The asymptotic core, nucleolus and Shapley value of smooth market games with symmetric large players

We examine the asymptotic nucleolus of a smooth and symmetric oligopoly with an atomless sector in a transferable utility (TU) market game. We provide sufficient conditions for the asymptotic core and the nucleolus to coincide with the unique TU competitive payoff distribution. This equivalence results from nucleolus of a finite TU market game belonging to its core, the core equivalence in a symmetric oligopoly with identical atoms and single-valuedness of the core in the limiting smooth game. In some cases (but not always), the asymptotic Shapley value is more favourable for the large traders than the nucleolus, in contrast to the monopoly case (Einy et al. in J Econ Theory 89(2):186–206, 1999), where the nucleolus allocation is larger than the Shapley value for the atom.

1. Recall, \(\frac{1}{r}\Pi _n\) contain players from all \(\{A_l\cap T_0,A_l\cap T_1\}\) such that \(\mu (\frac{1}{r}\Pi _n)=\frac{1}{r}\mu (\Pi _n)\).

2. Note it does not imply \(L=M\): with 3 commodities and \(L=\{1,2\}\), \(\omega _1=(1,1,0)\) and \(\omega _2=(0,1,1)\) are corner endowments of a non-trivial game.

3. The general results about existence of asymptotic Shapley value are systematised in Neyman (2002).

4. We wish to thank the anonymous referee for an idea leading to this construction.

Authors and Affiliations

1. Department of Economics, University of Haifa, Mount Carmel, 31905 , Haifa, Israel

   Avishay Aiche, Anna Rubinchik & Benyamin Shitovitz

Corresponding author

Correspondence to Benyamin Shitovitz.

A. Aiche’s contribution is part of his Ph.D. thesis being prepared at the University of Haifa under the supervision of Prof. B. Shitovitz.

Appendix A: Regular production functions

Proposition 5

For a collection \((F_l)_{l=1}^L\) of functions such that \(F_{l}\in {{\fancyscript{F}}}\) \(\forall l\), define \(f:{\hbox {I}\!\hbox {R}}_{+}^{M}\rightarrow {\hbox {I}\!\hbox {R}}_{+}\) by \(z\mapsto \underset{(y_{l})_l}{\max }\left\{ \sum _{l=1}^{L}F_{l}(y_{l}): \sum \nolimits _{l=1}^{L}y_{l}\le z\right\} \). Then \(f\in {{\fancyscript{F}}} \) and \(f^m(z)=F^m_{l}(y_{l})\) for \(m,l:(y_l)_m>0\), and for \((y_l)_l\) such that \(f(z)=\sum _{l=1}^{L}F_{l}(y_{l})\) and \(\sum \nolimits _{l=1}^{L}y_{l}\le z\).

Proof

Since the set \(\{( y_{l})_{l\in L}\in {\hbox {I}\!\hbox {R}}_{+}^{ML}:\sum \nolimits _{l=1}^{L}y_{l}\le z\}\) is compact and all \(F_{l}\) are continuous, the maximum is attained. Concavity of \(f\) follows from that of \(F_l\) and from the inequality \(f(\alpha z^1+(1-\alpha )z^2)\ge \max _{(y^1_l)_l,(y^2_l)_l:\sum y^1_l\le z^1,\sum y^2_l\le z^2}\sum _l F_l(\alpha y_l^1+(1-\alpha )y_l^2)\) for \(\alpha \in [0,1]\). Clearly, \(f(0)=0\) and \(f\) is non-negative and super-additive.

Next, we show continuity of \(f\) on \({\hbox {I}\!\hbox {R}}_{+}^{M}\).

