# How would one rescale the variance and standard deviation of a set of data?

I have a set of data with a certain mean, variance, and standard deviation. I centered the mean around the origin the standard way by subtracting it from the data.

Now how do I modify the data to make variance = 1?

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Just divide everything by the standard deviation.

Suppose first that you have a finite data set $x_1,x_2,x_3,\ldots,x_n$.

Let $\overline{x} = (x_1+\cdots+x_n)/n$ be the mean and $s=\sqrt{(1/n)\sum_{i=1}^n (x_i - \overline{x})^2}$ be the standard deviation. Let $y_i=x_i/s$, i.e. we're dividing everything by the standard deviation. Then $$\overline{y} = \frac{y_1+\cdots+y_n}{n} = \frac{\frac{x_1}{s}+ \cdots+\frac{x_n}{s}}{n} = \frac 1 s \cdot \frac{x_1+\cdots+x_n}{n} = \frac{\overline{x}}{s}.$$ And the standard deviation of the $y$-values is $$\sqrt{\frac 1 n ((y_1-\overline{y})^2 + \cdots+(y_n-\overline{y})^2)} = \sqrt{\frac 1 n \left( \left(\frac{x_1}{s}-\frac{\overline{x}}{s}\right)^2+ \cdots + \left(\frac{x_n}{s}-\frac{\overline{x}}{s}\right)^2\right)}$$ $$= \frac{1}{\sqrt{s^2}}\sqrt{\frac 1 n ((x_1-\overline{x})^2 + \cdots+(x_n-\overline{x})^2)} = \frac 1 s \cdot s = 1.$$

For discrete data sets with non-uniform distributions, including infinite discrete data sets, one replaces $1/n$ with appropriate weights $p_i$. For continuous data sets, one used integrals, but the argument is essentially the same.

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I knew that it would be something trivial like this. Can you show why this works? –  mathjacks May 21 '12 at 1:22
@flapjackery: Scaling the data by a value $a$ scales the standard deviation also by $a$, as you can see from the definition $\sigma = \sqrt{\frac1n\sum{x_i^2} - \big(\frac1n\sum x_i\big)^2}$. –  Rahul May 21 '12 at 1:48
@RahulNarain : That works if $a\ge 0$. In general multiplying everything by $a$ has the effect of multiplying the standard deviation by $|a|$. –  Michael Hardy May 21 '12 at 1:54
@MichaelHardy: You're right, of course. Thanks for the correction. –  Rahul May 21 '12 at 2:00