Question

Given that $\overline{X}$ is the mean and ${\sigma }^2$ is the variance of $n$ observations $X_1,X_2...X_n$. Prove that the mean and variance of the observations $a{X}_1,a{X}_2,a{X}_3....a{X}_n$ are $\overline{ax}$ and $a^2{\sigma }^2$ respectively $(a\ne 0)$.

Answer 1

The given $n$ observation are $x_1,x_2..x_n$
Mean = $\overline{x}$
variance = ${\sigma }^2$
$\therefore {\sigma }^2=\frac{1}{n}\sum _{i=l}^n{y}_i{(x_i-\overline{x})}^2$ ....(i)
If each observation is multiplied by a and the new observation are $y_i$ then
$y_i=a{x}_i,i.e,x_i=\frac{1}{a}{y}_i$
$\therefore \overline{y}=\frac{1}{n}\sum _{i=l}^n{y}_i=\frac{1}{n}\sum _{i=l}^{n}a{x}_i=\frac{a}{n}\sum _{i=l}^n{x}_i=\overline{ax},(\because \overline{x}=\frac{1}{n}\sum _{i=1}^n{x}_i)$
Therefore mean of the observation, $a{x}_1,a{x}_2....a{x}_n$ is $\overline{ax}$
Substituting the values of $x_i$ and $\overline{x}$ in (1) we obtain
${\sigma }^2=\frac{1}{2}\sum _{i=l}^n{(\frac{1}{a}{y}_i-\frac{1}{a}\overline{y})}^2$
$\Rightarrow a^2{\sigma }^2=\frac{1}{n}\sum _{i=l}^n{(y_i-\overline{y})}^2$
Thus the variance of the observation $a{x}_1,a{x}_2...a{x}_n$ is $a^2{\sigma }^2$