Since \(f\) is concave it is also continuous on \({\hbox {I}\!\hbox {R}}_{++}^{M}\). Pick a sequence \((z^{k})_k\) converging to a point on the boundary, \(z^{*}\in {\hbox {I}\!\hbox {R}}_{+}^{M}\). If \(z^*=0\), the conclusion follows immediately since \(F_l(0)=0\,\forall l\). Assume now \(z^*\ne 0\). Let \(\hat{z}^{k}\equiv z^{k}\wedge z^{*}\), i.e., \(\hat{z}_{m}^{k}\equiv \min \{z_{m}^{k},z_{m}^{*}\}\) for every \(m\in \{1,\dots ,M\}\). So, \(\hat{z}^{k}\underset{k\rightarrow \infty }{\rightarrow }z^{*}\wedge z^{*}=z^{*}\). Denote by \(z^*_+\) the vector composed of strictly positive components of \(z^*\), so \(z_+^*\in {\hbox {I}\!\hbox {R}}^n_{++}\), where \(n\) is the number of elements in the set \(\left\{ m:\text { }z_{m}^{*}>0\right\} \). Let \(\hat{f}\) be the restriction of \(f\) to \({\hbox {I}\!\hbox {R}}_{+}^{n}\), i.e., \(\hat{f}(x)=f(x,0)\), where \(x\in {\hbox {I}\!\hbox {R}}^n_+\) and \((x,0)\in {\hbox {I}\!\hbox {R}}^M_+\). Then \(\hat{f}\) too is concave and therefore continuous on \({\hbox {I}\!\hbox {R}}_{++}^{n}\). Similarly, let \(\hat{z}^k_+\) be the vector composed of strictly positive components of \(\hat{z}^k\). Since \(\hat{z}^k\rightarrow z^*\), for \(k\) large enough, \(\hat{z}^k_+\in {\hbox {I}\!\hbox {R}}^n_{++}\). But \(f\) is non-decreasing, since \(F_l\) are so by Remark 1, hence \(f(z^{k})\ge \hat{f}(\hat{z}^{k})\rightarrow \hat{f}(\hat{z}^{*})\equiv f(z^{*})\). We conclude that \(\liminf f(z_{k})\ge f(z^{*})\).

To complete the proof of continuity consider again the converging sequence, \(z^{k}\rightarrow z^{*}\) and let \((y_{l}^{k})_{l\in L}\) be such that \(z_{k}=\sum _{l=1}^{L}y_{l}^{k}\) with \(f(z^{k})=\sum F_{l}(y_{l}^{k})\). By construction, for every \(l\) the sequence \(y_l^k\) is bounded, hence is convergent, so \(y_{l}^{k}\rightarrow y_{l}\), extracting a subsequence if necessary. Continuity of \(F_l\) yields \(f(z^{k})=\sum F_{l}(y_{l}^{k})\rightarrow \sum F_{l}(y_{l})\), and since \(\sum y_{l}=z^{*}\), \(f(z^{*})\ge \sum F_{l}(y_{l})\), by construction of \(f\). Thus we obtain \(f(z^{*})\ge {\limsup }f(z^{k})\). Therefore \(f\) is continuous.

It is left to prove existence of partial derivatives \(f^m(z^*)\) with respect to strictly positive coordinates. Let \(z^{*}\in {\hbox {I}\!\hbox {R}}_{+}^{M}\) with \(z_{m}^{*}>0\). So we need to show that \(f^{m}(z^{*})\) exists.

First, note that by concavity of \(f\) it is enough to show that its partial derivative from the left \(\left( D_{m}^{-}f\right) (z^{*})\) equals the partial derivative from the right \(\left( D_{m}^{+}f\right) (z^{*})\), or, using the definitions that \(\underset{\varepsilon \rightarrow 0^{+}}{\lim }\frac{f(z^{*}+\varepsilon e_{m})-f(z^{*})}{\varepsilon }=\underset{\varepsilon \rightarrow 0^{+}}{\lim }\frac{f(z^{*})-f(z^{*}-\varepsilon e_{m})}{\varepsilon }\), where \(e_{m}\) is the \(m\)-th unit vector.

