Mechanisms for (Mis)allocating Scientific Credit

Abstract

Scientific communities confer many forms of credit on their successful members. The motivation provided by these forms of credit helps shaping a community’s collective attention toward different lines of research. The allocation of scientific credit, however, has also been the focus of long-documented pathologies: certain research questions are said to command more credit then they deserve; and certain researchers seem to receive a disproportionate share of the credit. Here we show that each of these pathologies can actually increase the collective productivity of a community. We consider a model for the allocation of credit, in which individuals pick a project among projects of varying importance and difficulty levels, and compete to receive credit with others who choose the same project. Under the most natural allocation mechanism, in which credit is divided equally among those who succeed at a project in proportion to the project’s importance, the resulting selection of projects by self-interested, credit-maximizing individuals will in general be socially sub-optimal. However, we show that there exist ways of allocating credit both out of proportion to the true importance of the projects and out of proportion to the relative contributions of the individuals, that lead credit-maximizing individuals to achieve social optimality. These results therefore suggest how well-known forms of misallocation of scientific credit can in fact serve to channel self-interested behavior into socially optimal outcomes.

Appendix

In this Appendix we prove Theorem 2.12 we show that, with ${\varepsilon }> 0$ sufficiently small and the re-weighting of players defined by $z_i = {\varepsilon }^i$, all Nash equilibria of the resulting game are socially optimal.

Even given the informal argument above, the proof is complicated by the fact that, with positive weights on all players, their strategic reasoning is more complex than it would be under an actual ordering. To prove Theorem 2.12, we consider the relationship between the actual utilities of the re-weighted players for a given strategy vector $\mathbf {a}$, denoted ${u_i}(\mathbf {a}; \mathbf {z})$:

$$\begin{aligned} {u_i}(\mathbf {a}; \mathbf {z}) = w_{a_i} q_{a_i} \sum _{S \subseteq \{K_{a_i}(\mathbf {a}) - i\}} \dfrac{z_i}{(\sum _{h \in S} z_h) + z_i} q_{a_i}^{|S|} (1-q_{a_i})^{k_{a_i}(\mathbf {a}) - |S| -1} \end{aligned}$$

and their “ideal” utility under the order we are trying to simulate, denoted $\widehat{u_i}(\mathbf {a}; \mathbf {z})$ which is formally defined next:

Definition 5.1

• $\pi _{\small< i}(S)=\{h \in S | z_i < z_h\}$ where $S \subseteq N$ – the set of players before player i.

• $\pi _{\small>i}(S)=\{h \in S | z_i > z_h\}$ where $S \subseteq N$ – the set of players after player i.

• ${\widehat{u}}_i(\mathbf {a}; \mathbf {z})= r_{a_i}(|\pi _{\small < i}(K_{a_i}(\mathbf {a}))|)$ – the marginal contribution of player i to the social welfare.

Recalling that the projects’ weights and success probabilities are rational, let d be their common denominator. We first show that if these two different utilities are close enough with respect to d, then our approximate simulation of an order using weights will succeed:

Claim 5.2

If for every player i, project j and strategy vector $\mathbf {a}$ we have

$$\begin{aligned} \widehat{u_i}(j,a_{-i}; \mathbf {z}) - \dfrac{1}{4d^{n+1}} \le {u_i}(j,a_{-i}; \mathbf {z}) \le \widehat{u_i} (j,a_{-i}; \mathbf {z}) + \dfrac{1}{4d^{n+1}}, \end{aligned}$$

then any Nash equilibrium in the game with the weights $\{z_i\}$ is also an optimal assignment.

