Award scheme in productivity-based contests

In this paper, we study the award scheme in the productivity-based contests in which the solvers’ performance is the product of the effort and their heterogeneous abilities. We consider the cases in which the seeker cares about the expected average quality (expected maximum quality) of solutions and the number of solvers is fixed. Our results show that the risk types of solvers, heterogeneous winning probability, and relative heterogeneity among the solvers can affect the award scheme and award allocation manner. Our study can provide some management insights for the seeker and enrich the heterogeneous contest theory.

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Authors and Affiliations

1. School of Management, Zhejiang University, Hangzhou, 310058, China

   Xu Tian

Corresponding author

Correspondence to Xu Tian.

Conflict of interest

None of the authors have a conflict of interest to disclose.

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Appendix A: Appendix

1.1 Appendix A.1: Proof of Proposition 1

(i) Suppose that all solvers except solver i have performance according to the best-response performance function \(q^{*}(a_{i})\), and \(q^{*}(a_{i})\) is continuously differentiable and increasing in the ability level \(a_{i}\). Then,

where \(q_{i}^{-1}(\cdot )\) and \(q^{*,-1}(\cdot )\) are the reverse functions of \(q_{i}(\cdot )\) and \(q^{*}(\cdot )\), respectively. Solver i with ability \(a_{i}\) will solve Problem (1) to determine her performance \(q_{i}\). In equilibrium, \(q_{i}(\cdot )=q^{*}(\cdot )\) and \(q^{*}(a_{i})=e^{*}(a_{i})\cdot a_{i}\). By the first-order condition of Problem (1), we obtain

where \(S_{j}^{n}(a_{i})=\frac{(n-1)!}{(j-1)!(n-j)!}F(a_{i})^{n-j-1}(1-F(a_{i}))^{j-2}f(a_{i})((n-j)(1-F(a_{i}))-(j-1)F(a_{i}))\). Since \(q^{*}(a_{i})\) is increasing, the least-ability solver has no chance of receiving the awards, which yields \(e^{*}(\underline{a})=0\). Thus, we can generate the equilibrium effort and performance as follows:

and

In the end, we check that the equilibrium performance \(q^{*}(a_{i})\) is continuously differentiable and increasing in the ability level \(a_{i}\). Note that the PDF \(f_{j}^{n}(\cdot )=\frac{n!}{(j-1)!(n-j)!}(1-F(\cdot ))^{j-1}{F(\cdot )}^{n-j}f(\cdot )\); then we can obtain:

Therefore, \((q^{*})^{\prime }(a_{i})>0\).

(ii) By the definition, the EAQ is given as

Using the Fubini Theorem, we can obtain the following:

Then, \(EAQ=\sum _{j=1}^{L}\int _{\underline{a}}^{\overline{a}}(U(V_{j})-U(V_{j+1}))f_{j}^{n-1}(a)\frac{a}{c}(1-F(a))da\).

(iii) Similar to (ii), \(EMQ=\sum _{j=1}^{L}\int _{\underline{a}}^{\overline{a}}(U(V_{j})-U(V_{j+1}))f_{j}^{n-1}(a)\frac{a}{c}(1-F_{1}^{n}(a))da\). \(\square \)

1.2 Appendix A.2: Proof of Theorem 1

(i) By the expression of EMQ, given any L, the optimization problem can be rewritten as

where \(l_{j}^{n}=\int _{\underline{a}}^{\overline{a}}f_{j}^{n-1}(a)a(1-F_{1}^{n}(a))da-\int _{\underline{a}}^{\overline{a}}f_{j-1}^{n-1}(a)a(1-F_{1}^{n}(a))da\) for \(2\le j\le n-1\), \(l_{n}^{n}=-\int _{\underline{a}}^{\overline{a}}f_{n-1}^{n-1}(a)a(1-F_{1}^{n}(a))da<0\), and \(l_{1}^{n}=\int _{\underline{a}}^{\overline{a}}f_{1}^{n-1}(a)a(1-F_{1}^{n}(a))da>0\). Next, we show that

when \(n\ge n_{L}\) for some critical value \(n_{L}\); and

Note that, this proof can be obtained in “Appendix A.3”.

