Axiomatization of reverse nested lottery contests

Abstract

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.

Appendix

1.1 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

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

(11)

Axiom R7 implies that

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

(12)

Define

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

(13)

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

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

(14)

where the second equality follows from

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

(15)

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 $\forall i,j\in \mathbf {N\backslash }\{1\}$,

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

(16)

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

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

(17)

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.,

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

(18)

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

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

(19)

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

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

(20)

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 $\tilde{y}_{j}>0$, $\forall j\in \mathbf {N} \backslash \mathbf {M}$ ,

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

(21)

Then by the result derived from the case with $|\mathbf {N}_{\mathbf {0}}(Y_{ \mathbf {N}})|=0$ and Axioms R4–R5,

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

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

1.2 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 R4–R7 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 R4–R7 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

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

(22)

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

(23)

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).