Binomialverteilung/Varianz/Fakt/Beweis

Beweis

Die zweite Aussage folgt unmittelbar aus der ersten durch Multiplikation mit ${}{\frac {1}{n^{2}}}$ . Es ist

{}{\begin{aligned}\sum _{(x_{1},x_{2},\ldots ,x_{n})\in \{0,1\}^{n}}{\left(\sum _{i=1}^{n}x_{i}-np\right)}^{2}P(x_{1},x_{2},\ldots ,x_{n})&=\sum _{(x_{1},x_{2},\ldots ,x_{n})\in \{0,1\}^{n}}{\left(x_{1}-p+x_{2}-p+\cdots +x_{n}-p\right)}^{2}P(x_{1},x_{2},\ldots ,x_{n})\\&=\sum _{(x_{1},x_{2},\ldots ,x_{n})\in \{0,1\}^{n}}{\left(\sum _{i=1}^{n}(x_{i}-p)^{2}+2\sum _{i<j}(x_{i}-p)(x_{j}-p)\right)}P(x_{1},x_{2},\ldots ,x_{n})\\&=\sum _{(x_{1},x_{2},\ldots ,x_{n})\in \{0,1\}^{n}}{\left(\sum _{i=1}^{n}(x_{i}-p)^{2}\right)}P(x_{1},x_{2},\ldots ,x_{n})+2\sum _{(x_{1},x_{2},\ldots ,x_{n})\in \{0,1\}^{n}}{\left(\sum _{i<j}(x_{i}-p)(x_{j}-p)\right)}P(x_{1},x_{2},\ldots ,x_{n}).\end{aligned}}

Für die vordere Summe ist

{}{\begin{aligned}\sum _{(x_{1},x_{2},\ldots ,x_{n})\in \{0,1\}^{n}}{\left(\sum _{i=1}^{n}(x_{i}-p)^{2}\right)}P(x_{1},x_{2},\ldots ,x_{n})&=\sum _{i=1}^{n}{\left(\sum _{(x_{1},x_{2},\ldots ,x_{n})\in \{0,1\}^{n}}(x_{i}-p)^{2}\cdot P(x_{1},x_{2},\ldots ,x_{n})\right)}\\&=\sum _{i=1}^{n}{\left(\sum _{(x_{1},\ldots ,x_{i-1},x_{i},\ldots ,x_{n})\in \{0,1\}^{n-1}}{\left(\sum _{x_{i}\in \{0,1\}}(x_{i}-p)^{2}\cdot P(x_{i})\right)}\prod _{k\neq i}P(x_{k})\right)}\\&=\sum _{i=1}^{n}{\left(\sum _{x_{i}\in \{0,1\}}(x_{i}-p)^{2}\cdot P(x_{i})\right)}\\&=\sum _{i=1}^{n}p(1-p)\\&=np(1-p)\end{aligned}}

Für die hintere Summe ist

{}{\begin{aligned}\sum _{(x_{1},x_{2},\ldots ,x_{n})\in \{0,1\}^{n}}{\left(\sum _{i<j}(x_{i}-p)(x_{j}-p)\right)}P(x_{1},x_{2},\ldots ,x_{n})&=\sum _{i<j}{\left(\sum _{(x_{1},x_{2},\ldots ,x_{n})\in \{0,1\}^{n}}(x_{i}-p)(x_{j}-p)P(x_{1},x_{2},\ldots ,x_{n})\right)}\\&=\sum _{i<j}{\left(\sum _{(x_{1},\ldots ,x_{i-1},x_{i+1},\ldots ,x_{j-1},x_{j+1},\ldots ,x_{n})\in \{0,1\}^{n-2}}{\left(\sum _{(x_{i},x_{j})\in \{0,1\}^{2}}(x_{i}-p)(x_{j}-p)P(x_{i})P(x_{j})\right)}\prod _{k\neq i,j}P(x_{k})\right)}\\&=0\end{aligned}}