Equity bargaining with common value

Abstract

We study a common-value bilateral bargaining model with equity offer. In particular, we consider a model in which players bargain over an equity share of a common-value stochastic pie (i.i.d. over time) and players receive private signals on the size of the pie each period. Efficient agreement is a stochastic rule: Delay is efficient if the expected size of today’s pie is small and the discount factor is high. Hence, information aggregation is crucial for efficiency. We derive the conditions under which an equilibrium that attains the efficient agreement exists. The key idea is that the proposer makes an offer in such a way that the responder will use her signal if the responder’s signal is crucial for an efficient agreement.

Appendices

Appendix 1: Proofs

1.1 A1.1 Proof of Lemma 2

Define, from the associated ICs,

$$\begin{aligned} \alpha _{1}=\frac{\Pr (h|h)\alpha (\ell )E[x|h,h]+\Pr (\ell |h)\delta W}{ E[x|h]},\ \ \ \alpha _{2}=\frac{\Pr (h|\ell )\alpha (\ell )E[x]+\Pr (\ell |\ell )\delta W}{E[x|\ell ]}. \end{aligned}$$

That is, $\alpha _{1}$ is the lowest $\alpha (h)$ satisfying (IC$_{h\ell }$) and $\alpha _{2}$ is the highest $\alpha (h)$ satisfying (IC$_{\ell h}$). To assure that there is $\alpha (h)\in [\alpha _{1},\alpha _{2}]$, we need $\alpha _{1}\le \alpha _{2}$. The condition $\alpha _{1}\le \alpha _{2}$ can be rewritten as

$$\begin{aligned}&\frac{\Pr (h|h)\alpha (\ell )E[x|h,h]+\Pr (\ell |h)\delta W}{E[x|h]}\le \frac{\Pr (h|\ell )\alpha (\ell )E[x]+\Pr \left( \ell |\ell \right) \delta W}{E[x|\ell ]}\\ \Leftrightarrow&[\Pr (h|h)\left( 1-\frac{\delta W}{E[x]}\right) E[x|h,h]+\Pr (\ell |h)\delta W]E[x|\ell ] \\&\le [\Pr (h|\ell )\left( 1-\frac{\delta W}{E[x]}\right) E[x]+\Pr (\ell |\ell )\delta W]E[x|h] \\ \Leftrightarrow&\Pr (h|h)E[x|h,h]E[x|\ell ]-\Pr (h|\ell )E[x]E[x|h] \\&\le \delta W[\{\Pr (h|h)\tfrac{E[x|h,h]}{E[x]}-\Pr (\ell |h)\}E[x|\ell ]+\{\Pr (\ell |\ell )-\Pr (h|\ell )\}E[x|h]] \\ \Leftrightarrow&\delta W\ge \frac{\Pr (h|h)E[x|h,h]E[x|\ell ]-\Pr (h|\ell )E[x]E[x|h]}{\{\Pr (h|h)\frac{E[x|h,h]}{E[x]}-\Pr (\ell |h)\}E[x|\ell ]+\{\Pr (\ell |\ell )-\Pr (h|\ell )\}E[x|h]} \end{aligned}$$

Define the RHS of the last inequality as A(q, L, H). Since $\delta W$ is increasing in $\delta ,$ $\delta W\ge A(q,L,H)$ defines a lower bound of $\delta$ for which $\alpha _{1}\le \alpha _{2}$ holds.²⁴

Simplifying the numerator of A(q, L, H) yields

$$\begin{aligned}&\Pr (h|h)E[x|h,h]E[x|\ell ]-\Pr (h|\ell )E[x]E[x|h] \\&\quad = \left( q^{2}H+(1-q)^{2}L\right) (qL+(1-q)H)-q(1-q)\left( H+L\right) (qH+(1-q)L) \\&\quad = \left( q^{2}(H+L)-(2q-1)L\right) (q(H+L)-(2q-1)H) \\&\qquad -q(1-q)\left( H+L\right) (q(H+L)-(2q-1)L) \\&\quad = (H+L)^{2}(q^{3}-q^{2}(1-q))+HL(2q-1)^{2}-(H+L)\\&\qquad \quad (2q-1)(qL+q^{2}H-q(1-q)L) \\&\quad = (H+L)^{2}q^{2}(2q-1)+HL(2q-1)^{2}-(H+L)^{2}q^{2}(2q-1) \\&\quad = HL(2q-1)^{2}. \end{aligned}$$

whereas simplifying the denominator of A(q, L, H) yields

$$\begin{aligned}&\left\{ \Pr (h|h)\frac{E[x|h,h]}{E[x]}-\Pr (\ell |h)\right\} E[x|\ell ]+\{\Pr (\ell |\ell )-\Pr (h|\ell )\}E[x|h] \\&\quad = \left[ \frac{2\left( q^{2}H+(1-q)^{2}L\right) }{H+L}-2q(1-q)\right] (qL+(1-q)H) \\&\qquad +\left[ (q^{2}+(1-q)^{2}-2q(1-q)\right] (qH+(1-q)L) \\&\quad = \left[ \frac{2\left( q^{2}(H+L)-(2q-1)L\right) }{H+L}-2q(1-q)\right] (q(H+L)-(2q-1)H) \\&\qquad +(2q-1)^{2}(q(H+L)-(2q-1)L) \\&\quad = \left[ (2q-1)^{2}+(2q-1)\frac{H-L}{H+L}\right] (q(H+L)-(2q-1)H) \\&\qquad +(2q-1)^{2}(q(H+L)-(2q-1)L) \\&\quad = (2q-1)^{2}(H+L)+(2q-1)\frac{H-L}{H+L}(q(H+L)-(2q-1)H) \\&\quad = (2q-1)(qH+(q-1)L)+2(2q-1)^{2}HL/(H+L) \end{aligned}$$

Hence

$$\begin{aligned} A(q,L,H)&=\frac{HL(2q-1)^{2}}{(2q-1)(qH+(q-1)L)+2(2q-1)^{2}HL/(H+L)} \\&=\left( \frac{q}{2q-1}\frac{1}{L}+\frac{q-1}{2q-1}\frac{1}{H}+\frac{1}{E[x] }\right) ^{-1} \end{aligned}$$

It is easy to see that A(q, L, H) is increasing in q.

