数理统计学导论参考答案第6~9章RV霍格.pdf

Solution to Exercises Chapter 6 Introduction to Statistical Inference Section 6.1 Point Estimation 6.1. Let n XXX, 21 represent a random sample from each of the distributions having the following probability density functions： (a） 0, 2, 1, 0, !/),(xxexf x , zero elsewhere, where . 1)0 , 0(f (b) 0 , 10,),( 1 xxxf, zero elsewhere. (c) 0,0, 1 ),( / xexf x , zero elsewhere. In each case find the m. l. e. of. Solution (a) The likelihood function of the sample is !/();( 21n nx xxxexL i Here !lnln);(ln ii xnxxL So we have . 0 )(ln n x d Ld i whose solution for is x which is the desired m .l. e. of the unknown parameter . (b) The likelihood function of the sample is 1 21 )();( n n xxxxL Here )ln)(1(ln);(ln i xnxL So we have . 0ln )(ln i x n d Ld whose solution for is i xnln/ which is the desired m. l. e. of the unknown parameter . (c) The likelihood function of the sample is / 1 );( i x n exL Here /ln);(ln i xnxL So we have . 0/ )(ln 2 i x n d Ld whose solution for is x which is the desired m. l. e. of the unknown parameter . 6.2. Let n XXX, 21 be i. i. d., each with the distribution having p. d. f. 211 / )( 221 0,)/1 (),;( 21 xexf x , zero elsewhere. Find the m. l. e. of 1 and 2 . Solution Given 2 , it is easily verify that the first order statistic can maximize the likelihood function, so the m. l. e. of 1 is the first order statistic 1 Y. The likelihood function of the sample is 211 / )( 221 0,)/1 (),;( 21 xexL i xn . 21221 / )(ln),;(ln i xnxL We observe that we may maximize by differentiation. We have 0 )(ln 2 2 1 22 i xnL whose solution is nYXi/ )( 1 which is the m. l. e. of the unknown parameter 2 . 6.3. Let n YYY 21 be the order statistics of a random sample from a distribution with p. d. f. , 2 1 2 1 , 1);(xxfzero elsewhere. Show that every statistic ),( 21n XXXu such that 2 1 ),( 2 1 21 nnn YXXXuY is a m. l. e. of . In particular, 6/ ) 142(and2/ )( , 6/ ) 124( 111 nnn YYYYYYare three such statistics. Thus the uniqueness is not in general a property of a m. l. e. Solution According to the definition of the order statistic, we have 2 1 ) 2 1 21 n YXY. From the inequality, we obtain 2 1 2 1 1 YYn which means that any statistic ),( 21n XXXu such that 2 1 ),( 2 1 121 YXXXuY nn is a m. l. e. of . Particularly, the statistics 2 1 and 2 1 1 YYn are both m. l. e. of . Furthermore, any weighty average of the two statistics is m. l. e. Since the statistics can be formulated as ). 2 1 ( 6 2 ) 2 1 ( 6 4 6/) 142( ), 2 1 ( 2 1 ) 2 1 ( 2 1 2/ )( ), 2 1 ( 6 4 ) 2 1 ( 6 2 6/ ) 124( 11 11 11 YYYY YYYY YYYY nn nn nn So 6/ ) 142(and2/ )( , 6/ ) 124( 111 nnn YYYYYYare three m. l. e. of . 6.4. Let 321 and,XXX have the multinomial distribution in which 4,25kn, and the unknown probabilities are 321 and,respectively. Here we can, for convenience, let 32143214 1and25XXXX. If the observed values of the random variables are, 7and,11, 4 321 xxx find the m. l. e. of 321 and,. Solution It is easily to understand that 3 , 2 , 1),25(ibX ii So the m. l. e. of the unknown parameters is 321 ,XXX, respectively. Thus the m. l. e. of 321 and,is 25 7 , 25 11 , 25 4 , respectively. 