How to rank rankings? Group performance in multiple-prize contests

Abstract

When groups of individuals compete in several multiple-prize contests, the performance of a group is a vector of ordered categories. As the prizes are aimed at ranking the participants, group performances are not trivially comparable. This note provides a theoretical discussion on how to rank group performances. In order to do so, I draw from the parallel that this problem has with the formally similar problem of measuring inequality. I describe three alternatives that generate partial orders for group performances. I define partial orders based on the first- and second-order dominance, two classes of performance measures, and two sets of basic transformations, and I prove equivalence theorems between them. I apply these theoretical results to discuss several sports ranking problems.

Appendix

Proof Theorem 1

The proof demonstrates the following chain of implications:

$$\begin{aligned} \mathbf {y} \ge _1 \mathbf {y'}\Rightarrow \mathbf {y} \ge _T \mathbf {y'} \Rightarrow \mathbf {y} \ge _I \mathbf {y'} \Rightarrow \mathbf {y} \ge _1 \mathbf {y'}. \end{aligned}$$

(i) $\mathbf {y} \ge _1 \mathbf {y'}\Rightarrow \mathbf {y} \ge _T \mathbf {y'}$.

Suppose $\mathbf {y} \ge _1 \mathbf {y'}$, I show how to get $\mathbf {y}$ from $\mathbf {y'}$ by T-transformation. I define recursively the n-tuple $\mathbf {y}(k)$ as

$$\begin{aligned} \mathbf {y}(k)=\mathbf {y}(k-1)+(Y_k-Y'_k)\mathbf {t}_k, \quad \forall k=1,..,n \end{aligned}$$

(A1)

with $\mathbf {y}(0)=\mathbf {y}'$. Notice that the n-tuple $\mathbf {y}(k)$ is not necessarily a performance. $\square$

Lemma 1

The cumulative of $\mathbf {y}(k)$, given by $Y(k)_i=\sum _{j=1}^{i} y(k)_j, \forall i,k=1,...,n$, is

$$\begin{aligned} Y(k)_i= {\left\{ \begin{array}{ll} Y_i, &{} \quad \text {if}\;\ i\le k \\ Y'_i, &{} \quad \text {if}\;\ i>k. \end{array}\right. } \end{aligned}$$

Proof

The proof is by induction. For $k=1$, I have $\mathbf {y}(1)=\mathbf {y}'+(Y_1-Y'_1)\mathbf {t}_1$. First, I have that $Y_1=y_1$ and $Y'_1=y'_1$, so $Y(1)_1=y(1)_1=Y_1$. Second, a transformation $\mathbf {t}_1$ has no effect on the cumulatives for $i\ge 2$. Consequently, $Y(1)_i=Y'_i$ for $i>1$. In the next step of induction, I assume that the relation is true for $k-1$ and I prove it for k. I have that $Y(k)_i=Y(k-1)_i=Y_i$ for $i<k$, where in the last equality I use the induction hypothesis. Given that a transformation $\mathbf {t}_k$ has no effect on the cumulatives for $i>k$, $Y(k)_i=Y(k-1)_i=Y'_i$ for $i>k$, where in the last equality I use the induction hypothesis. For k, I have that $Y(k)_k=Y(k-1)_k+(Y_k-Y'_k)=Y'_k$, where in the last equality I use the induction hypothesis. This completes the proof. $\square$

From the Lemma, $\mathbf {Y}(n)=\mathbf {Y}$, which implies that $\mathbf {y}(n)=\mathbf {y}$. As $\mathbf {y}(0)=\mathbf {y}'$, I use the Definition A1 to write $\mathbf {y}=\mathbf {y}'+\sum _{k=1}^n(Y_k-Y'_k)\mathbf {t}_k$. As $\mathbf {y} \ge _1 \mathbf {y'}$, $(Y_k-Y'_k)\ge 0,\forall k=1,..,n$. Hence $\mathbf {y}$ is obtained from $\mathbf {y}'$ through a T-transformation.

(ii) $\mathbf {y} \ge _T \mathbf {y'}\Rightarrow \mathbf {y} \ge _I \mathbf {y'}$.

Assume that $\mathbf {y}$ is obtained from $\mathbf {y'}$ through a T-transformation. Then there exists an index set $\mathcal {T}$ such that

$$\begin{aligned} \mathbf {y}= & {} \mathbf {y'}+\sum _{i\in \mathcal {T}}\mathbf {t_{i}} . \end{aligned}$$

As the performance measure is linear:

$$\begin{aligned} P(\mathbf {y;v})= & {} P(\mathbf {y';v})+\sum _{i\in \mathcal {T}} (v_i-v_{i+1}). \end{aligned}$$

For any $\mathbf {v}\in V_I$, $(v_i-v_{i+1})\ge 0,\forall i$. I conclude that $P(\mathbf {y;v})\ge P(\mathbf {y';v}), \forall \mathbf {v}\in V_I$.

(iii) $\mathbf {y} \ge _I \mathbf {y'}\Rightarrow \mathbf {y} \ge _1 \mathbf {y'}$.

