29
$\begingroup$

I am looking for a method to transform my dataset from its current mean and standard deviation to a target mean and a target standard deviation. Basically, I want to shrink/expand the dispersion and scale all numbers to a mean.

It doesn't work to do two separate linear transformations, one for standard deviation, and then one for mean. What method should I use?

$\endgroup$
3
  • 4
    $\begingroup$ If $Y = aX + b$, then $E(Y) = a E(X) + b$ and $Var(Y) = a^2 Var(X)$. Does this help? $\endgroup$
    – ocram
    Commented Dec 22, 2012 at 9:00
  • $\begingroup$ @ocram, I think that's an answer (and a good one)... $\endgroup$ Commented Dec 22, 2012 at 10:09
  • $\begingroup$ @PeterEllis: Thanks! I'll make it an answer then :-) $\endgroup$
    – ocram
    Commented Dec 22, 2012 at 10:20

2 Answers 2

48
$\begingroup$

Suppose you start $\{x_i\}$ with mean $m_1$ and non-zero standard deviation $s_1$ and you want to arrive at a similar set with mean $m_2$ and standard deviation $s_2$.

Then multiplying all your values by $\frac{s_2}{s_1}$ will give a set with mean $m_1 \times \frac{s_2}{s_1}$ and standard deviation $s_2$.

Now adding $m_2 - m_1 \times \frac{s_2}{s_1}$ will give a set with mean $m_2$ and standard deviation $s_2$.

So a new set $\{y_i\}$ with $$y_i= m_2+ (x_i- m_1) \times \frac{s_2}{s_1} $$ has mean $m_2$ and standard deviation $s_2$.

You would get the same result with the three steps: translate the mean to $0$, scale to the desired standard deviation; translate to the desired mean.

$\endgroup$
0
9
$\begingroup$

Let’s consider the z-score calculation of data $x_i$ with mean $\bar{x}$ and standard deviation $s_x$.

$$z_i = \dfrac{x_i-\bar{x}}{s_x}$$

This means that, given some data $(x_i)$, we can transform to data with a mean of $0$ and standard deviation of $1$.

Rearranging, we get:

$$x_i = z_i s_x+ \bar{x}$$

This gives us back our original data with the original mean $\bar{x}$ and standard deviation $s_x$. But we could’ve gone to data $y_i$ with any mean $\bar{y}$ and standard deviation $s_y$.

$$y_i = z_i s_y +\bar{y}$$

Now combine the two transformations, first to $z_i$ and then to $y_i$.

$$y_i = \dfrac{x_i-\bar{x}}{s_x}s_y + \bar{y}$$

This is the same as what Henry posted, but I do think it is helpful to see that we get there by first going to standardized data and then transforming to data with the mean and standard deviation values we desire.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.