Individual selection criteria for optimal team composition

Abstract

In this paper, we derive necessary and sufficient conditions on team based tasks in order for a selection criterion applied to individuals to produce optimal teams. We assume only that individuals have types and that a team’s performance depends on its size and the type composition of its members. We first derive the selection principle which states that if a selection criterion exists, it must rank types by homogeneous team performance, the performance of a team consisting only of that type. We then prove that a selection criterion exists if and only if replacing the team’s lowest ranked type, as measured by homogeneous team performance, with a higher ranked type  increases team performance. Finally, we show that the replace the lowest ranked property rules out most common types of team complementarities, including benefits to diverse types and types that fill structural holes.

Appendix

Proof of Theorem 1 (Selection Principle):

Let R be a selection criterion for $V_{K}.$ We need to show that for any $c\in C$ and $c^{\prime }\in C$, $R(c)\ge R(c^{\prime })$ iff $h_{K}(c)\ge h_{K}(c^{\prime }).$ First, for any $c\in C$ and $c^{\prime }\in C$ such that $R(c)\ge R(c^{\prime }),$ we show $h_{K}(c)\ge h_{K}(c^{\prime }).$ Let $A=\{\overset{K}{\overbrace{c,...,c}},\overset{K}{ \overbrace{c^{\prime },...,c^{\prime }}}\}$ be a set of candidates. Since $R(c)\ge R(c^{\prime })$, $\{\overset{K}{\overbrace{c,...,c}}\}$ is a K highest scoring team from A. Since R is a selection criterion for $V_{K},$ team $\{\overset{K}{\overbrace{c,...,c}}\}$ maximizes $V_{K}$ over all possible teams of size K from A which includes team $\{\overset{K}{ \overbrace{c^{\prime },...,c^{\prime }}}\}.$ Therefore, $V_{K}(c,...,c)\ge V_{K}(c^{\prime },...,c^{\prime }).$

We now consider the other direction. For any $c\in C$ and $c^{\prime }\in C$ such that $h_{K}(c)\ge h_{K}(c^{\prime }),$ we show $R(c)\ge R(c^{\prime }).$ We prove this by contradiction. Suppose $R(c)<R(c^{\prime })$ instead. Consider again $A=\{\overset{K}{\overbrace{c,...,c}},\overset{K}{\overbrace{ c^{\prime },...,c^{\prime }}}\}.$ Now team $\{\overset{K}{\overbrace{ c^{\prime },...,c^{\prime }}}\}$ is a K highest scoring team from set A while team $\{\overset{K}{\overbrace{c,...,c}}\}$ is not. Since R is a selection criterion for $V_{K}$, $\{\overset{K}{\overbrace{c^{\prime },...,c^{\prime }}}\}$ maximizes $V_{K}$ over all possible teams of size K from A while team $\{\overset{K}{\overbrace{c,...,c}}\}$ does not. This implies that $V_{K}(c^{\prime },...,c^{\prime })>$ $V_{K}(c,...,c)$ or $h_{K}(c^{\prime })>h_{K}(c)$. A contradiction.

Proof of Claim 1

First, we prove necessity. Given $(c_{1},...,c_{K})\in C^{K}.$ Without loss of generality (WLOG), assume that type $c_{K}$ has the lowest R score and it is being replaced by another type $c^{\prime }\in C$. We need to show that $R(c^{\prime })>R(c_{K})$ implies $V_{K}(c_{1},...,c_{K-1,}c^{\prime })>V_{K}(c_{1},...,c_{K})$ and $R(c^{\prime })=R(c_{K})$ implies $V_{K}(c_{1},...,c_{K-1,}c^{\prime })=V_{K}(c_{1},...,c_{K}).$ Consider the set of candidates $\{c_{1},...,c_{K-1},c_{K},c^{\prime }\}.$ If $R(c^{\prime })>R(c_{K}),$ then team $\{c_{1},...,c_{K-1,}c^{\prime }\}$ is a K highest R scoring team but team $\{c_{1},...,c_{K}\}$ is not. Since R is a selection criterion for $V_{K}$, we have $V_{K}(c_{1},...,c_{K-1,}c^{\prime })>V_{K}(c_{1},...,c_{K}).$ If $R(c^{\prime })=R(c_{K})$, then both teams $\{c_{1},...,c_{K-1,}c^{\prime }\}$ and $(c_{1},...,c_{K})$ are K highest R scoring teams. Again R being a selection criterion implies that they both maximize team performance. Therefore $V_{K}(c_{1},...,c_{K-1,}c^{\prime })=V_{K}(c_{1},...,c_{K}).$

