Cheap talk when the receiver has uncertain information sources

Abstract

In this paper we analyze a cheap-talk model with a partially informed receiver. In clear contrast to the previous literature, we find that there is a case where the receiver’s prior knowledge enhances the amount of information conveyed via cheap talk. The point of departure is our explicit focus on the “dual role” of the sender’s message in our context: When the receiver’s prior belief is subject to higher-order uncertainty, the sender’s message provides information not only about the true state but also about the reliability of the receiver’s private information. Building on this result, we argue that whether information acquisition and communication are complements or substitutes depends crucially on the extent of uncertainty regarding the information source.

Appendices

Appendix A: the proofs

Proof of Proposition 1

The proof is quite standard and closely follows that of Proposition 1 of Moreno de Barreda (2013). We relegate the proof to Online Appendix. $\square$

Proof of Proposition 2

The local incentive condition is necessary for an MPE but not sufficient. To show the existence of an MPE, we need to show that the sender has an incentive to follow the equilibrium strategy for all t.

First, consider $t\in T_n$. If $\lim _{ t \uparrow t _n } {\partial \Delta \over \partial t} \ge 0$, then ${\partial \Delta \over \partial t} \ge 0$ holds and the sender has no incentive to deviate for any $t\in T_n$ since ${\partial ^2 \Delta \over \partial t^2}\le 0$. Taking derivative of $\Delta$ with respect to t yields

$$\begin{aligned} {\partial \Delta \over \partial t} = 2 ( \beta _{ n + 1 } - \beta _{ n } ) - \frac{ 2 q ^2 \left( q + 2 ( 1 - q ) \tau _{ n } \right) }{ \left( q + ( 1 - q ) \tau _{ n } \right) ^2 } ( t - \beta _n ) - \frac{ 2 q ^2 }{ q + ( 1 - q ) \tau _{ n } } b, \end{aligned}$$

which presents an additional condition to be satisfied:

$$\begin{aligned} 2 ( \beta _{ n + 1 } - \beta _{ n } ) - \frac{ 2 q ^2 \left( q + 2 ( 1 - q ) \tau _{ n } \right) }{ \left( q + ( 1 - q ) \tau _{ n } \right) ^2 } ( t - \beta _n ) - \frac{ 2 q ^2 }{ q + ( 1 - q ) \tau _{ n } } b > 0. \end{aligned}$$

(3)

Second, for $t\in T_{n+1}$, taking derivative of $\Delta$ with respect to t yields

$$\begin{aligned} \frac{ \partial \Delta }{ \partial t } = 2 ( \beta _{ n + 1 } - \beta _{ n } ) + \frac{ 2 q ^2 \left( q + 2 ( 1 - q ) \tau _{ n + 1 } \right) }{ \left( q + ( 1 - q ) \tau _{ n + 1 } \right) ^2 } ( t - \beta _{ n + 1 } ) + \frac{ 2 q ^2 }{ q + ( 1 - q ) \tau _{ n + 1 } } b. \end{aligned}$$

Note that $\lim _{ t \downarrow t _n } \frac{\partial \Delta }{\partial t } > 0$. Then, the sender has no incentive to deviate for any $t \in T _{ n + 1 }$ since $\frac{\partial ^2 \Delta }{\partial t ^2 } \ge 0$.

Lastly, for $t \in T \backslash ( T _{ n } \cup T _{ n + 1 } )$, $\Delta$ can be written as

$$\begin{aligned} \Delta&( t ; t _{ n - 1 } , t _n , t _{ n + 1 } ) \\&= -\, ( \beta _{ n + 1 } - \beta _{ n } ) ( \beta _{ n + 1 } + \beta _{ n } - 2 t - 2 b ) \\&\quad - \frac{ q ^2 ( 1 - q ) }{ \left( q + ( 1 - q ) \tau _{ n + 1 } \right) ^2 } \int _{ r \in T _{ n + 1 } } ( r - \beta _{ n + 1 } ) ^2 f ( r )\, \hbox {d}r \\&\quad + \frac{ q ^2 ( 1 - q ) }{ \left( q + ( 1 - q ) \tau _{ n } \right) ^2 } \int _{ r \in T _{ n } } ( r - \beta _{ n } ) ^2 f ( r )\, \hbox {d} r. \end{aligned}$$

It is also verified that

$$\begin{aligned} \frac{ \partial \Delta }{ \partial t } = 2 ( \beta _{ n + 1 } - \beta _{ n + 1 } ) > 0 . \end{aligned}$$

Then, $\Delta ( t ; t _{ n - 1 } , t _{ n } , t _{ n + 1 } ) \le 0$ for $t \in T _{ n ' }$ and $n ' < n$ if and only if

$$\begin{aligned} \lim _{ t \uparrow t _{ n ' } } \Delta ( t ; t _{ n - 1 } , t _{ n } , t _{ n + 1 } ) \le 0. \end{aligned}$$

This holds if and only if

$$\begin{aligned} \lim _{ t \uparrow t _{ n - 1 } } \Delta ( t ; t _{ n - 1 } , t _{ n } , t _{ n + 1 } ) \le 0, \end{aligned}$$

since $\lim _{ t \uparrow t _{ n ' } } \Delta ( t ; t _{ n - 1 } , t _{ n } , t _{ n + 1 } ) = \lim _{ t \downarrow t _{ n ' } } \Delta ( t ; t _{ n - 1 } , t _{ n } , t _{ n + 1 } )$ for $n ' < n - 1$. This is guaranteed by $\underline{G}(t_{n-1}, t_n , t_{n+1})\le 0$ and (3).

