- #1
lantay77
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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!
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!