‘NEXT’ events: a cooperative game theoretic view to festivals

Abstract

During a cultural festival, artists and theaters act as a cartel by agreeing on pricing decisions that maximize the groups’ profit as a whole. We model the problem of sharing the profit created by a festival among organizing theaters as a cooperative game. In such a game, the worth of a coalition is defined as the theaters’ profit from the optimal fixation of prices. We show that this class of games is convex and we axiomatically characterize the Shapley value (Shapley 1953) for this class of games. We also provide an axiomatic basis for the downstream incremental solution. Finally, we apply this model to the NEXT festival, for which we have collected data. We propose an approach to derive the games’ vector from the data and we compute the different solutions.

Appendix

1.1 Proofs

1.1.1 Lemma 1

Clearly, by choosing $p_i=c_i$ we can secure $S_i(p_i)(p_i-c_i)=0.$ We also have $\frac{d}{d p}\left\{ S_i(p_i) (p_i-c_i) \right\} _{p_i=c_i}=S_i(c_i)>0.$ It follows that the price that maximizes the profit function must belong to $p_i \in ]c_i,\overline{p}_{S_i}].$

Now, if $p_i<0$ then $S_i(p_i)(p_i-c_i)=p_i-c_i<-c_i=S_i(0)(0-c_i)$ so we may restrict our attention to the case $p_i \ge 0.$ Moreover $S_i'(0)=0$ and $S_i(0)=1$ imply $\frac{d}{d p_i}\left\{ S_i(p_i) (p_i-c_i) \right\} _{p_i=0}=1>0$ so we now consider $p_i \in ]\max \{c_i,0\},\overline{p}_{S_i}].$ Over this interval we have

$$\begin{aligned} \frac{d^2}{d p_i^2}\left\{ S_i(p_i) (p_i-c) \right\} =S''_i(p_i) (p_i-c)+ 2S'_i(p_i)<0 \end{aligned}$$

which implies i).

Obviously if $c'<c_i$ then

$$\begin{aligned} S_i(p_i^m(c_i))(p_i^m(c_i)-c')>S_i(p_i^m(c_i))(p_i^m(c_i)-c_i) \end{aligned}$$

where $p_i^m(c)={{\,\mathrm{arg\,max}\,}}_{p_i \in \mathbb {R}}\; S_i(p_i) (p_i-c_i)$, which implies ii).

Finally, $d \pi _i^m (c') / dp_i = S'(p_i)(c_i - c')<0$ implies iii).

1.1.2 Lemma 2

Let us denote by $d_i=(e_i+\rho _{i}d_{i-1})S_i(p_i)$ the demand for the show of theater i. According to Definition 1, theater $i_T$ must solve

$$\begin{aligned} \max _{p_{i_T} \in \mathbb {R}}\; S_{i_T}(p_{i_T}) (p_{i_T}-c_{i_T})(e_{i_T}+\rho _{i_T}d_{-1+i_T}) \end{aligned}$$

where $d_{-1+i_T}$ results from the choices of all upward theaters. Clearly, theater $i_T$ must choose the monopoly price $p^m_{S_{i_T}}(c_{i_T})$ which is uniquely defined according to Lemma 1 and definition 1 since we assumed $S_{i_T}(c_{i_T})>0.$ To save space, we shall write $p^m_{i_T}$ for $p^m_{S_{i_T}}(c_{i_T}).$

Now proceeding upwards and consider theater $i_{T-1}.$ Fixing the pricing rules of all upward theaters up to index $i_{T-1}$ excluded, this theater solves

$$\begin{aligned} \max _{p_{i_{T-1}}\in \mathbb {R}}\; S_{i_{T-1}}(p_{i_{T-1}}) (p_{i_{T-1}}-c_{i_{T-1}})(e_{i_{T-1}}+\rho _{i_{T-1}}d_{-1+i_{T-1}})+ S_{i_T}(p^m_{i_T}) (p^m_{i_T}-c_{i_T})(e_{i_T}+\rho _{i_T}d_{-1+i_T}) \end{aligned}$$

which is equivalent to

$$\begin{aligned} \max _{p_{i_{T-1}}\in \mathbb {R}}\; S_{i_{T-1}}(p_{i_{T-1}}) (p_{i_{T-1}}-c_{i_{T-1}})(e_{i_{T-1}}+\rho _{i_{T-1}}d_{-1+i_{T-1}})+ S_{i_T}(p^m_{i_T}) (p^m_{i_T}-c_{i_T})\rho _{i_T}d_{-1+i_T} \end{aligned}$$

Now we have

$$\begin{aligned} d_{-1+i_T}=(e_{-1+i_T}+\rho _{-1+i_T}d_{-2+i_T})S_{-1+i_T}(p_{-1+i_T}) \end{aligned}$$