Now we prove sufficiency. We show that if $V_{K}$ satisfies the RTLR property in R,  then R is a selection criterion for $V_{K}.$ This is proven in two parts. First, we show that given any set of candidates, any team that maximizes team performance $V_{K}$ consists of K members whose types have the K highest R scores of the candidate set. Let $\{c_{1}^{b},...,c_{K}^{b}\}$ be a team that maximizes team performance, but some member’s type scores lower than all K highest R scores. Then there is a member $c^{\prime }$ who has one of the K highest scores. So $R(c^{\prime })>R(c_{K}^{b})$ if WLOG we assume $c_{K}^{b}$ has the lowest R score among $\{c_{1}^{b},...,c_{K}^{b}\}$. By the RTLR property, replacing $c_{K}^{b}$ with $c^{\prime }$ in team $\{c_{1}^{b},...,c_{K}^{b}\}$ will strictly improve team performance. A contradiction.

Second, we show that any team with the K highest scores maximizes team performance. Let $\{c_{1}^{h},...,c_{K}^{h}\}$ be a team with the K highest scores from a candidate set A, but it doesn’t maximize team performance. By what we’d just shown in the first part, if team $\{c_{1}^{b},...,c_{K}^{b}\}$ maximizes team performance, then $\{c_{1}^{b},...,c_{K}^{b}\}$ is also a K highest scoring team from A. By construction, the only members from team $\{c_{1}^{h},...,c_{K}^{h}\}$ that can be different from the members of team $\{c_{1}^{b},...,c_{K}^{b}\}$ are those that have the lowest R score among the K highest R scores of the candidate set. That means (assuming Kth team member is the lowest scoring one) $c_{K}^{h}$ and $c_{K}^{b}$ may be different types, but $R(c_{K}^{h})=R(c_{K}^{b})$. By the RTLR property, $V_{K}(c_{1}^{h},...,c_{K}^{h})=V_{K}(c_{1}^{b},...,c_{K}^{b})$. Therefore $\{c_{1}^{h},...,c_{K}^{h}\}$ maximizes team performance. A contradiction.

We now consider the strictly increasing property that is stronger than the RTLR property.

Definition 4

A team performance function $V_{K}$ is strictly increasing in $R:C\rightarrow \Re$ if (a) for any $(c_{1},...,c_{K})\in C^{K}$and $(c_{1}^{\prime },...,c_{K}^{\prime })\in C^{K}$ such that $R(c_{i})=R(c_{i})$ for all i, $V_{K}(c_{1},...,c_{K})=V_{K}(c_{1}^{\prime },...,c_{K}^{\prime })$ and (b) for any $(c_{1},...,c_{K})\in C^{K}$ and $(c_{1}^{\prime },...,c_{K}^{\prime })\in C^{K}$ such that $R(c_{i})\ge R(c_{i}^{\prime })$ for all i and the strict inequality holds for at least one i, $V_{K}(c_{1},...,c_{K})>V_{K}(c_{1}^{\prime },...,c_{K}^{\prime })$.

