Axiomatization of reverse nested lottery contests


The reverse nested lottery contest proposed by Fu et al. (2014) is the “mirror image” of the classical nested lottery contest of Clark and Riis (1996a), which has been axiomatized by Lu and Wang (2015). In this paper, we close the gap and provide an axiomatic underpinning for the reverse nested lottery contest by identifying a set of six necessary and sufficient axioms. These axioms proposed specify the properties of contestants’ probabilities of being ranked the lowest among all players or within subgroups, while the axiomatization of the classical nested lottery contest by Lu and Wang (2015) relies on axioms on contestants’ probabilities of being ranked the highest among all players or within subgroups.

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  1. 1.

    Namely, axioms: (A1) Imperfect Discrimination; (A2) Monotonicity; (A3) Anonymity; (A4) Sub-contest Consistency; and (A5) Independence from Irrelevant Contestants (IIC).

  2. 2.

    For instance, the model has been adopted in the studies of Clark and Riis (1998b), Amegashie (2000), Yates and Heckelman (2001), Szymanski and Valletti (2005), Fu and Lu (2009), Fu and Lu (2012a), Fu and Lu (2012b), Schweinzer and Segev (2012), etc.

  3. 3.

    Fu et al. (2014) also provide a micro foundation for the reverse nested lottery contest by showing that the model is uniquely underpinned by a noisy-performance-ranking model. This noisy performance model with Weibull (minimum) distribution for the shocks was first proposed by Hirshleifer and Riley (1992) in a two-player environment.

  4. 4.

    In the literature on perfectly discriminating contests, the most “classical” model on multi-prize contests might be the all-pay auction/contest model where players are ranked in a deterministic fashion. In this paper we focus on the literature on imperfectly discriminating contests where the nested lottery contest model is widely adopted—we call it the “classical” nested lottery contest model to distinguish it from the reverse nested lottery contest model.

  5. 5.

    In the noisy performance ranking contest, each contestant’s observable output is the sum of a deterministic component, which increases with his effort, and a random noise term. Clark and Riis (1996b) first point out the equivalence between a noisy-ranking-based discrete random choice model (McFadden 1973, 1974) and the single-prize lottery contest model. Fu and Lu (2012b) further establish a unique stochastic equivalence between a noisy performance ranking contest and the classical (multi-prize) nested lottery contest.

  6. 6.

    The performance of each experiment could follow any distribution.

  7. 7.

    Real-world examples include competitions in acrobatics, diving, and many skilled competitions across different professions.

  8. 8.

    As mentioned earlier, the existing literature has shown that the classical and reverse models can be interpreted as best-shot contests and worst-shot contests, respectively.

  9. 9.

    Before that, Fu et al. (2014) find that with a common single prize and identical impact function of power function form across the two models, the reverse model always induces more effort whenever a pure-strategy equilibrium exists. However, when the number of (common) prizes gets larger, the reverse contest loses its advantage gradually and eventually is dominated by the classical contest.

  10. 10.

    The lottery-form CSF is called the ratio-form CSF in Hwang (2012).

  11. 11.

    As shown in Fu et al. (2014), the classical and reverse nested lottery contests converge if and only if \(n=2\). In this paper we focus on the multi-prize case which requires at least three players.

  12. 12.

    Here “complete” means all players are ranked and “strict” means they are ranked without ties.

  13. 13.

    One can find that the two sets of axioms that underpin the classical and reverse contests [Lu and Wang (2015) and this paper] are parallel, so we use the same set of names of axioms for brevity. But one should bear in mind that those axioms in the two studies differ in whether they are concerned with the players’ probabilities of being ranked the highest or lowest in subgroups.

  14. 14.

    Some of the axioms may seem to be a bit stronger, this is because [differing from Skaperdas (1996)] the axioms in this paper explicitly allow zero effort in our framework.

  15. 15.

