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1、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 ,
2、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 th
3、e 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 He
4、re /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 . Solutio
5、n 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 di
6、fferentiation. 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 t
7、hat 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 ineq
8、uality, 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 form
9、ulated 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 unknow
10、n 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 paramete
11、rs 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 th
12、is 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 th
13、e 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 ev
14、ery 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
