# Invariance property of MLE: what is the MLE of $\theta^2$ of normal, $\bar{X}^2$?

Invariance property of MLE: if $\hat{\theta}$ is the MLE of $\theta$, then for any function $f(\theta)$, the MLE of $f(\theta)$ is $f(\hat{\theta})$.

Also, $f$ must be a one-to-one function.

The book says, "For example, to estimate ${\theta}^2$, the square of a normal mean, the mapping is not one-to-one." So, we can't use invariance property.

But then, it proves the property and says, "we now see that MLE of $\theta^2$, the square of a normal mean is $\bar{x}^2$".

This seems self-contradicting, we are squaring $\bar{x}$, but square of anything is not one-to-one, what am I reading wrong here? Thanks!

source: Casella & Berger "Statistical Inference"

• "Also, f must be a one-to-one function." -- must it? Why? Nov 25, 2013 at 4:25

## 2 Answers

That's not exactly what Casella and Berger say. They recognize (page 319) that when the transformation is one-to-one the proof of the invariance property is very simple. But then they extend the invariance property to arbitrary transformations of the parameters introducing an induced likelihood function on page 320. Theorem 7.2.10 on the same page gives the proof of the extended property. Hence, no contradiction here.

From page 350 of "Probability and Statistical Inference":

(Note: This Theorem can be found on pg. 320 and labeled as 7.2.10 in the second edition.)