Recall that E[x] is constant and $E[x|\ell ,\ell ]$ is decreasing in q, and that $E[x|\ell ,\ell ]=E[x]$ for $q=1/2.$ We can also see that

$$\begin{aligned} A(1,L,H)=\left( \frac{1}{L}+\frac{1}{E[x]}\right) ^{-1},\ \ \ A(1/2,L,H)=0, \end{aligned}$$

and

$$\begin{aligned} \left( \frac{1}{L}+\frac{1}{E[x]}\right) ^{-1}&<\left( \frac{2}{E[x]} \right) ^{-1}=\frac{E[x]}{2}, \\ \left( \frac{1}{L}+\frac{1}{E[x]}\right) ^{-1}&>\left( \frac{2}{L}\right) ^{-1}=\left. \frac{E[x|\ell ,\ell ]}{2}\right| _{q=1}. \end{aligned}$$

This shows that $E[x]>2A(q,L,H)$ for all q,  and there is a threshold $\tilde{q}\in (0,1)$ such that $E[x|\ell ,\ell ]>2A(q,L,H)$ for $q<\tilde{q}$ and $E[x|\ell ,\ell ]<2A(q,L,H)$ for $q>\tilde{q}.$

1.2 A1.2 Proof of Lemma 3

Consider $\alpha (h)$ satisfying (IC$_{h\ell }$) and (IC$_{\ell h}$). We first show that if $E[x]>2\delta W,$ we can take $\alpha (h)$ to be smaller than $\alpha (\ell ).$ It suffices to show that $\alpha _{1}$ in Lemma 2 is smaller than $\alpha (\ell ).$ Indeed

$$\begin{aligned} \alpha _{1}-\alpha (\ell )&=\frac{\Pr (h|h)\alpha (\ell )E[x|h,h]+\Pr (\ell |h)\delta W}{E[x|h]}-\left( 1-\frac{\delta W}{E[x]}\right) \\&=\left( \frac{\Pr (h|h)E[x|h,h]}{E[x|h]}-1\right) \left( 1-\frac{\delta W}{E[x]}\right) +\frac{\Pr (\ell |h)\delta W}{E[x|h]} \\&=\left( \frac{{-}\Pr (\ell |h)E[x]}{E[x|h]}\right) \left( 1{-}\frac{\delta W}{E[x]}\right) {+}\frac{\Pr (\ell |h)\delta W}{E[x|h]}\\&=\frac{\Pr (\ell |h)(-E[x]{+}2\delta W)}{E[x|h]}<0. \end{aligned}$$

Next we show that we can take $\alpha (h)$ such that $(1-\alpha (h))E[x|\ell ,\ell ]<\delta W\Leftrightarrow \alpha _{2}>1-\delta W/E[x|\ell ,\ell ].$ We verify this inequality by establishing $\alpha _{2}\ge 1-\delta W/E[x|\ell ]$ $(>1-\delta W/E[x|\ell ,\ell ]).$ Indeed

$$\begin{aligned} \alpha _{2}-\left( 1-\frac{\delta W}{E[x|\ell ]}\right)&=\frac{\Pr (h|\ell )\alpha (\ell )E[x]+\Pr (\ell |\ell )\delta W}{E[x|\ell ]}-\left( 1-\frac{\delta W}{ E[x|\ell ]}\right) \\&=\frac{\Pr (h|\ell )(E[x]-\delta W)+\Pr (\ell |\ell )\delta W}{E[x|\ell ]} -\left( 1-\frac{\delta W}{E[x|\ell ]}\right) \\&=\frac{\Pr (h|\ell )E[x]-E[x|\ell ]+2\Pr (\ell |\ell )\delta W}{E[x|\ell ]}\\&=\frac{\Pr (\ell |\ell )(-E[x|\ell ,\ell ]+2\delta W)}{E[x|\ell ]}\ge 0. \end{aligned}$$

1.3 A1.3 Proof of Lemma 4

Step 1 If an efficient equilibrium with $\alpha (h)>\alpha (\ell )$ exists, then $\alpha (h)=\alpha ^{*}$ i.e., $(1-\alpha (h))E[x]=\delta W.$

Proof Since the responder with $\ell$ signal must accept $\alpha (h)$ in efficient equilibrium, $(1-\alpha (h))E[x]\ge \delta W.$ Suppose $(1-\alpha (h))E[x]>\delta W$ and consider $\alpha \in (\alpha (h),\alpha ^{*}].$ Note that $(1-\alpha )E[x]>\delta W$ so that (i) if the responder believes the proposer has h signal, she accepts even if her signal is $\ell ,$ and (ii) if the responder believes the proposer has $\ell$ signal,  she accepts if and only if her signal is h. In other words, the responder’s reaction to $\alpha$ is the same as when $\alpha (h)$ is offered (case (i)) or when $\alpha (\ell )$ is offered (case (ii)). This implies that either the proposer with h signal or the one with $\ell$ signal has an incentive to deviate to offer $\alpha ,$ contradicting to equilibrium. Hence, $(1-\alpha (h))E[x]=\delta W.$

Step 2 If an efficient equilibrium with $\alpha (h)>\alpha (\ell )$ exists, then there is some $\alpha ^{h\prime }<\alpha ^{*}$ such that

In other words, the nonmimicry conditions for an efficient equilibrium hold.

