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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 $\widehat{Var}(\hat{\mu}, \hat{\sigma})$.

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    $\begingroup$ Can you show where you get stuck? Can you calculate $I$? $\endgroup$ – Glen_b Dec 9 '18 at 1:53
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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.

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  • $\begingroup$ 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. $\endgroup$ – mlofton Dec 9 '18 at 3:46

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