User daniel reck - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:40:20Z http://mathoverflow.net/feeds/user/30103 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116829/standard-errors-for-two-step-mle-procedure-for-computationally-intensive-likeliho Standard Errors for Two-Step MLE Procedure for Computationally Intensive Likelihood Functions Daniel Reck 2012-12-19T22:59:28Z 2012-12-19T22:59:28Z <p>Suppose we have a likelihood function, $L(\theta_{1},\theta_{2},\theta_{3};X_{1},X_{2})$ where $\theta_{1}...\theta_{3}$ are sets of parameters and $X_{1}$ and $X_{2}$ are data. The model is fully identified, but computational difficulties make it nearly impossible to maximize this function over $\theta_{1},$ $\theta_{2},$ and $\theta_{3}$ simultaneously, because the liklihood function is time-consuming to compute and can only be computed with error. So with too many dimensions it takes any appropirate algorithm a very long time to converge. However, we can restrict the model to estimate just some of the parameters consistently. More specifically, we can generate likelihood functions $L_{1}(\theta_{1};X_{1})$ and $L_{2}(\theta_{2};X_{2})$ to estimate $\theta_{1}$ and $\theta_{2}$ consistently. If doing so gives us estimates $\hat{\theta_{1}}$ and $\hat{\theta_{2}},$ then maximizing $L(\hat{\theta_{1}},\hat{\theta_{2}},\theta_{3};X_{1},X_{2})$ gives a consistent--if inefficient--estimate of $\theta_{3}$, which is not subject to the computational difficulties.</p> <p>My question concerns the standard errors for $\hat{\theta_{3}}$, or confidence intervals obtained via, say, the liklihood ratio test. Is it appropriate to uses the usual methods for obtaining standard errors in this case, or should I do something different to account for the odd way in which I obtained my estimate? If I do things the usual way, I should use approximations and/or evaluations involving the function $L(\theta_{1},\theta_{2},\theta_{3};X_{1},X_{2})$ and <em>not</em> $L(\hat{\theta_{1}},\hat{\theta_{2}},\theta_{3};X_{1},X_{2})$, right?</p>