If $-1+i_T=i_{T-1}$ theater $i_{T-1}$ must solve

$$\begin{aligned} \max _{p_{i_{T-1}}\in \mathbb {R}}\; S_{i_{T-1}}(p_{i_{T-1}}) (p_{i_{T-1}}-c_{i_{T-1}}+S_{i_T}(p^m_{i_T}) (p^m_{i_T}-c_{i_T})\rho _{i_T})(e_{i_{T-1}}+\rho _{i_{T-1}}d_{-1+i_{T-1}}) \end{aligned}$$

or

$$\begin{aligned} \max _{p_{i_{T-1}}\in \mathbb {R}}\; S_{i_{T-1}}(p_{i_{T-1}}) (p_{i_{T-1}}-c_{i_{T-1}}(T))(e_{i_{T-1}}+\rho _{i_{T-1}}d_{-1+i_{T-1}}) \end{aligned}$$

with $c_{i_{T-1}}(T)=c_{i_{T-1}}-S_{i_T}(p^m_{i_T}) (p^m_{i_T}-c_{i_T})\rho _{i_T}.$

Otherwise if $-1+i_T>i_{T-1}$ we know from definition 1 that $S_{-1+i_T}(p_{-1+i_T})=S_{-1+i_T}(p^m_{-1+i_T})=S_{-1+i_T}^m$. So that theater $i_{T-1}$ solves

$$\begin{aligned} \begin{array}{ll} \max _{p_{i_{T-1}}\in \mathbb {R}} &{} S_{i_{T-1}}(p_{i_{T-1}}) (p_{i_{T-1}}-c_{i_{T-1}})(e_{i_{T-1}}+\rho _{i_{T-1}}d_{-1+i_{T-1}})\\ {} &{}+ S_{i_T}(p^m_{i_T}) (p^m_{i_T}-c_{i_T})\rho _{i_T}(e_{-1+i_T}+\rho _{-1+i_T}d_{-2+i_T})S^m_{-1+i_T} \end{array} \end{aligned}$$

which is equivalent to

$$\begin{aligned} \begin{array}{ll} \max _{p_{i_{T-1}}\in \mathbb {R}} &{} S_{i_{T-1}}(p_{i_{T-1}}) (p_{i_{T-1}}-c_{i_{T-1}})(e_{i_{T-1}}+\rho _{i_{T-1}}d_{-1+i_{T-1}})\\ {} &{}+ S_{i_T}(p^m_{i_T}) (p^m_{i_T}-c_{i_T})\rho _{i_T}\rho _{-1+i_T}d_{-2+i_T}S^m_{-1+i_T} \end{array} \end{aligned}$$

and proceeding upwards the objective of theater $i_{T-1}$ is

$$\begin{aligned} \begin{array}{ll} \max _{p_{i_{T-1}}\in \mathbb {R}}&{} S_{i_{T-1}}(p_{i_{T-1}}) (p_{i_{T-1}}-c_{i_{T-1}})(e_{i_{T-1}}+\rho _{i_{T-1}}d_{-1+i_{T-1}})\\ &{} + S_{i_T}(p^m_{i_T}) (p^m_{i_T}-c_{i_T})\rho _{i_T}\left( \prod _{1+i_{T-1}< j \le i_T}\rho _{j}S^m_{j}\right) d_{i_{T-1}} \end{array} \end{aligned}$$

or, using $d_{i_{T-1}}=S_{i_{T-1}}(p_{i_{T-1}})(e_{i_{T-1}}+\rho _{i_{T-1}}d_{-1+i_{T-1}})$

$$\begin{aligned} \begin{array}{l} \max _{p_{i_{T-1}}\in \mathbb {R}} S_{i_{T-1}}(p_{i_{T-1}}) (p_{i_{T-1}}-c_{i_{T-1}})(e_{i_{T-1}}+\rho _{i_{T-1}}d_{-1+i_{T-1}})\\ + S_{i_T}(p^m_{i_T}) (p^m_{i_T}-c_{i_T})\rho _{i_T}\left( \prod _{1+i_{T-1} <j\le i_T}\rho _{j}S^m_{j}\right) S_{i_{T-1}}(p_{i_{T-1}})(e_{i_{T-1}}+\rho _{i_{T-1}}d_{-1+i_{T-1}}) \end{array} \end{aligned}$$

In any case, the objective of theater $i_{T-1}$ may be written as

$$\begin{aligned} \max _{p_{i_{T-1}}\in \mathbb {R}}\; S_{i_{T-1}}(p_{i_{T-1}}) (p_{i_{T-1}}-c_{i_{T-1}}(T))(e_{i_{T-1}}+\rho _{i_{T-1}}d_{-1+i_{T-1}}) \end{aligned}$$

where

$$\begin{aligned} c_{i_{T-1}}(T)=c_{i_{T-1}}-S_{i_T}(p^m_{i_T}) (p^m_{i_T}-c_{i_T})\rho _{i_T}\prod _{1+i_{T-1} < j\le i_T}\rho _{j}S^m_j \end{aligned}$$

with the convention $\prod _{j \in J}=1$ whenever J is an empty set.

