An equilibrium analysis of a discrete-time Markovian queue with endogenous abandonments

Abstract

This paper studies a Geo/Geo/1$+GI$ queue in which the abandonments are endogenous. One crucial feature of this model is that the abandonment behavior is affected by the system performance and vice versa. Our model captures this interaction by developing two closely related models: an abandonment model and a queueing model. In the abandonment model, customers take the virtual waiting time distribution as given. They receive a reward r from service and incur a cost c per period of waiting. Customers are forward-looking and maximize their expected discounted utilities by making wait or abandon decisions dynamically as they wait in the queue. The queueing model takes the customers’ abandonment time distribution as an input and studies the resulting virtual waiting time distribution. In equilibrium, the customers’ abandonment behavior and the system performance must be consistent across the two models. Therefore, combining the two models and imposing this consistency requirement, we show that there exists a unique equilibrium. Lastly, we provide a computational scheme to calculate the equilibrium numerically.

Appendices

Appendix 1: Proofs of Lemmas, Propositions, Corollary and Theorems in the paper

1.1 Proofs of results in Sect. 2

Proof of Proposition 1

It follows from Eq. (2) that V(t) is a Markov process. Since we only consider the underloaded case, i.e., $a<b$, there is a unique stationary distribution of V(t). Let v(w) denote the stationary distribution of V(t), i.e., $v(w) = \lim _{t\rightarrow \infty } \mathbb {P}(V(t) = w)$ for $w\ge 1$. The flow balance equations of the Markov process described in (2) are given as follows:

$$\begin{aligned} v(w)&= \left\{ \begin{array}{ll} \sum _{i=0}^w v(i)a\bar{G}(i) (1-b)^{w-i}b+ v(w+1)(1-a\bar{G}(w+1)),&{} w\ge 1,\\ v(0)(1-a\bar{G}(0)+ab\bar{G}(0)) + v(1)(1-a\bar{G}(1)), &{} w=0, \end{array}\right. \end{aligned}$$

(41)

where $\bar{G}(w) = 1-G(w)$ for $w\ge 1$. We simplify these equations by defining $r(w) = v(w)(1-b)^{-w}$. By substituting the definition of v(w) into (41) for $w=0$ and rearranging the terms, we have that

$$\begin{aligned} (1-a\bar{G}(1)) r(1) - a\bar{G}(0)r(0) = 0. \end{aligned}$$

(42)

By substituting r(w) into (41), we obtain the following: For $w\ge 1$,

$$\begin{aligned} \begin{aligned} r(w) =&(1\!-b)^{-w}\left[ \sum _{i=0}^w (1\!-b)^w r(i)ab\bar{G}(i){+} (1-b)^{w+1}r(w{+}1)(1\!-a\bar{G}(w{+}1))\right] \\ =&\sum _{i=0}^w r(i)ab\bar{G}(i)+ (1-b)r(w+1)(1-a\bar{G}(w+1)). \end{aligned} \end{aligned}$$

Rearranging the terms for $w=1$, we obtain that

$$\begin{aligned} (1-b)[(1-a\bar{G}(2)) r(2) - r(1)] =b[(1-a\bar{G}(1))r(1)-a\bar{G}(0)r(0)] = 0, \end{aligned}$$

(43)

where the last equality follows from (42). Subtracting r(w) from $r(w+1)$, we obtain that for $w\ge 1$,

$$\begin{aligned} \begin{aligned} r(w+1) -r(w) =&(1-b)(1-a\bar{G}(w+2))h(w+2) + r(w+1)ab\bar{G}(w+1) \\&- (1-b)(1-a\bar{G}(w+1))r(w+1)\\ =&(1-b)(1-a\bar{G}(w+2))r(w+2) \\&- (1-b-a\bar{G}(w+1))r(w+1). \end{aligned} \end{aligned}$$

Rearranging the terms, we have that for $w\ge 1$,

$$\begin{aligned}&(1-b)(1-a\bar{G}(w+2))r(w+2) - r(w+1) \\&\quad = [(1-a\bar{G}(w+1))r(w+1) - r(w)]. \end{aligned}$$

By substituting (43) into this equation recursively, we have that

$$\begin{aligned} (1-a\bar{G}(w+1))r(w+1) =r(w), \ w\ge 1. \end{aligned}$$

This gives that

$$\begin{aligned} r(w)&= r(1)\prod _{i = 2}^w (1-a\bar{G}(i))^{-1} , w\ge 2. \end{aligned}$$

By substituting $v(w) = (1-b)^w r(w)$ into this equation, we obtain the following:

$$\begin{aligned} v(w) = v(1) \prod _{i = 2}^w \frac{(1-b)}{(1-a\bar{G}(i))}, w\ge 2. \end{aligned}$$

Let $\beta (w)$ denote the probability of entering service in next period (in steady state) after waiting for w periods. Thus, $\beta (w)$ is given as follows: For $w\ge 1$,

$$\begin{aligned} \beta (w)= & {} \lim _{t\rightarrow \infty } \mathbb {P}(V(t)=w | V(t) \ge w) \\= & {} \frac{\displaystyle v(w)}{\displaystyle \sum _{t=w}^\infty v(w)}\\= & {} \frac{ \displaystyle v(1)\prod _{i = 2}^w (1-b)(1-a\bar{G}(i))^{-1}}{\displaystyle v(1)\sum _{t=w}^\infty \prod _{i = 2}^t (1-b)(1-a\bar{G}(i))^{-1}}\\= & {} \left( 1+\sum _{t=w+1}^\infty \prod _{i = w+1}^t (1-b)(1-a\bar{G}(i))^{-1}\right) ^{-1}. \end{aligned}$$

$\square$

Proof of Lemma 1

Substituting Eq. (5) into (9) yields the following: For $w\ge 1$,

$$\begin{aligned} \begin{aligned} q(w)&=\mathbb {P}(\varepsilon (1)-\varepsilon (0)\ge -c+\alpha \left\{ \beta (w)r+(1-\beta (w))J(w+1)\right\} )\\&=\bar{F}\left( -c+\alpha \left\{ \beta (w)r+(1-\beta (w))J(w+1)\right\} \right) . \end{aligned} \end{aligned}$$

We now derive the equation that characterizes $J(\cdot )$ given $q(\cdot )$. By substituting Eq. (10) into (5), the utility of staying in the queue can be written in terms of the abandonment probability, i.e.,

$$\begin{aligned} u(w,0,\varepsilon (w,0)) = \bar{F}^{-1}(q(w)) + \varepsilon (0). \end{aligned}$$

Substituting this into Eq. (6), we obtain that for $w\ge 1$,

$$\begin{aligned} \begin{aligned} J(w)&= \mathbb {E}_{\varepsilon } \left[ \max \{ u(w,0,\varepsilon (w,0)), u(w,1,\varepsilon (w,1))\}\right] \\&= \mathbb {E}_{\varepsilon } \left[ \max \{\bar{F}^{-1}(q(w)) + \varepsilon (0), \varepsilon (1)\}\right] \\&= \mathbb {E}_{\varepsilon } \left[ \bar{F}^{-1}(q(w)) + \varepsilon (0)-\varepsilon (1)\right] ^+ + \mathbb {E}_{\varepsilon }[\varepsilon (1)]\\&= \mathbb {E}_{\varepsilon } \left[ \bar{F}^{-1}(q(w)) + \varepsilon (0)-\varepsilon (1)\right] ^+. \end{aligned} \end{aligned}$$

The last line follows from Assumption 1, i.e., $\mathbb {E}_{\varepsilon }[\varepsilon (1)] = 0$. $\square$

Proof of Proposition 2

Fixing $\beta (\cdot )$, define $T_{\beta }:l^\infty \rightarrow l^\infty$ as follows: For $w\ge 1$,

$$\begin{aligned} \begin{aligned} T_{\beta }\, x(w)&=\mathbb {E}_{\varepsilon }[\max \{\varepsilon (1), -c+\alpha [\beta (w)r+(1 -\beta (w))x(w+1)]+\varepsilon (0)\}]\\&= \mathbb {E}_{\varepsilon }[ -c\!+\alpha [\beta (w)r\!+(1\!-\beta (w))x(w+1)]-(\varepsilon (1)-\varepsilon (0))]^{+} +\mathbb {E}_{\varepsilon }[\varepsilon (1)]\\&= \mathbb {E}_{\varepsilon }[ -c+\alpha [\beta (w)r+(1 -\beta (w))x(w+1)]-(\varepsilon (1)-\varepsilon (0))]^+, \end{aligned} \end{aligned}$$

(44)

where $l^\infty$ is the space of bounded sequences of real numbers. The last equality follows from Assumption 1 that $\mathbb {E}_{\varepsilon }[\varepsilon (1)] = 0$. Note that $T_{\beta }$ is the right-hand side of Eq. (8). Therefore, the expected discounted utility function J is the fixed point of operator $T_{\beta }$, i.e., $J = T_{\beta }\,J$.

We use Blackwell’s sufficient conditions for a contraction [39, Theorem 3.3] to show that the operator $T_{\beta }$ is a contraction⁶.

First, we check that if $J^1(w)\le J^2(w)$ (for all $w\ge 1$), then $T_{\beta }\,J^1_k(w) \le T_{\beta }\,J^2(w)$ for all $w\ge 1$. The following inequality holds: For all $w\ge 1$,

$$\begin{aligned} \begin{aligned}&(1-\beta (w))J^1_{(w+1)} \le (1-\beta (w))J^2(w+1), \end{aligned} \end{aligned}$$

because $1-\beta (w)\ge 0$ and $J^1(w+1) \le J^2(w+1)$. Since the inequality is preserved by the $\max$ operator and the expectation, it follows that $T_{\beta }\, J^1(w) \le T_{\beta }\, J^2(w)$ for all $w\ge 1$.

Then we show that $T_{\beta }\,(J+e)(w) \le T_{\beta }\,J(w)+\alpha e$ for all $e>0$ and $w\ge 1$. It follows that

$$\begin{aligned} \begin{aligned}&T_{\beta }\,(J+e)(w) \\&\quad = \mathbb {E}_{\varepsilon } \left[ \max \{\varepsilon (1), -c+\alpha [\beta (w)r + (1-\beta (w))[J(w+1)+e]]+\varepsilon (0)\}\right] \\&\quad \le \mathbb {E}_{\varepsilon } \left[ \max \{\varepsilon (1), -c+\alpha [\beta (w)r + (1-\beta (w))J(w+1)]+\varepsilon (0)+\alpha e\}\right] \\&\quad \le \mathbb {E}_{\varepsilon } \left[ \max \{\varepsilon (1)+\alpha e, -c+\alpha [\beta (w)r + (1-\beta (w))J(w+1)]+\varepsilon (0)+\alpha e\}\right] \\&\quad = \mathbb {E}_{\varepsilon } \left[ \max \{\varepsilon (1), -c+\alpha [\beta (w)r + (1-\beta (w))J(w+1)]+\varepsilon (0)\}\right] +\alpha e\\&\quad = T_{\beta }\,J(w)+\alpha e. \end{aligned} \end{aligned}$$

The first inequality holds because $(1-\beta (w))e \le e$. Hence, the two sufficient conditions of Blackwell are satisfied, i.e., $T_{\beta }$ is a contraction mapping. It follows from the Banach fixed point theorem that there exists a unique fixed point of $x = T_{\beta }\,x$. Since the solution to Eq. (8) is equivalent to the fixed point of $J = T_{\beta }\,J$, the solution is unique. $\square$

Proof of Corollary 1

Fix $\beta \in [0,1]^\infty$. Let $\mathscr {V} = \{\nu \in l^\infty : \nu (w) \in [0,r],\ w\ge 1\}$. In addition, let $T_{\beta }:l^\infty \rightarrow l^\infty$ be the operator defined in Eq. (44). For any $J^0\in \mathscr {V}$ and $w\ge 1$, the following inequality holds:

$$\begin{aligned} \begin{aligned} T_{\beta }\, J^0(w) =&\mathbb {E}_{\varepsilon } \left[ \max \{\varepsilon (1), -c+\alpha [\beta (w)r + (1-\beta (w))J^0_k(w+1)]+\varepsilon (0)\}\right] \\ \le&\mathbb {E}_{\varepsilon } \left[ \max \{\varepsilon (1), -c+\alpha r+\varepsilon (0)\}\right] \le r. \end{aligned} \end{aligned}$$

The first inequality holds because $J^0\in \mathscr {V}$. In particular, $J^0(w+1) \le r$. The second inequality follows from Assumption 2. In addition, it is immediate from (44) that $T_{\beta }\,J^0(w) \ge 0$ for all $w\ge 1$. Therefore, $T_{\beta }\, J^0\in \mathscr {V}$. Since $T_{\beta }$ is a contraction mapping and $\mathscr {V}$ is closed, $J = \lim _{n\rightarrow \infty } T_{\beta }^n\, J^0 \in \mathscr {V}$. In particular, $J(w)\le r$ for $w\ge 1$.

It follows from Eq. (10) to $J(w+1)\le r$ that

$$\begin{aligned} q(w) \ge \bar{F}(-c + \alpha r) \ge \bar{F}(r), \ w\ge 1. \end{aligned}$$

Thus, we have that $q(w) \ge \underline{q} = \bar{F}(r)$. Lastly, we have that $\underline{q}>0$ because r is in the interior of the support of $\bar{F}(\cdot )$. $\square$

1.2 Proofs of results in Sect. 3

Proof of Lemma 2

For a given $\beta \in [0,1]^\infty$, let $\tilde{\beta } = \varPhi (\varGamma (\beta ))$. Note that the mapping $\varPhi (\varGamma (\cdot ))$ is characterized by Eqs. (8), (10), (1) and (3).

