# Value that increases the Standard Deviation

#### JohnK

##### New Member
Hello,

I am puzzled by the following statement

" In order to increase the standard deviation of a set of numbers, you must add a value that is more than one standard deviation away from the mean"

What is the proof of that? I know of course how we define the standard deviation but that part I seem to miss somehow. Any comments? Thanks!

#### BGM

##### TS Contributor
For a given set of numbers $$\{x_1, x_2, \ldots, x_n\}$$, the sample variance is given by

$$s_n^2 = \frac {1} {n-1} \sum_{i=1}^n x_i^2 - \frac {1} {n(n-1)} \left(\sum_{i=1}^n x_i\right)^2$$

With a new additional number $$x_{n+1}$$, the sample variance becomes

$$s_{n+1}^2 = \frac {1} {n} \sum_{i=1}^{n+1} x_i^2 - \frac {1} {n(n+1)} \left(\sum_{i=1}^{n+1} x_i\right)^2$$

$$= \frac {1} {n} \sum_{i=1}^{n} x_i^2 + \frac {x_{n+1}^2} {n} - \frac {1} {n(n+1)} \left(\sum_{i=1}^n x_i\right)^2 - \frac {1} {n(n+1)} \left(2x_{n+1}\sum_{i=1}^n x_i + x_{n+1}^2\right)$$

Then the difference is

$$s_{n+1}^2 - s_n^2$$

$$= \frac {x_{n+1}^2} {n+1} - \frac {2x_{n+1}} {n(n+1)}\sum_{i=1}^n x_i -\frac {1} {n(n-1)} \sum_{i=1}^n x_i^2 + \frac {2} {n(n-1)(n+1)}\left(\sum_{i=1}^n x_i\right)^2$$

and you see this is a quadratic expression in $$x_{n+1}^2$$

To shorten our notation, let $$a = \sum_{i=1}^n x_i = n\bar{x}$$ and $$b = \sum_{i=1}^n x_i^2$$. Then we can solve the quadratic inequality

$$s_{n+1}^2 - s_n^2 > 0$$

$$\iff x_{n+1}^2 - 2\bar{x}x_{n+1} - \frac {n+1} {n(n-1)} b + \frac {2} {n(n-1)}a^2 > 0$$

$$\iff x_{n+1} < \bar{x} - \frac {\sqrt{bn^3 - (a^2 + 2b)n^2 + bn + a^2}} {n(n-1)} ~~\text{or}$$
$$x_{n+1} > \bar{x} + \frac {\sqrt{bn^3 - (a^2 + 2b)n^2 + bn + a^2}} {n(n-1)}$$

So the width is not exactly 1 standard deviation as

$$s_n = \sqrt{\frac {nb - a^2} {n(n-1)}}$$

Anyway the answer change if you, e.g. divide the sample variance by $$n$$ rather than $$n - 1$$

#### JohnK

##### New Member
Thanks BGM, that is indeed rigorous and to the point. But may I ask how you derive the last results? After completing the square and substituting for $$\bar{x}\ \text{with}\ \frac{a}{n}$$ what I get is:

$$\left( x_{n+1}-\bar{x} \right)^2 > \frac{a^2}{n^2} + \frac{ \left( n+1 \right)b -2a^2 } {n \left( n-1 \right) }$$ which is not exactly the same even after adding the fractions.