As we assumed in definition 1 that $\rho _i \ge 0$ and $S_{i_{T-1}}(c_{i_{T-1}})>0$ we have $S_{i_{T-1}}(c_{i_{T-1}}(T))>0$ and we then deduce the optimal pricing rule for theater $i_{T-1}$ is $p^m_{i_{T-1}}(c_{i_{T-1}}(T))$ with transparent notations.

Observe the objective functions of theaters $i_{T-1}$ and $i_T$ are now similar, we may proceed further upward using the same approach to derive the optimal pricing rule for all theaters in coalition T.

1.1.3 Lemma 4

Let us denote by $d_i=(e_i+\rho _{i}d_{i-1})S_i(p_i)$ the demand for the show of theater i. The result is trivial if the last theater leaves. If $n > 2$, consider $T \subset N,$ $1<i<n$ and $i \not \in T$ we have $p_i=p^m_i$ and $S_i(p_i)q_i=(e_i+\rho _{i}d_{i-1})S_i^m$ where $S_i^m=S_i(p^m_i).$

We may then write:

$$\begin{aligned} \begin{array}{l} d_1=e_1S_1(p_1) \\ d_2=(e_2+\rho _2d_1)S_2(p_2) \ldots \\ d_{i-1}=(e_{i-1}+\rho _{i-1}d_{i-2})S_{i-1}(p_{i-1}) \\ d_{i+1}=(e_{i+1}+\rho _{i+1}e_iS_i^m+\rho _{i+1}\rho _{i}S_i^md_{i-1})S_{i+1}(p_{i+1})\\ d_{i+2}=(e_{i+2}+\rho _{i+2}d_{i+1})S_{i+2}(p_{i+2}) \ldots \\ d_{n}=(e_{n}+\rho _{n}d_{n-1})S_{n}(p_{n}) \end{array} \end{aligned}$$

Finally if theater 1 leaves coalition T we have

$$\begin{aligned} \begin{array}{l} d_1=e_1S_1^m \\ d_{2}=(e_{2}+\rho _{2}e_1S_1^m) S_2(p_{2}) \\ d_{3}=(e_{3}+\rho _{3}d_2)S_3(p_{3}) \ldots \\ d_{n}=(e_{n}+\rho _{n}d_{n-1})S_{n}(p_{n}) \end{array} \end{aligned}$$

In any case, if theater i leaves, the remaining players enter a new festival game in which $e_{i+1}$ and $\rho _{i+1}$ are redefined.

1.1.4 Proposition 1

Using Lemma 3, we may consider the case $e=u_n.$

The case $n=2$ is easy since convexity amounts to show $v(\{1,2\})\ge v(\{1\})+v(\{2\})$. Now if 1 and 2 choose to keep their monopoly prices the total payoff of the grand coalition is exactly $v(\{1\})+v(\{2\}).$

Now assume the property is established for all festival games of size smaller or equal to $n-1$ and consider the case of game of size n. We have to show that $R \subset T$ and $i \not \in T$ imply

$$\begin{aligned} v(T \cup \{i\}) -v(T) \ge v(R \cup \{i\}) -v(R). \end{aligned}$$

But as showed in Lemma 4, if a theater leaves a festival, the remaining players are involved in a new festival with one player less. Hence, by induction we only need to establish that the previous inequality holds in the case $T \cup \{i\}=N$ and $R=N \backslash \{i,j\}.$ We then need to show

$$\begin{aligned} v(N)-v(N \backslash \{i\}) \ge v(N \backslash \{j\}) -v(N \backslash \{i,j\}). \end{aligned}$$

Notice that this inequality is equivalent to

$$\begin{aligned} v(N) \ge v(N \backslash \{i\}) + v(N \backslash \{j\})-v(N \backslash \{i,j\}), \end{aligned}$$

hence, by symmetry we may consider the case $j>i.$

First assume $i=1.$ Using lemma 2, we have

$$\begin{aligned} v(N)-v(N \backslash \{1\})=-v(N \backslash \{1\})+\max _{p_1}\left\{ S_1(p_1)(p_1-c_1)+ \frac{S_1(p_1)}{S_1^m}v(N \backslash \{1\})\right\} \end{aligned}$$

where $S_1^m=S_1(p^m_1).$ Similarly, we get (for $j >1$)

$$\begin{aligned} v(N \backslash \{j\})-v(N \backslash \{1,j\})=-v(N \backslash \{1,j\})+\max _{p_1}\left\{ S_1(p_1)(p_1-c_1)+ \frac{S_1(p_1)}{S_1^m}v(N \backslash \{1,j\})\right\} \end{aligned}$$

Consider the following function

$$\begin{aligned} f_1(x)=-x+\max _{p_1}\left\{ S_1(p_1)(p_1-c_1)+ \frac{S_1(p_1)}{S_1^m}x\right\} \end{aligned}$$

for positive values of x.

