# Quod Erat Demonstrandum

## 2012/01/14

### 續 Core Math 某題

Filed under: mathematics,NSS — johnmayhk @ 9:32 上午

I’m asked to generalize the solution of the previous post, okay, do it.

Let $n$ be a positive even integer.

Set 1: {$a_1,a_2,\dots,a_{n/2},a_{n/2+1},\dots,a_n$}
Set 2: {$a_1,a_2,\dots,a_{n/2},0,a_{n/2+1},\dots,a_n$}

where $a_1 < a_2 < \dots < a_{n/2} < 0 < a_{n/2+1} < \dots < a_n$.

Let $\sigma_1, \sigma_2$ be the standard deviations of Set 1 and Set 2 respectively.

Show that $\sigma_1 > \sigma_2$.

Solution (by ugly brute force)

$\sigma_1^2-\sigma_2^2$

$=\frac{\displaystyle\sum_{k=1}^na_k^2}{n}-(\frac{\displaystyle\sum_{k=1}^na_k}{n})^2-\frac{\displaystyle\sum_{k=1}^na_k^2}{n+1}+(\frac{\displaystyle\sum_{k=1}^na_k}{n+1})^2$

$=\frac{1}{n^2(n+1)^2}(n(n+1)\displaystyle\sum_{k=1}^na_k^2-(2n+1)(\displaystyle\sum_{k=1}^na_k)^2)$

$=\frac{1}{n^2(n+1)^2}((n^2-n-1)\displaystyle\sum_{k=1}^na_k^2-2(2n+1)\displaystyle\sum_{1 \le i < j \le n}a_ia_j)$

$=\frac{1}{n^2(n+1)^2}((n^2-n-1)\displaystyle\sum_{k=1}^na_k^2-2(2n+1)(\displaystyle\sum_{i=1}^{n/2}a_i)(\displaystyle\sum_{i=n/2+1}^{n}a_i)-2(2n+1)\displaystyle\sum_{1 \le i < j \le n/2}a_ia_j-2(2n+1)\displaystyle\sum_{n/2+1 \le i < j \le n}a_ia_j)$

Note that, since $a_1 < a_2 < \dots < a_{n/2} < 0 < a_{n/2+1} < \dots < a_n$,

$-2(2n+1)(\displaystyle\sum_{i=1}^{n/2}a_i)(\displaystyle\sum_{i=n/2+1}^{n}a_i)$ must be positive,

it suffices to prove

$(n^2-n-1)\displaystyle\sum_{k=1}^na_k^2-2(2n+1)\displaystyle\sum_{1 \le i < j \le n/2}a_ia_j-2(2n+1)\displaystyle\sum_{n/2+1 \le i < j \le n}a_ia_j> 0$

Consider

$(n^2-n-1)\displaystyle\sum_{k=1}^{n/2}a_k^2-2(2n+1)\displaystyle\sum_{1 \le i < j \le n/2}a_ia_j$

$=(n^2-n-1-n/2)\displaystyle\sum_{k=1}^{n/2}a_k^2-2(2n+1)\displaystyle\sum_{1 \le i < j \le n/2}a_ia_j+(n/2)\displaystyle\sum_{k=1}^{n/2}a_k^2$

$=(n/2-1)(2n+1)\displaystyle\sum_{k=1}^{n/2}a_k^2-2(2n+1)\displaystyle\sum_{1 \le i < j \le n/2}a_ia_j+(n/2)\displaystyle\sum_{k=1}^{n/2}a_k^2$

$=(2n+1)((n/2-1)\displaystyle\sum_{k=1}^{n/2}a_k^2-2\displaystyle\sum_{1 \le i < j \le n/2}a_ia_j)+(n/2)\displaystyle\sum_{k=1}^{n/2}a_k^2$

$=(2n+1)\displaystyle\sum_{1\le i < j\le n/2}(a_i-a_j)^2+(n/2)\displaystyle\sum_{k=1}^{n/2}a_k^2$

$> 0$

Similar, we can obtain

$(n^2-n-1)\displaystyle\sum_{k=n/2+1}^{n}a_k^2-2(2n+1)\displaystyle\sum_{n/2+1 \le i < j \le n}a_ia_j > 0$

Thus,

$(n^2-n-1)\displaystyle\sum_{k=1}^na_k^2-2(2n+1)\displaystyle\sum_{1 \le i < j \le n/2}a_ia_j-2(2n+1)\displaystyle\sum_{n/2+1 \le i < j \le n}a_ia_j> 0$

result follows.