= Normal mean proportional to standard deviation model
{title2=$X_i\sim N(\mu,\mu^2),\quad\mu>0$}
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 $A=\sum_ix_i$ and $B=\sum_ix_i^2$. 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> is
$$
\widehat\mu=\frac{-A+\sqrt{A^2+4nB}}{2n}.
$$
The 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 $\sqrt n(\widehat\mu-\mu)\xrightarrow dN(0,\mu^2/3)$. The <optimal asymptotic combination of a normal sample mean and standard deviation> attains the same first-order precision.
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