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From a description of DiD method of Borusyak,2020 , I saw that

pretrends(integer): if some value k>0 is specified, the command will performs a test for parallel trends, by a separate regression on nontreated observations only: of the outcome on the dummies for 1,...,k periods before treatment, in addition to all the FE and controls. The coefficients are reported as pre1,...,prek. The Wald statistic, pvalue, and degrees-of-freedom as reported in e(pre_chi2), e(pre_p), and e(pre_df) resp.

  • Use a reasonable number of pre-trends, do not use all of the available ones unless you have a really large never-treated group. With too many pre-trend coefficients, the power of the joint test will be lower.

I am wondering how to explain the bold sentence. In another word, why joint null test has less power when there are more coefficients involve?

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You may remember that the reason the sample variance is divided by (n-1) instead of n, is because the sample variance calculation makes use of a population mean estimate, namely the sample mean. Not correcting for this 'cost' yields a biased estimate of the population variance. You make a trade. You give up an independent point in your data for an accurate estimate of some quantity, the variance in this example.

Each parameter $k$ you estimate comes at a similar cost, it costs a datapoint (also referred to as degree of freedom) which you fix and thus no longer provides any useful information to you for other statistical quantities you may want to estimate. The power of the joint test depends on the number of independent data points $n-k$ you have to 'spend'. The larger your sample size $n$, the more accurate your test (power). The more pre-trend coefficients you need to account for, the larger $k$ is and thus the smaller $n-k$ is.

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  • $\begingroup$ Thanks a heap. I am still confused. The power of joint null test based on k parameters, not n-k, so the higher number of coefficients, the higher power? May I fall into a dilemma but it is my understanding then. We are talking about the joint null test itself. $\endgroup$
    – Louise
    Sep 3 at 13:18

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