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.

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