[讨论]如何调用数个模型的估计量?

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在构造一些统计量时,比如wald统计量,hausman统计量等,往往需要用到几个模型的估计结果.

这个问题在EVIEWS中较容易解决:

例如用到两个残差平方和, (eq1.@ssr-eq0.@ssr)就可以了,在STATA中应如何表示呢?

在STATA作回归时,甚至没有标明模型的序号.如 (reg y x)

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关键词：估计量 hausman EVIEWS ausman Stata 讨论 模型

请看张晓峒教授《计量经济学基础》第2版p.181中CES生产函数模型的示例。注意ereturn list命令的执行结果，可以实现EViews中(eq1.@ssr-eq0.@ssr)，例如估计一个模型以后将它们先保存，备用。

. reg y x1 x2

Source | SS df MS Number of obs = 17
-------------+------------------------------ F( 2, 14) = 5020.09
Model | 8.9868461 2 4.49342305 Prob > F = 0.0000
Residual | .012531224 14 .000895087 R-squared = 0.9986
-------------+------------------------------ Adj R-squared = 0.9984
Total | 8.99937733 16 .562461083 Root MSE = .02992

------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
x1 | 1.021123 .0294043 34.73 0.000 .9580574 1.084189
x2 | 1.471946 .2392908 6.15 0.000 .9587184 1.985174
_cons | -10.46387 1.287012 -8.13 0.000 -13.22424 -7.703506
------------------------------------------------------------------------------

. ereturn list

scalars:
e(N) = 17
e(df_m) = 2
e(df_r) = 14
e(F) = 5020.09387451462
e(r2) = .9986075453980505
e(rmse) = .0299180122244058
e(mss) = 8.986846104616193
e(rss) = .0125312243764357
e(r2_a) = .9984086233120577
e(ll) = 37.18637865362844
e(ll_0) = -18.71546244884161

macros:
e(title) : "Linear regression"
e(depvar) : "y"
e(cmd) : "regress"
e(properties) : "b V"
e(predict) : "regres_p"
e(model) : "ols"
e(estat_cmd) : "regress_estat"

matrices:
e(b) : 1 x 3
e(V) : 3 x 3

functions:
e(sample)

. reg y x1 x2 x3

Source | SS df MS Number of obs = 17
-------------+------------------------------ F( 3, 13) = 4023.22
Model | 8.98969469 3 2.9965649 Prob > F = 0.0000
Residual | .009682637 13 .000744818 R-squared = 0.9989
-------------+------------------------------ Adj R-squared = 0.9987
Total | 8.99937733 16 .562461083 Root MSE = .02729

------------------------------------------------------------------------------
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
x1 | 1.169302 .0803773 14.55 0.000 .9956573 1.342947
x2 | 1.029332 .3144375 3.27 0.006 .3500309 1.708633
x3 | -.0601934 .0307793 -1.96 0.072 -.1266881 .0063013
_cons | -8.714548 1.475957 -5.90 0.000 -11.90316 -5.525936
------------------------------------------------------------------------------

. matrix b=get(_b)

. mat list b

b[1,4]
x1 x2 x3 _cons
y1 1.1693019 1.0293318 -.0601934 -8.7145482

. * 广义技术进步
. scalar A=exp(b[1,4])

. dis "A="A
A=.00016418

. * 分配系数
. scalar delta=b[1,1]/(b[1,1]+b[1,2])

. dis "delta="delta
delta=.53183116

. * 与要素替代弹性相关的系数rho
. scalar rho=(-2*b[1,3]*(b[1,1]+b[1,2]))/(b[1,1]*b[1,2])

. dis "rho="rho
rho=.21991238

. * 规模报酬系数
. scalar m=b[1,1]+b[1,2]

. dis "m="m
m=2.1986338

. * 资本替代劳动的替代弹性系数
. scalar sigma=1/(1+rho)

. dis "sigma="sigma
sigma=.819731