Let \((y_{l})_{l\in L}\) be such that \(f(z^{*})=\sum F_{l}(y_{l})\) where \(\sum y_{l}\le z^{*}\). Take \(m,l^{*}\) such that \((y_{l^{*}})_{m}>0\), and hence \(z^*_m>0\). Obviously \(z^{*}+\varepsilon e_{m}=\sum _{l^{\prime }\ne l^{*}}y_{l^{\prime }}+(y_{l^{*}}+\varepsilon e_{m})\) and \(z^{*}-\varepsilon e_{m}=\sum _{l^{\prime }\ne l^{*}}y_{l^{\prime }}+(y_{l^{*}}-\varepsilon e_{m})\). Therefore, by definition of \(f\) we have: \(f(z^{*}+\varepsilon e_{m})\ge \sum _{l^{\prime }\ne l^{*}}F_{l^{\prime }}(y_{l^{\prime }})+F_{l^{*}}(y_{l^{*}}+\varepsilon e_{m})\) and \(f(z^{*}-\varepsilon e_{m})\ge \sum _{l^{\prime }\ne l^{*}}F_{l^{\prime }}(y_{l^{\prime }})+F_{l^{*}}(y_{l^{*}}-\varepsilon e_{m})\). This implies

Thus, \(\left( D_{m}^{+}f\right) (z^{*})\ge \left( D_{m}^{-}f\right) (z^{*})\). By the concavity of \(f\) we have \(\left( D_{m}^{+}f\right) (z^{*})\le \left( D_{m}^{-}f\right) (z^{*})\). Therefore \(f^m(z^*)=F^m_{l^*}(y_{l^*})\).

It follows that \(f\in {{\fancyscript{F}}} \).

\(\square \)

Appendix B: Existence of Shapley value for a simple duopoly game

Proposition 6

Let \(f:{\hbox {I}\!\hbox {R}}_+^M\rightarrow {\hbox {I}\!\hbox {R}}\) be concave, non-decreasing with \(f(0)=0\) and continuous at zero. Let \(T_0=[0,1]\), \(T_1=\{a_1,a_2\}\), \(\lambda _0\) be the Lebesgue measure on \([0,1]\), \(\lambda _1(a_i)=1\). For any \(S\in \Sigma \), let \(\mu (S)=\lambda _0([0,1]\cap S)\omega _0+\lambda _1(\{a_1,a_2\}\cap S)\omega _1\) such that \(\omega _0+2\omega _1\in {\hbox {I}\!\hbox {R}}^M_{++}\).

Then the game \(V=f\circ \mu \) has an asymptotic Shapley value.

Proof

Let ^{Footnote 4}

Let \(\sigma {\mathop {=}\limits ^{\mathrm{def}}}\lambda _0+2\lambda _1\). Note, that the difference between \(\sigma (S)\) and its integer part is \(\lambda _0\), while the integer part equals \(2\lambda _1\), whenever \(\lambda _0<1\). When \(\lambda _0=1\) then \(\sigma (S)\in \{1,3,5\}\) thus identifying the value of \(\lambda _1\in \{0,1,2\}\). Hence, the range of \(\sigma \) is \([0,1]\cup [2,3]\cup [4,5]\) and every element of the range uniquely identifies the pair \(\lambda _0,\lambda _1\). It follows then that \(g\circ \sigma \) is well-defined, and \(f\circ \mu =g\circ \sigma \) on \(\Sigma \). Since, by continuity of \(f\) at zero and construction of \(g\), \(\lim _{x\rightarrow 0}g(x)=\lim _{x\rightarrow 0} f(x\omega _0)=f(0)=g(0)=0\) and by concavity of \(f\) (hence its continuity in the interior), \(\lim _{x\rightarrow 5}g(x)=\lim _{x\rightarrow 5} f((x-4)\omega _0+2\omega _1)=f(\omega _0+2\omega _1)=g(5)\), \(g\) is continuous at the corners and is zero at zero. \(g\) is of bounded variation: indeed, by monotonicity of \(f\), \(g\) is monotonic on each of the intervals \(I_1,I_3,I_5\), respectively, while being zero elsewhere, so the total variation of \(g\) is \(2(f(1)+f(3))+f(5)<\infty \).

Following Hart (1973), define \(bv'\) as a set of real-valued functions of bounded variation defined on \([0,1]\) that are zero at zero and are continuous at \(0\) and \(1\), so, when normalised, \(g\in bv'\). Since also \(\sigma \in FL\), with \(FL\) the space of measures that can be represented as a sum of non-atomic measure and a measure with a finite carrier, \(g\circ \sigma \in bv'FL\). But since by Neyman (1979) (cf. Neyman 2002, thm. 4) \(bv'FL\subset ASYMP\), the conclusion follows.\(\square \)

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