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Thread: Value that increases the Standard Deviation

  1. #1
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    Value that increases the Standard Deviation




    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!

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    Re: Value that increases the Standard Deviation

    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

  3. #3
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    Re: Value that increases the Standard Deviation


    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.

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