# Standard deviation translation property proof - confused with property

1. Sep 7, 2014

### lantay77

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. Sep 7, 2014

### mal4mac

Plug that straight into the equation for standard deviation. Doesn't the proof then appear trivial?

3. Sep 7, 2014

### mathman

Simple equation: $x_i+c-mean(x+c) = x_i-mean(x)$.

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