In order to compute the mean and variance we need to recall Newton's binomial formula $$ \begin{equation*} (a+b)^m=\sum_{n=0}^m \left(\begin{array}{c} m \\ n\end{array}\right)a^nb^{m-n}, \end{equation*} $$ which can be used to show that $$ \begin{equation*} \sum_{x=0}^n\left(\begin{array}{c} n \\ x\end{array}\right)y^x(1-y)^{n-x} = (y+1-y)^n = 1, \end{equation*} $$ the PDF is normalized to one. The mean value is $$ \begin{equation*} \mu = \sum_{x=0}^n x\left(\begin{array}{c} n \\ x\end{array}\right)y^x(1-y)^{n-x} = \sum_{x=0}^n x\frac{n!}{x!(n-x)!}y^x(1-y)^{n-x}, \end{equation*} $$ resulting in $$ \begin{equation*} \mu = \sum_{x=0}^n x\frac{(n-1)!}{(x-1)!(n-1-(x-1))!}y^{x-1}(1-y)^{n-1-(x-1)}, \end{equation*} $$ which we rewrite as $$ \begin{equation*} \mu=ny\sum_{\nu=0}^n\left(\begin{array}{c} n-1 \\ \nu\end{array}\right)y^{\nu}(1-y)^{n-1-\nu} =ny(y+1-y)^{n-1}=ny. \end{equation*} $$