Claim 6

If $V_{K}$ is strictly increasing in a scoring rule R, then R is a selection criterion for $V_{K}.$

Proof of Claim 6:

Let $\{c_{1},...,c_{K}\}$ be a team consisting of K members with the highest R-scores from a given set of candidates A. Then for any other team of size K that contains at least one member whose R-score is not among the top K scores of A, $(c_{1}^{\prime },...,c_{K}^{\prime }),$ we have $R(c_{i})\ge R(c_{i}^{\prime })$ for all i and the strict inequality holds for at least one i. By Condition (b) of $V_{K}$ being strictly increasing in R, we have $V_{K}(c_{1},...,c_{K})>V_{K}(c_{1}^{\prime },...,c_{K}^{\prime }).$ This means that a team of K members that maximizes team performance must consist of K members with the highest R -scores from A. On the other hand, any two teams, each consisting of K members with the highest R-scores, must have the same team performance by Condition (a) of $V_{K}$ being strictly increasing in R. This means that any team consisting of K members with the highest R-scores must also maximize team performance. $\square$

Combining Claim 6 with the selection principle (Theorem 1), produces the following corollary:

Corollary 1

If $V_{K}$ is strictly increasing in homogeneous team performance $h_{K}$, then $h_{K}$ is a selection criterion.

To see why $V_{K}$ being strictly increasing in the homogeneous team performance $h_{K}$ is not necessary for a selection criterion, consider a special case of our coordinated additive performance function from Example 2 where $C=\{{1,2,3\}.}$ Let $K=2$. Since $V_{2}(3,1)=0<2=V_{2}(2,1),$ $V_{2}$ is not strictly increasing in homogeneous team performance ($h_{2}(c)=2c$) but given any population of possible types, the best team consists of the two types with the highest homogeneous team performances. In other words, $h_{K}$ is a selection criterion for this special case.

The key insight is as follows: A comparison between team performance of a team with types $\{1,3\}$ and a team with types $\{1,2\}$ is only required when all three types $\{1,2,3\}$ are possible candidates. In those cases, the optimal team consists of types $\{2,3\}$. Thus, the assumption of being strictly increasing in $h_{K}$ is too strong. What is necessary is that replacing the lowest ranked with a higher ranked type must increase team performance - the replace the lowest ranked (RTLR) property.

It is possible to construct any number of classes of examples in which a selection criterion exists but the strictly increasing property is violated. We present one class here. Assume that individuals possess knowledge sets and that team performance equals the product of the size of the union of the individuals’ sets and the size of intersection of those sets. Thus, holding total knowledge constant and increasing the intersection of knowledge increases team performance as does holding the intersection constant and adding knowledge to any one team member. These assumptions are consistent with a team that benefits from more knowledge but also benefits from common understandings.

Example 7

Communicative Knowledge: Individual i has knowledge set $c_{i}\subset \Omega$, a collection of relevant knowledge. The performance of a team of size K equals:

$$\begin{aligned} V_{K}(c_{1},c_{2},\dots ,c_{K})=\left| \bigcup _{i=1}^{K}c_{i}\right| \cdot \left| \bigcap _{i=1}^{K}c_{i}\right| \end{aligned}$$

If the types satisfy set containment, for any i and j, either $c_{i}\subset c_{j}$ or $c_{j}\subset c_{i}$, then individual performance is a selection criterion and $V_{K}$ is also strictly increasing in it. It is straightforward to construct examples that do not assume set containment in which a selection criterion exists and in which strictly increasing property is also violated. Before we present a general class of such examples, we consider one instance of that class. Let $\Omega =\{b_{1},b_{2},a_{1},...,a_{11}\}$ and $C=\{c_{1},c_{2},c_{3}\}$ where $c_{1}=\{b_{1},b_{2},a_{4},...,a_{11}\}$, $c_{2}= \{b_{1},b_{2},a_{2},...,a_{6}\}$, $c_{3}=\{b_{1},b_{2},a_{1},a_{2},a_{3}\}$. Let $K=2.$ The homogeneous team performance $h_{2}(c)=\left| c\right| ^{2}$ is a selection criterion and it ranks $c_{1}$, $c_{2}$ and $c_{3}$ in a strictly descending order, but team performance is not strictly increasing in $h_{2}$ because $V_{2}(c_{1},c_{3})=(13)(2)<(8)(4)=V_{2}(c_{2},c_{3}).$