Proof If an efficient equilibrium with $\alpha (h)=\alpha ^{*}>\alpha (\ell )$ exists, we have the following nonmimicry conditions

Since $\alpha ^{*}>\alpha (\ell ),$ the latter inequality implies

$$\begin{aligned} \alpha ^{*}E[x|\ell ]<\Pr (h|\ell )\alpha ^{*}E[x] +\Pr (\ell |\ell )\delta W. \end{aligned}$$

Note also that $\alpha ^{*}=1-\delta W/E[x]>1/2,$ since $E[x]>2\delta W.$ This implies

$$\begin{aligned} \alpha ^{*}E[x|h]= & {} \Pr (h|h)\alpha ^{*}E[x|h,h]+\Pr (\ell |h)\alpha ^{*}E[x] \\>&\Pr (h|h)\alpha ^{*}E[x|h,h]+\Pr (\ell |h)2\alpha ^{*}\delta W \\> & {} \Pr (h|h)\alpha ^{*}E[x|h,h]+\Pr (\ell |h)\delta W \end{aligned}$$

We then have the following inequalities:

$$\begin{aligned} \alpha ^{*}E[x|h]&>\Pr (h|h)\alpha ^{*}E[x|h,h]+\Pr (\ell |h)\delta W, \\ \alpha ^{*}E[x|\ell ]&<\Pr (h|\ell )\alpha ^{*}E[x] +\Pr (\ell |\ell )\delta W. \end{aligned}$$

Now we substitute the $\alpha ^{*}$ in the left-hand sides of the inequalities with $\alpha ^{h\prime }$. Since the inequalities are strict, slightly lowering $\alpha ^{h\prime }$ does not change the inequalities. The resulting inequalities are indeed the nonmimicry conditions to be verified.

Appendix 2: Equilibrium for no delay schedule

Suppose that the discount factor is such that No Delay Schedule is efficient. To attain this outcome in equilibrium, the proposer’s offer must be accepted regardless of his own and the responder’s signals. The proposer’s offer must be pooling, since otherwise he has an incentive to always make the more favorable claim of the possible offers. Let $\alpha ^{p}\$be the proposer’s pooling offer. The responder with $\ell$ signal accepts $\alpha ^{p}$ if

$$\begin{aligned} (1-\alpha ^{p})E[x|\ell ]\ge \delta W(\delta ,q,L,H)=\frac{\delta E[x]}{2}. \end{aligned}$$

Among such shares, the best one for the proposer is the highest offer. As we will see in the following, the main problem in constructing an efficient equilibrium is to deter the proposer from making an off-path offer. Note that the higher the on-path pooling offer, the weaker the proposer’s incentive to make an off-path offer. We thus focus on an equilibrium with the highest offer in order to weaken the condition to attain an efficient equilibrium:

$$\begin{aligned} (1-\alpha ^{p})E[x|\ell ]=\delta W\Leftrightarrow \alpha ^{p}=1-\frac{\delta E[x]}{2E[x|\ell ]}. \end{aligned}$$

We consider the following pessimistic off-path belief:

$$\begin{aligned} \beta (\alpha ,s_{r}{=}\ell ){=}\left\{ \begin{array}{ll} 2q(1-q) &{}\quad \text {if }\alpha \le \alpha ^{p} \\ 0 &{}\quad \text {otherwise} \end{array} \right. ,\ \ \ \beta (\alpha ,s_{r}{=}h){=}\left\{ \begin{array}{ll} q^{2}+(1-q)^{2} &{}\quad \text {if }\alpha \le \alpha ^{p} \\ 0 &{}\quad \text {otherwise} \end{array} \right. . \end{aligned}$$

Unlike the previous cases, the responder’s on-path belief depends on her own signal due to the correlation of the signals. It is important to set zero belief for higher $\alpha$ to deter the proposer’s deviation; as before, the responder needs to change her acceptance behavior for higher $\alpha .$ Note, however, that $(1-\alpha ^{p})E[x]>(1-\alpha ^{p})E[x|\ell ]=\delta W,$ so that the responder with h signal is willing to accept an $\alpha$ slightly higher than $\alpha ^{p}$ even though the belief is pessimistic. Accordingly, the responder’s optimal reaction to offer $\alpha$ is to accept always for $\alpha \le \alpha ^{p},$ to accept if the responder’s signal is h for $\alpha \in (\alpha ^{p},\alpha ^{**}]$ where $\left( 1-\alpha ^{**}\right) E[x]=\delta W,$ and to reject always for $\alpha >\alpha ^{**}$ (note that $\alpha ^{**}=1-\delta /2$ since $W=E[x]/2).$

Therefore, if an efficient equilibrium exists, it should take the following form:

• The proposer’s offer strategy is $\alpha (h)=\alpha (\ell )=\alpha ^{p}.$

• The responder’s strategy is

  $$\begin{aligned} \text {for }\alpha&>\alpha ^{*},\ \sigma (\alpha ,s_{r})=\text {reject for }s_{r}=h,\ell , \\ \text {for }\alpha&\in (\alpha ^{p},\alpha ^{**}],\, \sigma (\alpha ,s_{r})=\left\{ \begin{array}{ll} \text {accept} &{} \text {for }s_{r}=h, \\ \text {reject} &{} \text {for }s_{r}=\ell , \end{array} \right. \\ \text {for }\alpha&\le \alpha ^{p},\quad \sigma (\alpha ,s_{r})=\text {accept for }s_{r}=h,\ell . \end{aligned}$$

• The belief is the above pessimistic belief system.