Similarly, $\Delta ( t ; t _{ n - 1 } , t _{ n } , t _{ n + 1 } ) \ge 0$ for $t \in T_{ n ' }$ and $n ' > n + 1$ if and only if

$$\begin{aligned} \lim _{ t \downarrow t _{ n ' } } \Delta ( t ; t _{ n - 1 } , t _{ n } , t _{ n + 1 } ) \ge 0. \end{aligned}$$

This holds if and only if

$$\begin{aligned} \lim _{ t \downarrow t _{ n + 1 } } \Delta ( t ; t _{ n - 1 } , t _{ n } , t_{ n + 1 } ) \ge 0, \end{aligned}$$

since $\lim _{ t \downarrow t _{ n ' } } \Delta ( t ; t _{ n - 1 } , t _{ n } , t _{ n + 1 } ) = \lim _{ t \uparrow t _{ n ' } } \Delta ( t ; t _{ n - 1 } , t _{ n } , t _{ n + 1 } )$ for $n ' > n + 1$. This is guaranteed by $\overline{G}(t_{n-1}, t_n , t_{n+1})\ge 0$. $\square$

Proof of Proposition 4

We have shown that the incentive conditions in our setting are less stringent than those in CS. Moreover, if $\beta _{ n + 1 } + \beta _{ n } - 2 t _{ n } - 2 b = 0$ holds, then

$$\begin{aligned} \lim _{ t \uparrow t _{ n } } \frac{ \partial \Delta }{ \partial t } = \left[ 4 - \frac{ 2 q ^2 \left( q + 2 ( 1 - q ) \tau _{ n } \right) }{ \left( q + ( 1 - q ) \tau _{ n } \right) ^2 } \right] ( t _n - \beta _n ) + \left[ 4 - \frac{ 2 q ^2 }{ q + ( 1 - q ) \tau _{ n } } \right] b > 0 . \end{aligned}$$

This means that the equilibrium partition in our setting can be more flexible than that in CS. Thus, the first statement of the proposition is verified.

As for the second statement, fix any MPE-0 ${\mathbf {t} } = ( t _0 , \dots , t _N )$ for $N \ge 2$. Note that

$$\begin{aligned} \frac{ \partial V ^R }{ \partial t _1 } \Big \vert _{ q = 0 , \beta _2 + \beta _1 - 2 t _1 - 2 b = 0 } = 2 b ( \beta _2 - \beta _1 ) f ( t _1 ) > 0 . \end{aligned}$$

It then follows that there exists $\bar{\varepsilon } > 0$ such that for any $\varepsilon \in ( 0 , \bar{\varepsilon } )$ a partition $\bar{\mathbf {t}} ( \varepsilon )$ satisfying

$$\begin{aligned} \bar{t} _n ( \varepsilon ) = {\left\{ \begin{array}{ll} t _1 + \varepsilon \quad &{} \text {if} \ n = 1 \\ t _n \quad &{} \text {if} \ n \not = 1 \end{array}\right. } \end{aligned}$$

constitutes an MPE-q and $V ^R ( \bar{\mathbf {t}} ( \varepsilon ) , 0 ) > V ^R ( \mathbf {{t}} , 0 )$. Also, the latter implies that $V ^S ( \bar{\mathbf {t}} ( \varepsilon ) , 0 ) > V ^S ( \mathbf {t} , 0 )$ by (2). $\square$

Proof of Proposition 5

Throughout this proof we assume that the state is distributed over [0, 1] . In this case $\underline{G}$ and $\overline{G}$ are rewitten as follows:

$$\begin{aligned} 4 \underline{G} ( t_{ n - 1 } , t _n , t _{ n + 1 } ; q )&= -\,( \tau _{ n + 1 } + \tau _{ n } ) ( \tau _{ n + 1 } - \tau _{ n } - 4 b ) \\&\quad - \frac{ q ^2 ( 1 - q ) \tau _{ n + 1 } ^3 }{ 3 \left( q + ( 1 - q ) \tau _{ n + 1 } \right) ^2 } - \frac{ 2 q ^2 ( 1 - q ) \tau _{ n } ^3 }{ 3 \left( q + ( 1 - q ) \tau _{ n } \right) ^2 }\\&\quad - \frac{ q ^2 \tau _{ n } ( \tau _{ n } + 4 b ) }{ q + ( 1 - q ) \tau _{ n } } \\ 4 \overline{G} ( t_{ n - 1 } , t _n , t _{ n + 1 } ; q )&= -\,( \tau _{ n + 1 } + \tau _{ n } ) ( \tau _{ n + 1 } - \tau _{ n } - 4 b ) \\&\quad + \frac{ 2 q ^2 ( 1 - q ) \tau _{ n + 1 } ^3 }{ 3 \left( q + ( 1 - q ) \tau _{ n + 1 } \right) ^2 } + \frac{ q ^2 ( 1 - q ) \tau _{ n } ^3 }{ 3 \left( q + ( 1 - q ) \tau _{ n } \right) ^2 } \\&\quad + \frac{ q ^2 \tau _{ n + 1 } ( \tau _{ n + 1 } + 4 b ) }{ q + ( 1 - q ) \tau _{ n + 1 } } . \end{aligned}$$

Then, the equilibrium conditions are simplified and obtained as follows.