The envelope theorem asserts that $f_1'(x)=\frac{S_1(p_1(x))}{S_1^m}-1$ where

$$\begin{aligned} p_1(x)={{\,\mathrm{arg\,max}\,}}_{p_1} \left\{ S_1(p_1)(p_1-c_1)+ \frac{S_1(p_1)}{S_1^m}x\right\} = {{\,\mathrm{arg\,max}\,}}_{p_1} \left\{ S_1(p_1)\left( p_1-\left( c_1-\frac{x}{S_1^m}\right) \right) \right\} \end{aligned}$$

As x is a positive value, using Lemma 1 iii) and $S'()<0$ we get $f'_1(x)>0.$

Now we conclude, using $v(N \backslash \{1\}) \ge v(N \backslash \{1,j\}),$

$$\begin{aligned} f_1(v(N \backslash \{1\}))=v(N)-v(N \backslash \{1\}) \ge v(N \backslash \{j\}) -v(N \backslash \{1,j\})=f_1(v(N \backslash \{1,j\})) \end{aligned}$$

A similar argument may be used if $n \ge 3$ and $i>1.$ More precisely, let us write $v(N)-v(N \backslash \{i\})$ as

$$\begin{aligned}&\begin{array}{lll} \max _{p_i} \left\{ \right. &{}\max _{p_1,p_2,\ldots p_{i-1}} \left\{ \right. &{} S_1(p_1)(p_1-c_1)+\rho _2S_1(p_1)S_2(p_2)(p_2-c_2)+\ldots \\ &{}&{}\ldots + \rho _2\times \ldots \times \rho _{i-1}S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})(p_{i-1}-c_{i-1}) \\ &{}&{}+ \rho _2\times \ldots \times \rho _{i-1}\rho _i S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})\times S(p_i)(p_{i}-c_{i}) \\ &{}&{} + \frac{S_{i}(p_{i})}{S_{i}^m}\frac{S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})}{S_1^m\times \ldots \times S_{i-1}^m}v(N \backslash \{1,\ldots ,i\}) \\ &{}&{} \left. \right\} \\ &{}\left. \right\} \end{array}\\&\begin{array}{ll} - \max _{p_1,p_2,\ldots p_{i-1}} \left\{ \right. &{} S_1(p_1)(p_1-c_1)+\rho _2S_1(p_1)S_2(p_2)(p_2-c_2)+\ldots \\ &{}\ldots + \rho _2\times \ldots \times \rho _{i-1}S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})(p_{i-1}-c_{i-1}) \\ &{} + \frac{S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})}{S_1^m\times \ldots \times S_{i-1}^m}v(N \backslash \{1,\ldots ,i\}) \\ &{} \left. \right\} \\ \end{array} \end{aligned}$$

Accordingly, as we assumed $j>i$ we may write $v(N\backslash \{j\})-v(N \backslash \{i,j\})$ as

$$\begin{aligned}&\begin{array}{lll} \max _{p_i} \left\{ \right. &{}\max _{p_1,p_2,\ldots p_{i-1}} \left\{ \right. &{} S_1(p_1)(p_1-c_1)+\rho _2S_1(p_1)S_2(p_2)(p_2-c_2)+\ldots \\ &{}&{}\ldots + \rho _2\times \ldots \times \rho _{i-1}S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})(p_{i-1}-c_{i-1}) \\ &{}&{}+ \rho _2\times \ldots \times \rho _{i-1}\rho _i S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})\times S(p_i)(p_{i}-c_{i}) \\ &{}&{} + \frac{S_{i}(p_{i})}{S_{i}^m}\frac{S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})}{S_1^m\times \ldots \times S_{i-1}^m}v(N \backslash \{1,\ldots ,i,j\}) \\ &{}&{} \left. \right\} \\ &{}\left. \right\} \end{array}\\&\begin{array}{ll} - \max _{p_1,p_2,\ldots p_{i-1}} \left\{ \right. &{} S_1(p_1)(p_1-c_1)+\rho _2S_1(p_1)S_2(p_2)(p_2-c_2)+\ldots \\ &{}\ldots + \rho _2\times \ldots \times \rho _{i-1}S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})(p_{i-1}-c_{i-1}) \\ &{} + \frac{S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})}{S_1^m\times \ldots \times S_{i-1}^m}v(N \backslash \{1,\ldots ,i,j\}) \\ &{} \left. \right\} \\ \end{array} \end{aligned}$$

