There are many questions and answers (see here for example) related to standandardization of variables, carried out by taking a value, subtracting the mean (centering) and dividing by standard deviation.

I do this in an index that has 3 components (x, y, z), to adjust their scale and make the sum possible. See here: $$ R_{i} = \frac{x_{i}-\mu_{x}}{SD_{x}} + \frac{y_{i}-\mu_{y}}{SD_{y}} + \frac{z_{i}-\mu_{z}}{SD_{z}} $$

My variables are all greater or equal to zero.

Of course when values are below the mean I get a negative sum, which is a problem. So I am looking for a way to keep the final index always positive (or zero), as the original variables. But I don't want to use a min-max normalization.

If I avoid centering, but keep dividing by the SD, do I bias my index? Is the sum still correct? I mean that the variables have very different scales, so I am wondering if dividing them by SD (without substracting the mean) is enough to make them comparable (and summable). The result would be like this:

$$ R_{i}^{'} = \frac{x_{i}}{SD_{x}} + \frac{y_{i}}{SD_{y}} + \frac{z_{i}}{SD_{z}} $$


I think that $R_{i}^{'} $ keeps the ranking of observations as $R_{i}$ and doesn't bias the original index formulation, because mean and SD are constant values, so I would have that the two indices only differ by a constant. See my reasoning here: $$ k_{x} = \frac{\mu_{x}}{SD_{x}}, k_{y} = \frac{\mu_{y}}{SD_{y}}, k_{z} = \frac{\mu_{z}}{SD_{z}} $$ so that $$ R_{i} = \frac{x_{i}}{SD_{x}} - k_{x} + \frac{y_{i}}{SD_{y}} - k_{y} + \frac{z_{i}}{SD_{z}} - k_{z} $$ and $$ R_{i} = R_{i}^{'} - (k_{x}+k_{y}+k_{z}) $$ so I should not have problems using $R_{i}^{'} $ instead of $R_{i}$, as they only differ by a constant.

HOWEVER I THINK I AM WRONG: as in this way I am still summing variables on different scales (since I did not subtract the mean, see comments here), right? Could a solution be to simply divide by the mean?

  • $\begingroup$ You are not scaling $x_i+y_i+z_i$ by a constant, so may get different results. What is $R_i$ supposed to represent? $\endgroup$
    – Henry
    May 27, 2021 at 15:01
  • $\begingroup$ Thanks, I am not sure I get exactly the answer. Numbers will be different, but for example ranking of different observations (i) still concordant? You can imagine the index as an intelligence score, measured along 3 intelligence dimensions. $\endgroup$ May 27, 2021 at 15:05


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