"如何用为wald检验对参数矩阵进行了对称检验"

这标题写的。。。。。。。。。。。让人泪如雨下！

比较急，很抱歉，你见过相关的例子吗？

不知道你能看明白吗？如何用stata进行编程，是否有相关的例子

受楼主召唤，自插双目扶墙而进，免得看到那令人蛋疼的标题而爆瞎。

楼主，你确认你的意思都表达完整清楚了么？

俺股沟了下，实在没有什么东东能把stata, wald test, matrix, symmetric联系在一起。

。。。。。。。。。。。。。。鄙人扶墙而出先。


PS： 发现有个笔算wald test的栗子，希望能给楼主丁点启发。


. ****** (B) TEST H0: b3 = 0, b4 = 0. . . * (1) Wald test requires estimation of unrestricted model only.
poisson y x2 x3 x4
Iteration 0:   log likelihood = -238.77153  
Iteration 1:   log likelihood = -238.77153  
Poisson regression                                Number of obs   =        200                                                  
LR chi2(3)      =       8.30                                                  Prob > chi2     =     0.0401
Log likelihood = -238.77153                       Pseudo R2       =     0.0171
------------------------------------------------------------------------------         
y |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------         
x2 |  -.0275702   .0767909    -0.36   0.720    -.1780775    .1229371         
x3 |   .1630037   .0670848     2.43   0.015     .0315199    .2944874         
x4 |   .1026568   .0802139     1.28   0.201    -.0545595    .2598732      
_cons |  -.1653238   .0773479    -2.14   0.033     -.316923   -.0137246
------------------------------------------------------------------------------
.
.
* (1A) Stata Wald test command.
test (x3=0) (x4=0)
( 1)  [y]x3 = 0 ( 2)  [y]x4 = 0           
chi2(  2) =    8.57         Prob > chi2 =    0.0138
.
.
* (1B) Wald test done manually.
* Use h'[RVR]-inv*h.
.
* Details below will change for each example.
.
* In particular, for nonlinear restrictions more work in forming R.
* Note that Stata puts the intercept last, not first.
.
* So here the second and third elements of b are set to zero.
. matrix bfull = e(b)                    /* 1xq row vector */   
. matrix vfull = e(V)                    /* qxq matrix */   
. matrix h = (bfull[1,2]\bfull[1,3])     /* hx1 vector */
. matrix R = (0,1,0,0\0,0,1,0)           /* h x q matrix */
. matrix Wald = h'*syminv(R*vfull*R')*h  /* scalar */
. matrix list hh[2,1]           
c1
r1  .16300365
r2  .10265681
. matrix list
RR[2,4]   
c1  c2  c3  c4
r1   0   1   0   0
r2   0   0   1   0
. matrix list
Wald symmetric
Wald[1,1]           
  c1
c1 8.5701855

Table 2 gives Wald- and J-tests for various hypotheses concerning the model. Since our sample size is large (8 equations times 4,847 observations per equation), we use a 1 percent critical value for all tests. Consider price effects first. The Wald test of symmetry in the asymmetric model is marginally insignificant with a/7-value of 1.4 percent, and symmetry is not rejected for either the level of prices (A/ = A,' for all /) or for prices interacted with implicit utility y (B = B'). This lack of symmetry rejection is not due to the irrelevance of price effects. In particular, the direct price effects given by the A0 matrix are strongly significant, as are many of the interactions between prices and demographics in the Az matrices. The interaction of prices with income are less significant, and we cannot reject the hypothesis that the py interaction given by the B matrix is excludable.