# Quod Erat Demonstrandum

## 2015/01/12

### just a core math question involving variance

Filed under: mathematics,NSS — johnmayhk @ 10:35 下午

Just reply to a student about a core math question used in school exam.

Question

Let $m$, $v$ be the mean and variance of {$2x_1, 2x_2, 2x_3$} respectively.

Show that the mean and variance of the set

{$x_1+x_1,x_1+x_2,x_1+x_3,x_2+x_1,x_2+x_2,x_2+x_3,x_3+x_1,x_3+x_2,x_3+x_3$}

are $m$ and $\frac{v}{2}$ respectively.

Proof

(Mean) For the original set,

$\frac{2x_1+2x_2+2x_3}{3}=m\Rightarrow x_1+x_2+x_3=\frac{3}{2}m$

For the new set,

mean

$=\frac{1}{9}((x_1+x_1)+(x_1+x_2)+\dots+(x_3+x_3))$

$=\frac{1}{9}((3x_1+\frac{3}{2}m)+(3x_2+\frac{3}{2}m)+(3x_3+\frac{3}{2}m))$

$=\frac{1}{9}(3(\frac{3}{2}m)+3(\frac{3}{2}m))$

$=m$

(Variance) For the original set,

$\frac{(2x_1-m)^2+(2x_2-m)^2+(2x_3-m)^2}{3}=v$

For the new set,

variance

$=\frac{(x_1+x_1-m)^2+(x_1+x_2-m)^2+\dots+(x_3+x_3-m)^2}{9}$

$=\frac{(2x_1+2x_1-2m)^2+(2x_1+2x_2-2m)^2+\dots+(2x_3+2x_3-2m)^2}{36}$

$=\frac{(2x_1-m+2x_1-m)^2+(2x_1-m+2x_2-m)^2+\dots+(2x_3-m+2x_3-m)^2}{36}$

$=\frac{((2x_1-m)^2+(2x_1-m)^2+2(2x_1-m)(2x_1-m))+((2x_1-m)^2+(2x_2-m)^2+2(2x_1-m)(2x_2-m))+\dots+((2x_3-m)^2+(2x_3-m)^2+2(2x_3-m)(2x_3-m))}{36}$

$=\frac{(3(2x_1-m)^2+3v)+(3(2x_2-m)^2+3v)+(3(2x_3-m)^2+3v)}{36}$　　（note $2(2x_i-m)(2x_1+2x_2+2x_3-3m)=0$ for $i=1,2,3$

$=\frac{3(3v)+9v}{36}$

$=\frac{v}{2}$

Remark:
1. It is easy to present the solution using the summation sign $\sum$ which is out of the present syllabus.

2. Here is a solution of an M.C. question from the core math practice paper 2 issued on Jan 2012, for further reference

https://johnmayhk.wordpress.com/2012/01/12/just-a-core-math-question/

3. Here is a general version of the solution of question above:

https://johnmayhk.wordpress.com/2012/01/14/just-a-question-of-core-math-2/