Suppose I have an estimator $\widehat{\theta}$ that is asymptotically efficient (for example, it could be the MLE of the mean $\mu$ and variance $\sigma^2$ of $Normal(\mu,\sigma^2)$). Suppose I also have $\widehat{\tau}=h(\widehat{\theta})$, where $h(\cdot)$ is a smooth transformation (in the normal example, an example could be $\tau=\mu^2\sigma^6$). My question is, is $\widehat{\tau}$ an asymptotically efficient estimator for $\tau$? Thanks!


1 Answer 1


Refer to the lecture notes here (page 6 specifically) by Shao.

Assume certain regularity conditions imposed on the distribution parameterized by $\theta$, so that the information matrix $I(\theta)$ is well-defined and positive definite.

Suppose $\theta \in \mathbb R^p$ and $\hat\theta_n$ is an asymptotically efficient estimator of $\theta$, so that

$$\sqrt n(\hat\theta_n-\theta)\stackrel{d}\to N_p\left(0,(I(\theta))^{-1}\right)$$

Let $h:\mathbb R^p\to \mathbb R$ be a differentiable function. Then by delta-method,

$$\sqrt n\left(h(\hat\theta_n)-h(\theta)\right)\stackrel{d}\to N\left(0,\nabla h(\theta)^T(I(\theta))^{-1}\nabla h(\theta)\right)\,,$$

where $\nabla h(\theta)$ is the gradient of $h(\cdot)$ at $\theta$.

Now it is indeed the case that $\nabla h(\theta)^T(I(\theta))^{-1}\nabla h(\theta)$ is the inverse of Fisher information about $h(\theta)$ (see this or this):

$$(\tilde I(h(\theta))^{-1}=\nabla h(\theta)^T(I(\theta))^{-1}\nabla h(\theta)$$

This makes $h(\hat\theta_n)$ also asymptotically efficient.

According to the linked notes, if $p>1$, we take $h(\hat\theta_n)$ to be asymptotically efficient if and only if $\hat\theta_n$ is asymptotically efficient.

  • $\begingroup$ Thank you very much! This is very helpful. $\endgroup$ Commented Jun 12, 2023 at 1:18

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