We now construct a class of examples of $\Omega$ and C in which a selection criterion exists but the strictly increasing property is violated. The class of examples is based on the Fibonacci series. Construct a collection of disjoint sets $\Omega =\{B\}\cup \{S_{i}\}_{i=1}^{T+1}$. Let set $S_{1}$ contain one element, $a_{1}$, set $S_{2}$ contain two elements, $a_{2}$ and $a_{3}$. For $j>2$, let set $S_{j}$ contain as a number of elements equal to the sum of the two previous sets, $\mid S_{j}\mid =\mid S_{j-1}\mid +\mid S_{j-2}\mid$ Let set B contain the same number of elements as $S_{T}$. Let type $c_{T}$ consist of the set B, and the sets $S_{1}$and $S_{2}$, and type $c_{T-1}$ consist of the sets B, $S_{2}$, and $S_{3}$. Construct the higher ranked types in similar fashion, $c_{T-k}$ consists of sets B, $S_{k+1}$, and $S_{k+2}$. The set of possible types $C=\{c_{1},...,c_{T}\}.$

We show that for $K=2,$ the communicative knowledge team performance function $V_{2}$ on this class of examples satisfies the RTLR property. Let $f_{\ell }$ equal the cardinality of the set $S_{\ell }$. Let b denote the cardinality of the set B. Recall that $b=f_{T}$. To prove that the RTLR property holds, it suffice to show (i) and (ii).

• (i) $V_{2}(c_{j},c_{j})<V_{2}(c_{j-1},c_{j})$. By construction, $V_{2}(c_{j},c_{j})=(b+f_{T-j+1}+f_{T-j+2})^{2}$, and $V_{2}(c_{j-1},c_{j})=(b+f_{T-j+1}+f_{T-j+2}+f_{T-j+3})(b+f_{T-j+2})$. Some simplification leads to $V_{2}(c_{j-1},c_{j})-V_{2}(c_{j},c_{j})=bf_{T-j+2}+f_{T-j+2}^{2}-f_{T-j+1}^{2}>0$ because by construction, $f_{T-j+2}>f_{T-j+1}$.

• (ii) $V_{2}(c_{j},c_{j})<V_{2}(c_{i},c_{j})$ for $i<j-1$. As above, $V_{2}(c_{j},c_{j})=(b+f_{T-j+1}+f_{T-j+2})^{2}=(b+f_{T-j+3})^{2}$. For $i<j-1$, $V_{2}(c_{i},c_{j})=(b+f_{T-j+1}+f_{T-j+2}+f_{T-i+1}+f_{T-i+2})b=(b+f_{T-j+3}+f_{T-i+3})b$. Again with some algebra and note that $i<j-1$, we have $V_{2}(c_{i},c_{j})-V_{2}(c_{j},c_{j})>f_{T-j+2}b+f_{T-j+3}(b-f_{T-j+3}).$ Given $i<j-1$, it follows that $j\ge 3$, which implies $b\ge f_{T-j+3}$ for any $j\ge 3.$ So $V_{2}(c_{i},c_{j})-V_{2}(c_{j},c_{j})>0$.

To show that the strictly increasing property is violated, it suffices to show that $V_{2}(c_{2},c_{3})>V_{2}(c_{ 1},c_{3})$. $V_{2}(c_{2},c_{3})=(b+f_{T-2}+f_{T-1}+f_{T})(b+f_{T-1})=(b+2f_{T})(b+f_{T-1})$ and $V_{2}(c_{1},c_{3})=(b+f_{T-2}+f_{T-1}+f_{T}+f_{T+1})b=(b+2f_{T}+f_{T+1})b$. It can be simplified that $V_{2}(c_{2},c_{3})-V_{2}(c_{1},c_{3})=f_{T}(2f_{T-1}-b).$ Plug in $b=f_{T}$ and note that $f_{T}=$ $f_{T-1}+f_{T-2}.$ $V_{2}(c_{2},c_{3})-V_{2}(c_{1},c_{3})=f_{T}(f_{T-1}-f_{T-2})>0.$