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

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
Commented Jun 21, 2022 at 19:53
• Agreed, but why not always use something (LRT, score etc) that converge to chi square to exploit asymmetric CIs? Commented Jun 22, 2022 at 0:27
• 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. Commented Mar 12 at 16:37

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.