In this case, by Problem (2), we can verify that a single-award scheme is optimal if \(U(\cdot )\) is linear or convex, i.e., the solvers are risk-neutral or risk-preferred. If \(U(\cdot )\) is concave, suppose that \(l_{j}^{n}>0\) for any \(1\le j\le L\), we claim that there must be at least L awards at optimality. Suppose that there are \(L-1\) awards, i.e., \(r_{1}\ge r_{2}\ge \cdots \ge r_{L-1}>0\), and \(r_{L}=0\). Since \(l_{L-1}^{n}>l_{L}^{n}>0\), let \(\hat{r}_{L-1}\) and \(\hat{r}_{L}\) satisfy that

In this way, we can verify that the allocation is not optimal. Therefore, we have, if \(l_{j}^{n}>0\) for any \(1\le j\le L\), then \(r_{1}^*\ge r_{2}^*\ge \cdots \ge r_{L}^*>0\), i.e., \(L^*\ge L\). That is, a multiple-award scheme is optimal with the optimal number of awards being \(L^*\ge L\) if the solvers are risk-averse.

(ii) Also, if the EAQ is preferred, the optimization problem is given as

where \(s_{j}^{n}=\int _{\underline{a}}^{\overline{a}}f_{j}^{n-1}(a)a(1-F(a))da-\int _{\underline{a}}^{\overline{a}}f_{j-1}^{n-1}(a)a(1-F(a))da\) for \(2\le j\le n-1\), \(s_{n}^{n}=-\int _{\underline{a}}^{\overline{a}}f_{n-1}^{n-1}(a)a(1-F(a))da<0\), and \(s_{1}^{n}=\int _{\underline{a}}^{\overline{a}}f_{1}^{n-1}(a)a(1-F(a))da>0\). Similarly, if the EAQ is preferred, we also only need to show that

when \(n\ge n_{L}\) for some critical value \(n_{L}\); and

Note that, this proof can be obtained in “Appendix A.4”.

Then, we can verify that when the EAQ is preferred, a single-award scheme is optimal if the solvers are risk-neutral or risk-preferred; a multiple-award scheme is optimal with the optimal number of awards being \(L^*\ge L\) if the solvers are risk-averse. \(\square \)

1.3 Appendix A.3: Property of \(l_{j}^{n}\)

Using integration by parts, we have

Then, \(l_{j}^{n}=\int _{\underline{a}}^{\overline{a}}f_{j}^{n-1}(a)a(1-F_{1}^{n}(a))da-\int _{\underline{a}}^{\overline{a}}f_{j-1}^{n-1}(a)a(1-F_{1}^{n}(a))da=-\int _{\underline{a}}^{\overline{a}}\frac{(n-1)!}{(j-1)!(n-j)!}(1-F(a))^{j-1}{F(a)}^{n-j}[a(1-F_{1}^{n}(a))]^{\prime }da,~j\ge 2\), and \(l_{1}^{n}=\int _{\underline{a}}^{\overline{a}}f_{1}^{n-1}(a)a(1-F_{1}^{n}(a))da>0\). Recall that \(F(a)=a^k\), by Lemma 1 in “Appendix A.9”, we have

Since \(nk+1 \rightarrow \infty \) if \(n\rightarrow \infty \). Then, it is not hard to verify that there exists a critical value \(n^{j}\) such that \(n\ge n^{j}\), \(l_{j}^{n}>0\).

