, two"blues.3 The sum of the corresponding proba- bilities is )+ +(。)()+ 1.3.16 ≥, Solve for n 1.3. 23 Choose an integer no>maxfa-l(1-a)-). Then fa)=nmno(a-i,a+1) Hence by(1.3.10) P(a=li 1.4.2 P[(C1∩C2∩C3)nCa=PC4C1nC2∩C3]P(C1∩c2nC) and so forth. That is write the last factor as P(C1∩C2)nC3]=PC3|Ci1∩C2]P(C1∩C2) 1.4.5 48 +(2) [(4)G8)(3)(16)+(4)()/(3) 1.4.9Part(b): (1/2)(6/10) (1/2)(3/10)+(1/2)(6/10)3 1.4.10 (2/3)3/10) 32 P(CIC) (2/3)(3/10)+(1/3)(8/10 )-73=P(Ch 1.4.12Part(c) P(C1UC2)=1-P[(1UC)9=1-P(C1∩C2) 1-(0.4)(0.3)=0.88 1.414Part(d): 1-(03)2(0.)(6) 1.4. 1-P(TT)=1-(1/ 2)(1/ 2)=3/4, assuming independence and that h an T' are equilikely 1.4.19 Let C be the complement of the event; i.e., C equals at most 3 draws to get the first spade (a)P(O)=+是+() (b)P(C)=+部+3 1.2 The probability that a wins is=0()”喜=善 1.4.27 Let y denote the bulb is yellow and let Ti and T2 denote bags of the first and second types, respectively P(Y)=P(Y)P(Ti)+P(YT2)P(2)=26+ P(T1Y) P(YT1P(T1 P(Y 1.4. 30 Suppose without loss of generality that the prize is behind curtain 1 Condition on the event that the contestant switches if the contestant chooses curtain 2 then she wins, (In this case Monte cannot choose cur- tain 1. so he must choose curtain 3 and hence. the contestant switches to curtain 1). Likewise, in the case the contestant chooses curtain 3. If the contestant chooses curtain 1. she loses Therefore the conditional probability that she wins is 4 Probability and Distributions 1.4.31(1)The probability is 1-(6) (2) The probability is 1- (6)2+30 15. 2 Part c[(2/3)+(2/3)2+(2/3)3+… c(2/3) 2c=1 1-(2/3) 1/2 1.5.5 Part(a) P(a) 5 0 where 1.5.9 Part(a) ∑ 50(51)51 /5050 T=1 2(5050)202 1.5. 11 For Part(c): Let Cn=X Sn. Then Cn C Cn+1 and Un Cn =R. Hence, F(n Let e>0 be given. Choose no such that n no implies 1-F(n)<E. Then if x 270, 1-F(r)<1-F(no)<e 1.6.2Part(a): 10 =1,2,10. 1.63 m) 1,2,3. (b)∑ 1/6 (25/36)11 16.8D={1,23,3,……}. The pmf of y is y p(3y) ∈ 171If√<10then F(x)=P[X()=c2≤x=P(c≤√) 102=10 T 0<x<100 elsewhere 5 1.7.2 1→P(C2)≤P(Ci=1-(3/8)=5/8 1.7.4 Among other characteristics 1「丌 丌(142a=- arctan 2 2 1.7.6 Part(b) P(X2<9 P(-3<X<3) d 1g 182 +2 1.7. 8 Part(c) 0: hence, x= 2 is the mode because it maximizes f(r) 1.7.9 Part(b) 32- d hence,m3=2-1andm=(1/2)1/3 l.7.10 4a3dx=0.2 hence,52=02and5o2=0.21/4 1.7.13 =l is the mode because for 0< oo because +re=re ∫() ce-x+e-=0, and∫(1)=0. 1.7.16 Since△>0 X>z→Y=X+△ Hence, P(X>2)< P(Y>a) 1.7. 19 Since f(ar)is symmetric about 0, 5.25 <0. So we need to solve, 52 The solution is $.25 Probability and Distributions 1.720For0<y<27, 1/3 dx 1 y 2/3 g(y)= 7 1.7.22 丌 ny dy 1 +y4 <y< q(3 ∞<y< T 1+y 2 1.7.23 (y)=P(-2logX4≤y)=P(X≥e-8) 4 xo dx= 1 /2 y g(y)= G(y) e-y/20<y<∞o 0 elsewhere 1.724 (X2≤y)=P(√≤X≤√) 0≤y<1 /￥=+31≤y<4 0≤y< 9(y) 6√y 1<y<4 elsewhere 1.8.5 B(1/X)=∑ 50 x=51 The latter sum is bounded by the two integrals 101 dx and 1001 An appropriate approximation might be r1015 (log100.5-log50.5) 50.5 【实例截图】
【核心代码】
标签：
好例子网口号：伸出你的我的手 — 分享！
小贴士
感谢您为本站写下的评论，您的评论对其它用户来说具有重要的参考价值，所以请认真填写。
- 类似“顶”、“沙发”之类没有营养的文字，对勤劳贡献的楼主来说是令人沮丧的反馈信息。
- 相信您也不想看到一排文字/表情墙，所以请不要反馈意义不大的重复字符，也请尽量不要纯表情的回复。
- 提问之前请再仔细看一遍楼主的说明，或许是您遗漏了。
- 请勿到处挖坑绊人、招贴广告。既占空间让人厌烦，又没人会搭理，于人于己都无利。
关于好例子网
本站旨在为广大IT学习爱好者提供一个非营利性互相学习交流分享平台。本站所有资源都可以被免费获取学习研究。本站资源来自网友分享，对搜索内容的合法性不具有预见性、识别性、控制性，仅供学习研究，请务必在下载后24小时内给予删除，不得用于其他任何用途，否则后果自负。基于互联网的特殊性，平台无法对用户传输的作品、信息、内容的权属或合法性、安全性、合规性、真实性、科学性、完整权、有效性等进行实质审查；无论平台是否已进行审查，用户均应自行承担因其传输的作品、信息、内容而可能或已经产生的侵权或权属纠纷等法律责任。本站所有资源不代表本站的观点或立场，基于网友分享，根据中国法律《信息网络传播权保护条例》第二十二与二十三条之规定，若资源存在侵权或相关问题请联系本站客服人员，点此联系我们。关于更多版权及免责申明参见 版权及免责申明
网友评论
我要评论