1.1 A2.1 Proposer’s off-path incentive condition

Since the responder’s reaction is constant over each of the intervals, $[0,\alpha ^{p}],(\alpha ^{p},\alpha ^{**}],$ and $(\alpha ^{**},1],$ the offers other than $\alpha ^{p},\alpha ^{**},$ and 1 are (weakly) dominated. In addition, due to the fact that

$$\begin{aligned} \alpha ^{p}E[x|\ell ]=E[x|\ell ]-\delta W>E[x|\ell ,\ell ]-\delta W\ge 2\delta W-\delta W=\delta W, \end{aligned}$$

offer $\alpha =1$ is also dominated by $\alpha ^{p}.$

To deter deviation to $\alpha ^{**}$, the following conditions must hold:

The second constraint, (IC$_{\ell }$), is indeed equivalent to the condition under which No Delay Schedule is efficient:

$$\begin{aligned}&\left( 1-\frac{\delta W}{E[x|\ell ]}\right) E[x|\ell ]\ge \Pr (h|\ell )E[x]\left( 1-\frac{\delta W}{E[x]}\right) +\Pr (\ell |\ell )\delta W \\&\Leftrightarrow \delta W\left( -\Pr (h|\ell )+1+\Pr (\ell |\ell )\right) \le -\Pr (h|\ell )E[x]+E[x|\ell ]\Leftrightarrow 2\delta W\le E[x|\ell ,\ell ]. \end{aligned}$$

On the other hand, the first condition, (IC$_{h}$), does not always hold even when No Delay Schedule is efficient. The following technical lemma gives the requirement for (IC$_{h}$):

Lemma 5

(IC$_{h}$) holds iff Condition B is satisfied, i.e., $\delta W\le B(q,L,H),$ where

$$\begin{aligned} B(q,L,H)=\frac{H+L}{2}\left( 2+\frac{2q-1}{2q(1-q)}\frac{H-L}{H+L}\frac{ qH+(1-q)L}{qL+(1-q)H}\right) ^{-1}, \end{aligned}$$

with the following properties:

• 1. $2B(1,L,H)=0,$ and $2B(1/2,L,H)=E[x]=\left. E[x|\ell ,\ell ]\right| _{q=1/2}$

• 2. If $L>0,$ then there is a $\tilde{q}$ such that $2B(q,L,H)>E[x|\ell ,\ell ]$ for $q\in (1/2,\tilde{q}),$ and $2B(q,L,H)<E[x|\ell ,\ell ]$ for $q> \tilde{q}.$

• 3. If $L=0,$ $2B(q,L,H)\ge E[x|\ell ,\ell ]$ for all q (equality holds for $q=1/2$, 1).

Proof

Rearranging (IC$_{h}$), we have

$$\begin{aligned}&\left( 1-\delta W/E[x|\ell ]\right) E[x|h]\ge (1-\delta W/E[x])\Pr (h|h)E[x|h,h]+\Pr (\ell |h)\delta W \\&\Leftrightarrow \delta W\left( \frac{E[x|h]-\Pr (\ell |h)E[x]}{E[x]}-\frac{ E[x|h]}{E[x|\ell ]}-\Pr (\ell |h)\right) \ge -\Pr (\ell |h)E[x] \\&\Leftrightarrow \delta W\le \frac{\Pr (\ell |h)E[x]}{2\Pr (\ell |h)-E[x|h](1/E[x]-1/E[x|\ell ])} \end{aligned}$$

Define

$$\begin{aligned} B(q,L,H)&=\frac{\Pr (\ell |h)E[x]}{2\Pr (\ell |h)-E[x|h]\left( 1/E[x]-1/E[x|\ell ]\right) } \\&=\frac{q(1-q)(H+L)}{4q(1-q)-(qH+(1-q)L))(2/(H+L)-1/(qL+(1-q)H))} \\&=\frac{H+L}{2}\left( 2+\frac{2q-1}{2q(1-q)}\frac{H-L}{H+L}\frac{qH+(1-q)L}{ qL+(1-q)H}\right) ^{-1} \end{aligned}$$

It is easy to see

$$\begin{aligned} B(1,L,H)&=0 \\ B(1/2,L,H)&=\frac{E[x]}{2}=\left. \frac{E[x|\ell ,\ell ]}{2}\right| _{q=1/2}. \end{aligned}$$

By definition,

$$\begin{aligned}&2B(q,L,H)-E[x|\ell ,\ell ] \\&=\frac{2\Pr (\ell |h)E[x]}{2\Pr (\ell |h)-E[x|h]\left( 1/E[x]-1/E[x|\ell ]\right) }-E[x|\ell ,\ell ] \\&=\frac{2E[x]^{2}\Pr (\ell |h)E[x|\ell ]{-}E[x|\ell ,\ell ][2E[x]\Pr (\ell |h)E[x|\ell ]{-}E[x|h]\left( E[x|\ell ]{-}E[x]\right) ]}{2E[x]\Pr (\ell |h)E[x|\ell ]{-}E[x|h]\left( E[x|\ell ]{-}E[x]\right) } \end{aligned}$$

Note that the denominator is positive for all q. The numerator is

$$\begin{aligned}&2E[x]^{2}\Pr (\ell |h)E[x|\ell ]-E[x|\ell ,\ell ][2E[x]\Pr (\ell |h)E[x|\ell ]-E[x|h]\left( E[x|\ell ]-E[x]\right) ] \\&=2E[x]\Pr (\ell |h)E[x|\ell ][E[x]-E[x|\ell ,\ell ]]-E[x|\ell ,\ell ]E[x|h]\left( E[x]-E[x|\ell ]\right) ] \\&=(H+L)2q(1-q)(qL+(1-q)H)[\frac{H+L}{2}-\frac{q^{2}L+(1-q)^{2}H}{ q^{2}+(1-q)^{2}}] \\&\quad -\,\frac{q^{2}L+(1-q)^{2}H}{q^{2}+(1-q)^{2}}(qH+(1-q)L)[\frac{H+L}{2} -(qL+(1-q)H)] \\&=\frac{(2q-1)(H-L)}{2(q^{2}+(1-q)^{2})} [2(H+L)q(1-q)(qL+(1-q)H)-(q^{2}L+(1-q)^{2}H)\\&\qquad (qH+(1-q)L)] \end{aligned}$$