Proof

Consider some Nash equilibrium $\mathbf {a}$ in the game with the weights $\mathbf {z}$. Let $\mathbf {o}$ be an optimal assignment that is the result of running the greedy algorithm from Claim 2.2 with the same players’ order. In case several such optimal assignments exist, we pick $\mathbf {o}$ to be the one that shares the longest prefix with $\mathbf {a}$: $\max _{i} \forall i'\le i,a_{i'}=o_{i'}$. Assume towards a contradiction that $\mathbf {a}$ is not an optimal assignment. By definition, player i is the first player in the order that works on a different project in $\mathbf {a}$ (denote it by j) and in $\mathbf {o}$ (denote it by l). Since $\mathbf {a}$ is a Nash equilibrium with the weights $\mathbf {z}$ we have that ${u_i}(j,a_{-i}; \mathbf {z}) \ge {u_i}(l,a_{-i}; \mathbf {z})$. By applying the assumption of the claim we have that:

$$\begin{aligned} \widehat{u_i}(j,a_{-i}; \mathbf {z}) + \dfrac{1}{4d^{n+1}} \ge {u_i}(j,a_{-i}; \mathbf {z}) \ge {u_i}(l,a_{-i}; \mathbf {z}) \ge \widehat{u_i}(l,a_{-i}; \mathbf {z}) - \dfrac{1}{4d^{n+1}}. \end{aligned}$$

This implies that $\widehat{u_i}(j,a_{-i}; \mathbf {z}) + \dfrac{1}{2d^{n+1}} \ge \widehat{u_i}(l,a_{-i}; \mathbf {z})$. Also, since $\mathbf {o}$ is the result of a greedy algorithm with the same order, we have that ${\widehat{u}}_i(l,o_{-i}; \mathbf {z}) > {\widehat{u}}_i(j,o_{-i}; \mathbf {z})$. Since, $\mathbf {a}$ and $\mathbf {o}$ are identical for all players prior to player i in the order, we have that ${\widehat{u}}_i(l,a_{-i}; \mathbf {z}) > {\widehat{u}}_i(j,a_{-i}; \mathbf {z})$. Hence,

$$\begin{aligned} 0 < \widehat{ u_i}(j,a_{-i}; \mathbf {z}) - \widehat{u_i}(j,a_{-i}; \mathbf {z}) \le \dfrac{1}{2d^{n+1}} \end{aligned}$$

To complete the proof, recall that d is the common denominator of all success probabilities and weights. As both $\widehat{u_i}(j,a_{-i}; \mathbf {z})$ and $\widehat{u_i}(l,a_{-i}; \mathbf {z})$ are products of at most $n+1$ terms of common denominator d, and they are not equal, they must differ by at least $\dfrac{1}{d^{n+1}}$. Thus, it is impossible that in the Nash equilibrium $\mathbf {a}$ player i works on project j and in the optimal solution $\mathbf {o}$ on project l and a contradiction is reached. $\square$

Next, we show that it is possible to choose ${\varepsilon }$ sufficiently small such that for every player i and project j: $\widehat{u}_i(j,a_{-i}; \mathbf {z}) - \dfrac{1}{4d^{n+1}} \le {u_i}(j,a_{-i}; \mathbf {z}) \le \widehat{u_i}(j,a_{-i}; \mathbf {z}) + \dfrac{1}{4d^{n+1}}$ as required by Claim 5.2. We begin by presenting the following definitions that will be useful in simplifying the utility function:

Definition 5.3

For a strategy vector $\mathbf {a}$, a project j and $S \subseteq K_j(a_{-i})$ we define

$$\begin{aligned} \psi _i(j;S;\mathbf {a})=\dfrac{z_i}{z_i+\sum _{h\in S} z_h} \cdot q_j^{|S|+1}\cdot (1- q_j)^{ k_j(a_{-i}) - |S|}. \end{aligned}$$

This is the probability that the players in $S \cup \{i\}$ succeed at project j, the rest of the players working on j fail and player i gets the credit for succeeding at the project.

Using the definition we now have ${u_i}(j,a_{-i}; \mathbf {z}) = w_j \sum _{S \subseteq K_j(a_{-i})} \psi _i(j;S;\mathbf {a})$.

By using $\pi _{\small < i}(S)$ and $\pi _{\small >i}(S)$ we can break up the player’s utility in the following manner:

$$\begin{aligned} {u_i}(j, a_{-i}; \mathbf {z}) =w_j \cdot \Bigg ( \sum _{S \subseteq \pi _{\small >i}(K_j(\mathbf {a}))} \psi _i(j;S;\mathbf {a}) ~~~+\sum _{\begin{array}{c} S \subseteq K_j( a_{-i}) \\ S \cap \pi _{\small < i}(K_j(\mathbf {a}))\ne \emptyset \end{array} } \psi _i(j;S;\mathbf {a}) \Bigg ) \end{aligned}$$

(3)

This is a convenient representation of a player’s utility since it partitions the successful player sets into two types:

• 1. $S \subseteq \pi _{\small >i}(K_j(\mathbf {a}))$: for such a set S, player i’s weight is dominant in S, and hence she gets most of the utility.