Using integration by parts and Lemma 1 in Appendix A.9, we have

and

Similarly, we have \(l_{j-1}^{n}>l_{j}^{n}\) if the number of solvers is sufficiently large. Finally, it is not hard to verify that \(a(1-F_{1}^{n}(a))\) is unimodal in a, by the approach in “Appendix A.8”, we can verify that there exists i such that \(l_{j}^{n}>0\) for \(1\le j\le i\), and \(l_{j}^{n}\le 0\) for \(i+1\le j\le n\). \(\square \)

1.4 Appendix A.4: Property of \(s_{j}^{n}\)

Using integration by parts, we have

Then, \(s_{j}^{n}=\int _{\underline{a}}^{\overline{a}}f_{j}^{n-1}(a)a(1-F(a))da-\int _{\underline{a}}^{\overline{a}}f_{j-1}^{n-1}(a)a(1-F(a))da=-\int _{\underline{a}}^{\overline{a}}\frac{(n-1)!}{(j-1)!(n-j)!}(1-F(a))^{j-1}{F(a)}^{n-j}[a(1-F(a))]^{\prime }da,~j\ge 2\), and \(s_{1}^{n}=\int _{\underline{a}}^{\overline{a}}f_{1}^{n-1}(a)a(1-F(a))da>0\). In “Appendix A.7”, we show that, if \([a(1-F(a))]\) is unimodal in a (this can be easily verified), then there exists a critical value \(n_{L}\) such that \(n\ge n_{L}\), \(s_{j}^{n}>0\).

Using integration by parts, we have

It is not hard to verify that \(\frac{[a(1-F(a))]^{\prime }}{f(a)}\) is decreasing or unimodal in a. By the approach in Appendix A.7, we can also verify that \(s_{j-1}^{n}-s_{j}^{n}>0\) if the number of solvers is sufficiently large. In the end, in Appendix A.8, we have shown that there exists i such that \(s_{j}^{n}>0\) for \(1\le j\le i\), and \(s_{j}^{n}\le 0\) for \(i+1\le j\le n\). \(\square \)

1.5 Appendix A.5: Proof of Theorem 2

(i) If the EMQ is preferred and the number of awards is L, we can verify that the optimal award allocation satisfies \(\frac{r_{j}^{*}}{r_{i}^{*}}=(\frac{l_{j}^{n}}{l_{i}^{n}})^{\frac{1}{1-b}}\) for \(1\le i\le j\le L\), where \(l_{j}^{n}\) is defined in the proof of Theorem 1. Next, we show that

in this case, it is not hard to verify that \(r_{j-1}^{*}-r_{j}^{*}>r_{j}^{*}-r_{j+1}^{*}\), i.e., the award allocation manner is unequal and convex.

Using integration by parts, we have

In addition, by Lemma 1 in Appendix A.9, we have

Since \(\frac{(nk+1)(nk+1-k)(nk+1-2k)}{(1-k)(1-2k)} \rightarrow \pm \infty \) when \(n\rightarrow \infty \), we have there exists a critical value \(n^{L}\) such that \(n\ge n^{L}\), \(l_{j-1}^{n}-l_{j}^{n}>l_{j}^{n}-l_{j+1}^{n}\).

(ii) If the EAQ is preferred, the case is similar. We only need to verify that \(s_{j-1}^{n}-s_{j}^{n}>s_{j}^{n}-s_{j+1}^{n}\).

Using integration by parts, we have

We can verify that \(\frac{[\frac{[a(1-F_{1}^{n}(a))]^{\prime }}{f(a)}]^{\prime }}{f(a)}\) is decreasing or unimodal in a. By the approach in “Appendix A.7”, we can also verify that \(s_{j-1}^{n}-s_{j}^{n}>s_{j}^{n}-s_{j+1}^{n}\) if the number of solvers is sufficiently large. \(\square \)

1.6 Appendix A.6: Proof of Proposition 2

(i) When the EMQ is preferred, by the proof of Theorem 1, we can easily verify that \(r_{j}^{*}/r_{i}^{*}=({l}_{j}^{n}/{l}_{i}^{n})^{\frac{1}{1-b}}\) where \({l}_{j}^{n}\) is defined in the proof of Theorem 1. As a result, we have

