All Questions
1
question
1
vote
1
answer
29
views
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 $H_0$.
$U(\theta )={\frac {\partial \log L(\theta \mid x)}{\partial \theta }}.$...
