Heckman两步法中，第二步多重共线性如果处理？

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1.Heckman两步法中，第一步中得到的inverse Mills Ratio 在第二步回归中不显著。

2.同时第二步回归中加入inverse Mills Ratio导致了其中两个变量由显著变为不显着，经检查inverse Mills Ratio与其他几个解释变量存在共线性。

这时如果处理 啊！

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关键词：heckman两步法 heckman 多重共线性 多重共线 共线性 线性 heckman 步法

我也遇到了这个问题，顶一下！！！

也顶一下，大牛帮帮忙回答

不懂，但帮顶一下！

victorliou 发表于 2010-4-22 17:38 1.Heckman两步法中，第一步中得到的inverse Mills Ratio 在第二步回归中不显著

这已经表明选择偏误不强烈吧。

可以试试TWO-PART模型

1: Heckman两步法中，第一步中得到的inverse Mills Ratio 在第二步回归中不显著。

There are two ways to interpreting the results. First, if you believe that your first stage regression is good enough (by that I mean you believe that any endogeneity/selection issue can be sufficiently controlled for with your first stage model), then you argue that since the Inverse Mills' ratio is not significant in the second stage, you DON'T need to use the Heckman. However, one may argue that your first stage model is not sufficiently specified. Thus, second, you may consider other possible "identification variables" for the first stage or second stage. Actually, I suspect that the second is more likely in your case, because you mentioned that the Inverse Mills' ratio is highly correlated with two other variables. Let me discuss this point in the following response.

2.同时第二步回归中加入inverse Mills Ratio导致了其中两个变量由显著变为不显着，经检查inverse Mills Ratio与其他几个解释变量存在共线性。

Do you also include the two variables in the first stage? If you do, then it indicates that the "selection" or "identification" variable(s) you choose for the first stage - variable(s) appearing in the first stage but not the second stage - is not good enough. at the very least, such variable(s) should have a significant effect in the first stage.

hope this helps.

Heckman两步法本身对扰动项的分布有很强的假设，不显著不一定是说没有选择效应啊。你可以试试其他的方法。

有的文章里没有报告MIlls的显著性。

不汇报MIlls的显著性是极端不规范！