The fraction in the last line is positive for $q>1/2.$ If $q=1/2,$ the expression in the square bracket is

$$\begin{aligned} 2(H+L)\frac{1}{4}\frac{H+L}{2}-\frac{H+L}{4}\frac{H+L}{2}=\frac{(H+L)^{2}}{8} >0 \end{aligned}$$

The expression also can be rewritten as

$$\begin{aligned}&2(H+L)q(1-q)(qL+(1-q)H)-(q^{2}L+(1-q)^{2}H)(qH+(1-q)L) \\&=((1-q)H^{2}+qL^{2}+5HL)q(1-q)-HL \end{aligned}$$

Since $((1-q)H^{2}+qL^{2})$ and $q(1-q)$ are decreasing for $q\in (1/2,1),$ the expression is monotone decreasing in q and goes to $-HL.$ Therefore, as long as $L>0,$ there is $\tilde{q}\in (1/2,1)$ such that $2B(q,L,H)<E[x|\ell ,\ell ]$ iff $q>\tilde{q}.$ $\square$

The condition $\delta W\le B(q,L,H)$ and the properties of B(q, L, H) provide information about when No Delay Schedule is efficient but (IC$_{h}$) fails (see also Fig. 5 below). If $L=0,$ it never fails since $2B(q,L,H)\ge E[x|\ell ,\ell ]\ge 2\delta W$ always holds under No Delay Schedule being efficient. However, if $L>0,$ there is a $\tilde{q}$ such that $q>\tilde{q}\Rightarrow 2B(q,L,H)<E[x|\ell ,\ell ]$; hence, $2B(q,L,H)<2\delta W\le E[x|\ell ,\ell ]$ can arise.

The properties of B(q, L, H) imply that (IC$_{h}$) would be harder to hold as $\delta$ becomes higher, and the threshold of $\delta$ decreases as q increases. Intuitively, if the signal accuracy is very high, the proposer with h signal would expect that it is very likely that the responder has the same signal and accepts the most profitable deviation offer $\alpha ^{**}>\alpha .$ To deter such deviation, the difference between $\alpha ^{**}-\alpha ^{p}=\delta (E[x]/E[x|\ell ]-1)/2$ must be small enough. This is why $\delta$ needs to be small (this also makes forgoing till the next period more costly, which contributes to (IC$_{h}$) being satisfied). Moreover, the lemma confirms that if q is above a certain level, the threshold of $\delta$ is below the level for which No Delay Schedule is efficient. This immediately implies that a fully efficient equilibrium for No Delay Schedule fails to exist if the signal accuracy is relatively high and if the discount factor is not too low.

Proposition 2

(Part 3, Restatement) Suppose $\delta \in [0,\underline{\delta }(q,L/H)]$ (No Delay Schedule is efficient). The strategy profile and the belief system defined above compose an efficient equilibrium if and only if Condition B holds, i.e., $\delta W\le B(q,L,H)$.

We plot the diagram for $L/H=.6,$ in which $\delta W\le B(q,L,H)$ holds in the area below the dotted (red) curve (Condition B in Figure 5). An efficient No Delay equilibrium does not exist in the region below the dashed (blue) curve and above the dotted (red) curve. Therefore, efficiency is not always attainable. As the discount factor decreases and/or the signal accuracy increases, the benefit to the proposer with a high signal of using signal-responsive offers becomes larger. Hence, the deviation temptation from a pooling offer is too high to sustain the efficient pooling equilibrium.

Fig. 5

The Region for Efficient Equilibria for No Delay Schedule

Consider the perfect information case, $q=1,$ in which efficiency is obtained by separating offers conditional on the signals. Notice that this efficient equilibrium cannot be achieved in the limit of efficient equilibria as $q\rightarrow 1.$ Instead, a sequence of separating equilibria can be made arbitrarily close to the efficient equilibrium by the approximate efficiency result in the following.

Comparative Statics for No Delay Schedule

For an efficient equilibrium under No Delay Schedule, we only consider how the proposer’s offer $\alpha ^{p}$ is affected by signal accuracy (q) and the relative size of the pie (L / H). We do not consider the effect on time to agreement because the players always agree immediately in an efficient equilibrium under No Delay Schedule (remember that the discount factor was too low so the players prefer to agree immediately under No Delay Schedule). The equilibrium offer $\alpha ^{p}$ is determined by $(1-\alpha ^{p})E[x|\ell ]=\delta W$, which can be rewritten as

$$\begin{aligned} \alpha ^{p}=1-\frac{\delta E[x]}{2E[x|\ell ]}=1-\delta \frac{L/H+1}{ 2(qL/H+1-q)}. \end{aligned}$$

Hence, the equilibrium offer $\alpha ^{p}$ is decreasing in q and increasing in L / H. Continuation value W increases in the size of L given H,  and is constant in q,  whereas the expected pie size,  $E[x|\ell ],$ is increasing in the size of L given H,  and decreasing in q. This shows why $\alpha ^{p}$ is decreasing in q. The intuition for $\alpha ^{p}$ increasing in L / H is analogous to that for Medium Delay Schedule.