Following the previous argument let us define

$$\begin{aligned} \begin{array}{ll} g_i(p_i,x)= \max _{p_1,p_2,\ldots p_{i-1}} \left\{ \right. &{} S_1(p_1)(p_1-c_1)+\rho _2S_1(p_1)S_2(p_2)(p_2-c_2)+\ldots \\ &{} \ldots + \rho _2\times \ldots \times \rho _{i-1}S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})(p_{i-1}-c_{i-1}) \\ &{}+ \rho _2\times \ldots \times \rho _{i-1}\rho _i S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})\times S(p_i)(p_{i}-c_{i}) \\ &{} + \frac{S_{i}(p_{i})}{S_{i}^m}\frac{S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})}{S_1^m\times \ldots \times S_{i-1}^m}x \\ &{} \left. \right\} \\ h_i(x)=g_i(p^m_i,x)= \max _{p_1,p_2,\ldots p_{i-1}} \left\{ \right. &{} S_1(p_1)(p_1-c_1)+\rho _2S_1(p_1)S_2(p_2)(p_2-c_2)+\ldots \\ &{} \ldots + \rho _2\times \ldots \times \rho _{i-1}S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})(p_{i-1}-c_{i-1}) \\ &{} + \frac{S_1(p_1)\times \ldots \times S_{i-1}(p_{i-1})}{S_1^m\times \ldots \times S_{i-1}^m}x \\ &{} \left. \right\} \\ f_i(x)=\max _{p_i} \{g_i(p_i,x)\} -h_i(x) \end{array} \end{aligned}$$

We shall prove that $f_i'>0$ so the conclusion follows as in the case $i=1$. The envelope theorem applies, but one must be cautious for the sequences $p_1(x),\ldots ,p_{i-1}(x)$ for which the maximum of the function defining $g_i$ and $h_i$ are not the same. To this end, define for all given value of x the function $p_i(x)$ as the value of $p_i$ for which the function defining $g_i$ achieves its maximum and $p_1(p_i(x),x),\ldots ,p_{i-1}(p_i(x),x)$ the remaining terms of this sequence whereas $p_1(x),\ldots ,p_{i-1}(x)$ correspond to the sequence of prices associated with the definition of $h_i.$ We then have

$$\begin{aligned} \begin{array}{ll} f_i'(x)&=\frac{S_{i}(p_{i}(x))}{S_{i}^m}\frac{S_1(p_1(p_i(x),x))\times \ldots \times S_{i-1}(p_{i-1}(p_i(x),x))}{S_1^m\times \ldots \times S_{i-1}^m} -\frac{S_1(p_1(x))\times \ldots \times S_{i-1}(p_{i-1}(x))}{S_1^m\times \ldots \times S_{i-1}^m} \end{array} \end{aligned}$$

Now using again $x>0$ and Lemma 1 iii) we have $S_k(p_k(x))<S_k(p_k(p_i(x),x))$ for all $k \, \in \, \{1,i-1 \}$ and $\frac{S_{i}(p_{i}(x))}{S_{i}^m}>1$ hence $f_i'>0.$

1.1.5 Proposition 3

First consider the case $n=2.$ The two conditions on $\psi _1$ and $\psi _2$ write as

$$\begin{aligned} \begin{array}{ll} \psi _1(v)-v(\{1\})=\psi _2(v)-v(\{2\})\\ \psi _1(v)+\psi _2(v)=v(\{1,2\})\\ \end{array} \end{aligned}$$

Hence $\psi (v)$ is uniquely defined and the result holds for $n=2.$ Now consider $\psi$ coincides with the Shapley value for all $(N,v) \, \in \, \mathcal{F}$ , with $2\le m <n$ and consider $(N,v) \, \in \, \mathcal{F}.$ Using the recurrence hypothesis and the fact that removing one player leaves us with a festival game of size $n-1$, we have $\psi _i(v_{-j})=\phi _i(v_{-j})$ for all i, j where $\phi$ stands for the Shapley value. Now the linear system

$$\begin{aligned} \begin{array}{l} \psi _i(v)-\phi _i(v_{-j})=\psi _j(v)-\phi _j(v_{-i}) \, \forall \, i,j \, \in \, N\times N \\ \sum _{i \in N} \psi _i(v)=v(N) \end{array} \end{aligned}$$

admits a single solution hence $\psi$ coincides with the Shapley value on $\mathcal{F}.$

1.1.6 Proposition 4

The case $k=1$ is trivial since it leads to $\psi _1=v(N).$ Now let $T \subset N.$ By repeated uses for all $i \in T$ of the fact that removing one player leaves us with a (sub) festival game, we establish that $(N\backslash \{T\},v_{|N\backslash \{T\}}) \in \mathcal{F}$ (with straightforward notation).