Note from (1) that $\bar{G}(w) \ge 0$ for all $w\ge 1$. Thus, it follows from Eq. (3) that for $w\ge 1$,

$$\begin{aligned} \tilde{\beta }(w) \le \left( 1+\sum _{i=1}^\infty (1-b)^i\right) ^{-1} = b. \end{aligned}$$

This gives the upper bound of Eq. (12). In addition, it also follows from Eq. (3) that for $w\ge 1$,

$$\begin{aligned} \begin{aligned} \tilde{\beta }(w) =&\left( 1+\sum _{t=w+1}^\infty \prod _{i=w+1}^t \frac{1-b}{1-a\bar{G}(i+1)}\right) ^{-1}\\ \ge&\left( 1+\sum _{i=1}^\infty \left( \frac{1-b}{1-a\bar{G}(w+1)}\right) ^i\right) ^{-1} \\ =&1-\frac{1-b}{1-a\bar{G}(w+1)} \\ =&\frac{b-a\bar{G}(w+1)}{1-a\bar{G}(w+1)}. \end{aligned} \end{aligned}$$

(45)

The inequality follows from the fact that $\bar{G}(w)$ is non-increasing, i.e., $\bar{G}(i) \le \bar{G}(w+1)$ for all $i\ge w+1$; see (1) for its definition. It follows from Corollary 1 that $q_1(w)\ge \bar{q}>0$ for all $w\ge 1$. Thus, it follows from Eq. (1) that

$$\begin{aligned} \bar{G}(w) = \prod _{i=1}^w (1-q(i))\le (1-\underline{q})^w,\ w\ge 1. \end{aligned}$$

Substituting this inequality into Eq. (45), we have that for $w\ge 1$,

$$\begin{aligned} \begin{aligned} \tilde{\beta }(w) \ge&1- \frac{1-b}{1-a(1-\underline{q})^{w+1}} = \frac{b-a(1-\underline{q})^{w+1}}{1-a(1-\underline{q})^{w+1}}. \end{aligned} \end{aligned}$$

This shows the lower bound of $\tilde{\beta }(w)$ provided in Eq. (12). We end the proof by showing that $\tilde{\beta }(w)$ is non-decreasing in w. Rearranging the terms in Eq. (4), we have the following: For $w\ge 1$,

$$\begin{aligned} \frac{1}{\tilde{\beta }(w+1)} = \left( \frac{1}{\tilde{\beta }(w)} - 1\right) \left( 1+\frac{b-a\bar{G}(w+1)}{1-b}\right) . \end{aligned}$$

Substituting this equation into the following, we obtain that for $w\ge 1$,

$$\begin{aligned} \begin{aligned} \frac{1}{\tilde{\beta }(w+1)} - \frac{1}{\tilde{\beta }(w)}&= \left( \frac{1}{\tilde{\beta }(w)} - 1\right) \left( 1+\frac{b-a\bar{G}(w+1)}{1-b}\right) - \frac{1}{\tilde{\beta }(w)} \\&= \frac{1}{\tilde{\beta }(w)}\frac{b-a\bar{G}(w+1)}{1-b} - \frac{1-a\bar{G}(w+1)}{1-b}\\&\le \frac{1-a\bar{G}(w+1)}{b-a\bar{G}(w+1)}\frac{b-a\bar{G}(w+1)}{1-b}-\frac{1-a\bar{G}(w+1)}{1-b} = 0. \end{aligned} \end{aligned}$$

The last inequality follows from (45). Thus, we have that $\tilde{\beta }(w) \le \tilde{\beta }(w+1)$ for all $w\ge 1$, i.e., $\tilde{\beta }(w)$ is non-decreasing in w. $\square$

Proof of Lemma 3

We first show that $\mathscr {B}$ is a compact set. Define a sequence $x_w$ as follows:

$$\begin{aligned} x_w = b - \frac{b-a(1-\underline{q})^{w+1}}{1-a(1-\underline{q})^{w+1}}, \ w\ge 1. \end{aligned}$$

Thus, it is equivalent to writing $\mathscr {B}$ as

$$\begin{aligned} \mathscr {B} = \{\beta \in l^\infty : b - x_w \le \beta (w) \le b, w\ge 1\}. \end{aligned}$$

Note that $x_w \rightarrow 0$ as $w\rightarrow \infty$. Thus, for any $\epsilon >0$, there exists n such that $x_w < \epsilon$ for $w\ge n$. Define a set $\mathscr {B}_{n}$ as follows:

$$\begin{aligned} \mathscr {B}_{n} = \{\beta \in \mathbb {R}^{n}: b - x_w \le \beta (w) \le b, w=1,\ldots , n\}. \end{aligned}$$

Since $\mathscr {B}_{n}$ is a compact set in $\mathbb {R}^{n}$, it is totally bounded, i.e., it has a finite cover of open balls of radius $\epsilon$. In other words, there exist l and $\nu _1,\ldots , \nu _l \in \mathbb {R}^{n}$ such that $\mathscr {B}_{n} \subseteq \cup _{i=1}^l B_n(\nu _i, \epsilon )$, where $B_n(\nu _i, \epsilon )$ is the open ball in $\mathbb {R}^n$ centered at $\nu _i$ and with radius $\epsilon$. Let $w_i = (\nu _i, 0,\ldots )$, $i=1,\ldots ,l$. It is immediate that $\mathscr {B}$ is covered by $B(w_i,\epsilon )$, $i=1,\ldots , l$, where $B(w_i,\epsilon )$ is the open ball in $l^\infty$ that centers at $w_i$ and has a radius $\epsilon$. Since $\epsilon$ is arbitrary, $\mathscr {B}$ is totally bounded. Since $l^\infty$ is a complete metric space, the totally bounded subset $\mathscr {B}$ of $l^\infty$ is compact; see Theorem 3.28 in [5].

Next we show that $\varPhi (\varGamma (\cdot ))$ is continuous. Note that $\varPhi (\varGamma (\cdot ))$ is characterized by Eqs. (8), (10), (1) and (3). Let $\beta _n, \beta \in \mathscr {B}$ be sequences such that $\beta _n \rightarrow \beta$ (under the sup-norm). Let $J_n$, $q_n$, $G_n$ and $\tilde{\beta }_n$ and J, q, G and $\tilde{\beta }$ be the left-hand sides of Eqs. (8), (10), (1) and (3) by substituting $\beta _n$ and $\beta$ into $\varPhi (\varGamma (\cdot ))$, respectively. Thus, we have that $\tilde{\beta }_n =\varPhi (\varGamma (\beta _n))$ and $\tilde{\beta } = \varPhi (\varGamma (\beta ))$. We need to show that $\tilde{\beta }_n \rightarrow \tilde{\beta }$ under the sup-norm.

It follows from $\beta _n \rightarrow \beta$ that for any $\epsilon >0$, there exists $n_1$ such that $|\beta _n(w)-\beta (w)|<\epsilon$ for all $n\ge n_1$ and $w\ge 1$. It follows from (8) that for all $w\ge 1$ and $n\ge n_1$,

$$\begin{aligned} \begin{aligned}&|J_n(w)-J(w)| \\&\quad = |\mathbb {E}_{\varepsilon } \left[ \max \{\varepsilon (1), -c+\alpha [\beta _n(w)r + (1-\beta _n(w))J_n(w+1)]+\varepsilon (0)\}\right] \\&\qquad -\mathbb {E}_{\varepsilon } \left[ \max \{\varepsilon (1), -c+\alpha [\beta (w)r + (1-\beta (w))J(w+1)]+\varepsilon (0)\}\right] |\\&\quad = |\mathbb {E}_{\varepsilon } \left[ -c+\alpha [\beta _n(w)r + (1-\beta _n(w))J_n(w+1)]+\varepsilon (0)-\varepsilon (1)\right] ^+\\&\qquad -\mathbb {E}_{\varepsilon } \left[ -c+\alpha [\beta _n(w)r + (1-\beta (w))J(w+1)]+\varepsilon (0)-\varepsilon (1)\right] ^+|\\&\quad \le |\mathbb {E}_{\varepsilon } \left[ \alpha [\beta _n(w)r {+} (1\!-\beta _n(w))J_n(w+\!1)]-\alpha [\beta (w)r + (1\!-\beta (w))J(w+1)]\right] |\\&\quad = \alpha |(r-J_n(w+1))(\beta _n(w) -\beta (w)) - (1-\beta (w))(J_n(w+1)-J(w+1))|\\&\quad \le \alpha r |\beta _n(w) - \beta (w)| + \alpha |J_n(w+1) - J(w+1)|\\&\quad \le \alpha r \epsilon + \alpha |J_n(w+1) - J(w+1)|. \end{aligned} \end{aligned}$$

The equality in the third line follows from the fact that $\mathbb {E}[\varepsilon (1)] = 0$. The inequality in the fourth line follows from $|x_1^+ - x_2^+| \le |x_1-x_2|$ for all $x_1,x_2\in \mathbb {R}$. It follows from Corollary 1 that $J_n(w),J(w) \in [0,r]$ for all $w\ge 1$. Thus, $|J_n(w+1) - J(w+1)|$ is bounded for all $w\ge 1$. Thus, by applying this inequality recursively, we obtain that for all $w\ge 1$ and $n\ge n_1$,

$$\begin{aligned} |J_n(w)-J(w)| \le \alpha r \epsilon \sum _{i=0}^\infty \alpha ^i = \frac{\alpha r \epsilon }{1-\alpha }. \end{aligned}$$

(46)

Let $C_0 = \sup _{x\in [-c, r]} f(x)$. It follows from Assumption 1 that $f(\cdot )$ is continuous. Thus, $C_0<\infty$. Since $\bar{F}'(x) = -f(x)$, it holds that for any $x_1,x_2\in [-c,r]$,

$$\begin{aligned} |\bar{F}(x_1) - \bar{F}(x_2)| \le C_0 |x_1-x_2|. \end{aligned}$$

(47)

Since $J_n(w) \le r$ and $J(w)\le r$ for all $w\ge 1$, we have that (for $w\ge 1$)

$$\begin{aligned}&-c \le -c+\alpha (\beta _n(w)r + (1-\beta _n(w))J_n(w+1)) \le -c+\alpha r \le r,\\&-c \le -c+\alpha (\beta (w)r + (1-\beta (w))J(w+1)) \le -c+\alpha r \le r. \end{aligned}$$

Thus, it follows from (10) to (46) that for all $n\ge n_1$ and $w\ge 1$,

$$\begin{aligned} |q_n(w) - q(w)|\le & {} C_0 |\alpha [\beta _n(w) r + (1-\beta _n(w))J_n(w+1)]\\&- \alpha [\beta (w)+(1-\beta (w))J(w+1)]|\\= & {} C_0\alpha |(r-J_n(w+1))(\beta _n(w)-\beta (w)) \\&-(1-\beta (w))(J_n(w+1)-J(w+1))|\\\le & {} C_0\alpha r |\beta _n(w)-\beta (w)| + c\alpha |J_n(w+1) - J(w+1)|\\\le & {} C_0\alpha r \epsilon + c\alpha \frac{\alpha r \epsilon }{1-\alpha } \\= & {} \frac{C_0\alpha r\epsilon }{1-\alpha } = C_1\epsilon , \end{aligned}$$

where $C_1 = C_0\alpha r/(1-\alpha )$. The first inequality follows from (47) and the second one follows from $J_n(w+1)\in [0,r]$ and $\beta (w)\in [0,1]$. The last inequality follows from the assumption that $|\beta _n(w) - \beta (w)|<\epsilon$ for $n\ge n_1$ and (46). Substituting this inequality into (1), we have that for all $n\ge n_1$ and $w\ge 1$,

$$\begin{aligned} \begin{aligned} |G_n(w+1) - G(w+1)| =&|\bar{G}_n(w+1) - \bar{G}(w+1)|\\ =&|\bar{G}_n(w)(1-q_n(w+1)) - \bar{G}(w)(1-q(w+1))|\\ \le&\bar{G}_n(w) |q_n(w+1)-q(w+1)|\\&+ (1-q(w+1)) |\bar{G}_n(w) - \bar{G}(w)|\\ \le&(1-\underline{q})^w C_1\epsilon + (1-\underline{q}) |\bar{G}_n(w) - \bar{G}(w)|, \end{aligned} \end{aligned}$$

(48)

where the last inequality follows from Corollary 1 and Eq. (1) that

$$\begin{aligned} |1-q(w+1)|\le 1-\underline{q},\ 0\le \bar{G}_n(w)\le (1-\underline{q})^w \ \text{ and } \ 0\le \bar{G}(w) \le (1-\underline{q})^w,\ w\ge 1. \end{aligned}$$

(49)

Applying (48) recursively, we obtain that for all $w\ge 1$ and $n\ge n_1$,

$$\begin{aligned} \begin{aligned} |G_n(w) - G(w)| \le&\sum _{i=1}^{w-1} (1-\underline{q})^i C_1\epsilon + (1-\underline{q})^{w-1} |\bar{G}_n(1) - \bar{G}(1)|\\ \le&\sum _{i=1}^{\infty } (1-\underline{q})^i C_1\epsilon + (1-\underline{q})^{w-1} |q_n(1) - q(1)|\\ \le&\frac{C_1\epsilon }{\underline{q}} + C_1\epsilon = \left( 1+\frac{1}{\underline{q}}\right) C_1\epsilon . \end{aligned} \end{aligned}$$

