A generalization of the Harsanyi NTU value to games with incomplete information

Abstract

In this paper, we introduce a solution concept generalizing the Harsanyi non-transferable utility (NTU) value to cooperative games with incomplete information. The so-defined S-solution is characterized by virtual utility scales that extend the Harsanyi-Shapley fictitious weighted-utility transfer procedure. We construct a three-player cooperative game in which Myerson’s (Int J Game Theory 13(2):69–96, 1984a) generalization of the Shapley NTU value does not capture some “negative” externality generated by the adverse selection. However, when we explicitly compute the S-solution in this game, it turns out that it prescribes a more intuitive outcome which takes into account the above mentioned informational externality.

Appendix

1.1 Proof of Proposition 4

Let $\mu _N$ be an S-solution of ${\varGamma }_C$ supported by $\lambda$ and $\eta =(\mu _S)_{S\subseteq N}$. We verify recursively conditions (i)-(iv). Because ${\varGamma }_C$ has complete information, there are no incentive constraints, which is equivalent to set $\alpha =0$, so that virtual utility reduces to $\lambda$-weighted utility and the egalitarian criterion in (12) becomes

$$\begin{aligned} \lambda _i\left( u_i(\mu _S)-u_i(\mu _{S\setminus j})\right) =\lambda _j\left( u_j(\mu _S)-u_j(\mu _{S\setminus i})\right) ,\quad \forall i,j\in S. \end{aligned}$$

(26)

For coalitions consisting of a single player i, it is clear that $u_i(\mu _i)=0$. For all two-person coalitions containing player 4, $D_{\{i,4\}}=\{[d_i,d_4]\}$. Then, it follows immediately that $u_i(\mu _{\{i,4\}})=u_4(\mu _{\{i,4\}})=0$ for all $i\in N\setminus 4$. Similarly, $D_{\{1,3\}}=\{[d_1,d_3]\}$, thus $u_1(\mu _{\{1,3\}})=u_3(\mu _{\{1,3\}})=0$. Consider now coalition $\{1,2\}$. It can be easily verified that an optimal egalitarian threat for this coalition satisfies:

$$\begin{aligned} \left( u_1(\mu _{\{1,2\}}),u_2(\mu _{\{1,2\}})\right) =\left\{ \begin{array}{ll} (2,2), &{} \quad \text {if}\;\lambda _1=\lambda _2>0 \\ (y,y),\; y\in [0,2], &{} \quad \text {if}\; \lambda _1=\lambda _2=0 \\ (0,0), &{} \quad \text {if}\;\lambda _1\ne \lambda _2 \end{array} \right. \end{aligned}$$

Similarly,

$$\begin{aligned} \left( u_2(\mu _{\{2,3\}}),u_3(\mu _{\{2,3\}})\right) =\left\{ \begin{array}{ll} (1,1), &{} \quad \text {if}\;\lambda _2=\lambda _3>0 \\ (y,y),\; y\in [0,1], &{} \quad \text {if}\; \lambda _2=\lambda _3=0 \\ (0,0), &{} \quad \text {if}\;\lambda _2\ne \lambda _3 \end{array} \right. \end{aligned}$$

We proceed now by cases.

Case 1:$\lambda _1=\lambda _2=\lambda _3>0$ Condition (26) applied to $S=\{1,2,3\}$ leads to Eqs.  (24) and (25). We have already shown in Sect. 7 that these two equations are incompatible.

Case 2:$\lambda _1=\lambda _2>0$, $\lambda _2\ne \lambda _3$ Without loss of generality, we can set $\lambda _1=\lambda _2=1$. It can be easily verified that, (26) implies: $u_1(\mu _{\{1,2,3\}})=u_2(\mu _{\{1,2,3\}})=u_1(\mu _{\{1,2,4\}})=u_2(\mu _{\{1,2,4\}})=2$ and $u_3(\mu _{\{1,2,3\}})=u_4(\mu _{\{1,2,4\}})=u_2(\mu _{\{2,3,4\}})=u_3(\mu _{\{2,3,4\}})=u_4(\mu _{\{2,3,4\}})=0$. Then, condition (26) applied to N reduces to

$$\begin{aligned} u_1(\mu _N)= & {} u_2(\mu _N) \end{aligned}$$

(27a)

$$\begin{aligned} u_1(\mu _N)= & {} \lambda _3u_3(\mu _N)+2 \end{aligned}$$

(27b)

$$\begin{aligned} u_1(\mu _N)= & {} \lambda _4u_4(\mu _N)+2 \end{aligned}$$

(27c)

