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Why does the score test work for values longer in the tail that have a small log-likelihood derivative?

The score test says that we take the derivative of the log-likelihood at $H_0$ and divide it by the fisher information at $H_0$. $U(\theta )={\frac {\partial \log L(\theta \mid x)}{\partial \theta }}.$...
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1 vote
1 answer
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Different version of Wald test statistic formula

I came across two formulas for the Wald test statistic in a maximum likelihood framework: One has $(R\hat{\theta}-r)'(RI_n^{-1}R')^{-1}(R\hat{\theta}-r)$, where $I_n^{-1}$ is the inverse of the ...
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