    Note that this result implies that \(\eta _{\mathbf {N}}^{i}(\mathbf {0})=1/| \mathbf {N}|=1/n\), which is consistent with Axiom R1. One may find that Axioms R1 and R3 are not independent as the second requirement of part (iii) of Axiom R1, which demands \(\eta _{\mathbf {N}}^{i}(Y_{\mathbf {N }})=1\mathbf {/|N}_{\mathbf {0}}(Y_{\mathbf {N}})|\), easily follows from other parts of Axiom R1 combined with Axiom R3. However, we do not weaken part (iii) of Axiom R1 as in later analysis (Proposition 2) it is indeed needed when Axiom R3 is dropped.

  16. 16.

    For instance, in a contest with \(\mathbf {N}=\{1,2,3\}\) players and effort entries \(Y_{\mathbf {N}}=(y_{1},y_{2},y_{3})\), let \(\mathbf {M}=\{1,2\}\), if \( \eta _{\mathbf {N}}^{1}(Y_{\mathbf {N}})=1/6\) and \(\eta _{\mathbf {N}}^{2}(Y_{ \mathbf {N}})=1/3\), then by Axiom R4, we have \(\eta _{\mathbf {M}}^{1}(Y_{ \mathbf {N}})=(1/6)/(1/6+1/3)=1/3\) and \(\eta _{\mathbf {M}}^{2}(Y_{\mathbf {N} })=(1/3)/(1/6+1/3)=2/3\), thus \(\eta _{\mathbf {M}}^{1}(Y_{\mathbf {N}})/\eta _{ \mathbf {M}}^{2}(Y_{\mathbf {N}})=\eta _{\mathbf {N}}^{1}(Y_{\mathbf {N}})/\eta _{\mathbf {N}}^{2}(Y_{\mathbf {N}})=1/2.\)

  17. 17.

    It was initially called the “ratio-form CEF” in Sect. 2.1 of Fu et al. (2014).

  18. 18.

    The detailed proof can be obtained following a very similar pattern in Appendix A.3.2 of Lu and Wang (2015), which is omitted for brevity in this study.

  19. 19.

    This differs from the classical nested lottery contest model of Clark and Riis (1996a) where the draws for higher prizes are conducted earlier.

  20. 20.

    When \(\exists j\in \Omega _{k}\), \(y_{j}=0\), the probability that player i \( \in \Omega _{k}\) wins prize \(v_{k}\) is \(1/[\#(j|f(y_{j})=0,j\in \Omega _{k})] \), where \(\#(j|f(y_{j})=0,j\in \Omega _{k})\) is the count of zero efforts among \(\Omega _{k}\).

  21. 21.

    It is straightforward to extend the result to the cases where \(\mathbf {N}_{ \mathbf {0}}\ne \mathbf {\emptyset }\) by allowing efforts of some players to be zero. The proof is omitted to save space.

  22. 22.

    It can be shown that the reverse nested lottery contest model also satisfies Axioms 1–3 and 5 of Lu and Wang (2015). Similarly, Axioms 4 and 6 in Lu and Wang (2015) characterize distinctive properties that only the classical nested lottery contest model satisfies, among the three models compared.

  23. 23.

    Lu, Lu et al. (2016) show that the existing results on the optimal design of multi-stage elimination contests, which are obtained under classical nested lottery technology [e.g. Gradstein and Konrad (1999), and Fu and Lu (2012a)], may vary when being reexamined under the reverse nested lottery technology.


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Corresponding author

Correspondence to Jingfeng Lu.

Additional information

We are grateful to the editors in charge and two anonymous referees for their insightful comments and suggestions. We thank Dan Kovenock, Chengzhong Qin, Birendra Rai, Christian Riis, Santiago Sanchez-Pages, and Alberto Vesperoni for their kind comments and discussions. Jingfeng Lu gratefully acknowledges the financial support from Ministry of Education, Academic Research Fund, Singapore (R-122-000-217-112). Wang gratefully acknowledges financial support from Project 71501112 supported by NSFC, and from Qilu Young Scholars and Zhongying Young Scholars of Shandong University. All remaining errors are our own.