Lemma 1

For any $\tau _n \in (0, 1)$ and $q\in [0, 1)$,

• (i) There exists $\underline{\tau }(\tau _n, q)$ and $\overline{\tau }(\tau _n, q)$ such that $\overline{G}(t_{n-1},t_n,t_{n+1};q) \ge 0\ge \underline{G}(t_{n-1},t_n,t_{n+1};q)$ if and only if $\overline{\tau }(\tau _n, q) \ge \tau _{n+1}\ge \underline{\tau }(\tau _n, q)$;

• (ii) $\underline{G}(t_{n-1},t_n,t_n+\underline{\tau }(\tau _n,q);q) = 0$ and $\overline{G}(t_{n-1},t_n,t_n+\overline{\tau }(\tau _n,q);q) = 0$;

• (iii) $\overline{\tau }(\tau _n, q)\ge \tau _n + 4b\ge \underline{\tau }(\tau _n,q )> 2b$ where

  $$\begin{aligned} q = 0 \&\Leftrightarrow \ \overline{\tau }( \tau _n , q ) = \tau _n + 4 b = \underline{\tau }( \tau _n , q ) , \\ q> 0 \&\Leftrightarrow \ \overline{\tau }(\tau _n,q)> \tau _n + 4b > \underline{\tau }(\tau _n,q). \end{aligned}$$

Moreover, $\overline{G}(t_{n-1},t_n,t_{n+1};q) \ge 0\ge \underline{G}(t_{n-1},t_n,t_{n+1};q)$ for $n = 1, \dots , N - 1$ is both necessary and sufficient for an MPE with N intervals.

Proof

We first make the following observations by direct calculation. $\square$

Observation 1

For any $\tau _n \in [0, 1]$ and $\tau _{n+1}\ge 0$,

$$\begin{aligned} {\partial ^2\underline{G}\over \partial t_{n+1}^2} < 0,\ {\partial \underline{G}\over \partial t_{n+1}}\Big \vert _{t_{n+1}=t_n} > 0, {\partial \underline{G}\over \partial t_{n+1}}\Big \vert _{t_{n+1}=t_n+2b} \le 0. \end{aligned}$$

Moreover,

$$\begin{aligned} \underline{G}(t_{n-1},t_n,t_n)> 0,\ \underline{G}(t_{n-1},t_n,t_n+2b) > 0,\ \underline{G}(t_{n-1},t_n, 2 t_n-t_{n-1}+4b)\le 0. \end{aligned}$$

where the last condition holds with equality if and only if $q = 0$.

Observation 2

For any $\tau _n \in [0, 1]$ and $\tau _{n+1}\ge 0$,¹⁷

$$\begin{aligned} {\partial ^3 \overline{G}\over \partial t_{n+1}^3}\le 0 ,\ {\partial ^2\overline{G}\over \partial t_{n+1}^2}\Big \vert _{t_{n+1}=t_n+2b}< 0 ,\ {\partial \overline{G}\over \partial t_{n+1}}\Big \vert _{t_{n+1}=t_n + 2b} \ge 0 ,\ \lim _{ t _{ n + 1 } \rightarrow \infty } \frac{ \partial \overline{G}}{ \partial t_{ n + 1 } } < 0. \end{aligned}$$

Moreover,

$$\begin{aligned} \overline{G}(t_{n-1},t_n,t_n+2b) > 0,\ \overline{G}(t_{n-1},t_n, 2 t_n-t_{n-1}+4b) \ge 0, \end{aligned}$$

where the latter holds with equality if and only if $q = 0$.

It follows from Observation 1 that for any $\tau _n \in [0, 1]$ and $\tau _{n+1} \ge 0$, $\underline{G}(t_{n-1},t_n,t_{n+1})\le 0$ if and only if $\tau _{n+1}\ge \underline{\tau }(\tau _n, q)$ where $\underline{\tau }(\tau _n, q)$ is defined as the unique solution $\tilde{\tau }\in [0,\infty )$ of $\underline{G}(t_{n-1},t_n,t_n+\tilde{\tau }) = 0$. Moreover, $\tau _n + 4b\ge \underline{\tau }(\tau _n, q) > 2b$ where $\underline{\tau }(\tau _n,q) = \tau _n + 4b$ if and only if $q = 0$. Similarly, it follows from Observation 2 that for any $\tau _n\in [0,1]$ and $\tau _{n+1}\ge 2b$, $\overline{G}(t_{n-1},t_n,t_{n+1}) \ge 0$ if and only if $\tau _{n+1} \le \overline{\tau }(\tau _n , q)$ where $\overline{\tau }(\tau _n,q)$ is defined as the unique solution $\tilde{\tau }\in [2b,\infty )$ of $\overline{G}(t_{n-1},t_n,t_n+\tilde{\tau }) = 0$. Moreover, $\overline{\tau }(\tau _n,q) \ge \tau _n + 4b$ which holds with equality if and only if $q=0$.