Consider the case $k=2.$ Since $v_{|N\backslash \{T_1\}}$ and $v_{|N\backslash \{T_2\}}$ both belong to $\mathcal{F}$ we get

$$\begin{aligned} \begin{array}{l} \psi _1(v)-\psi _1(v_{|N\backslash \{T_2\}})=\psi _2(v)-\psi _2(v_{|N\backslash \{T_1\}}) \\ \psi _1(v)+\psi _2(v)=v(N) \end{array} \end{aligned}$$

which is equivalent to

$$\begin{aligned} \begin{array}{l} \psi _1(v)-v(T_2)=\psi _2(v)-v(T_1) \\ \psi _1(v)+\psi _2(v)=v(N) \end{array} \end{aligned}$$

and this system admits a unique solution. Finally assume the property has been established up to $k-1$ theaters and proceed as in the proof of Proposition 3 above.

1.1.7 Proposition 5

The proof follows van den Brink et al. (2011) and we provide it for completeness. First according to van den Brink et al. the downstream incremental solution satisfies the three above axioms. For sufficiency, consider (N, v) associated with $(e,c,\rho ,S)$ and let us start with theater 1. Consider the game $(N,v^{\star }_1)$ associated with $(e^{\star }, c^{\star }, \rho ^{\star }, S^{\star })$ where

$$\begin{aligned} \begin{array}{l} e^{\star }=(e_1,0,\ldots ,0)' \\ c^{\star }=c \\ \rho ^{\star }=(0,0,\ldots ,0)' \\ S^{\star }=S \end{array} \end{aligned}$$

By the no contribution property we have $\psi _i=0$ for all $i \ge 2$ and by efficiency we get $\psi _1=v(\{1\}).$ Now consider the property is fulfilled up to theater $i-1<n-1.$ Consider the game $(N,v^{\star }_i)$ associated with $(e^{\star }, c^{\star }, \rho ^{\star }, S^{\star })$ where

$$\begin{aligned} \begin{array}{l} e^{\star }=(e_1,e_2,\ldots ,e_{i-1},e_i,0,\ldots ,0)' \\ c^{\star }=c \\ \rho ^{\star }=(\rho _2, \rho _3,\ldots ,\rho _{i-2},\rho _{i},0,\ldots ,0)' \\ S^{\star }=S \end{array} \end{aligned}$$

By induction and the no contribution property we have $\psi _j=0$ for all $j >i.$ Now by induction and efficiency we get $\psi _i=v(N\backslash \{i+1,\ldots ,n\})-v(N\backslash \{i,\ldots ,n\}).$ Now for theater n by induction and efficiency we get $\psi _n=v(N)-v(N\backslash \{n\}).$

We now prove that Axioms 1 to 4 are logically independent. If all players receive nothing, this solution satisfied axioms 4 and 4 but not axiom 1. If we give v(N) to the last player, this rule satisfies axioms 1 and 4 but not axiom 3. Finally, the Shapley value satisfies axiom 1. As for axiom 3, remark the no-contribution property applies to player i if and only if $\max _{j \in N , j \le i}\{e_j\}=0.$ Now in this case, it is clear that the surpluses generated by player i are all zero, so the Shapley value also fulfills axiom 3. Consider the case $N=\{1,2\},$ player 1 gets $v(\{1\})+ \frac{1}{2}(v(N)-v(\{2\}))$ if we adopt the Shapley value. The quantity $v(\{1\})$ does not depend on $e_2.$ But, as a consequence of Lemma 3, $\frac{1}{2}(v(N)-v(\{2\}))$ is a linear function of $e_2$ and it is trivial to verify that this quantity strictly increases with $e_2$ whenever $\rho _2>0.$ We deduce the Shapley value does not verify axiom 4.