By letting $C_2 =( 1 + 1/\underline{q})C_1$, we have that $|G_n(w) - G(w)|\le C_2\epsilon$ for all $w\ge 1$ and $n\ge n_1$. It follows from (4) that for $w\ge 1$ and $n\ge n_1$,

$$\begin{aligned}&|\tilde{\beta }_n(w) - \tilde{\beta }(w)|\nonumber \\&\quad = \tilde{\beta }_n(w)\tilde{\beta }(w)\left| \frac{1}{\tilde{\beta }_n(w)} - \frac{1}{\tilde{\beta }(w)}\right| \nonumber \\&\quad \le b^2 \left| \frac{1-b}{(1-a\bar{G}_n(w+1))\tilde{\beta }_n(w+1)}-\frac{1-b}{(1-a\bar{G}(w+1))\tilde{\beta }(w+1)}\right| \nonumber \\&\quad = b^2(1-b)\frac{|(1-a\bar{G}_n(w+1))(\tilde{\beta }_n(w+1) - \tilde{\beta }(w+1))+a \tilde{\beta }(w+1)(\bar{G}_n(w+1)-\bar{G}(w+1))|}{(1-a\bar{G}_n(w+1))\tilde{\beta }_n(w+1)(1-a\bar{G}(w+1))\tilde{\beta }(w+1)}\nonumber \\&\quad \le b^2(1-b)\frac{(1-a\bar{G}_n(w+1))|\tilde{\beta }_n(w+1) - \tilde{\beta }(w+1)|+a \tilde{\beta }(w+1)|\bar{G}_n(w+1)-\bar{G}(w+1)|}{(1-a\bar{G}_n(w+1))\tilde{\beta }_n(w+1)(1-a\bar{G}(w+1))\tilde{\beta }(w+1)}\nonumber \\&\quad = \frac{b^2(1-b)|\tilde{\beta }_n(w+1) - \tilde{\beta }(w+1)|}{\tilde{\beta }_n(w+1)(1-a\bar{G}(w+1))\tilde{\beta }(w+1)}+\frac{ab^2(1-b)|\bar{G}_n(w+1)-\bar{G}(w+1)|}{(1-a\bar{G}_n(w+1))\tilde{\beta }_n(w+1)(1-a\bar{G}(w+1))},\nonumber \\ \end{aligned}$$

(50)

where the first inequality follows from Lemma 2 that $\tilde{\beta }_n(w) \le b$, $\tilde{\beta }(w) \le b$ and (4). Note that $(b-a(1-\underline{q})^{w})/(1-a(1-\underline{q})^{w}) \rightarrow b$ as $w\rightarrow \infty$. In addition, $(1-\underline{q})^w\rightarrow 0$ as $w\rightarrow \infty$. Thus, there exists $w_1$ such that for $w\ge w_1$,

$$\begin{aligned} \begin{aligned}&\left( \frac{b-a(1-\underline{q})^{w}}{1-a(1-\underline{q})^{w}}\right) ^2(1-a(1-(1-\underline{q})^{w})) \ge b^2\sqrt{1-b} ,\\&\frac{b-a(1-\underline{q})^{w}}{1-a(1-\underline{q})^{w}} (1-a(1-(1-\underline{q})^{w}))^2\ge b\sqrt{1-b}. \end{aligned} \end{aligned}$$

Substituting these two inequalities into (12) and (49), we have that for $w\ge w_1$ and $n\ge n_1$,

$$\begin{aligned} \begin{aligned}&\tilde{\beta }_n(w+1)(1-a\bar{G}(w+1))\tilde{\beta }_n(w+1) \ge b^2\sqrt{1-b}, \\&(1-a\bar{G}_n(w+1))\tilde{\beta }_n(w+1)(1-a\bar{G}(w+1))\ge b\sqrt{1-b}. \end{aligned} \end{aligned}$$

Substituting these two inequalities into (50) yields that for $w\ge w_1$ and $n\ge n_1$

$$\begin{aligned} \begin{aligned} |\tilde{\beta }_n(w) - \tilde{\beta }(w)| \le&\sqrt{1-b}|\beta _n(w+1) - \tilde{\beta }(w+1)|\\&+ab\sqrt{1-b}|\bar{G}_n(w+1)-\bar{G}(w+1)|\\ \le&\sqrt{1-b}|\tilde{\beta }_n(w+1) -\tilde{\beta }(w+1)|+ab\sqrt{1-b}C_2\epsilon . \end{aligned} \end{aligned}$$

Applying this inequality recursively, we have that for all $w\ge w_1$ and $n\ge n_1$,

$$\begin{aligned} |\tilde{\beta }_n(w) -\tilde{\beta }(w)| \le ab\sqrt{1-b}C_2e\sum _{i=1}^\infty \left( \sqrt{1-b}\right) ^{i-1} = \frac{ab\sqrt{1-b}C_2\epsilon }{1-\sqrt{1-b}} = C_3\epsilon , \end{aligned}$$

(51)

where $C_3 = ab\sqrt{1-b}C_2/(1-\sqrt{1-b})$. It follows from (50) that for $w< w_1$,

$$\begin{aligned} \begin{aligned}&|\tilde{\beta }_n(w) - \tilde{\beta }(w)| \\&\quad \le \frac{b^2(1-b)|\beta _n(w+1) - \beta '(w+1)|}{\tilde{\beta }_n(w+1)(1-a\bar{G}(w+1))\tilde{\beta }(w+1)}\\&\qquad +\frac{ab^2(1-b)|\bar{G}_n(w+1)-\bar{G}(w+1)|}{(1-a\bar{G}_n(w+1))\tilde{\beta }_n(w+1)(1-a\bar{G}(w+1))}\\&\quad \le \frac{b^2(1-b)|\tilde{\beta }_n(w+1) - \tilde{\beta }(w+1)|}{(b-a)^2(1-a)}+\frac{ab^2(1-b)C_2\epsilon }{(1-a)^2(b-a)}. \end{aligned} \end{aligned}$$

(52)

The last inequality follows from $\bar{G}_n(w+1) \le 1$ and $\bar{G}(w+1)\le 1$ and from Lemma 2 that

$$\begin{aligned} \tilde{\beta }_n(w)\ge \frac{b-a(1-\underline{q})^{w}}{1-a(1-\underline{q})^w} \ge b-a \ \text{ and } \ \tilde{\beta }(w) \ge b-a. \end{aligned}$$

By applying (52) recursively, we have that for all $w< w_1$ and $n\ge n_1$

$$\begin{aligned} |\tilde{\beta }_n(w) - \tilde{\beta }(w)|\le & {} \left( \frac{b^2(1-b)}{(b-a)^2(1-a)}\right) ^{w_1-w}|\tilde{\beta }_n(w_1) - \tilde{\beta }(w_1)|\nonumber \\&+\frac{ab^2(1-b)C_2\epsilon }{(1-a)^2(b-a)}\sum _{i=0}^{w_1-w-1} \left( \frac{b^2(1-b)}{(b-a)^2(1-a)}\right) ^{i}\nonumber \\\le & {} \left( \frac{b^2(1-b)}{(b-a)^2(1-a)}\right) ^{w_1-w}C_3\epsilon \nonumber \\&+\frac{ab^2(1-b)C_2\epsilon }{(1-a)^2(b-a)}\sum _{i=0}^{w_1-w-1} \left( \frac{b^2(1-b)}{(b-a)^2(1-a)}\right) ^{i} \le c_1\epsilon ,\nonumber \\ \end{aligned}$$

(53)

where

$$\begin{aligned} \begin{aligned} c_1 =&\sup _{1\le w \le w_1} \left( \frac{b^2(1-b)}{(b-a)^2(1-a)}\right) ^{w_1-w}C_3 \\&+ \frac{ab^2(1-b)C_2}{(1-a)^2(b-a)}\sum _{i=0}^{w_1-w-1} \left( \frac{b^2(1-b)}{(b-a)^2(1-a)}\right) ^{i}. \end{aligned} \end{aligned}$$

Note that $w_1$ is independent of $\epsilon$. Thus, the constant $c_1$ is independent of $\epsilon$ as well. Combining Eqs. (51) and (53), we have that

$$\begin{aligned} |\tilde{\beta }_n(w) - \tilde{\beta }(w)| \le c_2\epsilon ,\ w\ge 1\ \text{ and } n\ge n_1, \end{aligned}$$

where $c_2 = \max \{C_3, c_1\}$. By letting $\epsilon \rightarrow 0$, we have that $\tilde{\beta }(w) \rightarrow \tilde{\beta }$ uniformly. This gives the continuity of $\varPhi (\varGamma (\cdot ))$. $\square$

Proof of Corollary 3

Since $\beta ^*$ is the solution to the fixed point problem $\beta ^* = \varPhi (\varGamma (\beta ^*))$, it is immediate from Lemma 2 that $\beta ^*(w)$ is increasing in w and satisfies inequality (12) for all $w\ge 1$. Note that the left-hand side of Eq. (12) converges to b as w goes to infinity. Therefore, $\lim _{w\rightarrow \infty }\beta ^*(w) = b$. $\square$

Proof of Lemma 4

Let $e^*= (\beta ^*,q^*)$ be an equilibrium. Let $J^*$ be the expected utility associated with $q^*$. It follows from Proposition 2 that $J^* = T_{\beta ^*}\,J^*$, where $T_{\beta ^*}$ is given by (44). Let $\mathscr {V} = \{J\in l^\infty : J(w_1)\le J(w_2)\le r, 1\le w_1\le w_2\}$. We first show that for any $J_0\in \mathscr {V}$, $J = T_{\beta ^*}\, J^0 \in \mathscr {V}$. It follows from (44) that for $w\ge 1$,

$$\begin{aligned} \begin{aligned} J(w)&= \mathbb {E}_\varepsilon \max \{\varepsilon (1), -c+\alpha [\beta ^*(w)r + (1-\beta ^*(w))J^0(w+1)]+\varepsilon (0)\}\\&\le \mathbb {E}_\varepsilon \max \{\varepsilon (1), -c+\alpha [\beta ^*(w+1)r + (1-\beta ^*(w+1))J^0(w+1)]+\varepsilon (0)\}\\&\le \mathbb {E}_\varepsilon \max \{\varepsilon (1), -c+\alpha [\beta ^*(w+1)r + (1-\beta ^*(w+1))J^0(w+2)]+\varepsilon (0)\} \\&=J(w+1). \end{aligned} \end{aligned}$$

(54)

The first inequality follows from Corollary 3 and the assumption that $J_0\in \mathscr {V}$. In particular, $\beta ^*(w) \le \beta ^*(w+1)$ and $J_0(w+1)\le r$. The second inequality follows from the assumption that $J^0(w+1) \le J^0(w+2)$. In addition, the following holds: For $w\ge 1$,

$$\begin{aligned} \begin{aligned} J(w)&= \mathbb {E}_\varepsilon \max \{\varepsilon (1), -c+\alpha [\beta ^*(w)r + (1-\beta ^*(w))J^0(w+1)]+\varepsilon (0)\}\\&\le \mathbb {E}_\varepsilon \max \{\varepsilon (1), -c+\alpha r+\varepsilon (0)\}\le r, \end{aligned} \end{aligned}$$

(55)

where the first inequality follows from $J_0(w+1)\le r$ and the second inequality follows from Assumption 2. Therefore, it follows from (54) to (55) that $J\in \mathscr {V}$. We have shown in the proof of Proposition 2 that T is a contraction mapping. Since $\mathscr {V}$ is a closed set, we have that $J^* = \lim _{n\rightarrow \infty } T_{\beta ^*}^n J^0 \in \mathscr {V}$. In particular, $J^*(w)$ is increasing in w and bounded above by r.

In addition, it follows from (10) that for $w\ge 1$,

$$\begin{aligned} \begin{aligned} q^*(w) =&\bar{F}(-c+\alpha [\beta ^*(w)r + (1-\beta ^*(w))J^*(w+1)])\\ \ge&\bar{F}(-c+\alpha [\beta ^*(w+1)r + (1-\beta ^*(w+1))J^*(w+2)]) \\ =&q^*(w+1). \end{aligned} \end{aligned}$$

The inequality follows from Corollary 3 that $\beta ^*(w)\le \beta ^*(w+1)$ and the monotonicity of $J^*$, i.e., $J^*(w+1) \le J^*(w+2) \le r$. Thus, $q^*(w)$ is decreasing in w. $\square$

Proof of Corollary 4

It follows from Lemma 4 and Corollary 1 that $J^*(w)$ is increasing in w and bounded above by r. Thus, there exists $J'_\infty \le r$ such that $\lim _{w\rightarrow \infty } J^*(w) = J'_\infty$.

Note that the right-hand side of (14) equals $\kappa (b,x)$, where $\kappa (\cdot )$ is defined in (95). It follows from Lemma 19 that the fixed point of (14) is unique. Let $J_\infty = j(b)$ be the fixed point of (14), where $j(\cdot )$ is given in Lemma 19.

Next, we show that $J_\infty ' = J_\infty$. We first show that $J'_\infty \le J_\infty$. Let $\beta _1(w) = b$ and $J_1(w) = J_\infty$ for all $w\ge 1$. It is immediate that $J_1$ is a fixed point of $J = T_{\beta _1}\,J$, where the operator $T_{\beta _1}$ is given by (44). As shown in the proof of Proposition 2 that $T_{\beta _1}$ is a contraction mapping, this fixed point is unique. Substituting the inequalities $\beta ^*(w)\le \beta _1(w) = b$ and $J^*(w) \le r$ into (44), we have that

$$\begin{aligned} J^*(w)= T_{\beta ^*}J^*(w) \le T_{\beta _1}J^*(w),\ w\ge 1. \end{aligned}$$

Substituting this inequality recursively into (44), we have that $J^*(w) \le T_{\beta _1}^n\,J^*(w)$ for all n, w. Thus, the following holds:

$$\begin{aligned} J^*(w) \le \lim _{n\rightarrow \infty } T_{\beta _1}^n\,J^*(w) = J_1(w) = J_\infty ,\ w\ge 1. \end{aligned}$$

(56)

Letting w go to infinity, we have that $J'_\infty =\lim _{w\rightarrow \infty } J^*(w) \le J_\infty$.

