# Two step maximum likelihood

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?

• 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. Apr 13 at 11:24

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.