6.5. The Pareto distribution is frequently used as a model in study of incomes and has the distribution function 0and0where elsewhere,zero,)/(1),;( 21 1121 2 xxxF If n XXX, 21 is a random sample from this distribution, find the m. l. e. of 21 and. Solution The p. d. f. of the population is x x xf 1 1 12 21 ,),;( 2 2 . Obviously, the m. l. e. of 1 is the first order statistic 1 Y. The likelihood function of the sample is 0lnln ln ,),;( 1 22 11 2 1 1 12 21 222 2 i n nn xn nL xxx xL Thus we obtain the m. l. e. of 2 1 2 lnln YnX n i . 6.6. Let n Y be a statistic such that 0limand)(lim 2 n n Yn n YE. Prove that n Y is consistent estimator of . Proof Since 2222 )()()()( nYnnnn YEYEYEYEYE n , So, in accordance with Chebyshevs inequality, we have n YE YE Y nY n n n as, 0 )( )( )|Pr(| 2 22 2 2 for every 0. Thus according to the definition of consistent estimator, we complete the proof. 6.7. For each of the distributions in Exercise 6.1, find an estimator of by the method of moments and show that it is consistent. Solution (1) It is obvious that the population is Poisson distribution with parameter . So )(XE. Let X. We get the estimator of by the method of moments is the sample mean X. n XVXE )(,)(. For any 0, we have n n XV Xas, 0 )( )|Pr(| 22 Thus the sample mean X is a consistent estimator of the population mean . (2) The population mean is . 1 d)( 1 0 xxXE In accordance with the idea of method of moments, let X 1 We have X X 1 which is the moment estimator of . (3) In fact, the population is Gamma distribution with parameters 1 and. So)(XE. Thus the estimator of by method of moments is the sample meanX. It is easily to verify that the sample X converges in probability to the population mean, so X is a consistent estimator of. 6.8. If a random sample of size n is taken from a distribution having p. d. f. ,0,/2);( 2 xxxf zero elsewhere, find (a) The m. l. e. for. (b) The constant c so that) (cE. (c) The m. l. e. for the median of the distribution. Solution (a) The likelihood function of the sample is n n n xxxxL 2 21 /2);(. It is obvious that the thn order statistic n Y can maximize the likelihood function, so the m. l. e. for is the thn order statistic n Y. b) (b) since the p. d. f. of n Y is n n n nn ynyyf0 ,/2)( 2 12 , thus 12 2 )( n n YE n So n n c 2 12 . (c) Since ,2/,/2 2 1 22 0 2 mmdxx m In accordance with the invariant property of m. l. e. we have 2/ n Ym . 6.9. Let n XXX, 21 be i. i. d., each with a distribution with p. d. f. xexf x 0,)/1 ();( / , zero elsewhere. Find the m. l. e. of )2Pr(X. Solution It is not difficult to find that the m. l. e. of is the sample mean X. .1)/1 ()2Pr( /2/ 2 0 edxeX x In accordance with the invariance property of m. l. e., the m. l. e. of )2Pr(Xis .1 /2 X e 6.10. Let X have a binomial distribution with parameters nand p. The variance of nX / is npp/ )1 ( ; This is sometimes estimated by the m. l. e. n n X n X / )1 ( . Is this an unbiased estimator ofnpp/ )1 ( ? If not, can you construct one by multiplying this one by a constant? Solution Since . ) 1( /)()(/ )(/ )/(/ )1 ( 2 223232 n pqn npqnpnpnXEXVnpnXEnnXEn n X n X E So n n X n X / )1 ( multiplied by ) 1/( nnbecomes an unbiased estimator of npp/ )1 ( . 6.11. Let the table x 0 1 2 3 4 5 Frequency 6 10 14 13 6 1 represent a summary of a sample of size 50 from a binomial distribution having 5n. Find the m. l. e. of ) 3Pr(X. Solution Since the population ),(nbX.So the m. l. e. of ) 3Pr(X is (13+6+1)/50=2/5. 