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Prove: $\operatorname{std}(\frac{v}{\operatorname{std}(v)}))=1$ for any vector $v$ where std is the standard deviation as $\sqrt{\frac{1}{n-1}\sum_{i=1}^n (v_i-\bar{v})^2}$

I think I have to cancel out the $(n-1)$s somewhere to get 1 in the end because of this proof for a z score but I can't figure out how to get to that point.

Any help is appreciated!

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    $\begingroup$ Use the fact that $std(c A) = |c| std(A)$. $\endgroup$ – Math Lover Sep 21 '17 at 4:13
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Let $v=(v_1,\dots, v_n)$. One has $$std(\frac{v}{std(v)}) = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (\frac{v_i}{std(v)}-\bar{\frac{v}{std(v)}}})^2 = \frac{1}{std(v)}\sqrt{\frac{1}{n-1}\sum_{i=1}^n(v_i-\bar{v})^2} = \frac{std(v)}{std(v)} = 1.$$

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  • $\begingroup$ Stylistically, it may be easier to follow Math Lover's suggestion in the comments and perform the calculations using a generic scalar like $c$ (so that there are fewer instances of $std$ floating around). $\endgroup$ – Erick Wong Sep 21 '17 at 4:41

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