在Stata中,命令nlcom
使用delta方法来测试关于估计系数的非线性假设。该命令在结果窗口中显示标准错误,但遗憾的是不能将它们保存在任何地方。
估算后可用的是矩阵r(V)
,但我无法弄清楚如何使用它来计算标准误差。
答案 0 :(得分:3)
显然你需要取r(V)对角元素的平方根。这是一种将标准误差作为变量返回到一个观测数据集中的方法。
sysuse auto, clear
reg mpg weight turn
nlcom (v1: 1/_b[weight]) (v2: _b[weight]/_b[turn])
mata: se = sqrt(diagonal(st_matrix("r(V)")))'
clear
getmata (se1 se2 ) = se /* supply names as needed */
list
答案 1 :(得分:2)
您需要使用post
选项,如下所示:
. sysuse auto
(1978 Automobile Data)
. reg price mpg weight
Source | SS df MS Number of obs = 74
-------------+------------------------------ F( 2, 71) = 14.74
Model | 186321280 2 93160639.9 Prob > F = 0.0000
Residual | 448744116 71 6320339.67 R-squared = 0.2934
-------------+------------------------------ Adj R-squared = 0.2735
Total | 635065396 73 8699525.97 Root MSE = 2514
------------------------------------------------------------------------------
price | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
mpg | -49.51222 86.15604 -0.57 0.567 -221.3025 122.278
weight | 1.746559 .6413538 2.72 0.008 .467736 3.025382
_cons | 1946.069 3597.05 0.54 0.590 -5226.245 9118.382
------------------------------------------------------------------------------
. nlcom ratio: _b[mpg]/_b[weight], post
ratio: _b[mpg]/_b[weight]
------------------------------------------------------------------------------
price | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
ratio | -28.34844 58.05769 -0.49 0.625 -142.1394 85.44254
------------------------------------------------------------------------------
. di _se[ratio]
58.057686
此标准误差是方差矩阵r(V)的条目的平方根:
. matrix list r(V)
symmetric r(V)[1,1]
ratio
ratio 3370.6949
. di sqrt(3370.6949)
58.057686