In this section we will aggregate the scores on the six dimensions, denoted by x
1,x
2,...,x
6,Footnote 3 using
$$ A\left( x\right) :=\sum_{k=1}^{6}w_{k}\cdot x_{\left( k\right) } $$
(1)
where the w
k
’s are weights and \(x_{\left( k\right) }\) is the k-th largest score, so \(x_{\left( 1\right) }\geq x_{\left( 2\right) }\geq ...\geq x_{\left( 6\right) }.\) These OWA operators were introduced by Yager (1988, 1996, 1999). With equal weights we get the simple average, \(\overline{x}\). With all weight assigned to w
1 we take the best score to be representative, and we turn a blind eye to the other scores, no matter how low. This approach is surely too lenient. With all weight assigned to w
6 the scores are summarized in an unforgiving way by the worst score: there is no compensation whatsoever by any of the other scores, no matter how high. This approach is surely too demanding. A compromise that encourages good performance across the board, is obtained by assigning weights to all scores, with higher weights to lower scores. It is shown in Dijkstra (2008) that this is equivalent to concavity of A as a function of x. This implies in particular that when two countries with scores x and y respectively are valued equally, \(A\left( x\right) =A\left( y\right) ,\) a third country with scores in between (: \( {\frac12} \cdot x+ {\frac12} \cdot y\)) is valued higher. A related implication is that equal scores on all dimensions are valued more than a diverse set of scores with the same simple average: \(A\left( \overline{x},\overline{x},...,\overline{x}\right) \geq A(x_{1},x_{2},...,x_{6})\). In other words, below average scores are not simply compensated by equally large above average scores. This will be exemplified below.
Concavity appears to be a natural requirement, but aggregators satisfying it can still be overly demanding. The increase in the weights can be too large for comfort, after all, the minimum score is still one of the possibilities. We suggested in Dijkstra (2008) to add ‘reflection neutrality’ to ease things up. This concept can be explained as follows. Suppose we have scores 4, 8, and 9 on three equally important criteria. The scores are grades, with the Dutch interpretation: they range from 1 (worst) to 10 (best), with 5 (not passed, but close) and 6 (just passed). The simple mean of the given grades is 7. The differences between grades and mean are − 3, + 1 and + 2 respectively. If we reverse their signs, and add the ensuing numbers to 7, we get scores 5, 6 and 10, with of course the same mean. We will call the latter set of scores the mirror scores, they are obtained by reflecting the old scores in the mean, so to speak. Which sequence is better, (4, 8, 9) or (5, 6, 10)? The latter sequence has both a better worst score and a better best score, but it also has a decidedly lower median score. Without the lowest scores we have (8, 9) versus (6, 10), and without the highest scores, (4, 8) versus (5, 6). It appears that one could argue either way. We chose to impose equality for the aggregates of both scores and mirror scores, called reflection neutrality, in addition to concavity. More formally, we require that \(A\left( x\right) =A\left( y\right) \) for every x and y that mirror each other, meaning that y
k
, the observation on the k-th dimension, equals \(\overline{x}-\left( x_{k}- \overline{x}\right) \) whereas \(x_{k}=\overline{x}+\left( x_{k}-\overline{x} \right) \). It can be shown that the set of weights satisfying both concavity and reflection neutrality equals all possible weighted combinations of the rows of the following matrix E, say:
$$ E:= \begin{vmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 2 \\ 0 & 0 & 1 & 1 & 2 & 2 \\ 0 & 0 & 0 & 2 & 2 & 2 \end{vmatrix} \div 6. $$
(2)
(Note that E is constructed in a very simple way, valid for all numbers of criteria: start with a row of ones, take away the first ‘1’ and add it to the last ‘1’ to get the second row, then take from the second row the first ‘1’ and add it to the penultimate element to get the third row; repeat this until only one ‘1’ or no ‘1’ is left). The ensuing averages combine weighted averages covering the lower half of the scores and more. The minimum score is no longer an option. Now the most unforgiving aggregator, the least inclined to allow for compensation, is the average of the lowest three scores. Had we ignored reflection neutrality, we would have combined all averages of the form \(\overline{x}_{\left( k:6\right) }:=\left( x_{k}+...+x_{6}\right) \div \left( 7-k\right) .\) It is useful to note that when we average the rows of E we get the barycentre of the convex polytope generated by these rows: \(\left[ 1,2,3,5,6,7\right] \div 24\). It is the representative weight vector for concave reflection neutral weights. It yields almost the rank weighted average with weights \( \left[ 1,2,3,4,5,6\right] \div 21\)
. For uneven numbers of criteria the rank weighted average is representative. See Dijkstra (2008) for more general results.
We took 10.000 random combinations of the rows of E, for each combination we calculated overall scores for all of the countries involved and ranked them. (Here rank is defined as in sport competitions: one plus the number of countries with a better score.) Finally we noted how often each country occupied every one of the possible ranks,Footnote 4 and sorted the countries according to the average rank. This agrees with the ranking obtained using the representative weights. Table 1
Footnote 5 collects the results.
Table 1 Innocenti’s scores, sorted
Please observe that the scores are in descending order, so the identity of the dimensions is erased, in agreement with the assumption of symmetry or ‘equal importance’. Innocenti has normalized the z-scores in such a way that their average equals 100 and their standard deviation 10. We rounded the scores to the first decimal to ease the readability. As one can see, the ranking based on a more demanding way of aggregating differs from the one based on the simple average, indicated by the column headed by ‘\(\left( \overline{x}\right) \)’. In particular, Germany, with a slightly lower average than Italy, is placed above Italy. Germany has scores hovering around 100, whereas Italy is once well above 100 but also well below 100. The bad score is not simply balanced out by the good score.