= Solution
{c}
The specified <asymptotic relative efficiency> is
$$
\boxed{\frac{v(T_2)}{v(T_1)}
=\frac{\theta(\theta+2)}{(\theta+1)^2}
=1-\frac1{(\theta+1)^2}}.
$$
For every fixed $\theta>0$ this lies strictly between zero and one. \b[Prefer $T_2$ on grounds of asymptotic precision]: it has the smaller leading <asymptotic variance> and attains the inverse <Fisher information>. Both are <consistent estimators>, by the <law of large numbers> and the <continuous mapping theorem> applied respectively to $\bar X$ and $A/n$. The efficiency approaches zero as $\theta\downarrow0$ and one as $\theta\to\infty$.
This preference is an asymptotic comparison. In particular, it does not assert that $T_2$ is an <unbiased estimator>: since $A$ has a <gamma distribution> of shape $n$ and rate $\theta$, $\mathbb E T_2=n\theta/(n-1)$ for $n>1$.
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