Finally, we show that $\overline{G}(t_{n-1},t_n,t_{n+1};q) \ge 0\ge \underline{G}(t_{n-1},t_n,t_{n+1};q)$ for $n = 1, \dots , N - 1$ is both necessary and sufficient for an MPE with N intervals when the state is uniformly distributed. To see this, since $\beta _n={t_n+t_{n-1}\over 2}$ and $\tau _n=t_n-t_{n-1}$ under the uniformed distribution, condition (ii) of Proposition 2 can be written as

$$\begin{aligned} \tau _{n+1} + \tau _n - {q^2\over q+(1-q)\tau _n}\Biggl ({q + 2(1-q)\tau _n\over q+(1-q)\tau _n}\tau _n + 2b\Biggr ) > 0. \end{aligned}$$

This condition holds because $\tau _{n+1}>2b$.$\square$

Given this result, we prove the proposition by a series of lemmas.

Lemma 2

Given ${ \mathbf t } = ( t _0 , \dots t _N )$, ${ \mathbf t ' } = ( t ' _0 , \dots t ' _N )$, if there exists $\hat{n}$ such that

• $t_{\hat{n}} > t'_{\hat{n}}$,

• $t _n = t ' _n$ for $n \not = \hat{n}$, and

• $\tau _{ \hat{n} } < \tau _{ \hat{n} + 1 }$.

then $\mathbf{t}$ is more efficient than $\mathbf{t'}$ at $q = 0$, i.e., $V^i(\mathbf{t},0)>V^i(\mathbf{t'},0)$.

Proof

We denote

$$\begin{aligned} M ( \tau , q ) := - \frac{ ( q + \tau ) \tau ^3 }{ q + ( 1 - q ) \tau }. \end{aligned}$$

Then, since it is verified

$$\begin{aligned} M ( \tau _{ \hat{n} } , 0 ) + M ( \tau _{ \hat{n} + 1 } , 0 ) - M ( \tau ' _{ \hat{n} } , 0 ) - M ( \tau ' _{ \hat{n} + 1 } , 0 ) > 0 , \end{aligned}$$

we obtain the lemma. $\square$

Lemma 3

Given b and $q > 0$, if ${\mathbf {t}}$ is the most efficient MPE and $\tau _{ n } < \tau _{ n + 1 }$ for $n = 1 , \dots , N - 1$, then $\tau _{ n + 1 } = \underline{\tau }( \tau _n , q )$ for $n = 1 , \dots , N - 1$.

Proof

Suppose the contrary. Define $\hat{n}$ such that

$$\begin{aligned} \hat{n} :=\min \left\{ n \vert \tau _{ n + 1 } > \underline{\tau }( \tau _n , q ) \right\} . \end{aligned}$$

Then, $\hat{n} = 1$ or

$$\begin{aligned} \tau _{ \hat{n} } = \underline{\tau }( \tau _{ \hat{n} - 1 } , q ) < \overline{\tau }( \tau _{ \hat{n} - 1 } , q ) . \end{aligned}$$

Thus, we can find another MPE ${\mathbf {t}} '$ where $t ' _{ \hat{n} }$ is slightly higher than $t _{ \hat{n} }$, $t ' _{ n } = t _{ n }$ for $n \not = \hat{n}$, and the presupposition of Lemma 2 holds at $n = \hat{n}$. Then the lemma is obtained from Lemma 2. $\square$

Lemma 4

There exists $\hat{q} > 0$ such that $\underline{\tau }( \tau , q ) > \tau$ for any $\tau \in ( 0 , 1 )$ and $q \in [ 0 , \hat{q} ]$.

Proof

By Lemma 1, $\underline{\tau }( \tau , 0 ) = \tau + 4 b > \tau$ for any $\tau \in ( 0 , 1 )$. Then, there must exist $\hat{q} > 0$ such that $\hat{\tau }( \tau , q ) > \tau$ for any $\tau \in ( 0 , 1 )$ and $q \in [ 0 , \hat{q} ]$. $\square$

Lemma 5

There exists $\check{q} > 0$ such that $\displaystyle \frac{ \partial \underline{\tau }}{ \partial q } < 0$ for any $\tau \in ( 0 , 1 )$ and $q \in ( 0 , \check{q} ]$.

Proof

Since $\frac{\partial \underline{G}}{ \partial t _{n+1}} < 0$ for $\tau _{n+1} > 2b$ (see the proof of Lemma 1), $\frac{ \partial \underline{G}}{ \partial t_{n+1}}\vert _{ \tau _{ n + 1 } = \underline{\tau }( \tau _n ) } < 0$.