Proof of Proposition 2

It is straightforward to show that probabilities \(\eta _{\mathbf {M}}^{i}(Y_{ \mathbf {N}})\) satisfying the conditions of Proposition 2 must satisfy Axioms R1, R2, R4, R5 and R7. We next show that Axioms R1, R2, R4, R5 and R7 imply properties (i) and (ii) of Proposition 2. We first consider the case with \(|\mathbf {M}_{\mathbf {0}}(Y_{\mathbf {N}})|\ge 1\). In this case, \(|\mathbf {N}_{\mathbf {0}}(Y_{\mathbf {N}})|\ge 1\), and by applying Axioms R1 (iii), R3 and R4, Proposition 2 (ii) can be obtained.

Now look at the case with \(|\mathbf {M}_{\mathbf {0}}(Y_{\mathbf {N}})|=0\). We consider two subcases of \(|\mathbf {M}_{\mathbf {0}}(Y_{\mathbf {N}})|=0\): Case I with \(|\mathbf {M}_{\mathbf {0}}(Y_{\mathbf {N}})|=|\mathbf {N}_{\mathbf {0} }(Y_{\mathbf {N}})|=0\) and Case II with \(|\mathbf {M}_{\mathbf {0}}(Y_{\mathbf {N }})|=0\) but \(|\mathbf {N}_{\mathbf {0}}(Y_{\mathbf {N}})|\ge 1\). We begin with Case I. By Axiom R1 (ii), we have \(y_{i}>0\), \(\eta _{\mathbf {N} }^{i}(Y_{\mathbf {N}})>0\),\(\forall i\in \mathbf {N}\) as \(|\mathbf {M}_{\mathbf {0}}(Y_{\mathbf {N}})|=|\mathbf {N}_{\mathbf {0} }(Y_{\mathbf {N}})|=0\). In this case, assume player 1 is an arbitrary player in \(\mathbf {N}\), then we have \(y_{1}>0\). By Axioms R4 and R5, we have

$$\begin{aligned} \frac{\eta _{\{i,1\}}^{i}(y_{1},y_{i})}{\eta _{\{i,1\}}^{1}(y_{1},y_{i})}= \frac{\eta _{\mathbf {N}}^{i}(Y_{\mathbf {N}})}{\eta _{\mathbf {N}}^{1}(Y_{ \mathbf {N}})}. \end{aligned}$$

Axiom R7 implies that

$$\begin{aligned} \frac{\eta _{\{i,1\}}^{i}(y_{1},y_{i})}{\eta _{\{i,1\}}^{1}(y_{1},y_{i})}= \frac{\eta _{\{i,1\}}^{i}(1,\frac{y_{i}}{y_{1}})}{\eta _{\{i,1\}}^{1}(1, \frac{y_{i}}{y_{1}})}. \end{aligned}$$


$$\begin{aligned} h_{i}(x)\equiv \frac{\eta _{\{i,1\}}^{i}(1,x)}{\eta _{\{i,1\}}^{1}(1,x)}. \end{aligned}$$

Note that \(h_{i}(x)\) does not depend on \(Y_{\mathbf {N}\backslash \{1,i\}}\) from Axiom R5. Then (11), (12) and (13) imply

$$\begin{aligned} \eta _{\mathbf {N}}^{i}(Y_{\mathbf {N}})=\eta _{\mathbf {N}}^{1}(Y_{\mathbf {N} })h_{i}(\tfrac{y_{i}}{y_{1}})=\frac{h_{i}(\tfrac{y_{i}}{y_{1}})}{ 1+\sum _{k=2}^{n}h_{k}(\tfrac{y_{k}}{y_{1}})}, \quad 2\le i\le n, \end{aligned}$$

where the second equality follows from

$$\begin{aligned} 1=\sum \nolimits _{k=1}^{n}\eta _{\mathbf {N}}^{k}(Y_{\mathbf {N}})=\eta _{ \mathbf {N}}^{1}(Y_{\mathbf {N}})\left( 1+\sum \nolimits _{k=2}^{n}h_{k}(\tfrac{ y_{k}}{y_{1}})\right) . \end{aligned}$$

Axiom R4 implies that for \(2\le i\le n\), . From (14), we further have

The last equality obtains by letting \(y_{1}=1\) since is independent of player 1’s effort \(y_{1}\)by Axiom R5. We thus have \(\forall i,j\in \mathbf {N\backslash }\{1\}\),

$$\begin{aligned} \frac{h_{i}\left( \tfrac{y_{i}}{y_{1}}\right) }{h_{j}\left( \tfrac{y_{j}}{y_{1}}\right) }=\frac{ h_{i}(y_{i})}{h_{j}(y_{j})}. \end{aligned}$$

