Why not always use CI's from LRT since they don't require symmetry?

Question

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?

There are other options too, such as inverting a score test. — Ben, Jun 21 '22 at 19:53
Agreed, but why not always use something (LRT, score etc) that converge to chi square to exploit asymmetric CIs? — Casey, Jun 22 '22 at 00:27
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. — Durden, Mar 12 '24 at 16:37

Sextus Empiricus · Answer 1 · 2024-03-12T16:04:35.857

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.

Why not always use CI's from LRT since they don't require symmetry?

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