Next we show that $J'_\infty \ge J_\infty$. It follows from Corollary 3 that $\beta ^*(w) \rightarrow b$. Thus, fixing $\epsilon >0$, there exists $w_1$ such that $\beta ^*(w)\ge b-\epsilon$ for all $w\ge w_1$. Let $\beta _2(w) = \beta ^*(w+w_1)$ and $\beta _3(w) = b-\epsilon$ for $w\ge 1$. In addition, let $J_2(w) = J^*(w+w_1)$ and $J_3(w) = j(b-\epsilon )$, where $j(\cdot )$ is given in Lemma 19. It is immediate that $J_i$ is the unique fixed point of $J = T_{\beta _i}\, J$, $i=2,3$. Since $\beta _2(w)\ge \beta _3(w)$ for all $w\ge 1$, we can repeat the proof of (56) and show that $J_2(w)\ge J_3(w) = j(b-\epsilon )$ for all $w\ge 1$. Letting w go to infinity, we obtain that

$$\begin{aligned} J'_\infty = \lim _{w\rightarrow \infty } J^*(w) = \lim _{w\rightarrow \infty } J_2(w) \ge j(b-\epsilon ). \end{aligned}$$

By letting $\epsilon \rightarrow 0$, it follows from the continuity of $j(\cdot )$ (cf. Lemma 20) that $J'_\infty \ge j(b) = J_\infty$. Thus, we conclude that $\lim _{w\rightarrow \infty } J^*(w) = J'_\infty = J_\infty$.

It follows from Lemma 4 and Corollary 1 that $q^*(w)$ is decreasing in w and bounded above from $\underline{q}$. Thus, there exists a constant $q_\infty$ such that $\lim _{w\rightarrow \infty } q^*(w) = q_\infty$. In addition, it follows from (10) that

$$\begin{aligned} \begin{aligned} q_\infty = \lim _{w\rightarrow \infty } q^*(w) =&\lim _{w\rightarrow \infty } \bar{F}(-c+\alpha [\beta ^*(w)r + (1-\beta ^*(w))J^*(w+1)]) \\ =&\bar{F}(-c+\alpha (br+(1-b)J_\infty )). \end{aligned} \end{aligned}$$

The last inequality follows from the continuity of $\bar{F}(\cdot )$ and that $\beta ^*(w)\rightarrow b$ and $J^*(w)\rightarrow J_\infty$ as $w\rightarrow \infty$. $\square$

1.3 Proofs of the Proposition and the Lemma in Section 4

Proof of Lemma 9

Fixing N and comparing (37)–(40) and (70)–(72), we have that

$$\begin{aligned} (e_N(w), \bar{G}_N(w)) = h(e_N(w+1), \bar{G}_N(w+1)),\ w< N. \end{aligned}$$

(57)

Note that the truncation in (72) is immaterial in this case because $\bar{G}_N(w) \le 1$ for $w\ge 1$. Fixing $w = N$ and substituting $z(N) = (\beta _N(N),q_N(N),\bar{G}_N(N))$ into Eq. (83), we have that the resulting $z(1) = (\beta _N(1),q_N(1),\bar{G}_N(1))$ satisfies (84). In particular, $\bar{G}_N(1) = 1-q_N(1)$. Thus, it follows from the definition of the function $f_N(\cdot )$ that

$$\begin{aligned} \bar{G}_N(w) = f_w(e_N(w)), w\ge N. \end{aligned}$$

(58)

In particular, $\bar{G}_N(N) = f_N(e_N(N))$. In other words, the value of $\bar{G}_N(N)$ is uniquely determined. Since the truncated equilibrium is fully characterized by $\bar{G}_N(N)$, we conclude that the truncated equilibrium is unique. $\square$

Proof of Proposition 5

To facilitate the analysis to follow, we define a function $\tilde{h}=(\tilde{h}_1,\tilde{h}_2)$ as follows: For $w\ge 1$ and $(\beta ,q) \in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)\subseteq (0,b]\times [q_\infty ,1)$,

$$\begin{aligned}&\tilde{h}_i(\beta ,q;w) = h(\beta ,q,f_w(\beta ,q)),\ i=1,2, \end{aligned}$$

where the functions $h(\cdot )$ and $f_w(\cdot )$ are defined in (70)–(72) and (82)–(84) and $\mathscr {Z}_1(w) \times \mathscr {Z}_2(w)$ is given in (101). Define a matrix $D\tilde{h}(\beta _1,q_1,\beta _2,q_2;w)$ as follows: For $w\ge 1$ and $(\beta _1,q_1),(\beta _2,q_2) \in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)$,

$$\begin{aligned} D\tilde{h}(\beta _1,q_1,\beta _2,q_2;w) = \left[ \begin{array}{cc} Dh_{11}(\beta _1,q_1;w) &{} Dh_{12}(\beta _1,q_1;w) \\ Dh_{21}(\beta _2,q_2;w) &{} Dh_{22}(\beta _2,q_2;w)\end{array}\right] , \end{aligned}$$

where

$$\begin{aligned} D\tilde{h}_{11}(\beta ,q;w) = \frac{\partial h_1(\beta ,q,f_w(\beta ,q))}{\partial z_1} + \frac{\partial h_1(\beta ,q,f_w(\beta ,q))}{\partial z_3}\frac{\partial f_w(\beta ,q)}{\partial \beta },\\ D\tilde{h}_{12}(\beta ,q;w) = \frac{\partial h_1(\beta ,q,f_w(\beta ,q))}{\partial z_2} + \frac{\partial h_1(\beta ,q,f_w(\beta ,q))}{\partial z_3}\frac{\partial f_w(\beta ,q)}{\partial q},\\ D\tilde{h}_{21}(\beta ,q;w) = \frac{\partial h_2(\beta ,q,f_w(\beta ,q))}{\partial z_1} + \frac{\partial h_2(\beta ,q,f_w(\beta ,q))}{\partial z_3}\frac{\partial f_w(\beta ,q)}{\partial \beta },\\ D\tilde{h}_{22}(\beta ,q;w) = \frac{\partial h_2(\beta ,q,f_w(\beta ,q))}{\partial z_2} + \frac{\partial h_2(\beta ,q,f_w(\beta ,q))}{\partial z_3}\frac{\partial f_w(\beta ,q)}{\partial q}. \end{aligned}$$

It is immediate that $D\tilde{h}(\beta ,q,\beta ,q;w)$ is the Jacobian matrix of $\tilde{h}(\beta ,q)$. In addition, define a constant matrix $Dh_0$ as follows:

$$\begin{aligned} D\tilde{h}_0 =&\left[ \begin{array}{cc} \displaystyle \frac{\partial h_1(z_0)}{\partial z_1} &{}\displaystyle \frac{\partial h_1(z_0)}{\partial z_2} \\ \displaystyle \frac{\partial h_2(z_0)}{\partial z_1} &{}\displaystyle \frac{\partial h_2(z_0)}{\partial z_2}\end{array}\right] \\ =&\left[ \begin{array}{cc} 1-b &{}0 \\ -f(\bar{F}^{-1}(q_\infty ))\alpha (r-J_\infty )(1-b) &{}\alpha (1-q_\infty )(1-b)\end{array}\right] , \end{aligned}$$

where $z_0 = (\beta ,q_\infty ,0)$. It is immediate that the eigenvalues of $D\tilde{h}_0$ are $1-b$ and $\alpha (1-q_\infty )(1-b)$. Thus, there exists an invertible matrix S such that the following holds:

$$\begin{aligned} \left[ \begin{array}{cc} 1-b &{}0 \\ 0 &{}\alpha (1-q_\infty )(1-b)\end{array}\right] = S(D\tilde{h}_0) S^{-1}. \end{aligned}$$

Define a vector norm$||\cdot ||_S$ and a matrix norm $|||\cdot |||_S$ as follows: For $x\in \mathbb {R}^2$ and $M\in \mathscr {M}_2$,

$$\begin{aligned} ||x||_S = ||Sx||_\infty \ \text{ and } \ |||M|||_S = |||SMS^{-1}|||_\infty . \end{aligned}$$

It is immediate that $|||D\tilde{h}_0|||_S = 1-b$. Define a sequence $a_w$ as follows:

$$\begin{aligned} a_w \!= \sup \left\{ |||D\tilde{h}(\beta _1,q_1,\beta _2,q_2;w)|||_S:(\beta _1,q_1),(\beta _2,q_2){\in } \mathscr {Z}_1(w) {\times } \mathscr {Z}_2(w)\right\} ,\ w{\ge } 1. \end{aligned}$$

(59)

We then show that $a_w \rightarrow |||D\tilde{h}_0|||_S = 1-b$ as $w\rightarrow \infty$. It follows from Lemma 16 that

$$\begin{aligned} \sup \left\{ |f_w(\beta ,q)| :(\beta ,q) \in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)\subseteq (0,b]\times [q_\infty ,1)\right\} \rightarrow 0\ \text{ as } \ w\rightarrow \infty . \end{aligned}$$

It follows from Lemma 21 and Eq. (101) that

$$\begin{aligned} \sup \left\{ |\beta -b|+|q-q_\infty | :(\beta ,q) \in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)\right\} \rightarrow 0\ \text{ as } \ w\rightarrow \infty . \end{aligned}$$

Thus, it follows from the continuity of the partial derivatives of $h(\cdot )$ (see (73)–(81)) that for $i=1,2$,

$$\begin{aligned} \sup \nolimits _{(\beta ,q)\in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)} \left| \frac{\partial h_i(\beta ,q,f_w(\beta ,q))}{\partial z_j} - \frac{\partial h_i(b,q_\infty ,0)}{\partial z_j} \right| \rightarrow 0\ \text{ as } \ w\rightarrow \infty . \end{aligned}$$

(60)

In addition, it follows from Lemma 24 that as $w\rightarrow \infty$,

$$\begin{aligned} \sup \nolimits _{(\beta ,q)\in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)} \left| \frac{\partial f_w(\beta ,q)}{\partial \beta }\right| {\rightarrow } 0\ \text{ and } \ \sup \nolimits _{(\beta ,q)\in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)} \left| \frac{\partial f_w(\beta ,q)}{\partial q}\right| {\rightarrow } 0. \end{aligned}$$

(61)

Substituting (60)–(61) into $D\tilde{h}(\cdot )$, we have that

$$\begin{aligned} \sup \nolimits _{(\beta _1,q_1),(\beta _2,q_2)\in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)} \left| D\tilde{h}(\beta _1,q_1,\beta _2,q_2;w)-D\tilde{h}_0\right| \rightarrow 0\ \text{ as } \ w\rightarrow \infty . \end{aligned}$$

By the continuity of the norm $|||\cdot |||_S$, we have that

$$\begin{aligned} \begin{aligned} \lim _{w\rightarrow \infty }a_w =&\lim _{w\rightarrow \infty }\sup \left\{ |||Dh(\beta _1,q_1,\beta _2,q_2;w)|||_S:(\beta _1,q_1),\right. \\&\left. (\beta _2,q_2){\in } \mathscr {Z}_1(w) {\times } \mathscr {Z}_2(w)\right\} \\ =&|||D\tilde{h}_0|||_S =1-b. \end{aligned} \end{aligned}$$

Thus, there exists $w_1\ge 1$ such that

$$\begin{aligned} a_w \le 1-b/2 < 1,\ w\ge w_1. \end{aligned}$$

(62)

Next we show that $e^N\rightarrow e^*$ uniformly. Define the difference of the truncated equilibrium and the equilibrium as follows:

$$\begin{aligned} \delta _\beta ^N(w) = \beta _N(w) -\beta ^*(w)\ \text{ and } \ \delta _q^N(w) = q_N(w) -q^*(w),\ N,w\ge 1. \end{aligned}$$

To show that $e^N\rightarrow e^*$ uniformly, we need to show that

$$\begin{aligned} \sup _{w\ge 1} \delta _\beta ^N(w) \rightarrow 0 \ \text{ and } \ \sup _{w\ge 1} \delta _q^N(w)\rightarrow 0\ \text{ as } N\rightarrow \infty . \end{aligned}$$

This is equivalent to showing that

$$\begin{aligned} \sup _{w\ge 1} ||\delta ^N(w)||_S \rightarrow 0\ \text{ as } N\rightarrow \infty , \end{aligned}$$

(63)

where $\delta ^N(w) = [ \delta _\beta ^N(w), \delta _q^N(w)]^T$ for all $N,w\ge 1$. The rest of this proof shows that (63) holds. It follows from Corollaries 3–4 that $\beta ^*(w) \rightarrow b$ and $q^*(w) \rightarrow q_\infty$ as $w\rightarrow \infty$. Note that $\beta _N(w) = b$ and $q_N(w) = q_\infty$ for $w\ge N$. Thus, for any $\epsilon >0$, there exists $N_1\ge w_1$ such that

$$\begin{aligned} ||\delta ^N(w)||_S < \epsilon , \ w\ge N\ge N_1. \end{aligned}$$

(64)

It follows from (57) to (58), Lemma 25 and Corollary 5 that for $N>w\ge 1$,

$$\begin{aligned} (\beta _N(w),q_N(w))= & {} \tilde{h}(\beta _N(w+1),q_N(w+1))\ \text{ and } \\ (\beta ^*(w),q^*(w))= & {} \tilde{h}(\beta ^*(w+1),q^*(w+1)). \end{aligned}$$

Thus, it follows from the mean value theorem that for $N>w\ge 1$,

$$\begin{aligned} \begin{aligned} \delta ^N(w)&= \tilde{h}(\beta _N(w+1),q_N(w+1))-\tilde{h}(\beta _N(w+1),q_N(w+1))\\&= D\tilde{h}(w+1;\beta _1^N(w+1),q_1^N(w+1),\beta _2^N(w+1),q_2^N(w+1) ) \delta ^N(w+1), \end{aligned} \end{aligned}$$

(65)

where

$$\begin{aligned} \begin{aligned}&(\beta _i^N(w+1),q_i^N(w+1)) \\&\quad = c_i^N(w{+}1) (\beta _N(w{+}1),q_N(w{+}1))+(1-c_i^N(w+1)) (\beta ^*(w+1),q^*(w+1)) \end{aligned} \end{aligned}$$

for some $c_i^N(w+1) \in (0,1)$, $i=1,2$. Note that

$$\begin{aligned} (\beta _N(N), q_N(N)) = (b,q_\infty ) \in \mathscr {Z}_1(N)\times \mathscr {Z}_2(N). \end{aligned}$$