6.12. Let n YYY 21 be the order statistics of a random sample of size n from the uniform distribution of the continuous type over the closed interval ,pp. Find the maximum likelihood estimators of and p. Are these two unbiased estimators? Solution Since pYYYp n 21 The p. d. f. of the distribution is pxp p pxf, 2 1 ),;(, it is obvious that ),;(pxf is a decreasing function of the parameter p, however, pYYn2 1 , so the m. l. e. of p is . 2/ )( 1 YYn On the other hand, pYpYn 1 Obviously, the weighty average 2/ )( ) ( 2 1 ) ( 2 1 11 YYpYpY nn of pYnand pY 1 is a m. l. e. of the parameter . It is not difficult to compute the following ) 1/() 1(2/ )(,2/ )( 11 nnpYYEYYE nn . So 2/ )( 1 YYnis an unbiased estimator of while 2/ )( 1 YYnis not. Section 6.2 Confidence Intervals for Means 6.14. Let the observed value of the mean X of a random sample of size 20 from a distribution that is )80,(N be 81.2. Find a 95 percent confidence interval for. Solution Since )4 ,(NX, so ) 1 , 0( 2 N X .And the 97.5 percent quantile of the distribution from Table III in Appendix B is 1.96. Thus we have 95. 0)92. 392. 3Pr()96. 1 2 96. 1Pr( XX X . If the observed value of the mean X is 81.2, then a 95 percent confidence interval foris )12.85,28.77(3.9281.2,3.92-81.2）（. 6.15. Let X be the mean of a random sample of size nfrom a distribution that is )9 ,(N. Find n such that 90. 0) 11Pr(XX, approximately. Solution Since )/9 ,(nNX, we have 090) /9 1 /9/9 1 Pr() 11Pr() 11Pr( nn X n XXX . From Table III in Appendix B , we have 645. 1 /9 1 n , Approximately, we obtain 24nor 25n. 6.16. Let a random sample of size 17 from the normal distribution ),( 2 N yield 7 . 4xand 76. 5 2 s. Determine a 90 percent confidence interval for . Solution Since ) 1( )(1 nt S Xn , we get the 95 percent quantile from Table IV in Appendix A being 1.746. The events 746. 1 )(1 746. 1 S Xn and 1/746. 11/746. 1nSXnSX are equivalent to each other. So if 7 . 4xand 76. 5 2 s, a 90 percent confidence interval for is )7476. 5,6524. 3()117/746. 1, 117/746. 1(sxsx. 6.17. Let X denote the mean of a random sample of size n from a distribution that has mean and variance10 2 . Find n so that the probability is approximately 0.954 that the random interval ) 2 1 , 2 1 (XX includes. Solution In accordance with the central limit theorem, approximately, X is normally distributed with mean and variancen/10. So approximately, 954. 0) /102 1 /10/102 1 Pr() 2 1 2 1 (Pr nn X n XX . Let 2 /102 1 n , we obtain 160n. 6.18. Let n XXX, 21 be a random sample of size 9 from a distribution that is ),( 2 N. (a) If is known, find the length of a 95 percent confidence interval for if this interval is based on the random variable ./ )(9X (b) If is unknown, find the expected value of the length of a 95 percent confidence interval for if this interval is based on the random variable ./ )(8SX Solution (a) If is known, the statistic ) 1 , 0(/ )(9NX, so we have 0.95)9/1.969/1.96Pr()1.96/ )(91.96(PrXXX. The length of the interval is 3 . 13/96. 12)9/1.96(-9/1.96XX）（. (b) If is unknown, the statistic )8(/ )(8tSX, so we have 95. 0)8/306. 28/306. 2Pr()306. 2/ )(8306. 2(PSXSXSXr. The length of the interval is .6 . 18/306. 22)8/306. 2(-8/306. 2SSSXSX）（ 6.19. Let 121 , nn XXXX be a random sample of size1, 1nn, from a distribution that is ),( 2 N. Let n i i XX 1 and nXXS n i i / )( 1 2 .Find the constant c so that the statistic SXXc n / )( 1 has a ondistributit. If8n, determine k such that .80. 