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?

  • 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


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.


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