Question

If the mean and variance of the observations $x_1,x_2,x_3,\cdots ,x_n$ are $\overline{x}$ and ${\sigma }^2$ respectively and a be a nonzero real number, then show that the mean and variance of $a{x}_1,a{x}_2,a{x}_3,\cdots .a{x}_n$ are $\overline{ax}$ and $a^2{\sigma }^2$ respectively.

Answer 1

Let $\overline{x}$ be the mean of $x_1,x_2,x_3,\cdots ,x_n$ and a be a nonzero real number.

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

Let $y_i=a{x}_i$ for each i =1, 2, 3, ..., n. Then,

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

$=\frac{1}{n}(a{x}_1+a{x}_2+a{x}_3+\cdots +a{x}_n)=a.\frac{1}{n}(x_1+x_2+x_3+\cdots x_n)=a\overline{x}$

Thus, $\overline{y}=a\overline{x}$

Now, the variance of new observations is given by

variance (y) = $={\sigma }_1^2$

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

$=\frac{1}{n}.{\sum }_{i=1}^n(a{x}_i-a\overline{x}{)}^2$ [$\because y_i=a{x}_i$ for each i and $\overline{y}=a\overline{x}$]

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

$=a^2$. {variance (x)} $=a^2{\sigma }^2$.

$\therefore$ new variance $=a^2{\sigma }^2$

Remark ${\sigma }_1=\sqrt{a^2{\sigma }^2}=|a|.\sigma$