Say I have an i.i.d. sequence sequence $X_1,\ldots X_n \sim \text{Bernoulli}(p)$, and I am interested in estimating $p^2$. Let $T$ denote $\sum_{i=1}^n X_i$. It turns out that the mle $\bar X^2 = \frac{T^2}{n^2}$ is biased, but the following estimator is not: $$\frac{T(T-1)}{n(n-1)}\,.$$
Now, by other means I know that this estimator is the least variance unbiased estimator for $p^2$. (Briefly, the Bernoulli distribution is a member of the exponential family, and $T$ is complete sufficient statistic for $p$, and thus $\frac{T(T-1)}{n(n-1)}$ is the LVUE for its expected value, $p^2$.)
However, I tried to show this fact another way, by showing that the estimator is actually efficient. There is a corollary of the Rao-Cramer lower bound (I got this from Casella & Berger's Statistical Inference, p. 341):
Let $X_1,\ldots,X_n$ be iid $f(\mathbf{x}|\theta)$, where $f(\mathbf{x}|\theta)$ satisfies the conditions of the Cramer-Rao Theorem. Let $L(\theta|\mathbf{x})$ denote the likelihood function. If $W(\mathbf{x})$ is any unbiased estimator of $\tau(\theta)$, then $W(\mathbf{x})$ attains the Cramer-Rao Lower Bound iff $$a(\theta)[W(\mathbf{x})-\tau(\theta)] = \frac{\partial}{\partial\theta} \log L(\theta|\mathbf{x})$$ for some function $a(\theta)$.
Being a member of the exponential family, our Bernoulli random variable easily satisfies the conditions for the CRLB Theorem. However, when I try to apply this Corollary the following happens:
$$\log L(p|\mathbf{x}) = \sum_{i=1}^n x_i \log(p) + (1-x_i)\log(1-p) = T \log(p) + (n-T)\log(1-p)$$ $$\implies \frac{\partial}{\partial p} \log L(p|\mathbf{x}) = \frac{T-np}{p(1-p)}$$ So for the lower bound to hold, we need to have: $$a(p)\left[\frac{T(T-1)}{n(n-1)}-p^2\right] = \frac{T-np}{p(1-p)}$$
As far as I can tell, there is no function $a(p)$ that can make this hold as the left side is quadratic in $T$ but the left side only linear.
The only resolution to this I can see is that there does not exist an unbiased, efficient estimator for $p^2$. Is this the case?
