= Solution
Expanding the exponent, the <log-likelihood function> is
$$
\ell(\theta)=C-\frac{\sum_i y_i}{2\theta^2}+\frac n\theta,
$$
where $C$ is independent of $\theta$. Its <score function> and second derivative are
$$
\ell'(\theta)=\frac{\sum_i y_i-n\theta}{\theta^3},\qquad
\ell''(\theta)=-\frac{3\sum_i y_i}{\theta^4}+\frac{2n}{\theta^3}.
$$
The first is positive below $\bar y$ and negative above it, so the unique <maximum-likelihood estimator> is
$$
\boxed{\widehat\theta=\bar Y}.
$$
Using $\mathbb E\sum_iY_i=n\theta$, the one-observation <Fisher information> is $I(\theta)=-n^{-1}\mathbb E\ell''(\theta)=\theta^{-3}$. The support does not depend on the interior parameter $\theta>0$, and the regular <asymptotic normality of a maximum likelihood estimator> gives
$$
\boxed{\sqrt n(\widehat\theta-\theta)\xrightarrow d N(0,\theta^3)},\qquad
\widehat\theta\ \mathrel{\dot\sim}\ N(\theta,\theta^3/n).
$$
This is <Inverse Gaussian mean estimation> with known shape one. Equivalently, the <Inverse Gaussian distribution> has <variance> $\theta^3$, so the same result follows directly from the <central limit theorem> applied to its <sample mean>.
Back to article page