• 2. $S \subseteq K_j( a_{-i})$ and $S \cap \pi _{\small < i}(K_j(\mathbf {a}))\ne \emptyset$: for such a set S, player i’s weight is dominated, and hence she gets only a very small fraction of the utility.

In the next two lemmas we bound player i’s utility with respect to each of these types separately. The bounds are later combined to bound ${u_i}(j, a_{-i}; \mathbf {z})$.

Lemma 5.4

$\dfrac{1}{1+2 {\varepsilon }} {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) \le w_j \cdot \sum _{S \subseteq \pi _{\small >i}(K_j(\mathbf {a}))} \psi _i(j;S;\mathbf {a}) \le {\widehat{u}}_i(j, a_{-i}; \mathbf {z})$.

Proof

We first show that $w_j \cdot \sum _{S \subseteq \pi _{\small >i}(K_j(\mathbf {a}))} \psi _i(j;S;\mathbf {a}) \le {\widehat{u}}_i(j, a_{-i}; \mathbf {z})$. To do this, we write an alternative expression for ${\widehat{u}}_i(j, a_{-i}; \mathbf {z})$:

$$\begin{aligned} {\widehat{u}}_i(j, a_{-i}; \mathbf {z})&=\Bigg ( \underbrace{w_jq_j(1-q_j)^{|\pi _{\small< i}(K_j(\mathbf {a}))|}}_{=r_j(|\pi _{\small < i}(K_j(\mathbf {a}))|)} \Bigg ) \\&\cdot \Bigg ( \underbrace{\sum _{S \subseteq \pi _{\small>i}(K_j(\mathbf {a}))} q_j^{|S|}\cdot (1- q_j)^{ |\pi _{\small>i}(K_j(\mathbf {a}))| - |S|}}_{=1} \Bigg ) \\&= w_j \sum _{S \subseteq \pi _{\small >i}(K_j(\mathbf {a}))} q_j^{|S|+1}\cdot (1- q_j)^{k_j(a_{-i}) - |S|} \end{aligned}$$

The resulting expression is an upper bound on $w_j\sum _{S \subseteq \pi _{\small >i}(K_j(\mathbf {a}))} \psi _i(j;S;\mathbf {a})$ since it assumes that $z_i>0$ and the other weights are 0.

We now turn to show that $\dfrac{1}{1+2 {\varepsilon }} {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) \le w_j \cdot \sum _{S \subseteq \pi _{\small >i}(K_j(\mathbf {a}))} \psi _i(j;S;\mathbf {a})$. By the definition of the weights, we have that for all $h\in \pi _{\small >i}(K_j(\mathbf {a}))$, $z_h \le {\varepsilon }z_i$. This allows us to bound the weight coefficients $\dfrac{z_i}{z_i + \sum _{h \in S} z_h}$ in ${u_i}(j, a_{-i}; \mathbf {z})$:

$$\begin{aligned} \forall S \subseteq \pi _{\small>i}(K_j(\mathbf {a})):~\dfrac{z_i}{z_i +\sum _{h\in S} z_h}&\ge \dfrac{z_i}{z_i + \sum _{h\in \pi _{\small>i}(K_j(\mathbf {a}))} z_h} \\&\ge \dfrac{z_i}{z_i + \sum _ {h=1}^{ |\pi _{\small>i}(K_j(\mathbf {a}))|} {\varepsilon }^h z_i } \\&= \dfrac{1}{1 + \sum _{h=1}^{|\pi _{\small>i}(K_j(\mathbf {a}))|} {\varepsilon }^h} > \dfrac{1}{1+ 2{\varepsilon }} \end{aligned}$$

where the last inequality holds for ${\varepsilon }< 0.5$. Hence we have that $\dfrac{1}{1+2 {\varepsilon }} {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) \le w_j \cdot \sum _{S \subseteq \pi _{\small >i}(K_j(\mathbf {a}))} \psi _i(j;S;\mathbf {a})$. $\square$

Lemma 5.5

$\sum _{\begin{array}{c} S \subseteq K_j(a_{-i}) \\ S \cap \pi _{\small < i}(K_j(\mathbf {a}))\ne \emptyset \end{array} } \psi _i(j;S;\mathbf {a}) \le \dfrac{{\varepsilon }}{1+ {\varepsilon }}$.