Approximate Efficiency of No Delay Schedule

For the case of approximate efficiency in No Delay Schedule, consider a separating equilibrium in which the proposer offers $\alpha (h)>\alpha (\ell )$, and the responder accepts $\alpha (h)$ if she receives h signal and always accepts $\alpha (\ell )$. The ex ante value from this agreement schedule is

$$\begin{aligned} V_{s}=\frac{1}{2}(E[x]-\Pr (h,\ell )(E[x]-2\delta V_{s}))=\frac{1-\Pr (h,\ell )}{1-\delta \Pr (h,\ell )}\frac{E[x]}{2}. \end{aligned}$$

The responder’s incentive conditions are

$$\begin{aligned} (1-\alpha (h))E[x|h,h]&\ge \delta V_{s}\ge (1-\alpha (h))E[x], \\ (1-\alpha (\ell ))E[x|\ell ,\ell ]&\ge \delta V_{s}. \end{aligned}$$

Again we assume the pessimistic belief for off-path offers.

The proposer’s off-path incentive condition implies $(1-\alpha (\ell ))E[x|\ell ,\ell ]\le \delta V_{s};$ otherwise the proposer can deviate to offer $\alpha ^{\prime },$which is slightly higher than $\alpha (\ell )$ but still acceptable. Hence,

$$\begin{aligned} \alpha (\ell )=1-\frac{\delta V_{s}}{E[x|\ell ,\ell ]}. \end{aligned}$$

We consider the case in which the responder receives no rent, i.e.,

$$\begin{aligned} (1-\alpha (h))E[x|h,h]=\delta V_{s}\Leftrightarrow \alpha (h)=1-\frac{\delta V_{s}}{E[x|h,h]}. \end{aligned}$$

Consider the nonmimicry conditions:

$$\begin{aligned} \Pr (h|h)\alpha (h)E[x|h,h]+\Pr (\ell |h)\delta V_{s}&\ge \alpha (\ell )E[x|h] \\ \Pr (h|\ell )\alpha (h)E[x]+\Pr (\ell |\ell )\delta V_{s}&\le \alpha (\ell )E[x|\ell ] \end{aligned}$$

The first IC is redundant for a sufficiently large q, since $q\rightarrow 1$ implies

$$\begin{aligned} \alpha (h)E[x|h,h]&=(1-\frac{\delta V_{s}}{E[x|h,h]})E[x|h,h]=H-\delta \frac{H+L}{4} \\&>H-\delta \frac{H+L}{4L}H=(1-\frac{\delta V_{s}}{E[x|\ell ,\ell ]} )E[x|h]=\alpha (\ell )E[x|h]. \end{aligned}$$

The second one can be rewritten as

$$\begin{aligned} \delta \frac{H+L}{4}\le L-\delta \frac{H+L}{4}=\alpha (\ell )E[x|\ell ]\Leftrightarrow \delta \le \frac{L}{(H+L)/2}=\frac{E[x|\ell ,\ell ]}{E[x]}. \end{aligned}$$

An argument analogous to approximate efficiency for Medium Delay Schedule shows that as q goes to 1,  an approximately efficient equilibrium exists.

Appendix 3: Discussion of off-path beliefs

We examine whether the off-path beliefs of the above equilibria are reasonable in line with the D1 criterion of Banks and Sobel (1987). Two remarks are noted. First, the game is not a standard signaling game in which each player can move only once. However, the information in our model is not persistent, so we can still apply the refinement argument within a period (at least for a stationary equilibrium). Second, we are dealing with two-sided incomplete information with correlation. We need to consider a belief system conditional on the responder’s type. To incorporate correlation, we adopt the following “ two-stage” beliefs.

Let $\beta (\alpha )$ denote the belief that the proposer has h signal given offer $\alpha$ ignoring the responder’s signal (think of it as an outsider’s belief). Suppose $\alpha$ is offered with some probability. By Bayes’ rule

$$\begin{aligned} \beta (\alpha )=\frac{\Pr (\alpha |s_{p}=h)\Pr (s_{p}=h)}{\Pr (\alpha |s_{p}=h)\Pr (s_{p}=h)+\Pr (\alpha |s_{p}=\ell )\Pr (s_{p}=\ell )}=\frac{\Pr (\alpha |h)}{\Pr (\alpha |h)+\Pr (\alpha |\ell )}. \end{aligned}$$

For example, with a pooling offer $\alpha ,$ i.e., $\Pr (\alpha |h)=\Pr (\alpha |\ell )=1,$ we have $\beta (\alpha )=.5.$ With a separating offer $\alpha (h),\alpha (\ell ),$ i.e., $\Pr (\alpha (h)|h)=\Pr (\alpha (\ell )|\ell )=1,\$we have $\beta (\alpha (h))=1$ and $\beta (\alpha (\ell ))=0.$ Note that

$$\begin{aligned} \Pr (s_{p}=h,\alpha |s_{r}=h)&=\Pr (\alpha |s_{p}=h,s_{r}=h)\Pr (s_{p}=h|s_{r}=h)\ \text {(Bayes' rule)} \\&=\Pr (\alpha |s_{p}=h)\Pr (s_{p}=h|s_{r}=h)\text { (the choice of }\alpha \text { only depends on }s_{p}\text {)} \end{aligned}$$

Responder’s belief $\beta (\alpha ,s_{r})$ can be induced by $\beta (\alpha )$:

$$\begin{aligned} \beta (\alpha ,h)&=\frac{\Pr (h|\alpha ,s_{r}=h)}{\Pr (h|\alpha ,s_{r}=h)+\Pr (\ell |\alpha ,s_{r}=h)} \\&=\frac{\Pr (h,\alpha |s_{r}=h)}{\Pr (h,\alpha |s_{r}=h)+\Pr (\ell ,\alpha |s_{r}=h)}\text { (Bayes' rule + eliminating }\Pr (\alpha |s_{r}=h)) \\&=\frac{\Pr (h|s_{r}=h)\Pr (\alpha |s_{p}=h)}{\Pr (h|s_{r}=h)\Pr (\alpha |s_{p}=h)+\Pr (\ell |s_{r}=h)\Pr (\alpha |s_{p}=\ell )}\text { (see the above note) } \\&=\frac{\Pr (h|h)\beta (\alpha )}{\Pr (h|h)\beta (\alpha )+\Pr (\ell |h)(1-\beta (\alpha ))} \\ \beta (\alpha ,\ell )&=\frac{\Pr (h|\ell )\beta (\alpha )}{\Pr (h|\ell )\beta (\alpha )+\Pr (\ell |\ell )(1-\beta (\alpha ))} \end{aligned}$$