On the other hand,

$$\begin{aligned} - {4 \over q}{\partial \underline{G}\over \partial q}&= \frac{ \left( - q ^ 2 + ( 1 - q ) ( 2 - q ) \tau _{ n + 1 } \right) \tau _{ n + 1 } ^3 }{ 3 \left( q + ( 1 - q ) \tau _{ n + 1 } \right) ^3 } \\&\qquad + \frac{ 2 \left( - q ^ 2 + ( 1 - q ) ( 2 - q ) \tau _{ n } \right) \tau _{ n } ^3 }{ 3 \left( q + ( 1 - q ) \tau _{ n } \right) ^3 } \\&\qquad + \frac{ \left( q + ( 2 - q ) \tau _n \right) \tau _n ( \tau _n + 4 b ) }{ \left( q + ( 1 - q ) \tau _n \right) ^2 } . \end{aligned}$$

Then, Observation 1 (in the proof of Lemma 1) implies

$$\begin{aligned} - {4\over q}{\partial \underline{G}\over \partial q}\Big \vert _{ q = 0, \tau _{ n + 1 } = \underline{\tau }( \tau _n ) }&= \frac{2}{3} \underline{\tau }( \tau _n ) + \frac{4}{3} \tau _n + 2 ( \tau _n + 4 b ) \\&> \frac{2}{3} 2 b + 8 b \\&= \frac{ 28 }{ 3 } b . \end{aligned}$$

Therefore, there exists $\check{q}$ such that $\frac{\partial \underline{G}}{ \partial q } \vert _{ \tau _{ n + 1 } = \underline{\tau }( \tau _n ) } < 0$ for any $q \in ( 0 , \check{q} ]$.

We now return to the Proof of Proposition 5. Letting $\overline{q} := \min \{ \hat{q} , \check{q} \}$, the proposition then follows from Lemmas 2, 3, 4, and 5. $\square$

Appendix B: The discrete-state model and convergence

In this appendix, we establish that the equilibrium of the discrete-state model converges to that of the continuous-state model as $L\rightarrow \infty$. In the process, we also derive the equilibrium conditions for a sufficiently large L which allows us to isolate the confirmation effect.

In what follows, we use the following notations:

• $m _n$ is a typical element of the messages sent by $t \in T ^L _n$,

• $T ^L _n = \{ t _{ I _{ n - 1 } + 1 } , \dots , t _{ I _{ n } } \}$ for $n = 1 \dots , N$ where $I _0 = - 1$,

• $\gamma _n = \frac{ I _n + I _{ n - 1 } + 1 }{ 2 L }$,

• $\lambda _n = \frac{ I _{ n } - I _{ n - 1 } }{ L }$, and

• $\ell = \frac{ 1 }{ L }$.

Define

$$\begin{aligned} \Delta ^L_n ( t ) := {\mathbb {E} } _{ r } \left[ U ^S \left( t , \alpha ( m _{ n + 1 } , r ) \right) \mid t \right] - {\mathbb {E} } _{ r } \left[ U ^S \left( t , \alpha ( m _{ n } , r ) \right) \mid t \right] . \end{aligned}$$

A sufficient condition for an MPE with $\{ T ^L _n \} _{ n = 1 ,\dots , N }$ is that

$$\begin{aligned} \Delta ^L _n ( t ) {\left\{ \begin{array}{ll} \le 0 \quad &{} \text {for} \ t \le t _{ I _n } \\ \ge 0 \quad &{} \text {for} \ t \ge t _{ I _{ n } + 1 } \end{array}\right. } \end{aligned}$$

(4)

holds for any $n = 1 , \dots , N - 1$. Note that $\Delta ^L_n(t_{I_n}) \le 0$ and $\Delta ^L_n ( t _{ I _{ n } + 1 } ) \ge 0$ are sufficient for (4) whenever L is sufficiently large.

Proposition 6

There exists $\varLambda$ such that for any finite model with $L > \varLambda$, $\Delta ^L _n ( t _{ I _{ n } } ) \le 0$ and $\Delta ^L _n ( t _{ I _{ n } + 1 } ) \ge 0$ are sufficient conditions for an MPE with $\{ T ^L _n \} _{ n = 1 , \dots , N }$.

Proof

The receiver’s posterior is written as

$$\begin{aligned} \beta ( t \vert m _n , r ) = {\left\{ \begin{array}{ll} \frac{ q ( 1 + \ell ) + ( 1 - q ) \ell }{ q ( 1 + \ell ) + ( 1 - q ) \lambda _n } \quad &{} \text {if} \ t \in T ^L _n , \ r \in T ^L _n , \ r = t \\ \frac{ ( 1 - q ) \ell }{ q ( 1 + \ell ) + ( 1 - q ) \lambda _n } \quad &{} \text {if} \ t \in T ^L _n , \ r \in T ^L _n , \ r \not = t \\ \frac{ L }{ \lambda _n } \quad &{} \text {if} \ t \in T ^L _n , \ r \notin T ^L _n \\ 0 \quad &{} \text {if} \ t \notin T ^L _n . \end{array}\right. } \end{aligned}$$