On the other hand, we have that

$$\begin{aligned}&u_1(d_N)+u_2(d_N)+\lambda _3u_3(d_N)+\lambda _4u_4(d_N) \nonumber \\&\quad =\left\{ \begin{array}{ll} 2+3(\lambda _3+\lambda _4), &{} \quad \text {if}\; d_N=d^1_N,\,d^2_N\\ 6-\lambda _3+3\lambda _4, &{} \quad \text {if}\; d_N=d^3_N\\ 6+3\lambda _3-\lambda _4, &{} \quad \text {if}\; d_N=d^4_N \end{array} \right. \end{aligned}$$

(27d)

Subcase 2.1:$\lambda _3>1$, $\lambda _4>1$ Condition (i) implies that $\mu _N(d^3_N)=\mu _N(d^4_N)=0$. But then, (27a) requires that $\mu _N(d^1_N)=\mu _N(d^2_N)=1/2$. Hence, $u_1(\mu _N)=u_2(\mu _N)=1$ and $u_3(\mu _N)=u_4(\mu _N)=3$. With this, (27b) reduces to $\lambda _3=-1/3$, which is a contradiction.

Subcase 2.2:$\lambda _3>1$, $\lambda _4=1$ Condition (i) implies that $\mu _N(d^3_N)=0$. But then, $u_1(\mu _N)<\lambda _3u_3(\mu _N)+2$, which contradicts (27b).

Subcase 2.3:$\lambda _3>1$, $\lambda _4<1$ Condition (i) implies that $\mu _N(d^4_N)=1$. But then, (27c) reduces to $\lambda _4=-1$.

Subcase 2.4:$\lambda _3<1$, $\lambda _4\ge 1$ Condition (i) implies that $\mu _N(d^3_N)=1$. But then, $u_1(\mu _N)>\lambda _3u_3(\mu _N)+2$, which contradicts (27b).

Subcase 2.5:$\lambda _4<\lambda _3<1$ Condition (i) implies that $\mu _N(d^4_N)=1$. The same conclusion as in the case 2.3 is obtained.

Subcase 2.6:$\lambda _3<\lambda _4<1$ Condition (i) implies that $\mu _N(d^3_N)=1$. The same conclusion as in the case 2.4 is obtained.

Subcase 2.7:$0<\lambda _3=\lambda _4<1$ Condition (i) implies that $\mu _N(d^3_N)=1-\mu _N(d^4_N)=\beta$ with $\beta \in [0,1]$. Condition (27b) implies

$$\begin{aligned} \beta [1+\lambda _3]+(1-\beta )[1-3\lambda _3]=0\;\Rightarrow \; \beta =\frac{3\lambda _3-1}{4\lambda _3} \end{aligned}$$

Similarly, (27c) implies

$$\begin{aligned} \beta [1-3\lambda _3]+(1-\beta )[1+\lambda _3]=0\;\Rightarrow \; \beta =\frac{1+\lambda _3}{4\lambda _3} \end{aligned}$$

Therefore, $3\lambda _3-1=1+\lambda _3$, or equivalently, $\lambda _3=1$, which is a contradiction.

Subcase 2.8:$\lambda _3=\lambda _4=0$ As in the previous case, condition (i) implies that $\mu _N(d^3_N)=1-\mu _N(d^4_N)=\beta$ with $\beta \in [0,1]$. However, (27a) and (27b) imply that $u_1(\mu _N)=u_2(\mu _N)=2$, which cannot be satisfied by any such mechanism.

