Let $X_1, X_2, \dots, X_n$ be a random sample following the Geometric distribution.
$$ \prod\limits_{i=1}^{n} f(x_i|p) = (1-p)^{\sum\limits_{i=1}^n x_i-n}p^n $$
Since the pmf of the Geometric distribution is exponential family, the factorization theorem yields that the statistic $$ T = \sum\limits_{i=1}^n x_i $$ is sufficient and complete. Then, $$ E[T] = E\left[ \sum\limits_{i=1}^n x_i \right] = \sum\limits_{i=1}^n E[x_i] = \frac{n}{p} $$ Therefore, according to Rao-Blackwell, the Minimum Variance Unbiased Estimator of $\frac{1}{p}$ is $$ W = \frac{1}{n} \sum\limits_{i=1}^n x_i $$ Now, Cramer-Rao's Lower Bound:
$$ LB = \frac{\left[\left(\frac{1}{p}\right)'\right]^2}{nI(p)} = \frac{1}{n}\frac{1-p}{p} $$
Question: Is $V[W] = LB$ in this specific example? If so, is there a reason why $W$ has the lowest possible variance that has something to do with the geometric distribution?