Then, the receiver’s best response is written as

$$\begin{aligned} \alpha ( m _n , r ) = {\left\{ \begin{array}{ll} \gamma _n + A _n ( r - \gamma _n ) \quad &{} \text {if} \ r \in T ^L _n \\ \gamma _n \quad &{} \text {if} \ r \notin T ^L _n . \end{array}\right. } \end{aligned}$$

For $t \in T ^L _n$, we obtain

$$\begin{aligned} \Delta ^L _n ( t )&= -\,( \gamma _{ n + 1 } - \gamma _n ) ( \gamma _{ n + 1 } + \gamma _n - 2 t - 2 b ) \\&\qquad - \frac{ 1 - q }{ L + 1 } \left[ \sum _{ r \in T ^L _{ n + 1 } } A _{ n + 1 } ^2 ( r - \gamma _{ n + 1 } ) ^2 - \sum _{ r \in T ^L _n } A _n ^2 ( r - \gamma _n ) ^2 \right] \\&\qquad -\,q A _n ( t - \gamma _n ) \left[ ( 2 - A _n ) ( t - \gamma _n ) + 2 b \right] . \end{aligned}$$

For $t \in T ^L _{ n + 1 }$,

$$\begin{aligned} \Delta ^L _n ( t )&= -\,( \gamma _{ n + 1 } - \gamma _n ) ( \gamma _{ n + 1 } + \gamma _n - 2 t - 2 b ) \\&\qquad - \frac{ 1 - q }{ L + 1 } \left[ \sum _{ r \in T ^L _{ n + 1 } } A _{ n + 1 } ^2 ( r - \gamma _{ n + 1 } ) ^2 - \sum _{ r \in T ^L _n } A _n ^2 ( r - \gamma _n ) ^2 \right] \\&\qquad +\,q A _{ n + 1 } ( t - \gamma _{ n + 1 } ) \left[ ( 2 - A _{ n + 1 } ) ( t - \gamma _{ n + 1 } ) + 2 b \right] . \end{aligned}$$

For $t \notin T ^L _{ n } \cup T ^L _{ n + 1 }$,

$$\begin{aligned} \Delta ^L _n ( t )&= -\,( \gamma _{ n + 1 } - \gamma _n ) ( \gamma _{ n + 1 } + \gamma _n - 2 t - 2 b ) \\&\qquad - \frac{ 1 - q }{ L + 1 } \left[ \sum _{ r \in T ^L _{ n + 1 } } A _{ n + 1 } ^2 ( r - \gamma _{ n + 1 } ) ^2 - \sum _{ r \in T ^L _n } A _n ^2 ( r - \gamma _n ) ^2 \right] . \end{aligned}$$

We show that $\Delta ^L _n ( t _{ I _{ n } } ) \le 0$ implies $\lambda _{ n + 1 } > 2 b$ whenever L is sufficiently large. First, $\Delta ^L _n ( t _{ I _{ n } } )$ is rewritten as

$$\begin{aligned} 4 \Delta ^L _n ( t _{ I _{ n } } )&= -\,( \lambda _{ n + 1 } + \lambda _{ n } ) ( \lambda _{ n + 1 } - \lambda _{ n } - 4b + 2 \ell ) \\&\quad - \frac{ 1 - q }{ 3 ( 1 + \ell ) } A _{ n + 1 } ^2 \lambda _{ n + 1 } ( \lambda _{ n + 1 } ^2 - \ell ^2 ) + \frac{ 1 - q }{ 3 ( 1 + \ell ) } A _{ n } ^2 \lambda _{ n } ( \lambda _{ n } ^2 - \ell ^2 ) \\&\quad - q A _{ n } ( \lambda _n - \ell ) \left( ( 2 - A _ n ) ( \lambda _{ n } - \ell ) + 4 b \right) . \end{aligned}$$

When $\Delta ^L _n ( t _{ I _{ n } } )$ is considered as a function of $\lambda _{ n + 1 }$¹⁸, we obtain

$$\begin{aligned} \frac{ \partial ^2 \Delta ^L _n ( t _{ I _{ n } } ) }{ \partial \lambda _{ n + 1 } ^2 }< 0,\ \frac{ \partial \Delta ^L _n ( t _{ I _{ n } } ) }{ \partial \lambda _{ n + 1 } } \Big \vert _{ \lambda _{ n + 1 } = 2 b } < 0,\ \Delta ^L _n ( t _{ I _{ n } } ) \vert _{ \lambda _{ n + 1 } = 2 b } > 0 . \end{aligned}$$

It then follows that $\Delta ^L _n ( t _{ I _{ n } } ) \le 0$ implies $\lambda _{ n + 1 } > 2 b$ for any sufficiently large L.

We are in a position to verify that $\Delta ^L _n ( t _{ I _{ n } } ) \le 0$ and $\Delta ^L _n ( t _{ I _{ n } + 1 } ) \ge 0$ are sufficient. We do it by dividing the situation into the following cases.

