0
$\begingroup$

Suppose I have a likelihood maximisation problem

$$ \hat{\theta} = \max L_n(\theta;y) $$

where $\theta = [\theta_1, \theta_2, ...., \theta_k]^T$.

What if I would estimate instead estimate the maximisation problem leaving out a parameter

$$ \hat{\theta_{-k}} = \max L_n(\theta_{-k};y) $$

but a loop over each value of $\theta_k$ and pick the specification with the highest likelihood. Would this be identical to estimating the problem jointly?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ I think so. Why would you think otherwise? Evaluating each value of $\theta_{k}$ yields the profile likelihood, which is in fact handy to compute confidence intervals for a parameter (see e.g. link.springer.com/article/10.1007/s11222-021-10012-y ). In particular, the maximum of the profile likelihood is the MLE. $\endgroup$
    – Samufi
    Apr 13 at 11:24

1 Answer 1

Reset to default
2
$\begingroup$

Note that \begin{align} \max_{\theta} L_n(\theta;y) &= \max_{\theta_1,\dots,\theta_k} L_n(\theta_1,\dots,\theta_k;y) \\ &= \max_{\theta_1} \left[\max_{\theta_2}\left[\cdots \max_{\theta_k} L_n(\theta_1,\dots,\theta_k;y)\right]\right] \end{align} See this answer for why the last equality is true. Any permutation of the $\max_{\theta_i}$ operators would work too.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.