Let $X_{1},X_{2},\ldots,X_{n}$ be random sample from $X\sim\text{Binomial}(2,\theta)$.
(a) Find the least variance from the set of all unbiased estimators of $\theta$.
(b) Find a sufficient statistics of $\theta$.
(c) Obtain from this statistics an unbiased estimator to $\theta$.
(d) Verify if this estimator is efficient.
MY ATTEMPTS (EDITED)
(a) To begin with, I started with \begin{align*} f(x|\theta) & = {{2}\choose{x}}\theta^{x}(1-\theta)^{2-x} = \frac{2}{x!(2-x)!}\theta^{x}(1-\theta)^{2-x} \Rightarrow\\\\ \ln f(x|\theta) & = \ln(2) - \ln(x!) - \ln[(2-x)!] + x\ln(\theta) + (2-x)\ln(1-\theta)\Rightarrow\\\\ \frac{\partial\ln f(x|\theta)}{\partial\theta} & = \frac{x}{\theta} + \frac{x-2}{1-\theta} \Rightarrow \frac{\partial^{2}\ln f(x|\theta)}{\partial\theta^{2}} = -\frac{x}{\theta^{2}} + \frac{x-2}{(1-\theta)^{2}} \Rightarrow\\\\ \textbf{E}\left(\frac{\partial^{2}\ln f(x|\theta)}{\partial\theta^{2}}\right) & = -\frac{2}{\theta} - \frac{2}{1-\theta} = -\frac{2}{\theta(1-\theta)} \Rightarrow I_{F}(\theta) = \frac{2}{\theta(1-\theta)} \end{align*}
(b) Observe that we can factor the likelihood function as follows:
\begin{align*} L(\textbf{x}|\theta) & = \prod_{k=1}^{n}f(x_{k}|\theta) = \prod_{k=1}^{n}{{2}\choose{x_{k}}}\theta^{x_{k}}(1-\theta)^{1-x_{k}}\\\\ & = h(x_{1},x_{2},\ldots,x_{k})g_{\theta}(T(x_{1},x_{2},\ldots,x_{n})) \end{align*}
where
\begin{align*} g_{\theta}(T(x_{1},x_{2},\ldots,x_{n})) = \theta^{\sum_{k=1}^{n}x_{k}}(1-\theta)^{n-\sum_{k=1}^{n}x_{k}} \end{align*}
Thefore $T(\textbf{x}) = \sum_{k=1}^{n}x_{k}$ is a sufficient statistics, as requested.
(c) To begin with, note that $\textbf{E}(X) = 2\theta$. Since $\textbf{E}(T(\textbf{x})) = 2n\theta$, we can set $\hat{\theta} = T(\textbf{x})/2n$.
(d) Let us start by determining $\textbf{Var}(\hat{\theta})$:
\begin{align*} \textbf{Var}(\hat{\theta}) = \textbf{Var}\left(\frac{1}{2n}\sum_{k=1}^{n}X_{k}\right) = \frac{1}{4n^{2}}\sum_{k=1}^{n}\textbf{Var}(X_{k}) = \frac{2n\theta(1-\theta)}{4n^{2}} = \frac{\theta(1-\theta)}{2n} \end{align*}
Therefore the efficiency of $\hat{\theta}$ is given by: \begin{align*} e(\hat{\theta}) = \frac{1}{\textbf{Var}(\hat{\theta})nI_{F}(\theta)} = 1 \end{align*}
Thus the estimator $\hat{\theta}$ is efficient.