Case 1 For $i = 0 , \dots , I _{ n - 1 } - 1$, it is verified that

$$\begin{aligned} \Delta ^L _n ( t _{ i + 1 } ) - \Delta ^L _n ( t _{ i } ) = ( \lambda _{ n + 1 } + \lambda _{ n } ) \ell > 0 . \end{aligned}$$

Case 2 It is verified that

$$\begin{aligned} \Delta ^L _n&( t _{ I _{ n } } ) - \Delta ^L _n ( t _{ I _{ n - 1 } } ) \\&= \lambda _n ( \lambda _{ n + 1 } + \lambda _{ n } ) - q A _{ n } \frac{ \lambda _{ n } - \ell }{ 2 } \left( ( 2 - A _n ) \frac{ \lambda _{ n } - \ell }{ 2 } + 2 b \right) \\&> \lambda _n ( 2 b + \lambda _{ n } ) - q A _{ n } \frac{ \lambda _{ n } - \ell }{ 2 } \left( ( 2 - A _{ n } ) \frac{ \lambda _{ n } - \ell }{ 2 } + 2 b \right) \\&= \left( \lambda _{ n } ^ 2 - q A _{ n } ( 2 - A _{ n } ) \left( \frac{ \lambda _{ n } - \ell }{ 2 } \right) ^2 \right) + 2 \left( \lambda _{ n } - q A _{ n } \frac{ \lambda _{ n } - \ell }{ 2 } \right) b \\&> 0 . \end{aligned}$$

Case 3 For $i = I _{ n - 1 } + 1 , \dots , I _{ n } - 1$,

$$\begin{aligned} \Delta ^L _n&( t _{ i + 1 } ) - \Delta ^L _n ( t _{ i } ) \\&= ( \lambda _{ n + 1 } + \lambda _{ n } ) \ell + q A _{ n } ( 2 - A _{ n } ) ( 2 \gamma _{ n } - t _{ i + 1 } - t _{ i } ) \ell - 2 q A _{ n } b \ell \\&> ( 2 b + \lambda _{ n } ) \ell + q A _{ n } ( 2 - A _{ n } ) ( 2 \gamma _{ n } - t _{ I _{ n }} - t _{ I _{ n } - 1 } ) \ell - 2 q A _{ n } b \ell \\&= \left( 1 - q A _{ n } ( 2 - A _{ n } ) \right) \lambda _{ n } \ell + 2 q A _{ n }( 2 - A _{ n } ) \ell ^2 + 2 ( 1 - q A _n ) b \ell \\&> 0 . \end{aligned}$$

Case 4 For $i = I _{ n } + 1 , \dots , I _{ n + 1 } - 1$, it is verified that

$$\begin{aligned} \Delta ^L _n&( t _{ i + 1 } ) - \Delta ^L _n ( t _{ i } ) \\&= ( \lambda _{ n + 1 } + \lambda _{ n } ) \ell + q A _{ n + 1 } ( 2 - A _{ n + 1 } ) ( t _{ i + 1 } + t _{ i } - 2 \gamma _{ n + 1 } ) \ell + 2 q A _{ n + 1 } b \ell \\&\ge ( \lambda _{ n + 1 } + \lambda _{ n } ) \ell + q A _{ n + 1 } ( 2 - A _{ n + 1 } ) ( t _{ I _{ n } + 2 } + t _{ I _n + 1 } - 2 \gamma _{ n + 1 } ) \ell + 2 q A _{ n + 1 } b \ell \\&= \left( 1 - q A _{ n + 1 } ( 2 - A _{ n + 1 } ) \right) \lambda _{ n + 1 } \ell + \lambda _{ n } \ell + 2 q A _{ n + 1 } ( 2 - A _{ n + 1 } ) \ell ^2 + 2 q A _{ n + 1 } b \ell \\&> 0 . \end{aligned}$$

Case 5 It is verified that

$$\begin{aligned} \Delta ^L _n&( t _{ I _{ n + 1 } + 1 } ) - \Delta ^L _n ( t _{ I _{ n } + 1 } ) \\&= \lambda _{ n + 1 } ( \lambda _{ n + 1 } + \lambda _{ n } ) - q A _{ n + 1 } \frac{ \lambda _{ n + 1 } - \ell }{ 2 } \left( ( 2 - A _{ n + 1 } ) \frac{ \lambda _{ n + 1 } - \ell }{ 2 } - 2 b \right) \\&= \left( \lambda _{ n + 1 } ^2 - q A _{ n + 1 } ( 2 - A _{ n + 1 } ) \left( \frac{ \lambda _{ n + 1 } - \ell }{ 2 } \right) ^2 \right) + \lambda _{ n + 1 } \lambda _{ n } + q A _{ n + 1 } ( \lambda _{ n + 1 } - \ell ) b \\&> 0 . \end{aligned}$$

Case 6 For $i = I _{ n + 1 } + 1 , \dots , L - 1$, it is verified that

$$\begin{aligned} \Delta ^L _n ( t _{ i + 1 } ) - \Delta ^L _n ( t _{ i } ) = ( \lambda _{ n + 1 } + \lambda _{ n } ) \ell > 0 . \end{aligned}$$

