$\cdot$ の期待値を $E[\cdot]$ 分散を $V[\cdot]$ で表す。
\begin{align} E \left[ \sum_{i=1}^n \left( X_i - \bar{X} \right)^2 \right] &= \sigma^2 E \left[ \sum_{i=1}^n \left( \frac{X_i - \bar{X}}{\sigma} \right)^2 \right] \\ &= \sigma^2 (n-1) \end{align}
\begin{align} V \left[ \sum_{i=1}^n \left( X_i - \bar{X} \right)^2 \right] &= \sigma^4 V \left[ \sum_{i=1}^n \left( \frac{X_i - \bar{X}}{\sigma} \right)^2 \right] \\ &= 2 \sigma^4 (n-1) \end{align}
$ W_i-a = \left( W_i - E \left[ W_i \right] \right) + \left( E \left[ W_i \right] - a \right) $ の両辺を2乗すると、 \begin{align} \left( W_i - a \right)^2 = \left( W_i - E \left[ W_i \right] \right)^2 + \left( E \left[ W_i \right] - a \right)^2 + 2 \left( W_i - E \left[ W_i \right] \right) \left( E \left[ W_i \right] - a \right) \end{align} であり、 \begin{align} E \left[ \left( W_i - a \right)^2 \right] &= E \left[ \left( W_i - E \left[ W_i \right] \right)^2 \right] + \left( E \left[ W_i \right] - a \right)^2 + 2 \left( E \left[ W_i \right] - E \left[ W_i \right] \right) \left( E \left[ W_i \right] - a \right) \\ &= E \left[ \left( W_i - E \left[ W_i \right] \right)^2 \right] + \left( E \left[ W_i \right] - a \right)^2 \\ &= V \left[ W_i \right] + \left( E \left[ W_i \right] - a \right)^2 \end{align} を得る。
\begin{align} E \left[ \left\{ c \sum_{i=1}^n \left( X_i - \bar{X} \right)^2 - \sigma^2 \right\}^2 \right] &= V \left[ c \sum_{i=1}^n \left( X_i - \bar{X} \right)^2 \right] + \left( E \left[ c \sum_{i=1}^n \left( X_i - \bar{X} \right)^2 \right] - \sigma^2 \right)^2 \\ &= 2 c^2 \sigma^4 (n-1) + \left( c \sigma^2 (n-1) - \sigma^2 \right)^2 \\ &= \sigma^4 \left[ (n+1)(n-1)c^2 - 2(n-1)c + 1 \right] \end{align}
\begin{align} E \left[ \left\{ c \sum_{i=1}^n \left( X_i - \bar{X} \right)^2 - \sigma^2 \right\}^2 \right] &= \sigma^4 (n+1)(n-1) \left( c^2 - \frac{2}{n+1} c \right) + \sigma^4 \\ &= \sigma^4 (n+1)(n-1) \left( c - \frac{1}{n+1} \right)^2 - \sigma^4 \frac{n-1}{n+1} + \sigma^4 \end{align} なので、求める $c$ の値は $c=1/(n+1)$ である。