0
$\begingroup$

This is in In Casella's Statistical Inference,page 473, the approximation of the variance of MLE (Cramer-Rao Lower Bound). I really confused with the conclusion:

$Var_{\hat{\theta}}h(\hat{\theta})$ is a consistent estimator of $Var_{\theta}h(\hat{\theta}).$

I guess the logic is

$\dfrac{[h'(\theta)]^2}{I_n(\theta)}$ is asymptotic variance of $Var_{\theta}h(\hat{\theta})$ and $Var_{\hat{\theta}}h(\hat{\theta})$ is consistent estimator of $\dfrac{[h'(\theta)]^2}{I_n(\theta)}.$ Then $Var_{\hat{\theta}}h(\hat{\theta})$ is a consistent estimator of $Var_{\theta}h(\hat{\theta}).$

My questions are:

  1. Why $-\dfrac{1}{n}\dfrac{\partial^2}{\partial\theta^2}\log L(\theta|\bf{X})\Big|_{\theta=\hat{\theta}}$ is consistent estimator $I(\theta) = \dfrac{1}{n}E_{\theta}[-\dfrac{\partial^2}{\partial\theta^2}\log L(\theta|\bf{X})]?$ since $E_{\theta}[-\dfrac{\partial^2}{\partial\theta^2}\log L(\theta|\bf{X})] \neq -\dfrac{\partial^2}{\partial\theta^2}\log L(\theta|\bf{X}),$ we cannot use the consistency property of MLE.

  2. Why A is asymptotic variance of B and C is the consistent estimator of A => C is the consistent estimator of B?

enter image description here enter image description here

$\endgroup$
1
0
$\begingroup$

1.$-\dfrac{1}{n}\dfrac{\partial^2}{\partial\theta^2}\log L(\theta|\bf{X})\Big|_{\theta=\hat{\theta}}$ is a consistent estimator of $I(\theta)$ since it converges in probability to the true value $\dfrac{1}{n}E_{\theta}[-\dfrac{\partial^2}{\partial\theta^2}\log L(\theta|\bf{X})]$. What you have pointed out is the condition for an estimator to be unbiased not consistent. Consistency and unbiasdness of estimators are different concepts.

2.could you please elaborate more on this?

Hope this helps

$\endgroup$
4
  • $\begingroup$ thanks, 1. actually, it is low of large number and nothing about the consistency of MLE (Theorem 10.1.6) as author said, right? 2. it is noly my guess, maybe you can ignore it and tell me why $Var_{\hat{\theta}}h(\hat{\theta})$ is a consistent estimator of $Var_{\theta}h(\hat{\theta})$ in author's deduction. $\endgroup$ – user6703592 Feb 17 '20 at 5:30
  • $\begingroup$ is there any conclusion that if $v(\theta)$ is the asymptotic variance of $T_n$ i.e. $T_n - \mu \rightarrow n(0,v(\theta))$ by distribution, then $\hat{Var}$ is the consistent estimator of $v(\theta),$ then $\hat{Var}$ is also the consistent estimator of $Var_{\theta} T_n?$ Since we already know $Var_{\hat{\theta}}h(\hat{\theta})$ is the consistent estimator of $\dfrac{[h'(\theta)]^2}{I_n(\theta)}$ and $\dfrac{[h'(\theta)]^2}{I_n(\theta)}$ is the asymptotic variance of $h(\hat{\theta}).$ $\endgroup$ – user6703592 Feb 18 '20 at 8:34
  • $\begingroup$ It is related to consistency of MLE since the statement "convergence in probability" is the Weak Law of Large Numbers. Now coming to your point2, that is correct since we are estimating an approximation of an estimator that is eventually going to be used for the parameter being estimated. :) $\endgroup$ – Nizam Feb 18 '20 at 12:45
  • $\begingroup$ sorry could you say more detail for second point: if $v(\theta)$ is the asymptotic variance of $T_n$ i.e. $T_n - \mu\rightarrow n(0,v(\theta))$ by distribution and $\widehat{Var}$ is the consistent estimator of $v(\theta)$, then $\widehat{Var}$ is also the consistent estimator of $Var_{\theta} T_n?$ $\endgroup$ – user6703592 Feb 18 '20 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.