请问使用随机效应模型后，Prob > chi2 后面没有数字，只有一个小点是怎么回事呢

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如图，  Wald chi2(7) 是蓝色的，p值也不显示。有大神可以帮忙看看吗，谢谢~

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关键词：随机效应模型 随机效应 Rob Wald 不显示

  Wald chi2(7) 点开：

The F or chi2 model statistic has been reported as missing

    Your estimation results show an F or chi2 model statistic reported to be missing.  Stata has done that so as to not be misleading, not because there is something
    necessarily wrong with your model.


Are any standard errors missing?

    If any standard errors are reported as dots, something is wrong with your model:  one or more coefficients could not be estimated in the normal statistical sense.
    You need to address that problem and ignore the rest of this discussion.


Are you using bootstrap or jackknife?

    The VCE you have just estimated is not of sufficient rank to perform the model test.  This is most likely due to not having enough replications.

    The bootstrap command has a reps(#) option, and if # is less than the number of coefficients in the model, the VCE will have insufficient rank.  The solution is to
    rerun bootstrap with a much larger number of replications.

    The jackknife command estimates the VCE by refitting the model for each observation in the dataset, leaving the associated observation out of the estimation sample
    each time.  As with the conventional variance estimator, the VCE will be singular if the number of observations is less than the number of parameters.  See the
    following discussion if you supplied the cluster() option to jackknife.


Are you using a svy estimator or did you specify the vce(cluster clustvar) option?

    The VCE you have just estimated is not of sufficient rank to perform the model test.  As discussed in [R] test, the model test with clustered or survey data is
    distributed as F(k,d-k+1) or chi2(k), where k is the number of constraints and d=number of clusters or d=number of PSUs minus the number of strata.  Because the
    rank of the VCE is at most d and the model test reserves 1 degree of freedom for the constant, at most d-1 constraints can be tested, so k must be less than d.
    The model that you just fit does not meet this requirement.

    To simplify the remaining discussion, let's consider the case of clustered data.  This discussion applies to survey estimation in general by substituting, "PSUs -
    strata" for "clusters".

    There is no mechanical problem with your model, but you need to consider carefully whether any of the reported standard errors mean anything.  The theory that
    justifies the standard error calculation is asymptotic in the number of clusters, and we have just established that you are estimating at least as many parameters
    as you have clusters.

    That concern aside, the model test statistic issue is that you cannot simultaneously test that all coefficients are zero because there is not enough information.
    You could test a subset, but not all, and so Stata refuses to report the overall model test statistic.

    Here note the degrees of freedom reported for the chi2 or F.  You might see chi2(6) or F(6, 5).  If you were to count the number of coefficients that would be
    constrained to 0 in a model test in this case, you would find that number to be greater than 6.  You could find out what that number is by reestimating the model
    parameters without the vce(robust) and vce(cluster clustvar) options (or, for the survey commands, using the corresponding non-svy estimator).  In any case, the 6
    reported is the maximum number of coefficients that could be simultaneously tested.


Is there a regressor that is nonzero for only 1 observation or for one cluster?

    The VCE you have just estimated is not of sufficient rank to perform the model test.  This can happen if there is a variable in your model that is nonzero for only
    1 observation in the estimation sample.  Likewise, it can happen if a variable is nonzero for only one cluster when using the cluster-robust VCE.  In such cases
    the derivative of the sum-of-squares or likelihood function with respect to that variable's parameter is zero for all observations.  That implies that the
    outer-product-of-gradients (OPG) variance matrix is singular.  Because the OPG variance matrix is used in computing the robust variance matrix, the latter is
    therefore singular as well.

您好，请问您的问题解决了吗？我遇到了相同的情况，请问您是怎么解决的呢？

ajinli 发表于 2021-11-24 22:13
您好，请问您的问题解决了吗？我遇到了相同的情况，请问您是怎么解决的呢？

请问您的问题解决了吗

请问您的问题解决了吗？怎么解决的呢？