Equation (16) gives that \(\forall i,j\in \mathbf {N\backslash }\{1\}\),

$$\begin{aligned} \frac{h_{i}\left( \tfrac{y_{i}}{y_{1}}\right) }{h_{i}(y_{i})}=\frac{h_{j}\left( \tfrac{y_{j}}{ y_{1}}\right) }{h_{j}(y_{j})}. \end{aligned}$$

As the right hand side of (17) is independent of \(y_{i}\), it also implies that the term on the left hand side of (17) is independent of \(y_{i}\), i.e.,

$$\begin{aligned} \frac{h_{i}\left( \tfrac{y_{i}}{y_{1}}\right) }{h_{i}(y_{i})}=\frac{h_{i}\left( \tfrac{1}{y_{1}} \right) }{h_{i}(1)},\quad \forall y_{1},y_{i}>0. \end{aligned}$$

Let \(x=y_{i}>0\) and \(z=1/y_{1}>0\), we can express (18) as

$$\begin{aligned} h_{i}(zx)=\frac{h_{i}(z)h_{i}(x)}{h_{i}(1)}. \end{aligned}$$

Define the function \(F_{i}(w)=h_{i}(w)/h_{i}(1),w>0\), then we can write (19) as

$$\begin{aligned} F_{i}(zx)=F_{i}(z)F_{i}(x),\quad \forall z,x>0, \end{aligned}$$

which is one of Cauchy’s four fundamental functional equations. Using the definition in (13) and Axiom R2, it is apparent that \(h_{i}(\cdot )\) and \(F_{i}(\cdot )\) are strictly monotonic. Then by Aczel (1966; 41), the unique solution to (20) is \(F_{i}(x)=x^{r_{i}}\Rightarrow h_{i}(x)=\beta _{i}x^{r_{i}}\), where \(\beta _{i}=h_{i}(1)\). As (17) can only be satisfied if \(r_{i}=-r\), \(\forall i\in \mathbf {N}\), thus \( h_{i}(x)=\beta _{i}x^{-r}\). Note that due to Axiom R2, we must have \(-r<0\), i.e., \(r>0\). Let \(\beta _{1}=1\), (14), (15) and \( h_{i}(x)=\beta _{i}x^{-r}.\) lead to \(\eta _{\mathbf {N}}^{i}(Y_{\mathbf {N} })=\beta _{i}y_{i}^{-r}/(\sum \nolimits _{j\in \mathbf {N}}\beta _{j}y_{j}^{-r})\), \(\forall i\in \mathbf {N},\forall Y_{\mathbf {N}}\) with \(| \mathbf {N}_{\mathbf {0}}(Y_{\mathbf {N}})|=0\). Let \(\alpha _{i}=\beta _{i}/(\sum \nolimits _{j\in \mathbf {N}}\beta _{j})>0\), \(\forall i\). We have \( \eta _{\mathbf {N}}^{i}(Y_{\mathbf {N}})=\alpha _{i}y_{i}^{-r}/(\sum \nolimits _{j\in \mathbf {N}}\alpha _{j}y_{j}^{-r})\), \( \forall i\in \mathbf {N}\) , \(\forall Y_{\mathbf {N}}\) where \(y_{j}>0,\) \(\forall j\in \mathbf {N}\). By Axiom R4, we have

$$\begin{aligned} \eta _{\mathbf {M}}^{i}(Y_{\mathbf {N}})=\eta _{\mathbf {M}}^{i}(Y_{\mathbf {M} })=\frac{\alpha _{i}y_{i}^{-r}}{\sum \nolimits _{j\in \mathbf {M}}\alpha _{j}y_{j}^{-r}},\quad \forall i\in \mathbf {M}\text {, }\forall Y_{\mathbf {N}} \quad \text {where }y_{j}>0\text {, }\forall j\in \mathbf {N}, \end{aligned}$$

which is Proposition 2 (i). From the way \(\alpha _{i}\)s (\(i\in \mathbf {N}\)) are constructed, those parameters \(\alpha _{i}\)s do not depend on the efforts of \(Y_{\mathbf {N}}\), and \(\sum \nolimits _{i\in \mathbf {N} }\alpha _{i}=1\).

