The Hausman test and weak instruments
Hausman 检验与弱工具变量AI 解读 ↓
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AI 根据所提供的论文材料归类：摘要围绕内生解释变量的工具变量识别及工具外生性检验展开，直接涉及因果识别方法。
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Hausman 检验与弱工具变量
- 研究问题
- 当可信外生工具解释力较弱，而解释力较强的工具可能内生时，如何有效检验后者的外生性，并处理异方差？
- 方法与思路
- 作者在 Staiger–Stock 弱工具渐近框架下分析标准 Hausman 检验，提出弱工具下有效的变体，并说明异方差时的实施。
- 材料陈述的贡献
- 摘要称标准检验在弱工具下失效，新版本仍可有效检验且模拟表现相对较好；同时明确指出所提检验不具有一致性。
- 阅读关注点
- 应核对弱工具渐近序列、检验对象与统计量构造，区分有效性和一致性，并查清异方差实现及模拟比较的具体设置。
- 阅读边界
- 仅依据所给出版社摘要，不能验证有限样本优势；一致检验不可能存在是作者表达的判断，摘要未提供相应证明或完整条件。
原文摘要
We consider the following problem. There is a structural equation of interest that contains an explanatory variable that theory predicts is endogenous. There are one or more instrumental variables that credibly are exogenous with regard to this structural equation, but which have limited explanatory power for the endogenous variable. Further, there is one or more potentially 'strong' instruments, which has much more explanatory power but which may not be exogenous. Hausman (1978) provided a test for the exogeneity of the second instrument when none of the instruments are weak. Here, we focus on how the standard Hausman test does in the presence of weak instruments using the Staiger-Stock asymptotics. It is natural to conjecture that the standard version of the Hausman test would be invalid in the weak instrument case, which we confirm. However, we provide a version of the Hausman test that is valid even in the presence of weak IV and illustrate how to implement the test in the presence of heteroskedasticity. We show that the situation we analyze occurs in several important economic examples. Our Monte Carlo experiments show that our procedure works relatively well in finite samples. We should note that our test is not consistent, although we believe that it is impossible to construct a consistent test with weak instruments.