# Is the smooth transformation of an asymptotically efficient estimator still asymptotically efficient?

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!

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.

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