Calculating the balanced worth vector

This algorithm calculates the so-called balanced worth vector, introduced in Herrero & Villar (2016), “The Balanced Worth: A procedure to compare distributions involving ordered attributes” (mimeo), to evaluate group performance with categorical data.

The balanced worth vector provides an evaluation of the relative performance of a given number of groups whose outcomes are expressed in terms of a finite set of ordered characteristics. So the ingredients of the problem are: a population divided into a set of \(g\) groups whose outcomes are classified into \(s\) categories, ordered from best to worst.

The input you need to feed the application is the matrix \(A = \{ a_{ ij } \} \) of relative frequencies, where the generic entry \( a_{ij}, \) \( i = 1, 2, \dots, g, \) \(j = 1, 2, \dots, s,\) denotes the share of individuals of group \( i \) in category \(j\). That is, a matrix with \(g\) rows and \(s\) columns, all whose rows add up to one.

The output you get is a vector of \(g\) components, one for each group, which gives you the relative evaluation of the performance of the \(g\) groups (the balanced worth vector).

To understand what the balanced worth vector means consider the following ideas. Let \(p_{ij}\) denote the probability that a member of group \(i\) belongs to a higher category than a member of group \(j\) (This probability is calculated as follows: \(p_{ij}=a_{i1}(a_{j2}+...+a_{js})+a_{i2}(a_{j3}+...+a_{js})+a_{i,s-1}a_{js}\)). Similarly, \(e_{ij}\) denotes the probability that a member of group \(i\) belongs to the same category that a member of group \(j\). By definition, \(p_{ij}+p_{ji}+e_{ij}=1\).

We define the relative advantage of group i with respect to group j, \(RA_{ij}\), as follows:

\(RA_{ij} = \frac{p_{ij}+(e_{ij}/2)}{\sum_{k \neq i}{(p_{ki}+(e_{ij}/2))}}\)

The relative advantage of group \(i\) with respect to group \(j\) is nothing more than the probability that group \(i\) dominates group \(j\), plus an equal split of the probability of a tie, divided by the sum of the corresponding probabilities that group \(i\) be dominated by some other group and the equal split of the probabilities of ties.

To obtain an overall evaluation of group \(i\) in society, we take a weighted sum of its relative advantages with respect to all other groups. That is, the relative advantage of the group \(i\) is given by:

\(RA_{i} = \sum_{j \neq i}{\lambda_{j}RA_{ij}}\)

Since the weights \(\lambda_{j}\) reflect the relevance of the different groups, it is only natural to choose them consistently with their own evaluation, ie: taking \(\lambda_{j} = RA_{j}\). In this way, each group enters the evaluation of the relative advantage of the others with the weight corresponding to its own relative advantage. This implies that we have to find a vector \(b = (b_{1}, b_{2}, \dots, b_{g}) > 0\) such that:

\(b_{i} = \frac{\sum_{j \neq i}{(p_{ij}+(e_{ij}/2))b_{j}} }{\sum_{k \neq i}{(p_{ki}+(e_{ki}/2))}}\), i=1,2,...,g

By replicating the argument in Herrero & Villar (2013) we can prove that this vector always exists, is strictly positive and unique (once normalized and provided a technical condition, known as irreducibility of the domination probability matrix, is satisfied).

This vector can be easily calculated since it corresponds to the dominant eigenvector of the following matrix:

\(M = \begin{bmatrix} (g-1)-\sum_{i \neq 1}{[p_{i1}+(e_{i1}/2)]} & p_{12}+(e_{12}/2) & \dots & p_{1g}+(e_{1g}/2) \\ p_{21}+(e_{21}/2) & (g-1)-\sum_{i \neq 2}{[p_{i2}+(e_{i2}/2)]} & \dots & p_{2g}+(e_{2g}/2) \\ \dots & \dots & \dots & \dots \\ p_{g1}+(e_{g1}/2) & p_{g2}+(e_{g2}/2) & \dots & (g-1)-\sum_{i \neq g}{[p_{ig}+(e_{ig}/2)]} \end{bmatrix} \)

The off-diagonal elements of the M matrix are the pair-wise dominance probabilities under this protocol. The elements on the diagonal tell us the probability that a randomly chosen individual from group i belongs to a category that is not worse than a randomly chosen individual from any other group. Is easy to see that the matrix M is a Perron matrix whose columns add up to (g - 1). From this it follows the existence, positivity and uniqueness (when M is irreducible) of the balanced worth vector.

Step 1. Construct the matrix \(A\) of relative frequencies as an Excel table, with groups in rows and categories in columns, checking that all rows add up to one and ordering the categories from best (first column) to worst (last column). If you fail in any of those steps you will not get what you are looking for.

Step 2. Once you have accepted the “conditions of use”, copy the body of that table (i.e. omitting the column with the names of the groups or the row with the labels of the categories) and plug it into the space provided by the application, indicating the number of rows and columns, as requested.

Step 3. Chose “Solve” and you get instantly the balanced worth vector on the left hand side of the screen. Click on the “Copy outcomes” message and plug it back into the Excel spreadsheet. That’s all, you are done.

Remark.- Don’t forget that the balanced worth vector is the eigenvector of a matrix and, as such, has a degree of freedom. So you can normalize the vector as it best suits you. The normalization provided by default is such that the sum of the elements of the balanced worth vector is equal to \(g\) (the number of groups) so that the mean worth value is equal to 1.