3
$\begingroup$

My definition of profile likelihood is that given a vector of parameters $(\theta_1, \theta_2)$, with $\theta_1$ the parameter of interest, and $\theta_2$ a nuisance parameter -- If $L(\theta_1, \theta_2 ; x)$ is the likelihood constructed from the data $x$, the profile likelihood for $\theta_1$ is defined as $L_P(\theta_1 ; x) = L(\theta_1, \hat{\theta}_2(\theta_1) ; x)$ where $ \hat{\theta}_2(\theta_1)$ is the MLE of $\theta_2$ for a fixed value of $\theta_1$.

Now, in the normal distribution case where we have $X_1, \ldots, X_n \overset{iid}{\sim} N(\mu, \sigma^2)$, the MLE, $\hat{\mu}, \hat{\sigma^2}$ is found by first differentiating the likelihood function with respect to $\mu$, then setting to zero for the solution. Then, we differentiate the likelihood function with respect to $\sigma^2$, set to zero, and THEN plug in $\hat{\mu}$.

It seems to me that this standard approach is exactly what the profile likelihood is doing. However, whenever I read about profile likelihoods, books or resources never mention the Normal case.

Question: Is what we do to find the MLE of normally distributed data the same as the profile likelihood approach?

$\endgroup$

1 Answer 1

1
$\begingroup$

You don't need the profile likelihood approach for the normal case. Note that the restricted maximum likelihood estimator of $\mu$, given a value of the variance $\sigma^2$, $\hat{\mu}(\sigma^2) = \bar{x}$, so is independent of the variance.

And, in general, you don't need the profile likelihood to find the MLE. Note that for the interest parameter $\theta_1$, the same value maximizes the full likelihood $L$ and the profile likelihood $L_P$. But $L_P$ becomes interesting when constructing confidence intervals, for instance. But this is mostly useful when the simpler methods, as finding ancillary functions, cannot be used. But in the normal distribution case, we can do exact small-sample inference, so do not need the more complicated profile likelihood methods.

$\endgroup$

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.