Can somebody tell me, why the following is true:
\begin{equation} \sum_{\alpha,\varepsilon} (f_{r}(\alpha,\varepsilon)-E[f_{r}])^{2} =\sum_{\alpha,\varepsilon} f_{r}^2(\alpha,\varepsilon)-\sum_{\alpha,\varepsilon}E[f_{r}]^{2}, \end{equation}
where $f_{r}(\alpha,\epsilon)$ is a random variable on a sample space of $n_{(r)}2^r$ points, each having an equal probability.
Furthermore, we have $f_{r}(\alpha, \epsilon):=|\{v \in V:v \neq \alpha_{1},...,\alpha_{r}\text{ and } a(v,\alpha_{j})=\varepsilon_{j}, \forall\ 1\leq j\leq r\}|$ and $E[f_{r}(\alpha,\varepsilon)]=\frac{n-r}{2^r}$.