# the approximation of the variance of MLE (Cramer-Rai Lower Bound)

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?

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.
• 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. Feb 17, 2020 at 5:30
• 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 $h(\hat{\theta}).$ Feb 18, 2020 at 8:34
• 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 $Var_{\theta} T_n?$ Feb 18, 2020 at 13:58