1.2 Supplementary material 1 : Next Festival

1.2.1 Program

Performance or artist	Dates	Place	Type	Price
name	(from 18/11. to 3/12 )			(full, in €)
M. A. Demey /J. van Dormael	18,19,20	Tournai	Dance	20
T. Castellucci /D. Dell	19	Schowburg	Dance	9
M. Depauw	19	Budascoop	Theater	14
N. Penino	19,20,22,23	Budascoop	Theater	9
B. Lachambre	22	Phenix	Dance	14
C. De Smedt	21, 22	Espace Pasolini	Dance	9
D. Veronese	22 to 26	Théâtre du Nord	Theater	20
F. Jaâbi /J. Baccar	23	Phenix	Theater	14
O. Dubois	23,24	Rose des Vents	Dance	14
A.C. Vandalem	24,25	Tournai	Theater	20
Gob Squad	25,27	Budascoop	Theater	14
I. Van Hove	26,27	Schouwburg	Theater	20
N. Lucas /H. Heisig	28	Espace Pasolini	Dance	9
L. Rodrigues	29	Phenix	Dance	14
R. Castellucci	29 30	Rose des Vents	Theater	20
C. Loemij /M. Lorimer	30	Schouwburg	Dance	9
E. Joris	1 to 3	Wazemmes	Theater	9
Berlin	30 to 3	Transfo	Theater	14
T. Castellucci /D. Dell	1,2	Budascoop	Dance	9
W. Vandekeybus	2,3	Rose des Vents	Dance	20
O. Normand /Y. Barelli	2	Espace Pasolini	Dance	14
Syndrome Collective	2	Schouwburg	Theater	0 (free event)

1.2.2 Survey Questions

• 1. Place where the survey has been conducted (9 items)

• 2. Where in this place (3 items : queue, bar, other)

• 3. Gender (2 items : Female, Male)

• 4. Current occupation (14 items)

• 5. “Where are you coming from” (free answer)

• 6. ZIP code (free answer)

• 7. Transportation device (5 items: pers. veh., someone else’s veh., public trans., Next bus, other)

• 8. If previous “other” describe (free answer)

• 9. Age (14 items : 5 years intervals from less than 15 to more than 74)

• 10. What kind of performance do think you are about to attend (2 items yes/no per category : classic, experimental, international masterpiece, new generation, some artist I know)

• 11. How many cultural performance do you attend on a yearly basis (4 item once a year, 2 to 4, 5 to 12, more than 12 per category : general, dance, theater, classic, experimental)

• 12. Do you see a difference between Next and the usual program ( 2 items yes/no)

• 13. If previous “yes” describe (free answer)

• 14. Do you attend other festivals (2 items yes/no)

• 15. If previous “yes” which one(s) (free answer)

• 16. Do you go to other theaters (2 items yes/no)

• 17. If previous “yes” which one(s) (open question)

• 18. Which other Next performances do you intend to /have you already see(n) (open question)

• 19. Do you have a personal practice of theater/dance (3 items : yes as amateur, yes as a professional, no)

• 20. How do you came to known NEXT’s festival (6 items)

• 21. How do you came to known NEXT’s program (8 items)

• 22. If previous “other” describe (free answer)

• 23. What type of tariff do you have (4 items unit price, subscription, free, other)

• 24. If previous “other” describe (free answer)

• 25. In case you know it what price did you pay (free answer)

• 26. Do you known the following theaters (4 items : by name, already went to, will soon go to, subscriber per theater : 9 possible)

• 27. Have you already answer this survey (2 items yes/no)

• 28. If previous “yes” at which performance (free answer)

1.2.3 Map and travelling distances

See Fig. 1 and Table 7.

Fig. 1

(1: Budascoop, 2: Espace Pasolini, 3: Rose des Vents, 4: Maison de la culture de Tournai, 5: Maison Folie de Wazemmes, 6: Phenix, 7: Schouwburg, 8:Théâtre du Nord, 9: Transfo)

Table 7 Distance and travelling times (by car) between major relevant cities

1.3 Supplementary material 2: inferential strategy

1.3.1 Survival functions

Our estimation strategy rests on ’robust’ assumptions about the choices of full prices. First, the monopoly price must be larger than the observed price since even if the theater leaves the coalition, it is against its own interest to set a price larger than the monopoly price. For $i \in N$, the monopoly price for a linear cost $c_i$ is

$$\begin{aligned} \frac{1}{3} \left( c_i+\sqrt{3\alpha _i^2+c_i^2} \right) \end{aligned}$$

Now when theater i joins the coalition, it chooses a price as if it was in a monopoly position with a smaller linear cost. We assume this effort cannot exceed the case in which the corrected value for c is negative, for otherwise it would be optimal to charge a negative price. Hence we have for all theater $i \in N$

$$\begin{aligned} \alpha _i \ge p_i \ge \alpha _i/\sqrt{3} \,\Leftrightarrow \, p_i \ge \alpha _i \ge \sqrt{3}p_i. \end{aligned}$$

As the prices are observed, we may calibrate $\alpha _i$ by interval regression using a specification such as $\alpha _i=\beta 'X_i+\epsilon _i$ where $X_i$ are some covariates and $\epsilon _i$ is an i.i.d. sample in the $\mathcal{N}(0,\sigma ^2)$ distribution. At this stage, the free event has not been used since, for this event we have $S_i=1$ by assumption.