We now consider Case II where \(|\mathbf {M}_{\mathbf {0}}(Y_{\mathbf {N}})|=0\) and \(|\mathbf {N}_{\mathbf {0}}(Y_{\mathbf {N}})|\ge 1\). By Axiom R1 (iii), we have \(\eta _{\mathbf {N}}^{i}(Y_{\mathbf {N}})=0\), \(\forall i\in \mathbf {M}\) as no player in \(\mathbf {M}\) exerts zero effort but some player(s) outside \(\mathbf {M}\) does, i.e., \(\mathbf {\eta }_{\mathbf {M}}(Y_{ \mathbf {N}})=(\eta _{\mathbf {N}}^{i}(Y_{\mathbf {N}}))_{i\in \mathbf {M}}= \mathbf {0}\). From Axioms R4, R5 and R1 (ii) we derive: \(\forall \tilde{Y}_{\mathbf {N}\backslash \mathbf {M}}=(\tilde{y}_{j})_{j\in \mathbf {N} \backslash \mathbf {M}}\) where \(\tilde{y}_{j}>0\), \(\forall j\in \mathbf {N} \backslash \mathbf {M}\) ,

$$\begin{aligned} \eta _{\mathbf {M}}^{i}(Y_{\mathbf {N}})=\eta _{\mathbf {M}}^{i}(Y_{\mathbf {M}}, \tilde{Y}_{\mathbf {N}\backslash \mathbf {M}})=\frac{\eta _{\mathbf {N}}^{i}(Y_{ \mathbf {M}},\tilde{Y}_{\mathbf {N}\backslash \mathbf {M}})}{\sum _{j\in \mathbf { M}}\eta _{\mathbf {N}}^{j}(Y_{\mathbf {M}},\tilde{Y}_{\mathbf {N}\backslash \mathbf {M}})}. \end{aligned}$$

Then by the result derived from the case with \(|\mathbf {N}_{\mathbf {0}}(Y_{ \mathbf {N}})|=0\) and Axioms R4R5,

$$\begin{aligned} \eta _{\mathbf {M}}^{i}(Y_{\mathbf {N}})= & {} \eta _{\mathbf {M}}^{i}(Y_{\mathbf {M }})=\eta _{\mathbf {M}}^{i}(Y_{\mathbf {M}},\tilde{Y}_{\mathbf {N}\backslash \mathbf {M}}) \\= & {} \frac{\alpha _{i}y_{i}^{-r}}{\sum \nolimits _{h\in \mathbf {M}}\alpha _{h}y_{h}^{-r}+\sum \nolimits _{h\in \mathbf {N}\backslash \mathbf {M}}\alpha _{h}\tilde{y}_{h}^{-r}}/ \sum _{j\in \mathbf {M}}\left( \frac{\alpha _{j}y_{j}^{-r}}{\sum \nolimits _{h\in \mathbf {M}}\alpha _{h}y_{h}^{-r}+\sum \nolimits _{h\in \mathbf {N}\backslash \mathbf {M}}\alpha _{h}\tilde{y}_{h}^{-r}}\right) \\= & {} \frac{\alpha _{i}y_{i}^{-r}}{\sum _{j\in \mathbf {M}}\alpha _{j}y_{j}^{-r}} \text {,} \end{aligned}$$

which is exactly the same as Proposition 2 (i).