Proof

Observe that since for each of the sets S included in the sum, $S \cap \pi _{\small < i}(K_j(\mathbf {a}))\ne \emptyset$, then in each set S there exists at least a single player $h\in S$ such that $z_h>z_i$. This implies the following bound on the weight coefficients in $\psi _i(j;S;\mathbf {a})$:

$$\begin{aligned} \dfrac{z_i}{z_i+\sum _{h\in S} z_h} \le \dfrac{z_i}{z_i+\min _{\{h\in S | z_h>z_i\}} z_h} \le \dfrac{z_i}{z_i+ \dfrac{z_i}{{\varepsilon }}} = \dfrac{{\varepsilon }}{1+{\varepsilon }} \end{aligned}$$

The proof is completed by observing that $\sum _{\begin{array}{c} S \subseteq K_j(a_{-i}) \\ S \cap \pi _{\small< i}(K_j(\mathbf {a}))\ne \emptyset \end{array} } q_j^{|S|+1} \cdot (1-q_j)^{k_j(a_{-i}) - |S|} <1$. $\square$

It is not hard to see that given Eq. 3 and Lemmas 5.4 and 5.5 the following bounds hold:

Corollary 5.6

For every strategy vector $\mathbf {a}$, player i and project j:

$$\begin{aligned} \dfrac{1}{1+2 {\varepsilon }} {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) \le {u_i}(j, a_{-i; \mathbf {z}}) \le {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) +w_j \dfrac{{\varepsilon }}{1+ {\varepsilon }}. \end{aligned}$$

We can now use the previous corollary to compute a value of ${\varepsilon }$ for which the assumptions of Lemma 5.2 hold:

Lemma 5.7

Let l be a project with maximal weight ($l \in \arg \max _{j \in M} w_j$). For ${\varepsilon }\le \dfrac{1}{8d^{n+1}w_l}$ we have that: ${\widehat{u}}_i(j,a_{-i}; \mathbf {z}) -\dfrac{1}{4d^{n+1}} \le {u_i}(j,a_{-i}; \mathbf {z}) \le \widehat{u}_i(j,a_{-i}; \mathbf {z}) + \dfrac{1}{4d^{n+1}}$.

Proof

Note that by the lower bound on ${u_i}(j, a_{-i}; \mathbf {z})$ from Corollary 5.6 and the choice of l and the choice of ${\varepsilon }$ we have that

$$\begin{aligned} {u_i}(j, a_{-i}; \mathbf {z})&\ge \dfrac{1}{1+2 {\varepsilon }} {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) = {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) - \dfrac{2 {\varepsilon }}{1+2 {\varepsilon }} {\widehat{u}}_i(j, a_{-i}; \mathbf {z})\\&\ge {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) - 2 {\varepsilon }\cdot w_l. \\&\ge {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) - 2 \cdot \dfrac{1}{8d^{n+1}w_l} \cdot w_l \\&\ge {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) - \dfrac{1}{4d^{n+1}} \end{aligned}$$

Similarly, by the upper bound on ${u_i}(j, a_{-i}; \mathbf {z})$ from Corollary we have that ${u_i}(j, a_{-i}; \mathbf {z}) \le {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) +w_j \dfrac{{\varepsilon }}{1+ {\varepsilon }} \le {\widehat{u}}_i(j, a_{-i}; \mathbf {z}) + \dfrac{1}{4d^{n+1}}$. $\square$

This completes our proof of Theorem 2.12 showing that there exists ${\varepsilon }$ for which any equilibrium of the game with player-weights $\{z_i={\varepsilon }^i\}$ is an optimal solution in the unweighted game.