In addition, it follows from Lemma 26 that $(\beta ^*(N), q^*(N)) \in \mathscr {Z}_1(N)\times \mathscr {Z}_2(N)$. It follows from Lemma 22 and the convexity of $\mathscr {Z}_1(w)\times \mathscr {Z}_2(w)$ that

$$\begin{aligned} (\beta _i^N(w+1),q_i^N(w+1)) \in \mathscr {Z}_1(w+1)\times \mathscr {Z}_2(w+1), \ w=1,\ldots , N-1,\ \text{ and } \ i=1,2. \end{aligned}$$

Thus, it follows from (59) that for $w=1,\ldots , N-1$,

$$\begin{aligned} |||D\tilde{h}(\beta _1^N(w+1),q_1^N(w+1),\beta _2^N(w+1),q_2^N(w+1) ;w+1 )|||_S \le a_{w+1}. \end{aligned}$$

By taking the norm of the both sides of (65), we obtain that, for $w= 1,\ldots , N-1$,

$$\begin{aligned} \begin{aligned} ||\delta ^N(w)||_S&\le |||D\tilde{h}(w+1;\beta _1^N(w+1),q_1^N(w+1),\beta _2^N(w+1),\\&\quad \times q_2^N(w+1) )|||_S ||\delta ^N(w+1)||_S\\&\le a_{w+1} ||\delta ^N(w+1)||_S. \end{aligned} \end{aligned}$$

(66)

Substituting (62) and (64) into (66) yields that for $N\ge N_1$ and $w= w_1,\ldots ,N$,

$$\begin{aligned} ||\delta ^N(w)||_S \le \left( 1-\frac{1}{2}b\right) ||\delta ^N(w+1)||_S\le \cdots \le \left( 1-\frac{1}{2}b\right) ^{N-w} ||\delta ^N(N)||_S < \epsilon . \end{aligned}$$

(67)

In addition, it follows from (66) that for $N\ge N_1$ and $w = 1,\ldots , w_1-1$,

$$\begin{aligned} ||\delta ^N(w)||_S \le \left( \prod _{i=w+1}^{w_1} a_i\right) ||\delta ^N(w_1)||_S< \bar{a}\epsilon , w< w_1, \end{aligned}$$

(68)

where

$$\begin{aligned} \bar{a} =\max \left\{ \sup _{i\in \{1,\ldots ,w_1-1\}} \prod _{j = i+1}^{w_1} a_j,1\right\} . \end{aligned}$$

Thus, we conclude from (64), (67)–(68) that $||\delta ^N(w)||_S < \bar{a}\epsilon$ for all $N\ge N_1$ and $w\ge 1$. By letting $\epsilon \rightarrow 0$, Eq. (63) holds. $\square$

Proof of Lemma 10

We show by induction that for $w = 1,\ldots , N$,

$$\begin{aligned} \beta _N^1(w) \le \beta _N^2(w),\ \ q_N^1(w) \ge q_N^2(w)\ \text{ and } \ \bar{G}_N^1(w) >\bar{G}_N^2(w). \end{aligned}$$

(69)

This is true for $w=N$ by assumption. As the inductive assumption, suppose (69) is true for w, then we argue that it is also true for $w-1$. It follows from Eq. (37) and the inductive assumption that $\beta _N^1(w-1)\le \beta _N^2(w-1)$. Similarly, it follows from (38) to (40) that

$$\begin{aligned} q_N^1(w-1) \ge q_N^2(w-1)\ \text{ and } \ \bar{G}_N^1(w-1) >\bar{G}_N^2(w-1). \end{aligned}$$

In particular, both of the following must be true:

$$\begin{aligned} q_N^1(1) \ge q_N^2(1)\ \text{ and } \ \bar{G}_N^1(1) >\bar{G}_N^2(1). \end{aligned}$$

Thus, the following holds:

$$\begin{aligned} \bar{G}_N^1(0) = \frac{\bar{G}_N^1(1)}{1-q_N^1(1)} > \bar{G}_N^2(0) = \frac{\bar{G}_N^2(1)}{1-q_N^2(1)}. \end{aligned}$$

$\square$

Appendix 2: Technical lemmas characterizing the equilibrium quantities in discrete time

This section proves Lemma 5 that facilitates the proof of uniqueness of the equilibrium. To prove this result, we define an auxiliary function $f_w(\cdot )$ implicitly and study its properties (especially the monotonicity and convergence of its partial derivatives as w gets large). The function helps characterize $\bar{G}$ in terms of $\beta$ and q. We then apply the mean value theorem to $f_w(\cdot )$ to establish the result in Lemma 5.

1.1 Definition of the auxiliary function $f_w(\cdot )$

The function $f_w(\cdot )$ is constructed such that for an equilibrium, the following holds:

$$\begin{aligned} \bar{G}^*(w) = f_w(\beta ^*(w),q^*(w)),\ w\ge 1. \end{aligned}$$

We establish this relationship in “Characterizing the equilibrium quantities with $f_w(\cdot )$” in Appendix 2. As a preliminary, we first define a function $h(\cdot )$. The function $f_w(\cdot )$ is then defined implicitly as the value satisfying a set of equations characterized by $h(\cdot )$ recursively.

To facilitate the analysis to follow, we define a function $h = (h_1,h_2,h_3): Z\rightarrow \mathbb {R}^3$ as follows⁷:

$$\begin{aligned}&h_1(z) = \left( 1+\frac{1-b}{1-az_3}\frac{1}{z_1}\right) ^{-1}, \end{aligned}$$

(70)

$$\begin{aligned}&h_2(z) = \bar{F}\left( -c+\alpha \left[ h_1(z) r + (1-h_1(z))\int _{-\infty }^{\bar{F}^{-1}(z_2)} F(x)\,\mathrm {d}x\right] \right) , \end{aligned}$$

(71)

$$\begin{aligned}&h_3(z) = \min \left( \frac{z_3}{1-z_2},1\right) , \end{aligned}$$

(72)

where $z = (z_1,z_2,z_3)$ and $Z = (0,b]\times [q_\infty ,1)\times [0,1]\subseteq \mathbb {R}^3$, where $q_\infty$ is the constant defined in Corollary 4. The following lemma shows that $h(\cdot )$ maps Z to Z.

Lemma 11

We have that $h(z) \in Z$ for all $z\in \mathscr {Z}$. In addition, the following inequality holds for all $z\in Z$:

$$\begin{aligned} \int _{-\infty }^{\bar{F}^{-1}(z_2)} F(x)\,\mathrm {d}x = \mathbb {E}[\bar{F}^{-1}(z_2)-(\varepsilon (1)-\varepsilon (0))]^+ \le J_\infty \le r, \end{aligned}$$

where $J_\infty$ is the constant defined in Corollary 4.

Proof

For any $z\in Z$, it is straightforward that $h_1(z) >0$. Since $h_1(\cdot )$ is increasing in $z_1$ and decreasing in $z_3$, $h_1(z) \le (1+ (1-b)/b)^{-1} = b$. Thus, $h_1(z) \in (0,b]$.

Recall that the cdf $F(\cdot )$ is the distribution function of the difference of the idiosyncratic shocks $\varepsilon (1)-\varepsilon (0)$. It follows from integration by parts that

$$\begin{aligned} \begin{aligned} \mathbb {E}[\bar{F}^{-1}(z_2)-(\varepsilon (1)-\varepsilon (0))]^+ =&\int _{-\infty }^{\bar{F}^{-1}(z_2)} \left( \bar{F}^{-1}(z_2)-x\right) \,\mathrm {d}F(x)\\ =&(\bar{F}^{-1}(z_2)-x)F(x)|_{-\infty }^{\bar{F}^{-1}(z_2)} \\&- \int _{-\infty }^{\bar{F}^{-1}(z_2)} F(x)\,\mathrm {d} \left( \bar{F}^{-1}(z_2)-x\right) \\ =&\int _{-\infty }^{\bar{F}^{-1}(z_2)} F(x)\,\mathrm {d}x. \end{aligned} \end{aligned}$$

By substituting the definitions of $J_\infty$ and $q_\infty$ (in Corollary 4) into the right-hand side, the following inequality holds: For any $z_2 \in [q_\infty , 1)$,

$$\begin{aligned} \begin{aligned} \int _{-\infty }^{\bar{F}^{-1}(z_2)} F(x)\,\mathrm {d}x&\le \int _{-\infty }^{\bar{F}^{-1}(q_\infty )} F(x)\,\mathrm {d}x\\&= \mathbb {E}[\bar{F}^{-1}(q_\infty )-(\varepsilon (1)-\varepsilon (0))]^+\\&= \mathbb {E}[-c+\alpha [br+(1-b)J_\infty ]-(\varepsilon (1)-\varepsilon (0))]^+\\&= J_\infty \le r. \end{aligned} \end{aligned}$$

The second inequality holds by the definition of $q_\infty$, i.e., $q_\infty = \bar{F}(-c+\alpha [br+(1-b)J_\infty ])$. The third equality follows from Eq. (14). This proves that for any $z\in Z$,

$$\begin{aligned} \int _{-\infty }^{\bar{F}^{-1}(z_2)} F(x) \le J_\infty . \end{aligned}$$

Substituting this inequality into (71), we obtain that for $z\in \mathscr {Z}$,

$$\begin{aligned} \begin{aligned} h_2(z)&= \bar{F}\left( -c+\alpha \left[ h_1(z) r + (1-h_1(z))\int _{-\infty }^{\bar{F}^{-1}(z_2)} F(x)\,\mathrm {d}x \right] \right) \\&\ge \bar{F}\left( -c+\alpha (h_1(z) r + (1-h_1(z))J_\infty )\right) \\&\ge \bar{F}\left( -c+\alpha (b r + (1-b)J_\infty )\right) \\&= q_\infty , \end{aligned} \end{aligned}$$

where the last inequality follows from $h_1(z)\in (0,b]$. In addition, $h_2(z) \le \bar{F}(-c) <1$. Thus, $h_2(z) \in [q_\infty , 1)$ for any $z\in Z$. Since $z_3 \ge 0$ and $z_2< 1$, it follows from (72) that $h_3(z) \in [0,1]$. Thus, $h(z) \in Z$. $\square$

The following lemma shows the elements of the Jacobian matrix of $h(\cdot )$ and the sign of each element.

Lemma 12

The partial derivatives of $h(\cdot )$ are given as follows:

$$\begin{aligned}&\frac{\partial h_1}{\partial z_1} = \frac{h^2_1(z)}{z_1^2}\frac{1-b}{1-az_3}>0, \end{aligned}$$

(73)

$$\begin{aligned}&\frac{\partial h_1}{\partial z_2} = 0, \end{aligned}$$

(74)

$$\begin{aligned}&\frac{\partial h_1}{\partial z_3} = -\frac{ah^2_1(z)}{(1-az_3)^2}\frac{1-b}{z_1}<0, \end{aligned}$$

(75)

$$\begin{aligned}&\frac{\partial h_2}{\partial z_1} = -f(\bar{F}^{-1}(h_2(z)))\alpha \left( r-\int _{-\infty }^{\bar{F}^{-1}(z_2)} F(x)\,\mathrm {d}x \right) \frac{h^2_1(z)}{z_1^2}\frac{1-b}{1-az_3} < 0, \end{aligned}$$

(76)

$$\begin{aligned}&\frac{\partial h_2}{\partial z_2} = \alpha (1-h_1(z))(1-z_2)\frac{f(\bar{F}^{-1}(h_2(z)))}{f(\bar{F}^{-1}(z_2))}> 0, \end{aligned}$$

(77)

$$\begin{aligned}&\frac{\partial h_2}{\partial z_3} = f(\bar{F}^{-1}(h_2(z))\alpha \left( r-\int _{-\infty }^{\bar{F}^{-1}(z_2)} F(x)\,\mathrm {d}x \right) \frac{ah^2_1(z)}{(1-az_3)^2}\frac{1-b}{z_1}> 0 , \end{aligned}$$

(78)

$$\begin{aligned}&\frac{\partial h_3}{\partial z_1} = 0, \end{aligned}$$

(79)

$$\begin{aligned}&\frac{\partial h_3}{\partial z_2} = \left\{ \begin{array}{ll}\displaystyle \frac{z_3}{(1-z_2)^2}\ge 0,&{} \text{ if } \ z_3 < 1-z_2,\\ 0, &{} \text{ if } \ z_3 > 1-z_2,\end{array}\right. \end{aligned}$$

(80)

$$\begin{aligned}&\frac{\partial h_3}{\partial z_3} = \left\{ \begin{array}{ll}\displaystyle \frac{1}{1-z_2}> 1,&{} \text{ if } \ z_3 < 1-z_2,\\ 0, &{} \text{ if } \ z_3 > 1-z_2.\end{array}\right. \end{aligned}$$

(81)

Proof

Since $h(\cdot )$ is given explicitly by Eqs. (70)–(72), the partial derivatives of $h(\cdot )$ are immediate. The signs of Eqs. (76) and (78) follow from Lemma 11. To be specific,

$$\begin{aligned} r-\int _{-\infty }^{\bar{F}^{-1}(z_2)} F(x)\,\mathrm {d}x \ge 0, \ z_2 \in [q_\infty , 1). \end{aligned}$$

The signs of other equations are immediate.

In addition, the following lemma will be useful in the analysis to follow.

Lemma 13

We have that $h_3(z) \ge z_3$ for all $z \in Z$.