Case 3:$\lambda _1=\lambda _2=\lambda _3=0$ From (26) with $S=N$, $i=2$ and $j=4$, we get that $\lambda _4u_4(\mu _N)=0$ (since $u_4(\mu _{\{1,3,4\}})=0$). Hence, $\sum _{i\in N}\lambda _iu_i(\mu _N)=0$. However, condition (i) implies that $\sum _{i\in N}\lambda _iu_i(\mu _N)=3$, which is a contradiction.

Case 4:$\lambda _1=\lambda _2=0$, $\lambda _2\ne \lambda _3$ Condition (26) applied to $S=N$ with $i=2$ and $j=3,4$ gives $\lambda _3u_3(\mu _N)=\lambda _4u_4(\mu _N)=0$ (since $u_3(\mu _{\{1,3,4\}})=u_4(\mu _{\{1,3,4\}})=0$). Hence, $\sum _{i\in N}\lambda _iu_i(\mu _N)=0$. However, condition (i) implies that $\sum _{i\in N}\lambda _iu_i(\mu _N)>0$, which is a contradiction.

Case 5:$\lambda _1\ne \lambda _2$, $\lambda _2=\lambda _3>0$ Without loss of generality, we can set $\lambda _2=\lambda _3=1$. It can be easily verified that, (26) implies: $u_2(\mu _{\{1,2,3\}})=u_3(\mu _{\{1,2,3\}})=u_2(\mu _{\{2,3,4\}})=u_3(\mu _{\{2,3,4\}})=1$ and $u_1(\mu _{\{1,2,3\}})=u_1(\mu _{\{1,2,4\}})=u_2(\mu _{\{1,2,4\}})=u_4(\mu _{\{1,2,4\}})=u_4(\mu _{\{2,3,4\}})=0$. Then, condition (26) applied to N reduces to

$$\begin{aligned} \lambda _1u_1(\mu _N)= & {} u_2(\mu _N)-1 \end{aligned}$$

(28a)

$$\begin{aligned} \lambda _1u_1(\mu _N)= & {} u_3(\mu _N)-1 \end{aligned}$$

(28b)

$$\begin{aligned} \lambda _1u_1(\mu _N)= & {} \lambda _4u_4(\mu _N) \end{aligned}$$

(28c)

On the other hand, we have that

$$\begin{aligned}&\lambda _1u_1(d_N)+u_2(d_N)+u_3(d_N)+\lambda _4u_4(d_N) \nonumber \\&\quad =\left\{ \begin{array}{ll} 2+3(\lambda _1+\lambda _4), &{} \text {if}\; d_N=d^2_N,\,d^3_N\\ 6-\lambda _1+3\lambda _4, &{} \text {if}\; d_N=d^1_N\\ 6+3\lambda _1-\lambda _4, &{} \text {if}\; d_N=d^4_N \end{array} \right. \end{aligned}$$

(28d)

Subcase 5.1:$\lambda _1>1$, $\lambda _4>1$ Condition (i) implies that $\mu _N(d^2_N)=1-\mu _N(d^3_N)=\beta$ with $\beta \in [0,1]$. But then, since $u_2(\mu _N)=u_3(\mu _N)$ (by (28a) and (28b)), we must necessarily have that $\beta =1/2$. Therefore, $u_2(\mu _N)=u_3(\mu _N)=1$ and $u_1(\mu _N)=u_4(\mu _N)=3$. However, this together with (28a) imply that $\lambda _1=0$, which is a contradiction.

Subcase 5.2:$\lambda _1>1$, $\lambda _4=1$ Condition (i) implies that $\mu _N(d^1_N)=0$. Hence, (since $\lambda _1>2/3$) $\lambda _1u_1(\mu _N)>u_2(\mu _N)-1$, which contradicts (28a).

Subcase 5.3:$\lambda _1>1$, $\lambda _4<1$ Condition (i) implies that $\mu _N(d^4_N)=1$. With this, (28a) implies that $\lambda _1=2/3<1$, which is a contradiction.

Subcase 5.4:$\lambda _1<1$, $\lambda _4\ge 1$ Condition (i) implies that $\mu _N(d^1_N)=1$. This together with (28a) imply that $\lambda _1=-2$.