By the proof of Theorem 1, we have \({l}_{1}^{n}>{l}_{2}^{n}>\cdots>{l}_{L^*}^{n}>0\), then \(\log ({l}_{1}^{n})>\log ({l}_{j}^{n})\) for any \(2\le j\le L^*\). Thus, \(r_{1}^{*}\) increases with b. Similarly, we can obtain that \(r_{L^*}^{*}\) decreases with b. In addition, when the EAQ is preferred, by the proof of Theorem 1, we can also verify this result.

(ii) When the EMQ is preferred, by (i), we have \((r^*_{j-1}-r^*_{j})/(r^*_{j}-r^*_{j+1})=\big (({l}_{j-1}^{n})^{\frac{1}{1-b}}-({l}_{j}^{n})^{\frac{1}{1-b}}\big )/\big (({l}_{j}^{n})^{\frac{1}{1-b}}-({l}_{j+1}^{n})^{\frac{1}{1-b}}\big )\). Let \(a_{1}={l}_{j-1}^{n}\), \(a_{2}={l}_{j}^{n}\), \(a_{3}={l}_{j+1}^{n}\), then \(a_{1}>a_{2}>a_{3}\). Let \(H(x)=(a_{1}^{x}-a_{2}^{x})/(a_{2}^{x}-a_{3}^{x})\), we have

Since \(\lim _{x \rightarrow \infty }(a_{1}^{x}a_{2}^{x}-a_{1}^{x}a_{3}^{x})/(a_{1}^{x}a_{3}^{x}-a_{2}^{x}a_{3}^{x}) \rightarrow \infty \), then there exists a sufficiently large x such that \(\partial H(x)/\partial x>0\). Therefore, there exists a \(\tilde{b}\), if \(\tilde{b}<b<1\), then \((r^*_{j-1}-r^*_{j})/(r^*_{j}-r^*_{j+1})\) increases with b. In addition, when the EAQ is preferred, by the proof of Theorem 1, we can also verify this result. Note that, the proof of Proposition 2 is similar to the proofs of Propositions 1 and B.1 in [17]. \(\square \)

1.7 Appendix A.7: Complement A in the Proof of Theorem 1

Below, we show that there exists a critical value \(n_{L}\) such that \(n\ge n_{L}\), \(s_{j}^{n}>0\), where \(s_{j}^{n}\) is defined in the proof of Theorem 1. Since \(F(a)=a^{k}\), then \(a(1-F(a))=a-a^{k+1}\), \([a(1-F(a))]^{\prime }=1-(k+1)a^{k}\). We can easily obtain that \(a(1-F(a))\) is unimodal in a, i.e., there exists \(\hat{a}\in (\underline{a},\overline{a})\), such that if \(a\in (\underline{a},\hat{a})\), \(a(1-F(a))\) is increasing; if \(a\in (\hat{a},\overline{a})\), \(a(1-F(a))\) is decreasing. Let

we have

Then, \(\forall ~\epsilon >0\), we have

Thus, \(g_{j}^{\prime }(n+\epsilon )>F(\hat{a})^{\epsilon }g_{j}^{\prime }(n)\), which implies that if

It is easy to see that \(g_{1}(2)<0\), and \(g_{1}(\infty )=0\), then there exists \(n_{1}\ge 2\), such that

Then,

We can also obtain that \(g_{2}(\infty )=0\), then there exists \(n_{2}\), such that \(g_{2}^{\prime }(n_{2})\ge 0\), and for \(n\ge n_{2}\), \(g_{2}(n)<0\). Using this approach repeatedly, for any given L, we can obtain that there exists \(n_{L}\), such that \(g_{L}^{\prime }(n_{L})\ge 0\), and for