Proof

For all $z \in Z$, the following inequality holds:

$$\begin{aligned} h_3(z) =\min \left( \frac{z_3}{1-z_2},1\right) \ge \min \left( z_3,1\right) = z_3. \end{aligned}$$

For every $w\ge 1$, we define an implicit function $f_w(\beta ,q): (0,b]\times [q_\infty , 1) \rightarrow [0,1]$ through the set of equations immediately below. Namely, for $(\beta ,q)\in (0,b]\times [q_\infty ,1)$, $f_w(\beta ,q)$ and z(k) for $k = w,\ldots , 1$ are defined implicitly by Eqs. (82)–(84).

$$\begin{aligned}&z(w) = (\beta , q, f_w(\beta ,q)), \end{aligned}$$

(82)

$$\begin{aligned}&z(k-1) = h(z(k))\ \ \text{ for }\ \ k=w,\ldots , 2, \end{aligned}$$

(83)

$$\begin{aligned}&z_2(1) = 1-z_3(1). \end{aligned}$$

(84)

Intuitively, given $(\beta ,q)$ and an initial guess of $f_w(\beta ,q)$, z(w) is defined by (82) and z(k) (for $k=1,\ldots ,w-1$) are well-defined by (83) and Lemma 11. The essence of what the next lemma shows is that there is a unique value of $f_w(\cdot )$ such that the boundary condition (84) is satisfied.

Lemma 14

The function $f_w(\cdot )$ is well-defined for all $w\ge 1$. In addition, $f_w(\beta ,q) \in (0,1)$ for all $(\beta ,q)\in (0,b]\times [q_\infty , 1)$.

Proof

Note that given $z(w) = (\beta , q, f_w(\beta ,q))$, Eq. (83) defines z(k) for $k = 1,\ldots , w-1$. However, the resulting z values must also satisfy the boundary condition (84). In other words, we need to show that for any $(\beta ,q)\in (0,b]\times [q_\infty , 1)$, there exists a unique value $f_w(\beta ,q)$ such that the resulting z(k), $k = 1,\ldots , w,$ satisfy (82)–(84).

We first show that there exists a value of $f_w(\beta ,q) = \eta$ such that (82)–(84) are satisfied. To this end, we view z(k) for $k=1,\ldots , w-1$ as functions of $\eta$, denoted by $z(k;\eta )$ for $\eta \in [0,1]$. In addition, we define a function $\phi (\eta )$ as follows:

$$\begin{aligned} \phi (\eta ) = z_2(1;\eta ) - (1-z_3(1;\eta )). \end{aligned}$$

Note that $z_3(1;0) = 0$ which follows from Eq. (72) inductively. It follows from Lemma 11 by induction that $z_2(1;\eta ) \in [q_\infty , 1)$ for all $\eta$. In particular, $z_2(1;0)<1$, so $\phi (0)=z_2(1;0)-1 <0$. Next, we argue that $\phi (1) > 0$. To see this, note that for any $\eta \in [0,1]$,

$$\begin{aligned} 1\ge z_3(1;\eta ) \ge z_3(2;\eta )\ge \cdots \ge z_3(w;\eta ) = \eta \ge 0, \end{aligned}$$

(85)

which follows from Lemma 13 inductively. Note from Eq. (85) that $z_3(1;1) = 1$.

Thus, $\phi (1) = z_2(1;1)\ge q_\infty > 0$.

Since $h(\cdot )$ is continuous, $\phi (\cdot )$ is continuous as well. Combining the facts that $\phi (0) < 0$ and $\phi (1)> 0$, we conclude that there exists $\eta \in (0,1)$ such that $\phi (\eta ) = 0$, i.e., condition (84) is satisfied. For such $\eta$, the resulting vectors $z(k;\eta )$, $k = 1,\ldots , w$ satisfy conditions (82)–(84).

For $\eta$ such that $\phi (\eta ) = 0$, $z_3(1;\eta ) = 1-z_2(1;\eta ) < 1$. It follows from Eq. (85) that $z_3(k;\eta ) < 1$. Substituting the definition of $h_3(\cdot )$ into the inequality $z_3(k;\eta ) < 1$, we argue that for such $\eta$, the following holds:

$$\begin{aligned} z_3(k;\eta ) = \frac{z_3(k+1;\eta )}{1-z_2(k+1;\eta )}, \ k=1,\ldots ,w, \end{aligned}$$

(86)

because $z_3(k;\eta )$ cannot take the value 1. In other words, the truncation by 1 on the right-hand side of (72) is immaterial for solutions of $f_w(\beta ,q)$ and z(k) for $k=1,\ldots , w-1$ defined through (82)–(84).

We conclude the proof by showing that there exists a unique $\eta$ satisfying conditions (82)–(84). Suppose there are multiple values of $f_w(\beta ,q)$, say $\eta \not =\tilde{\eta }$, satisfying the conditions (82)–(84). Without loss of generality, assume $\eta >\tilde{\eta }$. Next, we show by induction that for $k = 1,\ldots , w$,

$$\begin{aligned} z_1(k;\eta ) \le z_1(k;\tilde{\eta }),\ \ z_2(k;\eta ) \ge z_2(k;\tilde{\eta })\ \text{ and } \ z_3(k;\eta ) >z_3(k;\tilde{\eta }). \end{aligned}$$

(87)

This is true for $k=w$ by assumption. Suppose it is true for k, then we argue that it is also true for $k-1$. Note that $z_1(k-1;\eta ) = h_1(z(k;\eta ))$ and $z_1(k-1;\tilde{\eta }) = h_1(z(k;\tilde{\eta }))$. Since $h_1(\cdot )$ is decreasing in its first argument, whereas increasing in its last two arguments, we conclude that $z_1(k-1;\eta ) \le z_1(k-1;\tilde{\eta })$. Similarly, because $h_2(\cdot )$ is decreasing in the first argument, whereas increasing in its last two arguments, we conclude that

$$\begin{aligned} z_2(k-1;\eta ) = h_2(z(k;\eta )) \ge h_2(z(k;\tilde{\eta })) =z_2(k-1;\tilde{\eta }). \end{aligned}$$

Also, it follows from Eq. (86) that

$$\begin{aligned} z_3(k-1;\eta ) = \frac{z_3(k;\eta )}{1-z_2(k;\eta )} >\frac{z_3(k;\tilde{\eta })}{1-z_2(k;\tilde{\eta })} = z_3(k-1;\tilde{\eta }). \end{aligned}$$

In particular, both of the following must be true:

$$\begin{aligned} z_3(1;\eta ) > z_3(1;\tilde{\eta })\ \text{ and } \ z_2(1;\eta ) \ge z_2(1;\tilde{\eta }). \end{aligned}$$

However, by Eq. (84), we conclude that

$$\begin{aligned} z_2(1;\eta ) = 1-z_3(1;\eta ) < 1-z_3(1;\tilde{\eta }) = \tilde{z}_2(1;\tilde{\eta }), \end{aligned}$$

which is a contraction. Therefore, there exists at most one value of $f_w(\beta , q)$ satisfying conditions (82)–(84).

1.2 Partial derivatives of the auxiliary function $f_w(\cdot )$

This subsection characterizes the partial derivatives of $f_w(\cdot )$ and establishes the monotonicity of $f_w(\cdot )$.

To facilitate the analysis to follow, fix $w\ge 1$ and denote by $z(k;w,\beta ,q)$ (for $k = 1,\ldots , w$) the z(k) defined by substituting $z(w) = (\beta ,q,f_w(\beta ,q))$ into Eq. (83). The following lemma shows that in this construction, the truncation by 1 in defining $h_3(\cdot )$ is immaterial, cf. Eq. (72).

Lemma 15

For $w\ge 1$ and $(\beta ,q)\in (0,b]\times [q_\infty , 1)$, the following holds:

$$\begin{aligned} z_3(k;w,\beta ,q) = h_3(z(k+1;w,\beta ,q)) = \frac{z_3(k+1;w,\beta ,q)}{1-z_2(k+1;w,\beta ,q)},\ k = 1,\ldots , w-1. \end{aligned}$$

Proof

It follows from Lemma 13 inductively that

$$\begin{aligned} z_3(1;w,\beta ,q) \ge z_3(2;w,\beta ,q)\ge \cdots \ge z_3(w;w,\beta ,q) = f_w(\beta ,q) >0. \end{aligned}$$

In addition, condition (84) ensures that $z_3(1;w,\beta ,q) = 1-z_2(1;w,\beta ,q) \le 1-q_\infty <1$. Combining these inequalities with Eq. (72) yields the following: For $k = 1,\ldots , w-1$,

$$\begin{aligned} 1> z_3(k;w,\beta ,q) = h_3(z(k+1;w,\beta ,q)) = \min \left( 1, \frac{z_3(k+1;w,\beta ,q)}{1-z_2(k+1;w,\beta ,q)}\right) . \end{aligned}$$

Since $z_3(k;w,\beta ,q)$ cannot take the value 1, the result follows.

The following lemma provides an upper bound of $z_3(k;w,\beta ,q)$ for $w\ge 1$.

Lemma 16

For $w\ge 1$ and $(\beta ,q) \in (0,b]\times [q_\infty , 1)$, we have the following inequality:

$$\begin{aligned} z_3(k;w,\beta ,q) \le (1-q_\infty )^k,\ k = 1,\ldots , w. \end{aligned}$$

In particular, $f_w(\beta ,q) \le (1-q_\infty )^w$ for $w\ge 1$.

Proof

We proceed by induction. On the induction basis, it follows from Eq. (84) that for $k = 1$,

$$\begin{aligned} z_3(1;w,\beta ,q) = (1-z_2(1;w,\beta ,q)) \le 1- q_\infty , \end{aligned}$$

where the inequality follows because $z_2(1;w,\beta ,q) \in [q_\infty , 1)$ by construction.

By the induction hypothesis, suppose that the statement is true for k. Then note from Lemma 15 that

$$\begin{aligned} z_3(k+1;w,\beta ,q) = z_3(k;w,\beta ,q)(1-z_2(k+1;w,\beta ,q)) \le (1-q_\infty )^{k+1}, \end{aligned}$$

where the inequality follows from the induction hypothesis and that $z_2(k+1;w,\beta ,q)\in [q_\infty , 1)$.

In addition, it follows from (82) that

$$\begin{aligned} f_w(\beta ,q) = z_3(w;w,\beta ,q) \le (1-q_\infty )^w. \end{aligned}$$

The following lemma characterizes the partial derivatives of $f_w(\cdot )$ recursively.

Lemma 17

The partial derivatives of $f_w(\cdot )$ with respect to $\beta$ and q for $w\ge 1$ and $(\beta ,q)\in (0,b]\times [q_\infty ,1)$ are given as follows: For $w = 1$, we have that

$$\begin{aligned} \frac{\partial f_{1}(\beta ,q)}{\partial \beta } = 0\ \text{ and } \ \frac{\partial f_{1}(\beta ,q)}{\partial q} = -1. \end{aligned}$$

(88)

In addition, we have the following recursive characterization for $w\ge 1$:

$$\begin{aligned}&\frac{\partial f_{w+1}(\beta ,q)}{\partial \beta } = -\frac{\displaystyle \frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_1}+\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_1}}{\displaystyle \frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_3}+\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_3}-\frac{\partial h_3(z(w+1))}{\partial z_3}}, \end{aligned}$$

(89)

$$\begin{aligned}&\frac{\partial f_{w+1}(\beta ,q)}{\partial q} = -\frac{\displaystyle \frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_2}-\frac{\partial h_3(z(w+1))}{\partial z_2}}{\displaystyle \frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_3}+\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_3}-\frac{\partial h_3(z(w+1))}{\partial z_3}}, \end{aligned}$$

(90)

where $z(k) = z(k;w+1,\beta ,q)$ for $k = w, w+1$, and $e(w) = (z_1(w),z_2(w))$.

Proof

Note that $f_1(\beta , q) = 1-q$. Hence, Eq. (88) is immediate.

Fixing $w\ge 1$ and $(\beta ,q) \in (0,b]\times [q_\infty ,1)$, we want to characterize the partial derivatives of $f_{w+1}(\cdot )$ with respect to $\beta$ and q. Recall that for $w+1$, $z(w;w+1,\beta ,q)$ is computed by substituting $z(w+1) = (\beta ,q,f_{w+1}(\beta ,q))$ into Eq. (83). In particular, we rewrite Eq. (83) that derives $z(w;w+1,\beta ,q)$ as follows:

$$\begin{aligned}&z_i(w;w+1,\beta ,q) = h_i(\beta ,q,f_{w+1}(\beta ,q)). \end{aligned}$$

(91)

Moreover, by substituting $z(w) = z(w;w+1,\beta ,q)$ into Eq. (83) for w, we find that condition (84) (for w) is satisfied. In other words, solutions of (82)–(84) for different w’s are consistent provided that $\beta ,q$’s are chosen consistently for each w. In particular, the following holds:

$$\begin{aligned} f_w( z_1(w;w+1,\beta ,q),z_2(w;w+1,\beta ,q))=z_3(w;w+1,\beta ,q). \end{aligned}$$

(92)

Substituting Eq. (91) into (92), we obtain the following identity:

$$\begin{aligned} \begin{aligned} f_w(h_1(\beta ,q,f_{w+1}(\beta ,q)), h_2(\beta ,q,f_{w+1}(\beta ,q))) = h_3(\beta ,q,f_{w+1}(\beta ,q)). \end{aligned} \end{aligned}$$

(93)

Both the left-hand side and the right-hand side of Eq. (93) are functions of $(\beta ,q)$. Since we focus our analysis on the derivation for $w+1$ with fixed initial values $(\beta ,q)$, we write $z(k) = z(k;w+1,\beta ,q)$ in short.

First, we take the partial derivative of both sides of Eq. (93) with respect to $\beta$ by the chain rule and evaluate the function at point $(\beta ,q)$. It follows from Eq. (92) that the partial derivatives of $f_w(\cdot )$ are evaluated at $(z_1(w),z_2(w))$. Since $z(w+1) = (\beta ,q,f_{w+1}(\beta ,q))$, the partial derivatives of $h_i(\cdot )$ are evaluated at $z(w+1)$ for $i = 1,2,3$ . Thus, we obtain the following equation:

$$\begin{aligned} \begin{aligned}&\frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_1} + \frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_3}\frac{\partial f_{w+1}(\beta ,q)}{\partial \beta }\\&\quad +\frac{\partial f_w(e(w))}{\partial q} \frac{\partial h_2(z(w+1))}{\partial z_1} +\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_3}\frac{\partial f_{w+1}(\beta ,q)}{\partial \beta }\\&\quad = \frac{\partial h_3(z(w+1))}{\partial z_1} +\frac{\partial h_3(z(w+1))}{\partial z_3}\frac{\partial f_{w+1}(\beta ,q)}{\partial \beta }, \end{aligned} \end{aligned}$$

where $e(w) = (z_1(w),z_2(w))$. Note that $\partial h_3/\partial z_1 = 0$ by (74). Thus, we can drop the first term on the right-hand side. Rearranging the terms yields Eq. (89).