This is consistent with the on-path belief of pooling, $\beta (\alpha ,h)=q^{2}+(1-q)^{2}.$ We apply the above formulas for off-path beliefs given an arbitrary $\beta (\alpha ).$ For notational simplicity, we denote $\beta$ for $\beta (\alpha ),$ $\beta ^{h}=\beta (\alpha ,h),$ and $\beta ^{\ell }=\beta (\alpha ,\ell )$. Straightforward algebra shows that $\beta ^{h}\ge \beta ^{\ell }$ for all $\beta \in [0,1],$ where equality holds at $\beta =0,1.$ There are three possible best responses by the responder given $\alpha$, $\beta$ and W: (i) if

$$\begin{aligned} (1-\alpha )\left[ \beta ^{\ell }E[x]+(1-\beta ^{\ell })E[x|\ell ,\ell ] \right] \ge \delta W, \end{aligned}$$

the responder always accepts, (ii) if

$$\begin{aligned} (1-\alpha )\left[ \beta ^{\ell }E[x]+(1-\beta ^{\ell })E[x|\ell ,\ell ] \right] <\delta W\le (1-\alpha )\left[ \beta ^{h}E[x|h,h]+(1-\beta ^{h})E[x] \right] \end{aligned}$$

only the responder with h signal accepts, and (iii) if

$$\begin{aligned} (1-\alpha )\left[ \beta ^{h}E[x|h,h]+(1-\beta ^{h})E[x]\right] <\delta W \end{aligned}$$

the responder always rejects.

Now we examine the plausibility of pessimistic belief. A version of D1 criterion for our model can be stated as follows (see Fudenberg and Tirole 1991, p.452): fix an equilibrium and off-path offer $\alpha .$ Suppose that for each $\beta \in [0,1]$ such that the proposer with h signal ($\ell$) is weakly better off by offering $\alpha$ than the equilibrium offer, $\alpha (h)$ ($\alpha (\ell )$),  the proposer with $\ell$ signal ( h ) is always strictly better off by offering $\alpha$ than by offering the equilibrium offer, $\alpha (\ell )$ ($\alpha (h)$). Suppose, in addition, that there is $\beta ^{\prime }$ for which only the proposer with $\ell$ signal ( h) is strictly better off, then the belief after observing $\alpha$ should be $\beta =0 \$($\beta =1$). An equilibrium passes the D1 test if all the off-path beliefs satisfy this criterion.

Efficient equilibrium for high patience (high delay schedule). First, note that in addition to the equilibrium discussed in Sect. 3.1.1., there are other equilibria: the strategy profile and the belief system are the same as in the equilibrium in Sect. 3.1.1 except for $\alpha (h)$. We can show that if $\alpha (h)$ satisfies

$$\begin{aligned} (1-\alpha (h))E[x|h,h]&\ge \delta W\ge (1-\alpha (h))E[x], \\ \alpha (h)E[x|h,h]&\ge \delta W\ge \alpha (h)E[x], \end{aligned}$$

the associated strategy profile and the belief system compose an equilibrium (note that these conditions hold for $\alpha (h)=1/2)$.²⁵

We show that for the highest $\alpha$ satisfying the above conditions, the associated equilibrium with such $\alpha$ passes the D1 test. The highest $\alpha$ satisfies either $(1-\alpha )E[x|h,h]=\delta W$ or $\alpha E[x]=\delta W.$ Consider the case where $(1-\alpha )E[x|h,h]=\delta W$ and $\alpha E[x]<\delta W.$ The proposer obtains all the rent from the efficient agreement, and therefore the proposer with h signal cannot be better off by any offer given that the responder receives at least the reservation payoff. This implies that any belief system passes the D1 test.

Consider the case where $\alpha E[x]=\delta W$ and $(1-\alpha )E[x|h,h]>\delta W.$ For offer $\alpha ^{\prime }$ such that $(1-\alpha ^{\prime })E[x|h,h]<\delta W,$ the responder rejects the offer for any belief; hence the D1 test does not impose any restriction on the belief for such offer.

Consider offer $\alpha ^{\prime }>\alpha$ for which $(1-\alpha ^{\prime })E[x|h,h]\ge \delta W$. Note that $\alpha ^{\prime }>\alpha$ implies $\delta W\ge (1-\alpha ^{\prime })E[x].$ If $\beta$ is high enough, the responder with h signal would accept $\alpha ^{\prime },$ and hence the proposer with h signal will be strictly better off. However, with such a belief, the proposer with $\ell$ signal would always be better off than the equilibrium payoff, since $\alpha ^{\prime }E[x]>\alpha E[x]=\delta W.$ Hence, $\beta =0$ is consistent with the D1 test.

Consider offer $\alpha ^{\prime }<\alpha .$ For such $\alpha ^{\prime },$ the payoff of the proposer with $\ell$ signal would be $\alpha ^{\prime }E[x|\ell ]<\alpha ^{\prime }E[x]<\alpha E[x]=\delta W$ if the responder always accepts, and $\alpha ^{\prime }\Pr (h|\ell )E[x]+\Pr (\ell |\ell )\delta W<\delta W$. This shows that the proposer with $\ell$ signal is always worse off. Thus $\beta =1$ for offer $\alpha ^{\prime }<\alpha$ is consistent with the D1 test.

