Let C be an estimator defined as $$C = \frac{1}{n}\sum_{k=0}^{n}1_{U_k>a}$$ where $a$ is a real number in $(0,1)$ , $1$ is an indicator function and $U_k$ are uniform random variable in $(0,1)$. I want to compute E[C^2]. Exploiting the independence of the r.v. i do $$C = \frac{1}{n^2}\sum_{k=0}^{n} \sum_{j=0}^{n}E[1_{U_k>a}]E[1_{U_j>a}]$$ and so eventually I get $E[C^2] = (1-a)^2$. However the results looks wrong.
Can someone help, many thanks
Your result was wrong just because when $k=j$, the summand is not $\mathbb{E}(\mathbf{1}_{\{ U_k>a \}})\mathbb{E}(\mathbf{1}_{\{U_j>a \}})$ but is simply $\mathbb{E}(\mathbf{1}_{\{U_k>a \}})$.
$C$ must be calculated as follows $$\begin{align} C &= \frac{1}{n^2}\sum_{k=1}^{n} \sum_{j=1}^{n}\mathbb{E}(\mathbf{1}_{\{U_k>a \}}\mathbf{1}_{\{U_j>a \}}) \\ &=\frac{1}{n^2} \left(\sum_{1\le k\color{red}{\ne}j\le n}\mathbb{E}(\mathbf{1}_{\{U_k>a \}}\mathbf{1}_{\{U_j>a \}}) + \sum_{k=1}^{n}\mathbb{E}(\mathbf{1}_{\{U_k>a \}})\right)\\ &=\frac{1}{n^2} \left(n(n-1)(1-a)^2 +n(1-a) \right) \end{align}$$