Proof of Theorem 2

(IF Part:) We first show that the IIC property (Axiom R5) is satisfied by the CRF described in Lemma 2. Recall \(\mathbf {N}_{\mathbf {0} }(Y_{\mathbf {N}})=\{i\in \mathbf {N|}y_{i}=0\},\) \(\forall Y_{\mathbf {N}}\). If \(|\mathbf {M}_{\mathbf {0}}(Y_{\mathbf {N}})|\ge 1\), then \(|\mathbf {N}_{ \mathbf {0}}(Y_{\mathbf {N}})|\ge 1\), it can be shown that Axiom R5 holds using (ii) of Lemma 2 and Definition 2. When \(|\mathbf {M}_{\mathbf { 0}}(Y_{\mathbf {N}})|\ge 1\), if \(i\in \mathbf {M}\backslash \mathbf {M}_{ \mathbf {0}}(Y_{\mathbf {N}})\), it can be shown that \(\eta _{\mathbf {M} }^{i}(Y_{\mathbf {N}})=\eta _{\mathbf {M}}^{i}(Y_{\mathbf {M}})=0\) using ( iii) of Lemma 2 and Definition 2; if \(i\in \mathbf {M}_{\mathbf {0} }(Y_{\mathbf {N}})\), by (ii) of Lemma 2 and Definition 2, we can derive \(\eta _{\mathbf {M}}^{i}(Y_{\mathbf {N}})=\eta _{\mathbf {M}}^{i}(Y_{ \mathbf {M}})=1/|\mathbf {M}_{\mathbf {0}}(Y_{\mathbf {N}})|\). When \(|\mathbf {M} _{\mathbf {0}}(Y_{\mathbf {N}})|=0\), we next show that

$$\begin{aligned} \eta _{\mathbf {M}}^{i}(Y_{\mathbf {N}})=\eta _{\mathbf {M}}^{i}(Y_{\mathbf {M} })=\frac{\alpha _{i}y_{i}^{-r}}{\sum \nolimits _{j\in \mathbf {M}}\alpha _{j}y_{j}^{-r}}. \end{aligned}$$

Consider the following random performance ranking model of Fu et al. (2014) with multiplicative noise: \(x_{i}=y_{i}^{r}\times \varepsilon _{i},\ \forall i\in \mathbf {N,}\) where \(x_{i}\) is the output of player i and \( y_{i}\) is his effort, the noise term \(\varepsilon _{i}\) follows a Weibull (minimum) distribution. Noises \(\varepsilon _{i}\) are independent across i . Prizes \(\mathbf {V}\) are allocated according to the rankings of observed outputs \(x_{i},\) \(i\in \mathbf {N}\). Each player wins one prize. Higher outputs win higher prizes. Ties are broken randomly with equal winning chances. Theorem 2 in Fu et al. (2014) implies that the above model generates exactly the same CRF as described by Lemma 2 if \(\mathbf {N}_{ \mathbf {0}}(Y_{\mathbf {N}})=\mathbf {\emptyset }\). It is straightforward to extend the result to the cases where \(\mathbf {N}_{\mathbf {0}}(Y_{\mathbf {N} })\ne \mathbf {\emptyset }\) by allowing efforts of some players to be zero, the proof is omitted. Therefore, the above noisy performance ranking model is stochastically equivalent to any model that generates the CRF of Lemma 2. We thus have that \(\eta _{\mathbf {M}}^{i}(Y_{\mathbf {N}})\) (Definition 2) derived from the CRF of Lemma 2 must coincide with the probability \(\Pr (x_{i}<x_{j},\forall j\in \mathbf {M|}Y_{\mathbf {N}})\). From Theorem 1 of Fu et al. (2014), we can derive that

$$\begin{aligned} \Pr (x_{i}<x_{j},\forall j\in \mathbf {M|}Y_{\mathbf {N}})=\Pr (x_{i}<x_{j},\quad \forall j\in \mathbf {M|}Y_{\mathbf {M}})=\frac{\beta _{i}y_{i}^{-r}}{\sum _{j\in \mathbf {M}}\beta _{j}y_{j}^{-r}} \end{aligned}$$

as \(y_{j}>0\), \(\forall j\in \mathbf {M}\). We thus have completed the proof for Axiom R5.

