# Constructing 95% confidence interval based on profile likelihood

In my elementary statistics course, I learnt how to construct 95% confidence interval such as population mean, mu based on asymptotic normality for "large" sample size. Apart from resampling methods, there is another approach based on "profile likelihood". Could someone elucidate this approach? Under what situations, the constructed 95% CI based on asymptotic normality and profile likelihood are comparable? I could not find any references on this topic, any suggested references, please? Why isn't it more widely used?

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In general, the confidence interval based on the standard error strongly depends on the assumption of normality for the estimator. The "profile likelihood confidence interval" provides an alternative.

I am pretty sure you can find documentation for this. For instance, here and references therein.

Here is a brief overview.

Let us say the data depend upon two (vectors of) parameters, $\theta$ and $\delta$, where $\theta$ is of interest and $\delta$ is a nuisance parameter.

The profile likelihood of $\theta$ is defined by

$L_p(\theta) = \max_{\delta} L(\theta, \delta)$

where $L(\theta, \delta)$ is the 'complete likelihood'. $L_p(\theta)$ does no longer depend on $\delta$ since it has been profiled out.

Let a null hypothesis be $H_0 : \theta = \theta_0$ and the likelihood ratio statistic be

$LR = 2 (\log L_p(\hat{\theta}) - \log L_p(\theta_0))$

where $\hat{\theta}$ is the value of $\theta$ that maximises the profile likelihood $L_p(\theta)$.

A "profile likelihood confidence interval" for $\theta$ consists of those values $\theta_0$ for which the test is not significant.

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@ocram- Thanks for the clarification. It seems the method requires some intensive computations, maximizing the profile likelihood. Just wondering why not simply resort to the bootstrap method if the estimator is not normally distributed. –  user2264 Apr 21 '11 at 17:46