Question

If each of the observations $x_1,x_2,x_3,\cdots ,x_n$ is increased by an amount a, where a is a negative or positive number, then show that the variance remains unchanged.

Answer 1

Let $\overline{x}$ be the mean of $x_1,x_2,x_3,\cdots ,x_n$. Then,

$\overline{x}=\frac{1}{n}(x_1+x_2+x_3+\cdots +x_n)$

Let $y_i=x_i+a$ for each i =1, 2, 3, ..., n.

Let $\overline{y}$ be the mean of $y_1,y_2,y_3,\cdots ,y_n$. Then,

$\overline{y}=\frac{1}{n}(y_1+y_2+y_3+\cdots +y_n)$

$=\frac{1}{n}(x_1+a+x_2+a+x_3+a+\cdots +x_n+a)$

$=\frac{1}{n}(x_1+x_2+x_3+\cdots +x_n)+\frac{1}{n}(a+a+a+\cdots n\text{times})$

$=\overline{x}+\frac{1}{n}\left(na\right)=(\overline{x}+a)$

$\therefore \overline{y}=(\overline{x}+a)$

Now, the variance of the new observations is given by

variance $\left(y\right)={\sigma }^2=\frac{1}{n}.{\sum }_{i=1}^n(y_i-\overline{y}{)}^2$

$=\frac{1}{n}.{\sum }_{i=1}^n{\left\{\right(x_i+a)-(\overline{x}+a\left)\right\}}^2$

$[\because y_i=(x_i+a)\text{and}\overline{y}=(\overline{x}+a\left)\right]$

$=\frac{1}{n}.{\sum }_{i=1}^n(x_1+a-\overline{x}-a{)}^2$

$=\frac{1}{n}.{\sum }_{i=1}^n(x_i-\overline{x}{)}^2$

= variance (x).

Hence, the variance of the new observations is the same as the variance of the original observations.