# Normalization of power-law distributed variables. Z-scores or Min-Max?

I need to make a composite index from the sum of three power-law distributed variables, which vary on different scales and have different variances. For each variable there are many observations with very low scores and few observations with high scores.

I need to normalize the variables to obtain a common scale, before summing them to obtain a single score of the final index. I'm considering two possibilities:

Min-Max Normalization

(Xi - min (X)) / (max(X) - min(X))

Standardization (Z-scores)

(Xi - mean(X)) / std(X)

Which solution is appropriate, given the power-law distribution of the three variables? Or are they both wrong? Why?

EDIT Please have a look to an example of the distribution I am referring to: I have three variables distributed like X and I need to normalized them before making a sum of the three.

• Why do you think the variables follow a power law when they are the sums of three exponential variates, which do not follow a power law? Have you left something out? – jbowman Dec 26 '17 at 23:23
• "many observations with very low scores and few observations with high scores" would suggest skewed, with a possibly unimodal, perhaps even monotonic density, but none of those things are sufficient for a power law, which says something more specific about the shape. If your variates were ordinary one-parameter exponential and you wanted to render them of comparable scale, dividing by the mean would be the most obvious thing to do. – Glen_b -Reinstate Monica Dec 27 '17 at 4:17
• I think the distribution is a power law. So I edited the question and added a graph. Moreover I never meant that the final index has a power-law distribution. I meant that the variables that I need to sum follow a power-law distribution. They change on different scales, and have different SD. Before summing them I want to normalize them, choosing the best way given their distribution. – Forinstance Dec 27 '17 at 8:51
• Whether one way of standardizing is superior to another one will likely depend on what you plan on doing with your sum score afterwards. – Stephan Kolassa Dec 27 '17 at 9:40
• @Stephan thank you. Could you explain more the difference and what you mean? I'm constructing an index for comparison among observations and ultimately for forecasting purposes. – Forinstance Dec 27 '17 at 11:24