I define $\mathbf {v}(k)=\mathbf {e_1}+...+\mathbf {e_k} \in V_I, \forall k$. I have that $P(\mathbf {y;v}(k))=\sum _{i=1}^k y_i=Y_k$. The implication is direct.

Proof Theorem 2

The proof demonstrates the following chain of implications:

$$\begin{aligned} \mathbf {y} \ge _2 \mathbf {y'}\Rightarrow \mathbf {y} \ge _S \mathbf {y'} \Rightarrow \mathbf {y} \ge _C \mathbf {y'} \Rightarrow \mathbf {y} \ge _2 \mathbf {y'}. \end{aligned}$$

(i) $\mathbf {y} \ge _2 \mathbf {y'}\Rightarrow \mathbf {y} \ge _S \mathbf {y'}$.

Suppose $\mathbf {y} \ge _2 \mathbf {y'}$, I show how to get $\mathbf {y}$ from $\mathbf {y'}$ by a sequence of S-transformations. For that, I notice that second-order dominance in performances is equivalent to first-order dominance in the cumulatives of the same performances, i.e., $\mathbf {y} \ge _2 \mathbf {y'}\Longleftrightarrow \mathbf {Y} \ge _1 \mathbf {Y'}$. The proof of this equivalence is direct by definition. From Theorem 1, I know that $\mathbf {Y}$ is obtained from $\mathbf {Y}'$ through a T-transformation, such that $\mathbf {Y}=\mathbf {Y'}+\sum _{i\in \mathcal {T}}\mathbf {t_{i}}$.

The proof consists in showing that a T-transformation for a cumulative performance $\mathbf {Y}$ is equivalent to an S-transformation of the underlying performance $\mathbf {y}$. For that, I prove the following Lemma. $\square$

Lemma 2

If $\mathbf {Y(i)}=\mathbf {Y'}+\mathbf {t_i}$ then the underlying performance $\mathbf {y(i)}$ can be written as

$$\begin{aligned} \mathbf {y(i)}= {\left\{ \begin{array}{ll} \mathbf {y'}+\mathbf {s_{ii+1}}, &{} \quad \text {if}\;\ i<n \\ \mathbf {y'}+\mathbf {t_n}, &{} \quad \text {if}\;\ i=n. \end{array}\right. } \end{aligned}$$

Proof

For $i<n$, $\mathbf {y'}+\mathbf {t_i}-\mathbf {t_{i+1}}$. The cumulative is thus $Y'_j$ for $j<i$, $Y(i)'_i=Y'_i+1$, $Y(i)'_{i+1}=Y'_i-1$, and $Y'_j$ for $j>i+1$. For $i=n$ the proof is direct. $\square$

From the lemma, if $\mathbf {Y}$ is obtained from $\mathbf {Y}'$ through a T-transformation, then $\mathbf {y}$ is obtained from $\mathbf {y}'$ through an S-transformation.

(ii) $\mathbf {y} \ge _S \mathbf {y'}\Rightarrow \mathbf {y} \ge _C \mathbf {y'}$.

Assume that $\mathbf {y}$ is obtained from $\mathbf {y'}$ through an S-transformation. Then there exist $\mathcal {T}$ and $\mathcal {S}$ index sets such that

$$\begin{aligned} \mathbf {y}= & {} \mathbf {y'}+\sum _{i\in \mathcal {T}}\mathbf {t_{i}}+\sum _{{ij}\in \mathcal {S}}\mathbf {s_{ij}}. \end{aligned}$$

The performance function is:

$$\begin{aligned} P(\mathbf {y;v})= & {} P(\mathbf {y';v})+\sum _{i\in \mathcal {T}} (v_i-v_{i+1})+\sum _{{ij}\in \mathcal {S}}[(v_i-v_{i+1})-(v_j-v_{j+1})]. \end{aligned}$$

For any $\mathbf {v}\in V_C$ I have the following inequalities. First, $(v_i-v_{i+1})\ge 0,\forall i$ since $\mathbf {v}\in V_C\Longrightarrow \mathbf {v}\in V_I$. Second, $[(v_i-v_{i+1})-(v_j-v_{j+1})]\ge 0, \forall i<j$ since $\mathbf {v}\in V_C\Longrightarrow (v_i-v_{i+1})\ge (v_{i+1}-v_{i+2})\ge ...\ge (v_j-v_{j+1})$. I conclude that $P(\mathbf {y;v})\ge P(\mathbf {y';v}), \forall \mathbf {v}\in V_C$.

(iii) $\mathbf {y} \ge _C \mathbf {y'}\Rightarrow \mathbf {y} \ge _2 \mathbf {y'}$.

I define $\mathbf {v}(k) \in V_C, \forall k$ as

$$\begin{aligned} \mathbf {v}(k)_i= {\left\{ \begin{array}{ll} k+1-i, &{} \text {if}\ i\le k \\ 0, &{} \text {if}\ i>k. \end{array}\right. } \end{aligned}$$

I have that $P(\mathbf {y;v}(k))=\sum _{i=1}^k (k+1-i)y_i=\sum _{i=1}^k Y_i=Z_k$. The implication is direct.