= Nonregular identification at a flat mean map
\b[Nonregular identification at a flat mean map] occurs when different parameter values give different <probability distributions>, yet the first derivative of the observable mean vanishes at a parameter of interest. <Identifiability> is a property of the entire parameter-to-distribution map; a first-order sensitivity calculation can miss it.
Consider an original <statistical model> with independent observations
$$
Y_i\sim N(\theta^3,1),\qquad \theta\in\mathbb R,\quad i=1,\ldots,n.
$$
The cubic mean map is a <strictly increasing function>, so equality of the <normal distributions> implies equality of $\theta$. The model is globally identifiable, including at $\theta=0$. Nevertheless, the single-observation <score function> and <Fisher information> are
$$
U_\theta(Y)=3\theta^2(Y-\theta^3),\qquad I(\theta)=9\theta^4.
$$
Thus \b[$I(0)=0$ does not imply nonidentifiability]. The rank is zero at the origin and one at every nonzero nearby point. This is precisely why a regular-point assumption matters: Rothenberg's local identification criterion assumes constant information-matrix rank in a neighborhood.<Identification in Parametric Models by Thomas J. Rothenberg>[p.579, Definition 4 and Theorem 1]{ref}
<Maximum likelihood estimation> gives $\widehat\theta=\operatorname{cbrt}(\overline Y)$, using the <real cube root>. At the true value zero, write $\overline Y=Z/\sqrt n$ with a <standard normal random variable> $Z$. Then
$$
\boxed{n^{1/6}\widehat\theta\ \overset d=\ \operatorname{cbrt}(Z).}
$$
This is an exact <sampling distribution> identity for every $n$, not merely a <convergence in distribution> statement. The estimation error has scale $n^{-1/6}$ and a nonnormal shape. At a fixed nonzero $\theta$, the usual <delta method> instead gives asymptotic <variance> $1/(9n\theta^4)$; that approximation cannot be extended uniformly through zero.
An <exact confidence interval through a cubic mean map> follows by applying the <strictly increasing function> $\operatorname{cbrt}$ to both endpoints of the ordinary known-variance <confidence interval> for the mean:
$$
\left[\operatorname{cbrt}\!\left(\overline Y-\frac{z_{1-\alpha/2}}{\sqrt n}\right),\quad
\operatorname{cbrt}\!\left(\overline Y+\frac{z_{1-\alpha/2}}{\sqrt n}\right)\right].
$$
Here $z_p$ is the <quantile function> of the <standard normal distribution>. The transformed interval has exact coverage $1-\alpha$ for every $\theta$. When $n=64$ and the observed <sample mean> is zero, the 95% interval is approximately \b[$[-0.626,0.626]$]. Multiplying $n$ by 64 halves its width at that observed mean. A first-order interval based on dividing by the zero mean-map derivative would fail instead.
\Image[/flat-mean-map-confidence.png]
{title=Cubic mean map and exact confidence-interval scaling at an observed sample mean of zero; original normal-observation model}
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