# Estimating the variance of a function of MLE

Say I have the following likelihood :

$$l(\alpha, \lambda) = n(\log \alpha + \log \lambda) + (\alpha -1 )\sum x_i - \lambda \sum x_i^\alpha$$

which is that of Weibull distribution.

The question is,

Estimate the variance of MLE of $$(\mu, \sigma) = (-(\log\lambda) / \alpha, 1/\alpha)$$

Try

I can estimate the MLE of $$(\alpha, \lambda)$$, $$(\hat{\alpha}, \hat{\lambda})$$, by solving

$$\left(\frac{\partial l}{\partial \alpha}, \frac{\partial l}{\partial \lambda} \right) \overset{set}{=} (0,0)$$

and I can estimate the variance, $$\widehat{Var}(\hat{\alpha}, \hat{\lambda}) = I^{-1}(\hat{\alpha}, \hat{\lambda})$$, i.e. plug-in estimator.

And by the invariance property of MLE, we have

$$(\hat{\mu}, \hat{\sigma}) = \left(-(\log\hat{\lambda}) / \hat{\alpha}, 1/\hat{\alpha} \right)$$

But I'm stuck at finding $$\widehat{Var}(\hat{\mu}, \hat{\sigma})$$.

• Can you show where you get stuck? Can you calculate $I$? Dec 9, 2018 at 1:53

Since your parameters of interest are non-linear functions of your estimates, I would suggest you apply the delta method. See Hayashi, pp. 93-94.

• You can calculate the hessian of the likelihood and invert it. the elements of that ( I think you have to divide it by thesample size. check that ) will be the variance and the covariance estimates asymptotically. Dec 9, 2018 at 3:46