Subcase 5.5:$\lambda _1<\lambda _4<1$ Condition (i) implies that $\mu _N(d^1_N)=1$. The same conclusion as in case 5.4 is obtained.

Subcase 5.6:$\lambda _4<\lambda _1<1$ Condition (i) implies that $\mu _N(d^4_N)=1$. Hence, (28c) implies that $3\lambda _1=-\lambda _4$. Therefore, (since $\lambda _1,\lambda _2\ge 0$) $\lambda _1=\lambda _2=0$, which is a contradiction.

Subcase 5.7:$0<\lambda _4=\lambda _1<1$ Condition (i) implies that $\mu _N(d^1_N)=1-\mu _N(d^4_N)=\beta$, with $\beta \in [0,1]$. On the other hand, (28c) implies that $u_1(\mu _N)=u_4(\mu _N)$. Hence, we must have that $\beta =1/2$. But then, (28a) implies that $\lambda _1=2$, which is a contradiction.

Subcase 5.8:$0=\lambda _4=\lambda _1$ As in the previous case, condition (i) implies that $\mu _N(d^1_N)=1-\mu _N(d^4_N)=\beta$, with $\beta \in [0,1]$. Hence, $u_2(\mu _N)=u_3(\mu _N)=3$. However, by (28a) and (28b), $u_2(\mu _N)=u_3(\mu _N)=1$, which contradicts the previous fact.

Case 6:$\lambda _1\ne \lambda _2$, $\lambda _2=\lambda _3=0$ Condition (26) applied to $S=N$ with $i=2$ and $j=1,4$ gives $\lambda _1u_1(\mu _N)=\lambda _4u_4(\mu _N)=0$ (since $u_1(\mu _{\{1,3,4\}})=u_4(\mu _{\{1,3,4\}})=0$). Hence, $\sum _{i\in N}\lambda _iu_i(\mu _N)=0$. However, condition (i) implies that $\sum _{i\in N}\lambda _iu_i(\mu _N)=3(1+\lambda _4)>0$, which is a contradiction.

Case 7:$0=\lambda _2\ne \lambda _1$, $\lambda _2\ne \lambda _3$ Condition (26) applied to $S=N$ with $i=2$ and $j=1,3,4$ gives $\lambda _1u_1(\mu _N)=\lambda _3u_3(\mu _N)=\lambda _4u_4(\mu _N)=0$ (since $u_1(\mu _{\{1,3,4\}})=u_3(\mu _{\{1,3,4\}})=u_4(\mu _{\{1,3,4\}})=0$). Hence, $\sum _{i\in N}\lambda _iu_i(\mu _N)=0$. However, condition (i) implies that $\sum _{i\in N}\lambda _iu_i(\mu _N)=3(\lambda _1+\lambda _3+\lambda _4)>0$ (since we must have $\lambda _1>0$ and $\lambda _3>0$). But this is a contradiction.

Case 8:$\lambda _2>\lambda _1$, $\lambda _2\ne \lambda _3$ Condition (26) applied to $S=N$ with $i=1$ and $j=2$ gives $\lambda _1u_1(\mu _N)=\lambda _2u_2(\mu _N)$ (since $u_1(\mu _{\{1,3,4\}})=u_2(\mu _{\{2,3,4\}})=0$). On the other hand, condition (i) implies that $\mu _N(d^2_N)=0$. But then, this implies that $\lambda _1u_1(\mu _N)<\lambda _2u_2(\mu _N)$, which is a contradiction.

Case 9:$\lambda _2<\lambda _1$, $\lambda _2\ne \lambda _3$ As in the previous case, we have that $\lambda _1u_1(\mu _N)=\lambda _2u_2(\mu _N)$. We also have that condition (i) implies $\mu _N(d^1_N)=0$. But then, this implies that $\lambda _1u_1(\mu _N)>\lambda _2u_2(\mu _N)$, which is a contradiction.

We conclude that ${\varGamma }_C$ has no S-solution. This completes the proof.