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发布时间 : 星期五 文章计量经济学习题参考答案更新完毕开始阅读4ebcc3fe4afe04a1b171de06

108972.4 49213.2 39769.4 17665.1 2324.7 120350.3 52571.3 45565.0 19119.9 3094.1 136398.8 56834.4 55963.0 20615.1 2986.3 160280.4 63833.5 69168.4 23199.4 4079.1 188692.1 71217.5 80646.3 26605.2 10223.1 221170.5 80120.5 94103.2 30292.7 16654.1 先判断模型的识别性(略)

利用两阶段最小二乘法可得参数估计结果如下:

System: UNTITLED

Estimation Method: Two-Stage Least Squares Date: 10/24/08 Time: 12:04 Sample: 1979 2006

Included observations: 28

Total system (balanced) observations 56 Coefficie

Std. t-StatistProb. nt Error ic C(1) 935.4488 330.7224 2.828501 0.0067 C(2) 0.081721 0.025100 3.255844 0.0020 C(3) 0.866883 0.070477 12.30015 0.0000 C(4) -1374.284 394.7544 -3.481364 0.0010 C(5) 1.050314 0.088376 11.88463 0.0000 C(6) -0.727237 0.101261 -7.181776 0.0000 Determinant residual 6.79E+11

covariance Equation: CT=C(1)+C(2)*YT+C(3)*CT(-1) Instruments: CT(-1) YT(-1) GT XT C Observations: 28 R-squared 0.997869 Mean dependent 25754.3

var 6

Adjusted 0.997698 S.D. dependent 23980.0R-squared var 1 S.E. of 1150.465 Sum squared 3308923regression resid 1 Durbin-Watson 0.536937 stat Equation: IT=C(4)+C(5)*YT+C(6)*YT(-1) Instruments: CT(-1) YT(-1) GT XT C Observations: 28 R-squared 0.996936 Mean dependent 23723.1

var 5

Adjusted 0.996691 S.D. dependent 25706.2R-squared var 4 S.E. of 1478.676 Sum squared 5466204regression resid 7

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2001 2002 2003 2004 2005 2006 解答

Durbin-Watson 0.559530 stat 其中,C(1)、 C(2) 、C(3)、 C(4)、 C(5) 、C(6)顺次表示模型中的参数 2?0、?1、?、?、、?1?20参数估计量不是无偏估计量,但满足渐近无偏性。

12.利用表6-4中的数据,对下列供需均衡模型进行参数估计,说明参数估计量的性质。

?Dt??0??1Pt??2Yt??1t??St??0??1Pt??2Wt??2t ?D?S?Qtt?t其中,Dt、St、Pt分别为需求量、供给量和价格,是模型的内生变量;Yt、Wt分别为

收入和气候条件,是模型的外生变量。

t 1 2 3 4 5 6 7 8 9 10 Qt 11 16 11 14 13 17 14 15 12 18 表6-4 样本数据 Yt Pt 20 8.1 18 8.4 12 8.5 21 8.5 27 8.8 28 9.0 25 8.9 27 9.4 30 9.5 28 9.9 Wt 42 58 35 46 41 56 48 50 39 52 解答 先判断模型的识别性(略)

利用两阶段最小二乘法可得参数估计结果如下:

System: UNTITLED

Estimation Method: Two-Stage Least Squares Date: 10/24/08 Time: 12:17 Sample: 1 10

Included observations: 10

Total system (balanced) observations 20 C(1) C(2) C(3) C(4) C(5) C(6) Coefficie

nt 112.5397 2.719132 -18.27092 -2.409108 0.215527 0.244597 Std. t-StatistError ic 244.4152 0.460445 5.438436 0.499984 41.56901 -0.439532 3.354744 -0.718120 0.114937 1.875180 0.065811 3.716676 30.47657

Prob. 0.6523 0.6249 0.6670 0.4845 0.0818 0.0023

Determinant residual covariance Equation: QT=C(1)+C(2)*PT+C(3)*YT

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