Solutions_Part3

Solutions_Part3 - Chapter 18 The Generalized Method of...

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Chapter 18  The Generalized Method of Moments 1 . For the normal distribution µ 2 k = σ 2 k (2 k )!/( k !2 k ) and 2 k +1 = 0, k = 0, 1, . . . . Use this result to analyze the two estimators 3 4 12 3/2 2 22 and m m bb mm == . where 1 1 ( n ki i n x = ) k x m . The following result will be useful: 211 11 .[ , ] j k j k j k jk k j Asy Cov nm nm jk j k µµµ +− + =− + . + Use the delta method to obtain the asymptotic variances and covariance of these two functions assuming the data are drawn from a normal distribution with mean and variance 2 . (Hint: Under the assumptions, the sample mean is a consistent estimator of , so for purposes of deriving asymptotic results, the difference between x and may be ignored. As such, no generality is lost by assuming the mean is zero, and proceeding from there. Obtain V , the 3 × 3 covariance matrix for the three moments, then use the delta method to show that the covariance matrix for the two estimators is 60 02 4  ′ =   JVJ where J is the 2 × 3 matrix of derivatives. The elements of J are 23 5/2 (3 /2 ) 32 2 0 m −− ∂∂ 1 4 b m = (2 ) 42 0 m b m 2 4 2 = Using the formula given for the moments, we obtain, µ 2 = σ 2 , µ 3 = 0, µ 4 = 3 σ 4 . Insert these in the derivatives above to obtain 3 00 24 σσ = J . Since the rows of J are orthogonal, we know that the off diagonal term in JVJ will be zero, which simplifies things a bit. Taking the parts directly, we can see that the asymptotic variance of 1 b will be σ -6 Asy.Var[m 3 ], which will be Asy.Var[ 1 b ] = σ -6 ( µ 6 - µ 3 2 + 9 µ 2 3 - 3 µ 2 µ 4 - 3 µ 2 µ 4 ). The parts needed, using the general result given earlier, are µ 6 = 15 σ 6 , µ 3 = 0, µ 2 = σ 2 , µ 4 = 3 σ 4 . Inserting these in the parentheses and multiplying it out and collecting terms produces the upper left element of JVJ equal to 6, which is the desired result. The lower right element will be Asy.Var[b 2 ] = 36 σ -4 Asy.Var[m 2 ] + σ -8 Asy.Var[m 4 ] - 2(6) σ -6 Asy.Cov[m 2 ,m 4 ]. The needed parts are Asy.Var[m 2 ] = 2 σ 4 93
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Asy.Var[m 4 ] = µ 8 - µ 4 2 = 105 σ 8 - (3 σ 4 ) 2 Asy.Cov[m 2 ,m 4 ] = µ 6 - µ 2 µ 4 = 15 σ 6 - σ 2 (3 σ 4 ). Inserting these parts in the expansion, multiplying it out and collecting terms produces the lower right element equal to 24, as expected. 2 . Using the results in Example 18.7, estimate the asymptotic covariance matrix of the method of moments estimators of P and λ based on and [Note: You will need to use the data in Example C.1 to estimate V .] 1 m 2 . m The necessary data are given in Examples 18.5 and 18.7. The two moments are m =31.278 and =1453.96. Based on the theoretical results m 1 2 . m 1 = P/ λ and m 2 = P(P+1)/ λ 2 , the solutions are P = µ 1 2 /( µ 2 - µ 1 2 ) and λ = µ 1 /( µ 2 - µ 1 2 ). Using the sample moments produces estimates P = 2.05682 and λ = 0.065759. The matrix of derivatives is 2 1/ / 15.207 475.648 .
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This note was uploaded on 02/28/2010 for the course ECO 211 taught by Professor Gilo during the Spring '10 term at Young Harris.

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Solutions_Part3 - Chapter 18 The Generalized Method of...

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