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Unformatted text preview: PROFESSOR HONG FALL 2007 TA: Jaeho Yun Economics 319: Selected Suggested Solutions for 2006 Final 4. [10 pts] (a) The sample mean is de...ned as follows: Pn Xi X n = i=1 : n The expectation of sample mean can be derived as follows: Pn Pn E[Xi ] i=1 Xi E(X n ) = E = i=1 n n Pn i=1 = n = : Therefore, X n is unbiased for : 2 (b) For this, we have to ...gure out whether E(X n ) = 2 : We know that that V ar(X n ) = 2 E[(X n )2 ] = n , since Xi is iid. Let' use this result. s
2 E[(X n E[(X n ; n 2 )2 ] = E[X n )] = = = = =
2 E[X n ] 2 E[X n ] 2 E[X n ] 2 E[X n ] 2 2 Xn + 2 2 E[2 X n ] + E[ 2 ] 2 E[X n ] + 2
2 2 + 2 :
2 2 Therefore, we have E[X n ] biased for 2 : 5. [10 pts] 2 2 = 2 n , which implies E[X n ] = + 2 n 6= 2 : So, X n is 2 1 (a) (1 )100% con...dence interval when Xn p z n
=2 = 5% is as follows: < Xn + p z n
=2 ; < for every n 1, where z =2 is the uppertail critical value of N (0; 1) at level =2 (z2:5% = 1:96). Note that we use standard normal distribuion since is known. (Otherwise, we should have used studentt distribution) Now, we have that n is 4, is 3 (Note that standard deviation should be used rather than variance), and X n = 5+10+6+7 = 7: Hence, 4 con...dence interval can be computed as follows: 7 3 p 4 1:96 < 4:06 < 3 <7+ p 4 < 9:94: 1:96; (b) The con...dence level is the probability that will lie in the interval between X n p z =2 and X n + p z =2 : The higher 1 , the more likely that will lie in the interval. n n In other words, the probability that will fall on the con...dence interval is (1 )100%: 6. [10 pts] We will use the tstatistics since 2 is unknown. Note that the degree of freedom is 1, since the sample size is 2. Compute the tstatistic as follows: t= X p 0: S= n If jtj < t1;0:005 ; then H0 is accepted, and HA is rejected. Otherwise, H0 is rejected, and HA is accepted. In order to implement this hypothesis test, we need the following: Sample mean: X= Sample variance: S
2 5 + 11 = 8: 2 2 1 = 9 + 9 = 18: = P2 i=1 (Xi X)2 = (X1 X)2 + (X2 X)2 2 Sample standard deviation: S= p S2 = p p 18 = 3 2: t1;0:005 = 63:656 from the studentt distribution table.
8 p p 4 3 2= 2 As a result, we have t = accept H0 ; and reject HA . 7. [15 pts] 2 It is known that nb21 this, we can know that E = 4=3: jtj is clearly less than t1;0:005 . Therefore, we 2 n (Note that this is wrongly stated in the question). From nb2 1
2 nb2 1
2 = n; and V ar = 2n: (a) Check whether b2 and b2 i unbiased for 2 : 1 h 22 are From the fact that E nb21 = n; we can know that E b2 = 2 : Therefore, b2 is 1 1 2 unbiased for : In order to check the unbiasedness of b2 ; we are going to use a wellknown property 2 of the sample variance for iid random sample. Pn X)2 2 i=1 (Xi ; S = n 1 E(S 2 ) = 2 ; 2 4 : V ar(S 2 ) = n 1 Note that b2 = 2
n 1 2 S : n Therefore, we have that n n n n
2 V ar[b2 ] = 2 Since E b2 = 2
n 1 n E b2 2 = 1 E S2 = 1
2 n n 1 2 ; 1 n
2 2 V ar[S 2 ] = n 2 4 n 1 4 = 2 : n 1 n2 6= 2 (b) Note that MSE=bias2 +Variance. , b2 is biased for 2 . 3 result, M SE(b2 ) is 2n : 1 1 Next, consider b2 : We know that this is biased. The bias is n 2 since E b2 = 2 2 n 1 2 from part (a). The variance is nn21 2 4 (also from part (a)). Therefore, we have n 2 1 1 that M SE(b2 ) = n2 4 + nn21 2 4 = 2n 2 1 4 = n 4 n2 4 : 2 n Because M SE(b2 ) > M SE(b2 ); we can conclude that b2 is more e cient than b2 . 1 2 2 1 8. [20 pts] (a) We have to ...nd the condition under which E [b] = : "
n X i=1 First, consider b2 :hThis is unbiased estimator, so bias term is zero. The variance can 1 i 4 nb2 1 be derived from V ar 2 = 2n in part (a). From this, we have V ar b2 = 2n : As a 1
4 E [b] = E = ci Xi =
n X i=1 # P Hence, in order for b to be unbiased, we should have n ci = 1. i=1 (b) For the most e cient unbiased estimator, we have to solve the following minimization problem: " n # X M in V ar ci Xi
ci i=1 n X i=1 n X i=1 ci E[Xi ] ci = ci : s:t: Using the indepence of Xi ; we have the following: " n # n X X V ar ci Xi = c2 V ar(Xi ) i
i=1 i=1 n X i=1 ci = 1 = n X i=1 2 c2 i i2 : Let' construct the Lagrangian equation to solve the above maximization problem: s L=
n X i=1 2 c2 2 i i + (1 n X i=1 ci ): 4 The ...rst order contition is as follows:
2 @L = 2 2 ci = 0; @ci i n X @L (ii) = 1 ci = 0: @ i=1 (i) for all i0 s; From (i), we can derive 2 2 ci = i2 for all i0 s: Let' sum up these equations for all i: s Then, we have the following: 2
2 n X i=1 ci = = n(n + 1)(2n + 1) 6 n X i=1 i2 (from the hint (there is a typo!)) If we use (ii), then we have the following: = 12 2 : n(n + 1)(2n + 1) Plug the above into (i), then we have: ci = = 1 2 1
2 i2 12 2 2 n(n + 1)(2n + 1) 6i2 = n(n + 1)(2n + 1) i2 2 Finally, we have the following best unbiased estimator: b =
n X i=1 6i2 Xi : n(n + 1)(2n + 1) (c) We know that b is the best unbiased estimator, and that the sample mean, X n is also unbiased. Therefore, b is clearly better than X n in terms of relative e ciency. 2 Note that, for b ; the weight, ci is increasing as the variance, i2 decreases with i. That is to put more weight on less noisy observation. This improves e ciency. However, the sample mean put the same weights on all the observations although they have dierent levels of information. This fact makes b more e cient than X n : 5 ...
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This note was uploaded on 12/05/2007 for the course ECON 3190 taught by Professor Hong during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 HONG
 Economics

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