Rank-1/2:一种改进尾部指数OLS估计的简单方法

X. Gabaix, R. Ibragimov
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引用次数: 105

摘要

尽管有更复杂的方法可用,但估计帕累托指数的一种流行方法仍然是运行OLS回归:log(Rank)=a-b log(Size),并将b作为帕累托指数的估计。这种流行的原因可以说是这种方法的简单性和健壮性。不幸的是,这种方法在小样本中存在严重偏差。我们为这种偏差提供了一个简单实用的补救措施,并建议,如果想使用OLS回归,应该使用Rank-1/2,并且运行日志(Rank-1/2)=a-b日志(大小)。1/2的移位是最优的,并将偏差减少到领先的顺序。帕累托指数的标准误差不是OLS标准误差,而是渐近的(2/n)^(1/2)数值结果表明,该方法优于标准OLS估计方法,并表明该方法在偏离幂律的依赖重尾过程中表现良好。使用美国城市规模分布的Zipf定律的经验应用说明了所考虑的估计程序。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Rank-1/2: A Simple Way to Improve the OLS Estimation of Tail Exponents
Despite the availability of more sophisticated methods, a popular way to estimate a Pareto exponent is still to run an OLS regression: log(Rank)=a-b log(Size), and take b as an estimate of the Pareto exponent. The reason for this popularity is arguably the simplicity and robustness of this method. Unfortunately, this procedure is strongly biased in small samples. We provide a simple practical remedy for this bias, and propose that, if one wants to use an OLS regression, one should use the Rank-1/2, and run log(Rank-1/2)=a-b log(Size). The shift of 1/2 is optimal, and reduces the bias to a leading order. The standard error on the Pareto exponent zeta is not the OLS standard error, but is asymptotically (2/n)^(1/2) zeta. Numerical results demonstrate the advantage of the proposed approach over the standard OLS estimation procedures and indicate that it performs well under dependent heavy-tailed processes exhibiting deviations from power laws. The estimation procedures considered are illustrated using an empirical application to Zipf's law for the U.S. city size distribution.
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