各位大神，求教个问题，在用广义倾向得分匹配（GPS）分析时，老是出现两个问题：1. The assumption of Normality is not statistically satisfied at .05 level
It is advisable to try a different trasformation of the treatment variable
2. 1 group found, 2 required

对于第一个问题，我依次尝试了ln, lnskew0, bcskew0,boxcox几种不同形式，还是出现这种提示。
求教对这两个问题应该如何解决，万分感谢！！

回归结果如下：
qui generate cut=20 if cen<=20

. . qui replace cut=40 if cen>20 & cen<=40

. . qui replace cut=60 if cen>40 & cen<=60

. . qui replace cut=80 if cen>60

. . qui replace cut=100 if cen>80

. matrix define tp=(10\20\30\40\50\60\70\80\90\100)

. doseresponse lnta , outcome(lntotal) t(cen) gpscore(pscore) predict(hat_treat) sigma(sd) cutpoints(cut) index(p50) nq_gps(5) t _transf(boxcox) dose_response(dose_response) tpoints(tp) delta(1) reg_type_t(quadratic) reg_type_gps(quadratic) interaction(1) bootstrap(yes) boot_reps(100) filename("output") analysis(yes) graph("graph_output") detail

********************************************
ESTIMATE OF THE GENERALIZED PROPENSITY SCORE
********************************************

Generalized Propensity Score

******************************************************
Algorithm to estimate the generalized propensity score
******************************************************



Estimation of the propensity score

The BoxCox transformation of the treatment variable cen is used

T
-------------------------------------------------------------
Percentiles Smallest
1% 1.166742 1.166742
5% 1.179652 1.166742
10% 1.179652 1.166742 Obs 22,231
25% 1.219498 1.166742 Sum of Wgt. 22,231

50% 1.252957 Mean 1.26956
Largest Std. Dev. .0656851
75% 1.340244 1.384281
90% 1.360923 1.384766 Variance .0043145
95% 1.368533 1.384929 Skewness .2122693
99% 1.376733 1.384948 Kurtosis 1.628219

initial: log likelihood = -<inf>(could not be evaluated)
feasible: log likelihood = -31542.819
rescale: log likelihood = -21284.562
rescale eq: log likelihood = -3300.4439
Iteration 0: log likelihood = -3300.4439(not concave)
Iteration 1: log likelihood =14393.478(not concave)
Iteration 2: log likelihood =27857.065
Iteration 3: log likelihood =28702.593
Iteration 4: log likelihood =29035.453
Iteration 5: log likelihood =29039.052
Iteration 6: log likelihood =29039.053

Number of obs = 22,231
Wald chi2(1) = 101.38
Log likelihood =29039.053 Prob > chi2 = 0.0000

------------------------------------------------------------------------------
T | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
eq1 |
lnta | .0032904 .0003268 10.07 0.000 .0026499 .0039309
_cons | 1.198379 .0070833 169.18 0.000 1.184496 1.212261
-------------+----------------------------------------------------------------
eq2 |
_cons | .0655344 .0003108 210.86 0.000 .0649253 .0661436
------------------------------------------------------------------------------

Test for normality of the disturbances

Kolmogorov-Smirnov equality-of-distributions test
Normal Distribution of the disturbances

One-sample Kolmogorov-Smirnov test against theoretical distribution
normal((res_etreat - r(mean))/sqrt(r(Var)))

Smaller group D P-valueCorrected
----------------------------------------------
res_etreat: 0.1579 0.000
Cumulative: -0.1224 0.000
Combined K-S: 0.1579 0.000 0.000

Note: Ties exist in dataset;
there are 22224 unique values out of 22231 observations.

The assumption of Normality is not statistically satisfied at .05 level
It is advisable to try a different trasformation of the treatment variable

Estimated generalized propensity score
-------------------------------------------------------------
Percentiles Smallest
1% 1.569283 1.146247
5% 1.821294 1.226061
10% 2.006727 1.228207 Obs 22,231
25% 2.664825 1.238576 Sum of Wgt. 22,231

50% 4.133582 Mean 3.962133
Largest Std. Dev. 1.381247
75% 5.111096 6.087499
90% 5.848306 6.087522 Variance 1.907842
95% 5.929101 6.087524 Skewness -.1122427
99% 6.018518 6.087524 Kurtosis 1.696721

********************************************
End of the algorithm to estimate the gpscore
********************************************

******************************************************************************
The set of the potential treatment values is divided into 6 intervals

The values of the gpscore evaluated at the representative point of each
treatment interval are divided into 5 intervals
******************************************************************************

***********************************************************
Summary statistics of the distribution of the GPS evaluated
at the representative point of each treatment interval
***********************************************************

Variable | Obs Mean Std. Dev. Min Max
-------------+---------------------------------------------------------
gps_1 | 22,231 4.543274 .237251 2.885289 5.826589

Variable | Obs Mean Std. Dev. Min Max
-------------+---------------------------------------------------------
gps_2 | 22,231 4.774775 .2135613 3.095474 5.918881

Variable | Obs Mean Std. Dev. Min Max
-------------+---------------------------------------------------------
gps_3 | 22,231 3.037522 .2429328 1.566484 4.691811

Variable | Obs Mean Std. Dev. Min Max
-------------+---------------------------------------------------------
gps_4 | 22,231 2.071809 .2117767 .9311559 3.648537

Variable | Obs Mean Std. Dev. Min Max
-------------+---------------------------------------------------------
gps_5 | 22,231 1.608863 .1853206 .6689561 3.050296

Variable | Obs Mean Std. Dev. Min Max
-------------+---------------------------------------------------------
gps_6 | 22,231 1.301614 .163156 .5098148 2.61117


************************************************************************************
Test that the conditional mean of the pre-treatment variables given the generalized
propensity score is not different between units who belong to a particular treatment
interval and units who belong to all other treatment intervals
************************************************************************************
1 group found, 2 required