0)Pr( 9 kSXXkSXThe observed interval ),(ksxksxis often called an 80 percent prediction interval for 9 X. Solution Since)(/),/,( 2222 nnSnNX, so ) 1( 1 ), 1 , 0( 12 1 nt nS XX n n NXX n n , So c=1/1n. If 8n, ，80. 0)31.41531.415Pr( 9 SXXSXthus 4.248.31.415k 6.20. Let Y be ),300(pb.If the observed value of Yis75y, find an approximate 90 percent confidence interval for p. Solution In accordance with the central limit theorem, we approximately have ) 1 , 0( )/1)(/( N nYnYn npY . Thus approximately, 9 . 0) )/1)(/(645. 1)/1)(/(645. 1 Pr()645. 1 )/1)(/( 645. 1Pr( n nYnYnY p n nYnYnY nYnYn npY If the observed value of Yis75y, an approximate 90 percent confidence interval for p is ）（2911. 0,0.2088. 6.24. Let xbe the observed mean of a random sample of size nfrom a distribution having mean and known variance 2 . Find nso that 4/x to 4/x is an approximate 95 percent confidence interval for. Solution In accordance with the central limit theorem, we approximately have ) 1 , 0( )( N Xn . Thus approximately, .95. 0)/96. 1/96. 1Pr()96. 1 )( 96. 1Pr( nXnX Xn Let 4/1/96. 1n, we have 61n or 62n. 6.25. Assume a binomial model for a certain random variable. If we desire a 90 percent confidence interval for p that is at most 02. 0 in length, findn. Solution In accordance with the central limit theorem, we approximately have ) 1 , 0( )/1)(/( N nYnYn npY . and 2/1)2/11 (2/1)/1 (/nyny. According to the Exercise 6.20 in the preceding, A 90 percent confidence interval for p in length is n n nYnYn 645. 1 )/1)(/(645. 12 , Let ,02. 0 645. 1 n we obtain6766n. 6.26. It is known that a random variable X has a Poisson distribution with parameter. A sample of 200 observations from this population has a mean equal to4 . 3. Compute an approximate 90 percent confidence interval for. Solution 6.27. Let n YYY 21 denote the order statistics of a random sample of size n from a distribution that has p.d.f. xxxf0,/3)( 32 , zero elsewhere. (a) Show that 1c0where,1) 1/Pr( 3 n n cYc . (b) If n is 4 and if the observed value of n Y is3 . 2, what is a 95 percent confidence interval for ? Solution (a) The distribution function of the population is xtttxF x 0,/d/3)( 33 0 32 , So the p.d.f. of the nth order statistic is y ny yfyFnyf n n n 0, 3 )()()( 3 13 1 . Thus .1d 3 )Pr() 1/Pr( 3 3 13 n c n n nn cy ny YcYc (b) In accordance with the preceding discussion, let 95. 01 12 c, we have 12 05. 0c, thus a 95 percent confidence interval for is )/3 . 2, 3 . 2()/,( 44 ccyy. 6.28. Let n XXX, 21 be a random sample from ),( 2 N, where both parameters and 2 are unknown. A confidence interval for 2 can be found as follows. We know that 22 /nSis ) 1( 2 n. Thus we can find constants aand bso that 975. 0)/Pr( 22 bnS and 95. 0)/Pr( 22 bnSa. (a) Show that this second probability statement can be written as 95. 0)/Pr( 222 anSbnS. (b) If 9n and63. 7 2 S, find a 95 percent confidence interval for 2 . (c) If is known, how would you modify the preceding procedure for finding a confidence interval for 2 ? Solution (a) Since the events bnSa 22 / and anSbnS/ 222 are equivalent. So we have 95. 0)/Pr()/Pr( 2222222 anSbnSbnSa. (b) ) If 9n and63. 7 2 S, we can be get 5 .17,18. 