Therefore, we obtain that, there exists a critical value \(n_{L}\) such that

\(\square \)

1.8 Appendix A.8: Complement B in the Proof of Theorem 1

In this part, we show that there exists i such that \(s_{j}^{n}>0\) for \(1\le j\le i\), and \(s_{j}^{n}\le 0\) for \(i+1\le j\le n\). Let

where \(g_{j}(n)\) is defined in “Appendix A.7”. Then, by a similar approach in “Appendix A.7”, we can obtain

which implies that \(G_{j+1}^{n}>(1-F(\hat{a}))G_{j}^{n}\). This suggests that if \(g_{j+1}(n)>g_{j}(n)\), then

Also, it is not difficult to verify that \(g_{1}(n)<0\) and \(g_{n}(n)>0\) for a fixed n. Via simple calculation, we can obtain that

This equals that there exists i such that \(s_{j}^{n}>0\) for \(1\le j\le i\), and \(s_{j}^{n}\le 0\) for \(i+1\le j\le n\). \(\square \)

1.9 Appendix A.9: A supporting lemma

Lemma 1

Let \(B(P,Q)=\int _{0}^{1}x^{P-1}(1-x)^{Q-1}dx\) for any \(P\ge 1,~Q\ge 1\), then, \(B(P,Q)=\frac{(Q-1)!}{P\cdot (P+1)\cdots (P+Q-1)}\).

This lemma can be easily checked, we thus omit it.

1.10 Appendix A.10: The utility function \(U(V)=V^{b}\) can be generalized

First, Theorem 1 holds if the utility function \(U(\cdot )\) satisfies \(U^{\prime }(0)=+\infty \) when \(U^{\prime \prime }(\cdot )<0\) and no constraint when \(U^{\prime \prime }(\cdot )\ge 0\), which can be easily verified by the proof of Theorem 1. Note that, for the utility function \(U(\cdot )\) satisfying \(U^{\prime }(0)=+\infty \) when \(U^{\prime \prime }(\cdot )<0\), it is not hard to verify that in this case, \(r_{j}^*>0\) if \(l_{j}^{n}>0\) (or \(s_{j}^{n}>0\)), which yields Theorem 1. Second, by the proof of Theorem 2, we can check that Theorem 2 holds when \(U(V)=\log (V)\) (see [9]). These imply that our results hold for general solvers’ utility functions.

1.11 Appendix A.11: Some numerical study

As discussed above, the award scheme deeply depends on \(l_{j}^{n}\) and \(s_{j}^{n}\), which are defined above. By the proof of Theorems 1 and 2, if \(l_{j}^{n}>0\) (\(s_{j}^{n}>0\)) and decreases in j for any \(1\le j\le L\) and \(l_{j-1}^{n}-l_{j}^{n}>l_{j}^{n}-l_{j+1}^{n}\) (\(s_{j-1}^{n}-s_{j}^{n}>s_{j}^{n}-s_{j+1}^{n}\)) for any \(2\le j\le L-1\), then when the solvers are risk-averse, a multiple-award scheme is optimal, and the award allocation manner is unequal and convex. Below, we give some numerical studies. Using Monte Carlo simulation in R software, we can obtain \(l_{j}^{n}\) and \(s_{j}^{n}\), which are listed in Tables 3 and 4. By the above analysis, for \(k=1\) (\(F(a)=a^k\)), an unequal and convex allocation method can hold when \(n\ge 10\) in the EMQ case; an unequal and convex allocation method can hold when \(n\ge 8\) in the EAQ case.

In addition, we numerically give the \(l_{j}^{n}\) and \(s_{j}^{n}\) based on other distributions of solvers (i.e., f(a)), which are listed in Tables 5, 6, 7 and 8. By the above analysis, we have for general cases, if the EAQ or EMQ is considered, when the solvers are risk-averse, a multiple-award scheme is optimal, and the award allocation manner is unequal and convex.

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