1.3.2 Costs

Clearly, since the monopoly price is a given function of $\alpha _i$ and $c_i$ we may recover the cost from the previous estimates of $\alpha _i$ and some knowledge about the monopoly prices.

To this end, we used again a robust bracketing argument. We know that the observed price cannot be larger than the monopoly price. Now, the effort when joining a coalition is maximized when the theater assumes that the audiences of downward shows result only of spillovers from its own audience. Moreover we know in this case that the effect on the linear cost is exactly equal to the per theater sum of profit of downward firms. Hence the difference between $p^m_i-p_i$ cannot be larger than

$$\begin{aligned} \frac{1}{3}\sum _{j>i} \pi _j/S_j(p_j)q_j \end{aligned}$$

We do not observe directly this profit (and it is unlikely that the theaters could do so before the end of festival) but it cannot be larger than the sum of observed (and committed) observed prices.¹⁹ We then get the following brackets

$$\begin{aligned} p_i+ \frac{1}{3}\sum _{j>i} p_j \ge p^m_i \ge p_i \end{aligned}$$

We used this bracket to perform an estimation of the monopoly price assuming again a linear gaussian specification for the logarithm of the monopoly price. Using the forecast values of $p^m_i$ and $\alpha _i$ we may recover the estimated cost $c_i.$

1.3.3 Queues

Using the estimates for $\alpha _i$ as well as the actual prices we may recover the survival rate, which, together with observed attendances leads us to the estimated version of the queue $q_i$. At this stage we face another issue since 13 events appeared to be sold out. For these events the computation of the queue as the ratio of the observed attendance over the survival rate $1-(p_i/\alpha _i)^2$ is a right-censored version of the actual queue. To overcome this problem, we specified a Tobit model to estimate a corrected version of the queue.

The ‘sold out’ outcome may result from very different factors. Of course, notoriety is expected to play a major role, but the capacity of the theater, the presence of competing cultural events within or outside the festival may also be important. Unfortunately the sample is rather small and we have no information about competing events outside the festival.

We performed a Principal Component Analysis on the available exogenous variables and then used the factors as explanatory variables in the Tobit equation. Significant factors are 1,3,6 and 7.

1.3.4 Spillovers

Now, we need a strategy to decompose the estimated queue between newcomers and spillovers effects. According to Lemma 2 prices should be not correlated with newcomers. We may then use prices as instruments to decompose $q_i=e_i+\Delta _i.$

The best model we came up with (in terms of significance of coefficients and relevance of instruments) is a two stages least squares with capacity and factor 5 of the PCA as (possibly endogenous) explanatory variables together with actual and three previous prices as instruments. We are able to recover the spillovers up to some constant (see below how we did calibrate the intercept parameter).

1.3.5 Dispatching

Finally for each show i we must decompose $\Delta _i$ over all possible previous shows. Unfortunately, we do not have direct access to this information in the data. We assumed:

• spillover from more than 1 day are negligible;

• nobody attends the same show twice in a row;²⁰

• the transition probabilities are as in Table 1.

Consider for instance the seventh performance that took place on november the 20-th at Budascoop. This was the last performance of the play ‘Frustrating Picture Book for Adults’ by N. Penino and the audience was 58 people. First, the estimation of the queue tells us that $q_7=89$ people were initially interested by the show. The decomposition of the queue tells us that $\Delta _7=24$—out of 89—people have attended a show in the NEXT festival the day before. On November the 19-th four plays were displayed: ‘Kiss and Cry’ at Maison de la Culture de Tournai, ‘Cinquanta urlanti, quaranta ruggenti, sessanta stridenti’ at Schouwburg, ‘Eden Central’ at Budascoop and the first performance of ‘Frustrating Picture Book for Adults’ also at Budascoop. The spillover $\Delta _7$ has then be dispatched as follows. The probability for someone arriving at Budascoop from Maison de la Culture de Tournai conditionnally on the fact that he/she either come from Maison de la Culture de Tournai, Budascoop or Schouwburg may be computed from Table 1 and it is 12.5%. So the estimated spillover from ‘Kiss and Cry’ on 19-th to ‘Frustrating Picture Book for Adults’ on the 20-th is $\Delta _7 \times 0.125 = 3$ persons.

A final detail is in order. Recall we are only able to estimate spillover (and consequently, the newcomers) up to some constant. This constant may be bounded from above, as the number of newcomers cannot be negative. Also, we assumed that there is no word-of-mouth effect. This last assumption means that the total amount of spillover that may be attributed to a given performance cannot exceed the total audience of this performance. Also the constant may be bounded from below as spillover must be positive. To assess the sensitivity of our computations to this parameter, we computed the Shapley value using three different magnitudes (high, middle, low) but the differences are negligible.