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

Invariance property of MLE: if $\hat$ is the MLE of $\theta$, then for any function $f(\theta)$, the MLE of $f(\theta)$ is $f(\hat)$. Also, $f$ must be a one-to-one function. The book says, "For example, to estimate $^2$, 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 $\bar^2$". This seems self-contradicting, we are squaring $\bar$, but square of anything is not one-to-one, what am I reading wrong here? Thanks! source: Casella & Berger "Statistical Inference"

asked Nov 25, 2013 at 4:06 956 4 4 gold badges 12 12 silver badges 21 21 bronze badges $\begingroup$ "Also, f must be a one-to-one function." -- must it? Why? $\endgroup$ Commented Nov 25, 2013 at 4:25

2 Answers 2

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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.

answered Nov 25, 2013 at 4:26 24.9k 4 4 gold badges 84 84 silver badges 125 125 bronze badges $\begingroup$

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