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That is, say I have a distribution with parameter $\theta$.

If I re-write it with a parameter $a$ such that $a^3=\theta$, is it possible that doing maximum likelihood estimation on a will yield an estimate $\hat a$ such that $\hat a^3 \neq \hat \theta$?

Could it be the case for another function different than $x^3$?

If so, what are some criteria to choose a parametrization?

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The Invariance Property of Maximum Likelihood Estimators (MLEs) says, if $\hat{\theta}$ is the MLE of $\theta$, then for any function $\tau(\theta)$ the MLE of $\tau(\theta)$ is $\tau(\hat{\theta})$.

So, if you define $a^3=\theta$, once you obtained your MLE for $\theta$, $\hat{\theta}$, you can apply the inverse function by taking the cubed root of $\hat{\theta}$ and obtain the MLE of $a$ (i.e. $\hat{a}=\hat{\theta}^{1\over{3}}$)

Update:

I've added the proof mentioned by Thomas Lumley in the comments:

Let $\hat{\eta}$ denote the value that maximizes $L^*(\eta|\textbf{x})$. We must show that $L^*(\hat{\eta}|\textbf{x})$=$L^*(\tau(\hat{\theta})|\textbf{x})$. The maxima of $L$ and $L^*$ coincide, so we have

\begin{eqnarray*} L^{*}(\hat{\eta}|\textbf{x}) & = & \underset{\eta}{\text{sup}}\underset{\{\theta:\tau(\theta)=\eta\}}{\text{sup}}\,L(\theta|\textbf{x})\\ & = & \underset{\theta}{\text{sup}}L(\theta|\textbf{x})\\ & = & L(\hat{\theta}|\textbf{x}), \end{eqnarray*}

The first and third equalities hold by definition of $L^{*}$ and $\hat{\theta}$ respectively, and the second equality holds because the iterated maximization is equal to the unconditional maximization over $\theta$, obtained at $\hat{\theta}$. Further,

\begin{eqnarray*} L(\hat{\theta}|\textbf{x}) & = & \underset{\{\theta:\tau(\theta)=\tau(\hat{\theta})\}}{\text{sup}}L(\theta|\textbf{x})\\ & = & L^{*}\left[\tau(\hat{\theta})|\textbf{x}\right]. \end{eqnarray*}

Hence, the string of equalities shows that $L^{*}(\hat{\eta}|\textbf{x})=L^{*}\left[\tau(\hat{\theta})|\textbf{x}\right]$ and that $\tau(\hat{\theta})$ is the MLE of $\tau(\theta)$. $\blacksquare$

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    $\begingroup$ Perhaps worth noting that the equivariance property holds even if the transformation isn't invertible. $\endgroup$ – Thomas Lumley Aug 17 '20 at 0:40
  • $\begingroup$ Good point, @ThomasLumley. I don't think I made that clear and, in fact, may have obfuscated that point with my example (or the OP's example). $\endgroup$ – StatsStudent Aug 17 '20 at 0:45
  • $\begingroup$ Wow. Very cool (and a bit surprising!). No conditions on tau? Do you happen to have a good statement of the theorem $\endgroup$ – josinalvo Aug 17 '20 at 2:58
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    $\begingroup$ The proof is much more straightforward for invertible transformations. This question math.stackexchange.com/questions/3246587/… gives a reference to the text by Casella and Berger for the general result. $\endgroup$ – Thomas Lumley Aug 17 '20 at 5:12
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    $\begingroup$ @ThomasLumley, I've added proof under the Update heading in case the OP does not have access to the book. $\endgroup$ – StatsStudent Aug 17 '20 at 5:32

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