= Solution
{c}
Write $A=-\sum_i\log X_i>0$, with $\log$ the <natural logarithm>. The <likelihood function> and <log-likelihood function> are
$$
L(\theta)=\theta^n\exp\{-(\theta-1)A\},\qquad
\ell(\theta)=n\log\theta-(\theta-1)A.
$$
The <score function> is $\ell'(\theta)=n/\theta-A$ and the second <derivative> is $\ell''(\theta)=-n/\theta^2<0$. Thus the <log-likelihood function> is strictly <concave>, its stationary point is the unique global maximum, and the <maximum-likelihood estimator> is
$$
\boxed{T_2=\widehat\theta=\frac nA=-\frac n{\sum_i\log X_i}}.
$$
The one-observation <Fisher information> is $I(\theta)=1/\theta^2$. The <support of a probability distribution> here is independent of $\theta>0$, so the regular <asymptotic normality of a maximum likelihood estimator> gives
$$
\boxed{\sqrt n(T_2-\theta)\xrightarrow dN(0,\theta^2)},\qquad
T_2\ \mathrel{\dot\sim}\ N(\theta,\theta^2/n),\qquad v(T_2)=\theta^2/n.
$$
There is also a direct check: $U_i=-\log X_i$ has an <exponential distribution> with rate $\theta$, since $\mathbb P(U_i>u)=\mathbb P(X_i<e^{-u})=e^{-\theta u}$ for $u>0$. Its <mean> and <variance> are $1/\theta$ and $1/\theta^2$. Apply the <central limit theorem> to $A/n$ and the <delta method> to $h(u)=1/u$ to obtain the same result.
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