Taking the partial derivative of both sides of Eq. (93) with respect to q and evaluating the function at value $(\beta ,q)$, we obtain the following equation:

$$\begin{aligned} \begin{aligned}&\frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_2}+ \frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_3}\frac{\partial f_{w+1}(\beta ,q)}{\partial q}\\&\quad +\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_2} +\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_3}\frac{\partial f_{w+1}(\beta ,q)}{\partial q}\\&\quad = \frac{\partial h_3(z(w+1))}{\partial z_2}+\frac{\partial h_3(z(w+1))}{\partial z_3}\frac{\partial f_{w+1}(\beta ,q)}{\partial q}. \end{aligned} \end{aligned}$$

It follows from (79) that $\partial h_1/\partial z_2 = 0$. Thus, we can drop the first term on the left-hand side. Rearranging the terms yields Eq. (90).

The following lemma shows the monotonicity of $f_w(\cdot )$.

Lemma 18

$f_w$ is non-decreasing in $\beta$ and non-increasing in q. That is, for all $w\ge 1$ and $(\beta ,q) \in (0,b]\times [q_\infty ,1)$,

$$\begin{aligned} \frac{\partial f_{w}(\beta ,q)}{\partial \beta } \ge 0\ \text{ and } \ \frac{\partial f_{w}(\beta ,q)}{\partial q} \le 0. \end{aligned}$$

(94)

Proof

Recall that $f_1(\beta ,q) = 1-q$. Thus, (94) is immediate for $w=1$.

We proceed by induction: Suppose (94) holds for $k = 1,\ldots , w$, and we next show that it holds for $k = w+1$. Note from Eqs. (73) to (81) that

$$\begin{aligned} \frac{\partial h_1}{\partial z_1}, \ \frac{\partial h_2}{\partial z_3}, \ \frac{\partial h_3}{\partial z_3}> 0\ \text{ and } \ \frac{\partial h_2}{\partial z_1},\ \frac{\partial h_1}{\partial z_3}<0. \end{aligned}$$

Consider the formula for $\partial f_{w+1}/\partial \beta$ given in Eq. (89). Every term in the numerator is positive, whereas every term in the denominator is negative so that

$$\begin{aligned} \frac{\partial f_{w+1}(\beta ,q)}{\partial \beta } \ge 0. \end{aligned}$$

Next, consider $\partial f_{w+1}/\partial q$. It follows from Eqs. (73) to (81) that

$$\begin{aligned} \frac{\partial h_2}{\partial z_2} >0\ \text{ and } \ \frac{\partial h_3}{\partial z_2} \ge 0. \end{aligned}$$

Every term in both the numerator and the denominator of Eq. (90) is negative. Thus, we conclude that

$$\begin{aligned} \frac{\partial f_{w+1}(\beta ,q)}{\partial q}\le 0. \end{aligned}$$

1.3 Properties of $f_w(\cdot )$ on a restricted set

This subsection studies the partial derivatives of $f_w(\beta ,q)$ as w gets large. To facilitate this analysis, we define subsets $\mathscr {Z}_1(w)\times \mathscr {Z}_2(w)$ for $w\ge 1$ such that for any potential equilibrium $(\beta ^*(w),q^*(w))\in \mathscr {Z}_1(w)\times \mathscr {Z}_2(w)$ for all $w\ge 1$. Restricting our analysis to the case where $(\beta ,q)\in \mathscr {Z}_1(w)\times \mathscr {Z}_2(w) \subseteq (0,b]\times [q_\infty ,1)$, we establish the desired convergence results for the partial derivatives of $f_w(\beta ,q)$. (Note from Corollaries 3 and 4 that $(\beta ^*(w),q^*(w))\rightarrow (b,q_\infty )$ as $w\rightarrow \infty$ for any potential equilibrium. We define $\mathscr {Z}_1(w)\times \mathscr {Z}_2(w)$ such that they shrink to the point $(b,q_\infty )$ as $w\rightarrow \infty$. )

To facilitate the analysis to follow, we first define a function $\kappa (x,y):[0,1]\times [0,r]\rightarrow \mathbb {R}$ as follows:

$$\begin{aligned} \kappa (x,y) = \mathbb {E}[-c+\alpha [ xr+(1-x)y]-(\varepsilon (1)-\varepsilon (0))]^+. \end{aligned}$$

(95)

The following lemma shows the properties of $\kappa$.

Lemma 19

The function $\kappa (x,y)$ has the following properties:

• (i) For any $(x,y)\in [0,1]\times [0,r]$, $\kappa (x,y) \in [0,r)$.

• (ii) For any fixed $x\in [0,1]$, $\kappa (x,y)$ is a contraction mapping. In particular, $|\partial \kappa (x,y)/\partial y| \le \alpha <1$ for all $x\in [0,1]$.

• (iii) For any fixed $x\in [0,1]$, there exists a unique $j(x)\in [0,r)$ satisfying $j(x) = \kappa (x,j(x))$.

Proof

The following inequality shows that (i) holds: For any $(x,y)\in [0,1]\times [0,r]$,

$$\begin{aligned} \begin{aligned} \kappa (x,y)&= \mathbb {E}[-c+\alpha [ xr+(1-x)y]-(\varepsilon (1)-\varepsilon (0))]^+\\&\le \mathbb {E}[-c+\alpha [ xr+(1-x)r]-(\varepsilon (1)-\varepsilon (0))]^+\\&=\mathbb {E}[-c+\alpha r-(\varepsilon (1)-\varepsilon (0))]^+\\&=\mathbb {E}[\max \{\varepsilon (1),-c+\alpha r+\varepsilon (0)\}] < r. \end{aligned} \end{aligned}$$

(96)

The first inequality follows from $y\le r$. The equality in the fourth line holds because $\mathbb {E}[\varepsilon (1)] = 0$. The last equality follows from Assumption 2. In addition, it is immediate that $\kappa (x,y)\ge 0$ for all $(x,y)\in [0,1]\times [0,r]$. Thus, we have that $\kappa (x,y)\in [0,r)$.

We can write $\kappa (x,y)$ in integral form and use integration by parts to arrive at the following:

$$\begin{aligned} \begin{aligned} \kappa (x,y)&= \mathbb {E}[-c+\alpha [ xr+(1-x)y]-(\varepsilon (1)-\varepsilon (0))]^+\\&= \int _{-\infty }^{-c+\alpha [ xr+(1-x)y]} \left( -c+\alpha [ xr+(1-x)y] -u\right) \,\mathrm {d}F(u)\\&= \int _{-\infty }^{-c+\alpha [ xr+(1-x)y]} F(u)\,\mathrm {d} u, \end{aligned} \end{aligned}$$

(97)

where the last inequality follows from integration by parts. Thus, the partial derivative of $\kappa (x,y)$ with respect to y is given as follows:

$$\begin{aligned}\begin{aligned} \frac{\partial \kappa (x,y)}{\partial y}&= \frac{\partial (\int _{-\infty }^{-c+\alpha [ xr+(1-x)y]} F(u)\,\mathrm {d} u)}{\partial y} \\&=\alpha (1-x) F(-c+\alpha [ xr+(1-x)y]) \in [0,\alpha ]. \end{aligned} \end{aligned}$$

Therefore, for any fixed $x\in [0,1]$, the following inequality holds:

$$\begin{aligned} \left| \frac{\partial \kappa (x,y)}{\partial y}\right| \le \alpha <1,\ y\in [0,r]. \end{aligned}$$

Thus, (ii) holds, i.e., $\kappa (x,y)$ is a contraction mapping for any fixed $x\in [0,1]$.

It follows from properties (i)–(ii) that for any fixed $x\in [0,1]$, $\kappa (x,y)$ is a contraction mapping from [0, r] to [0, r). By the Banach fixed point theorem, there exists a unique fixed point $j(x)\in [0,r]$ such that $j(x)=\kappa (x,j(x))$. It follows from (96) that $\kappa (x,r) < r$. Thus, $j(x)\not =r$, which leads to $j(x)\in [0,r)$. In other words, property (iii) holds.

The following lemma shows useful properties of the function $j(\cdot )$.

Lemma 20

The function $j(\cdot )$ is increasing and differentiable.

Proof

The function j(x) is defined implicitly as follows:

$$\begin{aligned} \kappa (x,j(x)) - j(x) = 0,\ x\in [0,1]. \end{aligned}$$

(98)

Note that $\kappa$ is continuously differentiable by Eq. (97). By the implicit function theorem, j(x) is differentiable; see Theorem 9.28 of [36]. It follows from (97) that

$$\begin{aligned} \begin{aligned} \frac{\partial \kappa (x,y)}{\partial x} =&\alpha (r-y)F(-c+\alpha [ xr+(1-x)y]) \\ \ge&\alpha (r-y)F(-c) > 0,\ (x,y)\in [0,1]\times [0,r), \end{aligned} \end{aligned}$$

where the last inequality holds because $y<r$ and $-c$ is in the interior of the support of $F(\cdot )$ by Assumption 1. For any fixed $x\in [0,1]$, taking the derivative of both sides of the equation $j(x) = \kappa (x,j(x))$ yields the following equation:

$$\begin{aligned} j'(x) = \frac{\partial \kappa (x,j(x))}{\partial x} + \frac{\partial \kappa (x,j(x))}{\partial y}j'(x). \end{aligned}$$

Rearranging the terms, we have that

$$\begin{aligned} j'(x) = \frac{\partial \kappa (x,j(x))}{\partial x} \Bigg /\left( 1-\frac{\partial \kappa (x,j(x))}{\partial y}\right) > 0,\ x\in [0,1], \end{aligned}$$

where the inequality follows from the fact that $\partial \kappa (x,y)/\partial x>0$ and property (ii) in Lemma 19. Therefore, j(x) is increasing.

To facilitate the definition of $\mathscr {Z}_1$ and $\mathscr {Z}_2$, the sequence $\underline{\beta }(w)$ is defined as follows:

$$\begin{aligned} \underline{\beta }(w) = \frac{b -a(1-q_\infty )^w}{1-a(1-q_\infty )^w},\ w\ge 1. \end{aligned}$$

Since $b> a$, we have that $\underline{\beta }(w)>0$ for all $w\ge 1$. Then we define $\underline{J}(w) = j(\underline{\beta }(w))$. By substituting Eqs. (95) and (98) into the definition of $\underline{J}(w)$, we have that

$$\begin{aligned} \underline{J}(w) = \mathbb {E}\left[ -c+\alpha [\underline{\beta }(w)r + (1-\underline{\beta }(w)) \underline{J}(w)]-(\varepsilon (1) - \varepsilon (0))\right] ^+,\ w\ge 1. \end{aligned}$$

(99)

In addition, define⁸

$$\begin{aligned} \bar{q}(w) = \bar{F}(-c+\alpha [\underline{\beta }(w)r+(1-\underline{\beta }(w))\underline{J}(w)]), \ w\ge 1. \end{aligned}$$

(100)

The following lemma shows the properties of the sequences $\underline{\beta }(w)$, $\underline{J}(w)$ and $\bar{q}(w)$ for $w\ge 1$.

Lemma 21

The sequences $\underline{\beta }(w)$, $\underline{J}(w)$ and $\bar{q}(w)$ for $w\ge 1$ have the following properties:

• (i) $\underline{\beta }(w)$ and $\underline{J}(w)$ are increasing, whereas $\bar{q}(w)$ is decreasing in w.

• (ii) $\lim _{w\rightarrow \infty } \underline{\beta }(w) = b$, $\lim _{w\rightarrow \infty } \underline{J}(w) = J_\infty$ and $\lim _{w\rightarrow \infty } \bar{q}(w) =q_\infty$, where $J_\infty$ and $q_\infty$ are constants defined in Corollary 4.

• (iii) $\underline{J}(w) = \int _{-\infty }^{\bar{F}^{-1}(\bar{q}(w))} F(x)\,\mathrm {d}x$.

Proof

We first show (i). It is immediate that $\underline{\beta }(w)$ is increasing. Thus, $\underline{J}(w)$ is increasing in w by Lemma 20. It follows from property (iii) of Lemma 19 that $j(x)\in [0,r)$ for all $x\in [0,1]$. Thus, $\underline{J}(w)< r$ for $w\ge 1$. It follows from Eq. (100) that $\bar{q}(w)$ is decreasing in w because $\underline{\beta }(w)$ and $\underline{J}(w)$ are non-increasing in w and $\underline{J}(w) < r$.

Next we show that (ii) holds. Clearly, $\lim _{w\rightarrow \infty } \underline{\beta }(w) = b$. It follows from Lemma 19 that j(x) is a differentiable function and thus is continuous. Therefore, the following equation holds:

$$\begin{aligned} \lim _{w\rightarrow \infty } \underline{J}(w) = \lim _{w\rightarrow \infty } j\left( \underline{\beta }(w)\right) = j\left( \lim _{w\rightarrow \infty }\underline{\beta }(w) \right) = j(b) = J_\infty , \end{aligned}$$

where the last inequality follows from (14) that $J_\infty = j(b)$. It follows from the continuity of $\bar{F}$ that

$$\begin{aligned} \begin{aligned} \lim _{w\rightarrow \infty } \bar{q}(w) =&\lim _{w\rightarrow \infty } \bar{F} \left( -c+\alpha [\underline{\beta }(w)r+(1-\underline{\beta }(w))\underline{J}(w)]\right) \\ =&\bar{F}\left( -c+\alpha [br+(1-b)J_\infty ]\right) = q_\infty . \end{aligned} \end{aligned}$$

The last equality follows from the definition of $q_\infty$ in Corollary 4.