2bafrom the Table II in the Appendix A. Thus According to the part (a), we have a 95 percent confidence interval for 2 is the interval)5 .31,924. 3(. 6.29. Let n XXX, 21 be a random sample from a gamma distribution with known parameter 3 and unknown0. Discuss the construction of a confidence interval for. Solution It is easy to verify the sample mean)/,3(nnX. Section 6.3 Confidence Intervals for differences for Means 6.30. Let two independent random samples, each of size 10, from two normal distributions ),( 2 1 N and ),( 2 2 N yield88. 7, 6 . 5,64. 8, 8 . 4 2 2 2 1 sysx. Find a 95 confidence interval for 21 . Solution From the Table IV in the Appendix A we get101. 2b. And the observed value of R in the text of section 6.3 is355. 1R. Thus a 95 confidence interval for 21 can be )047. 2 ,646. 3()355. 1101. 2,355. 1101. 2(yxyx. 6.31. Let two independent random variables 1 Yand 2 Y, with binomial distributions that have parameters 100 21 nn, 1 pand 2 p,respectively, be observed to be equal to 40and50 21 yy. Determine an approximate 90 percent confidence interval for 21 pp . Solution In accordance with the central limiting theorem, we approximately have 95. 0)645. 1645. 1Pr( 2 2 1 1 21 2 2 1 1 U n Y n Y ppU n Y n Y . From the given data, the observed value of the statistic Uis .07. 0/ )/1 (/ )/1 (/ 2222211111 nnynynnyny Thus an approximate 90 percent confidence interval for 21 pp is )24. 0 ,04. 0()24 . 05 . 0,24 . 05 . 0(uu. 6.32. Discuss the problem of finding a confidence interval for the difference 21 between the two means of two normal distribution if the variances 2 1 and 2 2 are known but not necessarily equal. Solution If the variances 2 1 and 2 2 are known but not necessarily equal, the sampling theorem of the sample mean is ),(),( 2 2 2 2 1 2 1 1 n NY n NX , So we have ),( 2 2 2 1 2 1 21 nn NYX . Thus for given confidence level , we can obtain the number a from the Table III in the Appendix A such that )()Pr( 2 2 2 1 2 1 21 2 2 2 1 2 1 nn bYX nn bYX . 6.33. Discuss Exercise 6.32 when it is assumed that the variances are unknown and unequal. This is a very difficult problem, and the discussion should point out exactly where the difficulty lies. If, however, the variances are unknown but their ratio 2 2 2 1 /is a known constantk, then a statistic that is a Trandom variable can again be used . Why? Solution According to the sampling theorem, we have ) 1 , 0( )()( 1 2 2 1 2 1 21 N nn YX , and )2( 21 2 2 2 2 22 2 1 2 11 nn SnSn and these two are independent each other, furthermore k 2 2 2 1 /, that is 2 2 2 1 kthus we can construct the statistic ).2( )2()()( 21 21 2121 2 22 2 11 21 nnt knn nnnkn SknSn YX T . Then we can apply the static T to obtain the confidence interval of 21 . 6.34. As an illustration of Exercise 6.33, one can let 921 ,XXX and 1221 ,YYYrepresent two independent random samples from the respective normal distributions ),( 2 11 N and ),( 2 22 N. It is given that 2 2 2 1 3, but 2 2 is unknown. Define a random variable which has a ondistributitthat can be used to find a 95 percent interval for 21 . Solution Since ).2( )2()()( 21 21 2121 2 22 2 11 21 nnt knn nnnkn SknSn YX T From the Table IV in the Appendix A, 093. 2b. Thus a 95 percent interval for 21 is ) )2( )( 093. 2)( , )2( )( 093. 2)( 2121 21 2 22 2 11 2121 21 2 22 2 11 nnnkn knnSknSn YX