= Solution
{c}
Using the <raw moments> from part (i), the <variance> of one observation is
$$
\operatorname{Var}_\theta(X)=\frac{\theta}{\theta+2}-\frac{\theta^2}{(\theta+1)^2}
=\frac{\theta}{(\theta+1)^2(\theta+2)}.
$$
The <central limit theorem> for the independent <statistical sample> gives
$$
\sqrt n(\bar X-\mu)\xrightarrow d
N\left(0,\frac{\theta}{(\theta+1)^2(\theta+2)}\right).
$$
For $g(x)=x/(1-x)$, its <derivative> at the <mean> is $g'(\mu)=(1-\mu)^{-2}=(\theta+1)^2$. The <delta method> therefore gives
$$
\boxed{\sqrt n(T_1-\theta)\xrightarrow d
N\left(0,\frac{\theta(\theta+1)^2}{\theta+2}\right)}.
$$
Equivalently, the large-sample <normal approximation> is
$$
T_1\ \mathrel{\dot\sim}\ N\left(\theta,\frac{\theta(\theta+1)^2}{n(\theta+2)}\right),
\qquad v(T_1)=\frac{\theta(\theta+1)^2}{n(\theta+2)}.
$$
Here $v$ denotes the leading <asymptotic variance>, rather than an assertion about an exact finite-sample <variance>.
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