离差平方和

题目

原有 \(n\) 个元素，现在假如一个元素 \(a\)，求证，加入之后离差平方和 \(d_{2}^{2} \geq\) 加入之后离差平方和 \(d_{1}^{2}\)

解答

原先平均数为 \(\overline{x}_1\)，有 \(n\) 个元素
加入 \(a\) 后平均数为 \(\overline{x}_2 = \frac{n \overline{x}_1 + a}{n + 1}\)，有 \(n + 1\) 个元素

原来的离差平方和

\[\begin{aligned} d_{1}^{2} &= (x_1 - \overline{x}_1)^2 + (x_2 - \overline{x}_1)^2 + (x_3 - \overline{x}_1)^2 + \cdots + (x_{n} - \overline{x}_1)^2 \\ &= x_1^2 - 2\overline{x}_1 x_1 + \overline{x}_1^{2} + x_2^2 - 2\overline{x}_1 x_2 + \overline{x}_1^{2} + \cdots + x_n^2 - 2x_n\overline{x}_1 + \overline{x}_1^2 \\ &= x_1^2+x_2^2 + \cdots +x_n^2 + n\overline{x}_1^{2} - 2\overline{x}_1(x_1+x_2+ \cdots + x_n)\\ &= x_1^2+x_2^2 + \cdots +x_n^2 + n\overline{x}_1^{2} - 2n\overline{x}_1^{2}\\ &= x_1^2+x_2^2 + \cdots +x_n^2 - n\overline{x}_1^{2} \end{aligned} \]

加入 \(a\) 之后

\[\begin{aligned} d_{2}^{2} &= (x_1 - \overline{x}_2)^2 + (x_2 - \overline{x}_2)^2 + (x_3 - \overline{x}_2)^2 + \cdots + (x_{n} - \overline{x}_2)^2 + (a - \overline{x}_2)^2\\ &= x_1^2 - 2\overline{x}_2x_1 + \overline{x}^{2}_2 + x_2^2 - 2\overline{x}_2 x_2 + \overline{x}_2^{2} + \cdots + x_n^2 - 2x_n\overline{x}_2 + \overline{x}^2_2 + a^2 - 2a\overline{x}_2 + \overline{x}^2_2\\ &= x_1^2+x_2^2 + \cdots + x_n^2 + a^2 + (n+1)\overline{x}^{2}_2 - 2\overline{x}_2(x_1+x_2+ \cdots+ x_n + a)\\ &= x_1^2+x_2^2 + \cdots + x_n^2 + a^2 + (n+1)\overline{x}^{2}_2 - 2(n+1)\overline{x}^{2}_2\\ &= x_1^2+x_2^2 + \cdots + x_n^2 + a^2 - (n+1)\overline{x}^{2}_2 \end{aligned} \]

作差

\[\begin{aligned} d_{2}^{2} - d_{1}^{2} &= x_1^2+x_2^2 + \cdots + x_n^2 + a^2 - (n+1)\overline{x}^{2}_2 - (x_1^2+x_2^2 + \cdots +x_n^2 - n\overline{x}_1^{2})\\ &= n\overline{x}_1^{2} + a^2 - (n+1)\overline{x}^{2}_2 \end{aligned} \]

把 \(\overline{x}_2 = \frac{n \overline{x}_1 + a}{n + 1}\) 代入

\[\begin{aligned} d_{2}^{2} - d_{1}^{2} &= n\overline{x}_1^{2} + a^2 -(n+1)(\frac{n \overline{x}_1 + a}{n + 1})^2\\ &= n\overline{x}_1^{2} + a^2 -(n+1)(\frac{(n \overline{x}_1 + a)^2}{(n + 1)^2}\\ &= n\overline{x}_1^{2} + a^2 -\frac{(n \overline{x}_1 + a)^2}{(n + 1)}\\ &= \frac{(n\overline{x}_1^{2} + a^2)(n+1)-(n \overline{x}_1 + a)^2}{(n + 1)}\\ &= \frac{n^{2}\overline{x}_1^{2}+n\overline{x}_1^{2}+a^2n + a^2 - n^{2}\overline{x}_1^{2} - 2n\overline{x}_1a-a^2}{n+1}\\ &= \frac{n\overline{x}_1^{2}+a^2n - 2n\overline{x}_1a}{n+1}\\ &= \frac{n(\overline{x}_1^{2}+a^2 - 2\overline{x}_1a)}{n+1}\\ &= \frac{n(\overline{x}_1^{2}-a)^2}{n+1} \geq 0 \end{aligned} \]

得证