From Cases 1–6, if $\Delta ^L _n ( t _{ I _{ n } } ) \le 0$, $\Delta ^L _n ( t _{ I _{ n } + 1 } ) \ge 0$, and L is sufficiently large, then we have

$$\begin{aligned}&0 \ge \Delta ^L _n ( t _{ I _{ n } } )> \Delta ^L _n ( t _{ I _{ n - 1 } } )> \Delta ^L _n ( t _{ I _{ n - 1 } - 1 } )> \dots> \Delta ^L _n ( t _{ 0 } ) \\&0 \ge \Delta ^L _n ( t _{ I _{ n } } )> \Delta ^L _n ( t _{ I _{ n } - 1 } )> \dots > \Delta ^L _n ( t _{ I _{ n - 1 } + 1 } ) \\&0 \le \Delta ^L _n ( t _{ I _{ n } + 1 } )< \Delta ^L _n ( t _{ I _{ n } + 2 } )< \dots< \Delta ^L _n ( t _{ I _{ n + 1 } } ) \\&0 \le \Delta ^L _n ( t _{ I _{ n } + 1 } )< \Delta ^L _n ( t _{ I _{ n + 1 } + 1 } )< \Delta ^L _n ( t _{ I _{ n + 1 } + 2 } )< \dots < \Delta ^L _n ( t _{ L } ) . \end{aligned}$$

Therefore, we conclude (4). $\square$

Given this, we now explicitly derive the equilibrium conditions. With some computation, we obtain

$$\begin{aligned} \Delta ^L _n ( t _{ I _{ n } } )&\quad = -\,( \gamma _{ n + 1 } - \gamma _n ) ( \gamma _{ n + 1 } + \gamma _n - 2 t _{ I _{ n } } - 2 b ) \\&\qquad - \frac{ 1 - q }{ L + 1 } \left[ \sum _{ r \in T ^L _{ n + 1 } } A _{ n + 1 } ^2 ( r - \gamma _{ n + 1 } ) ^2 - \sum _{ r \in T ^L _n } A _n ^2 ( r - \gamma _n ) ^2 \right] \\&\qquad -\,q A _n ( t _{ I _{ n } } - \gamma _n ) \left[ ( 2 - A _n ) ( t _{ I _{ n } } - \gamma _n ) + 2 b \right] , \end{aligned}$$

$$\begin{aligned} \Delta ^L _n ( t _{ I _{ n } + 1 } )&\quad = -\,( \gamma _{ n + 1 } - \gamma _n ) ( \gamma _{ n + 1 } + \gamma _n - 2 t _{ I _{ n } + 1 } - 2 b ) \\&\qquad - \frac{ 1 - q }{ L + 1 } \left[ \sum _{ r \in T ^L _{ n + 1 } } A _{ n + 1 } ^2 ( r - \gamma _{ n + 1 } ) ^2 - \sum _{ r \in T ^L _n } A _n ^2 ( r - \gamma _n ) ^2 \right] \\&\qquad +\,q A _{ n + 1 } ( t _{ I _{ n } + 1 } - \gamma _{ n + 1 } ) \left[ ( 2 - A _{ n + 1 } ) ( t _{ I _{ n } + 1 } - \gamma _{ n + 1 } ) + 2 b \right] . \end{aligned}$$

where $A _n := \frac{ q ( 1 + \ell ) }{ q ( 1 + \ell ) + ( 1 - q ) \lambda _n }$. The third line of each equation represents the confirmation effect, which tends to relax the equilibrium conditions.

Finally, to motivate the use of the discrete-state model, we confirm that our baseline model is indeed the limit of a sequence of discrete-state models as $L \rightarrow \infty$. More precisely, we define $T^L_n = T_n \cap T^L$ for $n = 1 , \dots , N$. It is then verified that

$$\begin{aligned} \Delta ^L _n ( t _{ I _{ n } } ) \underset{ L \rightarrow \infty }{ \rightarrow } \underline{G} ( t _{ n - 1 } , t _{ n } , t _{ n + 1 } ) , \\ \Delta ^L _n ( t _{ I _{ n } + 1 } ) \underset{ L \rightarrow \infty }{ \rightarrow } \overline{G} ( t _{ n - 1 } , t _{ n } , t _{ n + 1 } ), \end{aligned}$$

leading to the following statement (the proof omitted).

Corollary 1

Consider the baseline model where the state is uniformly distributed. Given any MPE with $\{ T _n \} _{ n = 1 , \dots , N }$ of the baseline model, which satisfies the equilibrium conditions with strict inequalities, i.e., $\overline{G}(t_{n-1}, t, t_{n+1})> 0 > \underline{G}(t_{n-1}, t, t_{n+1})$ for $n = 1 , \dots , N - 1$, then there exists $\varLambda$ such that for any discrete-state model with $L > \varLambda$, there exists an MPE with $\{ T ^L _n \} _{ n = 1 , \dots , N }$ such that $T ^L _n \subset T _n$ for any n.