Are the following two definitions of Borda winner equivalent?

Question

The Borda count is a method used to determine the winner object where people rank objects. For instance, imagine each person ranking 3 objects. The highest ranked object gets 2 points, the second gets 1 and the last gets 0 points. Generalize to $m$ objects. The Borda winner is that object which has the highest summation of points.

I've seen another definition in the context of pairwise comparisons. Here again, people rank the objects, but we don't get to see their rankings. For any pair of objects A and B, we can only know the fraction of people $P_{AB}$ that rank object A over object B. The Borda winner is defined as that object i for which $\sum\limits_{j\neq i}P_{ij}$ is highest.

Are the above two definitions equivalent? I'm not able to prove their equivalence, and neither can I find a counterexample where they aren't the same.

Answer 1

Yes the two definitions are equivalent.

Let the set of objects be $T := \{t_1, \dots, t_m\}$ and the set of individuals be $N := \{1,\dots,n\}$.

Also for any agent $i\in N$, let $P_i$ represent $i$'s preference relation so that $t_k~P_i~t_j$ means that object $k$ is ranked higher by $i$ than object $j$.

If you follow the first definition, you can write the score of any object $t_j$ as

$S(t_j) := \sum_{i=1}^n \#\{t_k \in T~|~ t_j~P_i~t_k\}$,

Where $\# A$ denotes the cardinality of any set $A$. You can rewrite this as

$S(t_j) = \sum_{i=1}^n \sum_{k\neq j} \begin{cases} 1, ~~~~\text{ if }~~~~~ t_j~P_i~t_k,\\ 0, ~~~~~\text{ if }~~~~ t_k~P_i~t_j \end{cases}$

Inverting the order of summation yields

$S(t_j) = \sum_{k\neq j} \sum_{i=1}^n \begin{cases} 1, ~~~~\text{ if }~~~~~ t_j~P_i~t_k,\\ 0, ~~~~~\text{ if }~~~~ t_k~P_i~t_j \end{cases}$

Now you see that the last term $\sum_{i=1}^n \begin{cases} 1, ~~~~\text{ if }~~~~~ t_j~P_i~t_k,\\ 0, ~~~~~\text{ if }~~~~ t_k~P_i~t_j \end{cases}$ is the number of people who prefer $j$ to $k$. Given your definition of $P_{jk}$, this is precisely $n*P_{jk}$. Thus we have

$S(t_j) = \sum_{k\neq j} n*P_{jk}$.

and the two definitions are equivalent because

\begin{align} S(t_a) > S(t_b) \text{ for all } t_b\neq t_a ~~~~~ \Leftrightarrow ~~~~~ \sum_{j\neq a} n*P_{jk} > \sum_{j\neq b} n*P_{jk} ~~~~ \Leftrightarrow ~~~~~ \sum_{j\neq a} P_{jk} > \sum_{j\neq b} P_{jk} \end{align}