= Solution
Put $S=\sum_iX_i$. <Independence> and the exponential sampling model give the <gamma distribution> with density
$$
f_S(s)=\frac{\lambda^n}{\Gamma(n)}s^{n-1}e^{-\lambda s},\qquad s>0.
$$
For $n>k$, substitution $u=\lambda s$ yields
$$
\mathbb ES^{-k}=\lambda^k\frac{\Gamma(n-k)}{\Gamma(n)}.
$$
At $k=1$, $n\geq2$, this gives $\mathbb E(\bar X^{-1})=n\lambda/(n-1)$. Therefore
$$
\boxed{\widehat\lambda=\frac{n-1}{S}=\frac{n-1}{n\bar X}}
$$
is an <unbiased estimator> of the rate. For $n>2$, the same formula with $k=2$ gives
$$
\mathbb E\widehat\lambda^2=\lambda^2\frac{n-1}{n-2},\qquad
\boxed{\operatorname{Var}(\widehat\lambda)=\frac{\lambda^2}{n-2}}.
$$
For the source-allowed value $n=2$, the first moment is still finite, but the second moment contains an integral proportional to $\int_0^1s^{-1}e^{-\lambda s}ds$ and diverges. Hence \b[the <estimator> has infinite <variance> when $n=2$]; the finite formula applies only to $n>2$.
The sample score is $n/\lambda-S$, and its <Fisher information> is $I_n(\lambda)=n/\lambda^2$. The <Cramér-Rao lower bound> for unbiased rate estimation is thus $\lambda^2/n$. For every $n>2$, the actual <variance> is strictly larger by the factor $n/(n-2)$. The ratio approaches one as $n$ grows, but equality is not attained at finite $n$. At $n=2$ the finite-<variance> form of the bound is inapplicable to this <estimator>; its infinite <variance> provides no efficiency equality.
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