Based on the stochastic equivalence pointed out as above, the IIR property (Axiom R6) can be shown by Theorem 2 of Fu et al. (2014). Subcontest Consistency property (Axiom R4) and Homogeneity property (Axiom R7) must hold as we have established above that \(\eta _{\mathbf {M}}^{i}(Y_{\mathbf {N} })=\beta _{i}y_{i}^{-r}/(\sum _{j\in \mathbf {M}}\beta _{j}y_{j}^{-r})\), \( \forall \mathbf {M\subseteq N}\). Notice that \(\eta _{\mathbf {N}}^{i}(Y_{ \mathbf {N}})=\beta _{i}y_{i}^{-r}/(\sum _{j\in \mathbf {N}}\beta _{j}y_{j}^{-r})\) if \(\mathbf {N}_{\mathbf {0}}(Y_{\mathbf {N}})=\mathbf { \emptyset }\), \(\eta _{\mathbf {N}}^{i}(Y_{\mathbf {N}})=1/|\mathbf {N}_{\mathbf { 0}}(Y_{\mathbf {N}})|\) if \(i\in \mathbf {N}_{\mathbf {0}}(Y_{\mathbf {N}})\) and \( \mathbf {N}_{\mathbf {0}}(Y_{\mathbf {N}})\ne \mathbf {\emptyset }\), \(\eta _{ \mathbf {N}}^{i}(Y_{\mathbf {N}})=0\) if \(i\in \mathbf {N}\backslash \mathbf {N}_{ \mathbf {0}}(Y_{\mathbf {N}})\) and \(\mathbf {N}_{\mathbf {0}}(Y_{\mathbf {N} })\ne \mathbf {\emptyset }\), Imperfect Discrimination property (Axiom R1) and Monotonicity property (Axiom R2) must therefore hold.

(ONLY IF Part:) Next, we are going to show that Axioms R1, R2 and R4R7 must render the CRF of Lemma 2. In the following analysis we focus on the case with \(|\mathbf {N}_{\mathbf {0}}(Y_{\mathbf {N}})|=0\), i.e., \( y_{i}>0,\forall i\in \mathbf {N}\), the case with \(|\mathbf {N}_{\mathbf {0}}(Y_{ \mathbf {N}})|\ge 1\) can be derived following a similar procedure, which we omit for brevity.

By Proposition 2 (i), we derive that when \(|\mathbf {N}_{\mathbf {0} }(Y_{\mathbf {N}})|=0\), Axioms R1, R2 and R4R7 lead to \(\eta _{\mathbf {N} }^{i}(Y_{\mathbf {N}})=\alpha _{i}y_{i}^{-r}/(\sum _{j\in \mathbf {N}}\alpha _{j}y_{j}^{-r})\). Substituting it into (1) in Axiom R4 yields

$$\begin{aligned} \eta _{\mathbf {M}}^{i}(Y_{\mathbf {N}})=\frac{\alpha _{i}y_{i}^{-r}}{ \sum _{j^{\prime }\in \mathbf {N}}\alpha _{j^{\prime }}y_{j^{\prime }}^{-r}} \bigg /\sum _{j\in \mathbf {M}}\left( \frac{\alpha _{j}y_{j}^{-r}}{\sum _{j^{\prime }\in \mathbf {N}}\alpha _{j^{\prime }}y_{j^{\prime }}^{-r}}\right) =\frac{ \alpha _{i}y_{i}^{-r}}{\sum _{j\in \mathbf {M}}\alpha _{j}y_{j}^{-r}}. \end{aligned}$$

By Axiom R6, the probability of a complete ranking \(\mathbf {r}_{\mathbf {N} }=(i_{k})_{k=1}^{n}\) can be expressed as


Substituting (22) into (23), we have

$$\begin{aligned} p(\mathbf {r}_{\mathbf {N}};Y_{\mathbf {N}})=\Pi _{k=1}^{n}\frac{\alpha _{i_{k}}y_{i_{k}}^{-r}}{\sum \nolimits _{k^{\prime }=1}^{k}\alpha _{i_{k^{\prime }}}y_{i_{k^{\prime }}}^{-r}}, \end{aligned}$$

which is exactly the form of CRF of Lemma 2 (i).

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Lu, J., Wang, Z. Axiomatization of reverse nested lottery contests. Soc Choice Welf 47, 939–957 (2016).

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  • Ranking Outcome
  • Reverse Model
  • Contest Success Function
  • Impact Function
  • Effort Entry