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I'm confused on why anyone would appeal to asymptotic normality of mle,

$$\hat{\theta} - \theta_0 \rightarrow^D N(0,I^{-1}(\theta))$$

When we can simply invert the likelihood ratio test

$$L(\hat{\theta}) - L(\theta_0) \rightarrow^D \chi^2_1$$

to obtain a $(1-\alpha)$ CI. Is there a situation where this is not a good idea?

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    $\begingroup$ There are other options too, such as inverting a score test. $\endgroup$
    – Ben
    Commented Jun 21, 2022 at 19:53
  • $\begingroup$ Agreed, but why not always use something (LRT, score etc) that converge to chi square to exploit asymmetric CIs? $\endgroup$
    – Casey
    Commented Jun 22, 2022 at 0:27
  • $\begingroup$ The $\chi^2$ distribution of the likelihood-ratio test is an asymptotic result, too, so I don't see what conceptual difference your approach would make. $\endgroup$
    – Durden
    Commented Mar 12 at 16:37

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The asymptotic distribution

$$\hat{\theta} - \theta_a \rightarrow^D N(0,I^{-1}(\theta))$$

can be rewritten as

$$\hat{\theta} \rightarrow^D N(\theta_a,I^{-1}(\theta))$$

and it becomes easy to compute p-values for different hypothetical values $\theta_a$ and associated confidence intervals.

This expression $\hat{\theta} - \theta_a$ is a simple translation. This is not the case for $L(\hat{\theta}) - L(\theta_a)$.


A complication is $I^{-1}(\theta)$ which is probably gonna need to be $I^{-1}(\theta_a)$ and changing the value of $\theta_a$ might be not the same as a simple translation (but also change the variance). An example is the Wilson score interval for a binomial proportion.

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