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I'm trying to compute the Gini index on the SO reputation distribution using SO Data Explorer. The equation I'm trying to implement is this: $$ G(S)=\frac{1}{n-1}\left(n+1-2\left(\frac{\sum^n_{i=1}(n+1-i)y_i}{\sum^n_{i=1}y_i}\right)\right) $$ Where: $n$ = number of users on the site; $i$ = user serial id (1 - 1,225,000); $y_i$ = reputation of user $i$.

This is how I implemented it (copied from here):

DECLARE @numUsers int
SELECT @numUsers = COUNT(*) FROM Users
DECLARE @totalRep float
SELECT @totalRep = SUM(Users.Reputation) FROM Users
DECLARE @giniNominator float
SELECT @giniNominator = SUM( (@numUsers + 1 - CAST(Users.Id as Float)) * 
                              CAST(Users.Reputation as Float)) FROM Users
DECLARE @giniCalc float
SELECT @giniCalc = (@numUsers + 1 - 2*(@giniNominator / @totalRep)) / @numUsers
SELECT @giniCalc

My result is (currently) -0.53, but it makes no sense: I'm not sure even how it could have become negative, and even in abs value, I would have expected the inequality to be much closer to 1, given how reputation grows the more you have it.

Am I unknowingly ignoring some assumption about the distribution of the reputation/users?

What do I do wrong?

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You're right, but I'm not sure I see why this should effect the calculation? – yossale Aug 25 '12 at 12:28
2  
I'm guessing that your question is about the nature & calculation of the Gini index, & not about how to implement that in SQL (correct me if I'm wrong). If the latter, we should migrate this to SO. Continuing w/ my assumption, I have copied your code from the SE data site, but it might help if you can also rewrite it in pseudo-code for those who may not read SQL well. – gung Aug 25 '12 at 13:36
@gung thanks - I do ask about the calculation, not the SQL implementation. I'll re write it in pseudo code – yossale Aug 25 '12 at 14:56

2 Answers

up vote 8 down vote accepted

I can't read the SQL code very easily, but if it helps, if I were going to calculate the Gini coefficient, this is what I would do (in plain English).

  1. Figure out the $n$ of $x$ (ie. the number of people with rep on SO)
  2. Sort $x$ from lowest to highest
  3. Sum each $x$ multiplied by its order in the rank (ie. if there are 10 people, the rep for the person with the lowest rep gets multiplied by 1 and the rep of the person with the highest rep gets multiplied by 10)
  4. Take that value and divide it by the product of $n$ and the sum of $x$ (ie. $n \times \sum $ rep) and then multiply that result by 2
  5. Take that result and subtract the value of $1-(1/n)$ from it.
  6. Voila!

I took those steps from the remarkably straight-forward code in the R function (in the ineq package) for calculating the Gini coefficient. For the record, here's that code:

> ineq::Gini
function (x) 
{
    n <- length(x)
    x <- sort(x)
    G <- sum(x * 1:n)
    G <- 2 * G/(n * sum(x))
    G - 1 - (1/n)
}
<environment: namespace:ineq>

It looks somewhat similar to your SQL code, but like I said, I can't really read that very easily!

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Thanks you very much! I missed the sorting part! that explains a lot... – yossale Aug 25 '12 at 15:21
Super. I'm interested in knowing what the value is so maybe leave a comment when you've made the calculation! – smillig Aug 25 '12 at 16:04
Well, When I aggregated the values (i.e if there are 10 people, with either 1,3, or 5 points, then i have just 3 ranks : 1:3,2:5,3:10) and multiplied the (how many with that score)*score*(rank of score) I got -0.98 , which would have made sense if not for the wrong sign. But I'm not sure how my little shortcut effects the gini scale – yossale Aug 25 '12 at 16:16

There are, I believe, four equivalent formulations of the Gini index. To me, the most natural one is a U-statistic: $$ G = \frac 2{\mu n(n-1)}\sum_{i\neq j} |x_i - x_j| $$ where $\mu$ is the mean of $x$'s. You can double-check your computations with this formula. Obviously, the result must be non-negative. For what I know about Gini indices, the reputation distribution on CV should have the Gini index above 0.9; whether 0.98 makes a lot of sense or not, I can't say though.

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