# Is the standard deviation of a data set invariant to translation?

Dec 13, 2016

Yes it is. See explanation for a proof.

#### Explanation:

Let $S$ be a data set:

$S = \left\{{x}_{1} , {x}_{2} , \ldots , {x}_{n}\right\}$
Its mean and standard deviation:

$\overline{x} = \frac{1}{n} \times {\Sigma}_{i = 1}^{i = n} \left({x}_{i}\right)$

$\sigma = \sqrt{{\Sigma}_{i = 1}^{n} {\left({x}_{i} - \overline{x}\right)}^{2}}$

Let ${S}_{1}$ be a data set $S$ translated by $a$:

${S}_{1} = \left\{{x}_{1} + a , {x}_{2} + a , \ldots , {x}_{n} + a\right\}$

Its mean would equal:

$\overline{{x}_{1}} = \frac{{x}_{1} + a + {x}_{2} + a + \ldots + {x}_{n} + a}{n} =$
$= \frac{{x}_{1} + {x}_{2} + \ldots + {x}_{n}}{n} + \frac{n a}{n} = \overline{x} + a$

The standard deviation would be:

${\sigma}_{1} = \sqrt{{\Sigma}_{i = 1}^{n} {\left({x}_{i} + a - \left(\overline{x} + a\right)\right)}^{2}} =$

 =sqrt(Sigma_{i=1}^{n}(x_i+a-bar(x)-a))^2)=

 =sqrt(Sigma_{i=1}^{n}(x_icancel(+a)-bar(x) cancel(-a)))^2)=

$= \sqrt{{\Sigma}_{i = 1}^{n} {\left({x}_{i} - \overline{x}\right)}^{2}} = \sigma$

The standard deviation of the new set is equal to the deviation of the set before translation.

QED