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This one-parameter normal distribution model has positive mean equal to its standard deviation. It is a curved exponential family: both the linear and quadratic natural parameters depend on the same scalar statistical parameter. It combines information about the center and spread, so the sample mean alone does not use all available Fisher information.

For a statistical sample of size $n$, let $A={\sum }_i{x}_i$ and $B={\sum }_i{x}_i^2$. Up to an additive constant the log-likelihood function is $\ell (\mu )=-n\log \mu -B/(2{\mu }^2)+A/\mu$. Its score function is $(B-A\mu -n{\mu }^2)/{\mu }^3$. When $B>0$, the numerator changes from positive to negative at its unique positive root, so the global maximum-likelihood estimator isThe all-zero observation vector has zero probability and is an exceptional case with an unbounded likelihood function near zero. Since $\mathbb{E}A=n\mu$ and $\mathbb{E}B=2n{\mu }^2$, the expected Fisher information is $I_n(\mu )=3n/{\mu }^2$. Thus regular asymptotic normality of a maximum likelihood estimator gives $\sqrt{n}(\hat{\mu }-\mu )\stackrel{d}{\to }N(0,{\mu }^2/3)$. The optimal asymptotic combination of a normal sample mean and standard deviation attains the same first-order precision.

Write $S^2=Q/(n-1)$ and $S=\sqrt{S^2}$, with $n\ge 2$. Two exact facts about a normal sample areand independence of the normal sample mean and sample variance holds. A chi-squared distribution with $n-1$ statistical degrees of freedom is a sum of $n-1$ independent ${\chi }_1^2$ variables, each with mean one and variance two. The central limit theorem therefore givesFor $g(x)=\sqrt{x}$, the derivative at ${\mu }^2$ is $1/(2\mu )$ because $\mu >0$. The delta method gives the asymptotic normality of a normal sample standard deviation:The exact independence of $\overline{X}$ and $S$ makes their limiting normal distributions independent. Consequently,ThusThis has the same leading asymptotic variance as $\hat{\mu }$. The weights $1/3$ and $2/3$ are those in the optimal asymptotic combination of a normal sample mean and standard deviation. This does not require $S$ to be an exactly unbiased estimator of $\mu$.