Efficient equilibrium for medium patience (medium delay schedule). We show that the equilibrium passes the D1 test iff (IC $_{\ell h}$) is binding.

Consider offer $\alpha ^{\prime }(<\alpha (h))$ for which the equilibrium belief is $\beta =1$ and the responder always accepts. With any belief, the proposer is never better off by offering $\alpha ^{\prime };$ if the belief is so high that the responder always accepts, this is the same action as in the equilibrium and the off-path ICs imply that offering $\alpha ^{\prime }$ is never profitable. If the belief is such that only the responder with h signal accepts, this offer is dominated by offering $\alpha (\ell ),$ and therefore the proposer is never better off. This shows $\beta =1$ for $\alpha ^{\prime }<\alpha$ is consistent with the D1 test.

Consider offer $\alpha ^{\prime }>\alpha (\ell ).$ If $(1-\alpha ^{\prime })E[x|h,h]<\delta W,$the responder never accepts the offer, regardless of the beliefs; hence, any beliefs are consistent with the D1 test. Now suppose $(1-\alpha ^{\prime })E[x|h,h]\ge \delta W.$ If $\beta$ is sufficiently high, the responder with h signal accepts it; hence, the proposer with $\ell$ signal strictly prefers such $\alpha ^{\prime }$ to $\alpha (\ell )$, whereas the proposer with h signal may not. Therefore, $\beta =0$ for $\alpha ^{\prime }$ is consistent with the D1 test.

Consider offer $\alpha ^{\prime }$ such that $\alpha (h)<\alpha ^{\prime }<\alpha (\ell ).$ Since $(1-\alpha (\ell ))E[x]=\delta W,$ we have $(1-\alpha ^{\prime })E[x]>\delta W,$ implying that the responder always accepts $\alpha ^{\prime }$ if $\beta$ is sufficiently high. Consider an equilibrium for which (IC$_{\ell h})$ binds (it is easy to see that such an equilibrium exists if there is an $\alpha (h)$ satisfying both (IC$_{h\ell }$ ) and (IC$_{\ell h}$)), namely

$$\begin{aligned} \alpha (h)E[x|\ell ]=\Pr (h|\ell )\alpha (\ell )E[x]+\Pr (\ell |\ell )\delta W. \end{aligned}$$

Given this condition, the proposer regardless of the signal is strictly better off by offering $\alpha ^{\prime }$ if offer $\alpha ^{\prime }$ is always accepted, while he is worse off if offer $\alpha ^{\prime }$ is not always accepted (by the off-path incentive conditions). Hence, $\beta =0$ for $\alpha ^{\prime }$ is consistent with the D1 test. On the other hand, for an equilibrium with (IC$_{\ell h})$ being slack, there is an $\alpha ^{\prime \prime }$ close to $\alpha (h)$ such that the proposer with h signal is better off by offering $\alpha ^{\prime \prime }$ if offer $\alpha ^{\prime \prime }$ is always accepted, while the proposer with $\ell$ signal is not, implying that $\beta =1$ for offer $\alpha ^{\prime \prime }$ under the D1 criterion. Hence the equilibrium in which (IC$_{\ell h}$) binds only survives the D1 test.

Efficient equilibrium for low patience (no delay schedule). Indeed, some (but not all) efficient equilibrium fail to pass the D1 test. If off-path offer $\alpha ^{\prime }>\alpha$ is always accepted, the proposer wishes to deviate regardless of its signal. Consider the case where $\alpha ^{\prime }>\alpha$ is accepted only by the responder with h signal . We check whether the proposer with h signal is better off by offering $\alpha ^{\prime }$ than with the equilibrium offer.

Consider offer $\alpha ^{\prime }$ for which $(1-\alpha ^{\prime })E[x|h,h]=\delta W.$ Note that a higher offer than $\alpha ^{\prime }$ is always rejected. If $\beta =1,$ the proposer with h signal is better off by offering $\alpha ^{\prime }$ than with the equilibrium offer $\alpha$ iff (recalling that $\alpha =1-\delta W/E[x|\ell ]$)

$$\begin{aligned}&\alpha E[x|h]\le \alpha ^{\prime }\Pr (h|h)E[x|h,h]+\Pr (\ell |h)\delta W\\&\Leftrightarrow \left( 1-\delta W/E[x|\ell ]\right) E[x|h]\le \Pr (h|h)E[x|h,h]+(2\Pr (\ell |h)-1)\delta W \\&\Leftrightarrow \Pr (\ell |h)E[x]\le \left( E[x|h]/E[x|\ell ]+2\Pr (\ell |h)-1\right) \delta W \\&\Leftrightarrow \delta W\ge \frac{\Pr (\ell |h)E[x]}{2\Pr (\ell |h)+E[x|h](1/E[x|\ell ]-1/E[x|h])}. \end{aligned}$$

If this does not hold, $\beta =0$ for $\alpha ^{\prime }$ is consistent with the D1 test.

Recall (IC$_{h}$) for this equilibrium:

$$\begin{aligned} \delta W\le \frac{\Pr (\ell |h)E[x]}{2\Pr (\ell |h)+E[x|h](1/E[x|\ell ]-1/E[x])} \end{aligned}$$

Since $-1/E[x]<-1/E[x|h],$ the above two inequalities are both satisfied in some parameter region; hence, there is some region in which the associated efficient equilibrium fails to pass the D1 test. Note also that if q is close to 1, the proposer with $\ell$ signal is nearly certain that the responder receives $\ell$ signal, so offer $\alpha ^{\prime }$ will be rejected even if $\beta =1.$ Hence, the proposer with $\ell$ signal is worse off by offering $\alpha ^{\prime }$ rather than by $\alpha$ (we could proceed to the tedious details for this argument).