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Standard deviation translation property proof - confused with property

  1. Sep 7, 2014 #1
    I need to prove that $$std(x+c) = std(x)$$

    I have been trying to use the properties of the mean such as $$mean(x+c) = mean(x) + c$$

    I am confused on the following property of the mean, is this statement correct?

    $$\sum\limits_{i=1}^{i=N}(x_i - mean(\{x\}))^2 = mean(\{x\}) \\$$

    If that is correct, then by the definition of standard deviation, is the following correct?

    $$std(x) = \sqrt{\frac{1}{N}\sum\limits_{i=1}^{i=N}(x_i - mean(\{x\}))^2} = \sqrt{\frac{1}{N}mean(\{x\})} \\$$

    This does not seem right to me.

    Here is my previous attempt at the proof, I think I am doing something wrong, this is why I am now attempting with the above property.

    $$std(\{x + c\}) = \sqrt{mean(\{((x+c) - mean(\{x\}) + c)^2\})}$$
    $$= \sqrt{mean(\{(x - mean(\{x\}) + 2c)^2\})}$$
    $$= \sqrt{\frac{((x_1 - mean(\{x\}) + 2c))^2) + ((x_2 - mean(\{x\}) + 2c))^2) + ... + ((x_n - mean(\{x\}) + 2c))^2)}{N}}$$

    Thanks!
     
  2. jcsd
  3. Sep 7, 2014 #2
    Plug that straight into the equation for standard deviation. Doesn't the proof then appear trivial?
     
  4. Sep 7, 2014 #3

    mathman

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    Simple equation: [itex]x_i+c-mean(x+c) = x_i-mean(x)[/itex].

    Plug into the definition of standard deviation to get desired result.
     
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