 Nonregular identification at a flat mean map

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 observationsThe 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 areThus $I(0)=0$ 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.^[ref]

Maximum likelihood estimation gives $\hat{\theta }=\mathrm{cbrt}(\overline{Y})$, using the real cube root. At the true value zero, write $\overline{Y}=Z/\sqrt{n}$ with a standard normal random variable $Z$. ThenThis 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 $\mathrm{cbrt}$ to both endpoints of the ordinary known-variance confidence interval for the mean: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 $[-0.626,0.626]$. 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.