Lastly, we show that (iii) holds. It follows from Eqs. (99) to (100) that for $w\ge 1$,

$$\begin{aligned} \begin{aligned} \underline{J}(w) =&\mathbb {E}\left[ -c+\alpha [\underline{\beta }(w)r + (1-\underline{\beta }(w)) \underline{J}(w)]-(\varepsilon (1) - \varepsilon (0))\right] ^+\\ =&\mathbb {E}\left[ \bar{F}^{-1}(\bar{q}(w))-(\varepsilon (1)-\varepsilon (0)) \right] ^+\\ =&\int _{-\infty }^{\bar{F}^{-1}(\bar{q}(w))} F(x)\,\mathrm {d}x. \end{aligned} \end{aligned}$$

To facilitate the analysis, define

$$\begin{aligned} \mathscr {Z}_1(w) = [\underline{\beta }(w), b]\ \text{ and } \ \mathscr {Z}_2(w) = [q_\infty , \bar{q}(w)],\ w\ge 1. \end{aligned}$$

(101)

It follows from properties (i)–(ii) of Lemma 21 that $\underline{\beta }(w) <b$ and $\bar{q}(w) \ge q_\infty$. Thus, both $\mathscr {Z}_1(w)$ and $\mathscr {Z}_2(w)$ are non-empty for all $w\ge 1$. Since we only consider the underloaded case, i.e., $b>a$, we have that $\underline{\beta }(w)>0$ for all $w\ge 1$. In addition, it follows from (100) that $\bar{q}(w)<\bar{F}(r) \le 1$, which gives that

$$\begin{aligned} \mathscr {Z}_1(w)\times \mathscr {Z}_2(w)\subseteq (0,b]\times [q_\infty ,1),\ w\ge 1. \end{aligned}$$

Next, we study the properties of $f_w(\beta ,q)$ when $(\beta ,q)\in \mathscr {Z}_1(w)\times \mathscr {Z}_2(w)$.

Lemma 22

For any $w\ge 1$, if $(\beta ,q) \in \mathscr {Z}_1(w)\times \mathscr {Z}_2(w)$, then

$$\begin{aligned} (z_1(k;w,\beta ,q), z_2(k;w,\beta ,q)) \in \mathscr {Z}_1(k)\times \mathscr {Z}_2(k),\ k = 1,\ldots , w. \end{aligned}$$

(102)

Proof

Fix $w\ge 1$. We proceed by induction. For $k = w$, (102) holds by assumption.

By the induction hypothesis, suppose (102) holds for $l = k + 1,\ldots ,w-1,w$. That is,

$$\begin{aligned} \underline{\beta }(l) \le z_1(l;w,\beta ,q) \le b \ \text{ and } \ q_\infty \le z_2(l;w,\beta ,q) \le \bar{q}(l),\ l=k+1,\ldots , w. \end{aligned}$$

Next, we show that (102) holds for $l=k$. The following holds:

$$\begin{aligned} \begin{aligned} z_1(k;w,\beta ,q)&= h_1(z(k+1;w,\beta ,q))\\&= \left( 1+\frac{1-b}{1-az_3(k+1;w,\beta ,q)}\frac{1}{z_2(k+1;w,\beta ,q)}\right) ^{-1}\\&> \left( 1+\frac{1-b}{1-a(1-q_\infty )^{k+1}}\frac{1-a(1-q_\infty )^{k+1}}{b -a(1-q_\infty )^{k+1}}\right) ^{-1}\\&= \left( 1+\frac{1-b}{b -a(1-q_\infty )^{k+1}}\right) ^{-1}\\&> \left( 1+\frac{1-b}{b -a(1-q_\infty )^k}\right) ^{-1} \\&= \frac{b-a(1-q_\infty )^k}{1-a(1-q_\infty )^k} = \underline{\beta }(k). \end{aligned} \end{aligned}$$

The first inequality follows from Lemma 16 and that $z_2(k+1;w,\beta ,q) >\beta (k+1)$. Thus, $z_1(k;w,\beta ,q) \in \mathscr {Z}_1(k)$. Combining the two cases, $z_1(k;w,\beta ,q) \in \mathscr {Z}_1(k)$.

Moreover, it follows from Eq. (71) that

$$\begin{aligned} z_2(k;w,\beta ,q)= & {} h_2(z(k+1;w,\beta ,q))\\= & {} \bar{F}\left[ -c + \alpha \left( z_1(k;w,\beta ,q)r + (1-z_1(k;w,\beta ,q))\right. \right. \\&\left. \left. \int _{-\infty }^{\bar{F}^{-1}(z_2(k+1;w,\beta ,q))} F(x)\,\mathrm {d}x\right) \right] \\\le & {} \bar{F}\left[ -c + \alpha \left( \underline{\beta }(k)r + (1-\underline{\beta }(k))\int _{-\infty }^{\bar{F}^{-1}(\bar{q}(k+1))} F(x)\,\mathrm {d}x\right) \right] \\\le & {} \bar{F}\left[ -c + \alpha \left( \underline{\beta }(k)r + (1-\underline{\beta }(k))\int _{-\infty }^{\bar{F}^{-1}(\bar{q}(k))} F(x)\,\mathrm {d}x\right) \right] \\= & {} \bar{F}\left[ -c + \alpha \left( \underline{\beta }(k)r + (1-\underline{\beta }(k))\underline{J}(k)\right) \right] =\bar{q}(k). \end{aligned}$$

The first inequality follows from $z_1(k;w,\beta ,q) \ge \underline{\beta }(k)$ and the assumption that $z_2(k+1;w,\beta ,q) \le \bar{q}(k+1)$ and Lemma 11 that the integral is less than r. The second inequality follows from Lemma 21 that $\bar{q}(k+1) <\bar{q}(k)$ and the last two equalities follow from property (iii) of Lemma 21 and (100). Since $z_2(k;w,\beta ,q) \ge q_\infty$ by construction, $z_2(k;w,\beta ,q) \in \mathscr {Z}_2(k)$.

The following lemma shows the properties of the partial derivatives of $h(\cdot )$ for values in the set $\mathscr {Z}_1(w+1)\times \mathscr {Z}_2(w+1)$.

Lemma 23

For any $\epsilon >0$, there exist $w_0,M\ge 0$ such that the following holds for $w\ge w_0$ and $(\beta ,q) \in \mathscr {Z}_1(w+1)\times \mathscr {Z}_2(w+1)$:

$$\begin{aligned}&\frac{\partial h_2(z)}{\partial z_2} \le 1,\ \frac{\partial h_3(z)}{\partial z_2} \le \frac{\epsilon }{2} q_\infty ,\ \frac{\partial h_1(z)}{\partial z_1} \le 1\ \text{ and } \ -\frac{\partial h_2(z)}{\partial z_1} \le M, \end{aligned}$$

(103)

where $z= z(w+1;w+1,\beta ,q)=(\beta ,q,f_{w+1}(\beta ,q))$.

Proof

We show that the four inequalities hold for w large enough one by one.

We first show that $\partial h_2/\partial z_2 \le 1$ for w large. Recall from Eq. (77) that for $z\in \mathscr {Z}$,

$$\begin{aligned} \begin{aligned} \frac{\partial h_2(z)}{\partial z_2} =&\alpha (1-\beta )(1-q)\frac{f[\bar{F}^{-1}(h_2(z))]}{f(\bar{F}^{-1}(q))} \le \alpha (1-q_\infty )\frac{f[\bar{F}^{-1}(h_2(z))]}{f(\bar{F}^{-1}(q))}, \end{aligned} \end{aligned}$$

(104)

where the inequality follows because $q \ge q_\infty$ and $\beta \in \mathscr {Z}_1(w+1) \subseteq (0,1]$.

By continuity of $f(\cdot )$ and $\bar{F}^{-1}(\cdot )$ at $q_\infty$, there exists $\delta _1>0$ such that for all x such that $|x-q_\infty |<\delta _1$, the following holds:

$$\begin{aligned} f(\bar{F}^{-1}(q_\infty ))\sqrt{1-q_\infty } \le f(\bar{F}^{-1}(x)) \le \frac{1}{\sqrt{1-q_\infty }} f(\bar{F}^{-1}(q_\infty )). \end{aligned}$$

(105)

It follows from Lemma 21 that $\bar{q}(w) \rightarrow q_\infty$ as $w\rightarrow \infty$. Thus, there exists $w_1\ge 1$ such that $|\bar{q}(w)-q_\infty |<\delta _1$ for $w\ge w_1$. In particular,

$$\begin{aligned} |\bar{q}(w+1) -q_\infty |< \delta _1\ \text{ and } \ |\bar{q}(w) - q_\infty | < \delta _1,\ w\ge w_1. \end{aligned}$$

(106)

It follows from (83) that $h_2(z) = z_2(w;w+1,\beta ,q)$. We have that $h_2(z) \in \mathscr {Z}_2(w)=[q_\infty , \bar{q}(w)]$ by Lemma 22. In addition, by assumption, $q\in \mathscr {Z}_2(w+1)=[q_\infty , \bar{q}(w+1)]$. Thus, it follows from (106) that for $w\ge w_1$ and $(\beta ,q) \in \mathscr {Z}_1(w+1)\times \mathscr {Z}_2(w+1)$,

$$\begin{aligned} |q -q_\infty |< \delta _1\ \text{ and } \ |h_2(z)- q_\infty | < \delta _1. \end{aligned}$$

By Eq. (105), we have that for $w\ge w_1$ and $(\beta ,q) \in \mathscr {Z}_1(w+1)\times \mathscr {Z}_2(w+1)$,

$$\begin{aligned} \frac{f[\bar{F}^{-1}(h_2(z))]}{f(\bar{F}^{-1}(q))} \le \frac{1}{1-q_\infty }. \end{aligned}$$

Substituting this inequality into Eq. (104), we obtain the following:

$$\begin{aligned} \frac{\partial h_2(z)}{\partial z_2} \le \alpha \le 1\ \text{ for } \ w\ge w_1\ \text{ and } \ (\beta ,q) \in \mathscr {Z}_1(w+1)\times \mathscr {Z}_2(w+1), \end{aligned}$$

where $z = (\beta ,q,f_{w+1}(\beta ,q)) = z(w+1;w+1,\beta ,q)$.

Next we show that $\partial h_3/\partial z_2 \le \epsilon q_\infty /2$ for w large. It follows from Eq. (80) that

$$\begin{aligned} \frac{\partial h_3(z)}{\partial z_2}= \frac{z_3(w+1;w+1,\beta ,q)}{(1-q)^2} \le \frac{(1-q_\infty )^{w+1}}{(1-q)^2} \le \frac{(1-q_\infty )^w}{(1-\bar{q}(w_1))^2}, \end{aligned}$$

(107)

where the first inequality follows from Lemma 16. Recall from Lemma 21 that $\bar{q}(w)$ is decreasing. Thus, the second inequality holds because

$$\begin{aligned} q \le \bar{q}(w+1)\le \bar{q}(w_1),\ w\ge w_1. \end{aligned}$$

Since $1-q_\infty < 1$, there exists a constant $w_2\ge w_1$ such that

$$\begin{aligned} (1-q_\infty )^w \le \frac{\epsilon }{2} q_\infty (1-\bar{q}({w_1}))^2,\ w\ge w_2. \end{aligned}$$

Substituting this inequality into (107), we have the following:

$$\begin{aligned} \frac{\partial h_3(z)}{\partial z_2} \le \frac{\epsilon }{2} q_\infty \ \text{ for } \ w\ge w_2\ \text{ and } \ (\beta ,q) \in \mathscr {Z}_1(w+1)\times \mathscr {Z}_2(w+1). \end{aligned}$$

We then show that $\partial h_1/z_1\le 1$ for w large enough. Recall from Eq. (73) that for $w\ge w_2$,

$$\begin{aligned} \begin{aligned} 0\le \frac{\partial h_1(z)}{\partial z_1}= \frac{h_1^2(z)}{\beta ^2}\frac{1-b}{1-az_3(w+1;w+1,\beta ,q)}\le \frac{h_1^2(z)}{\beta ^2}\frac{1-b}{1-a(1-q_ \infty )^{w+1}}, \end{aligned} \end{aligned}$$

(108)

where the inequality follows from Lemma 16. Note by assumption that $\beta \in \mathscr {Z}_1(w+1)=[\underline{\beta }(w+1),b]$ and Lemma 22 that $h_1(z) = z_1(w;w+1,\beta ,q) \in \mathscr {Z}_1(w)=[\underline{\beta }(w),b]$. In other words, the following holds:

$$\begin{aligned} \underline{\beta }(w+1)\le \beta \le b\ \text{ and } \ \underline{\beta }(w)\le h_1(z) \le b. \end{aligned}$$

It follows from Lemma 21 that $\underline{\beta }(w) \rightarrow b$ as $w\rightarrow \infty$. In addition, $1-a(1-q_\infty )^{w+1}\rightarrow 1$ as $w\rightarrow \infty$. Thus, there exists $w_3\ge w_2$ such that

$$\begin{aligned} \frac{b}{\underline{\beta }(w+1)} \le \frac{1}{\root 4 \of {1-b}}\ \text{ and } \ 1-a(1-q_\infty )^{w+1} > \root 4 \of {1-b},\ w\ge w_3. \end{aligned}$$

Substituting these two inequalities into (108), we have that for $w\ge w_3$ and $(\beta ,q) \in \mathscr {Z}_1(w+1) \times \mathscr {Z}_2(w+1)$,

$$\begin{aligned} 0\le & {} \frac{\partial h_1(z)}{\partial z_1}\le \frac{h_1^2(z)}{\beta ^2}\frac{1-b}{1-a(1-q_\infty )^{w+1}} \\\le & {} \frac{b^2}{\underline{\beta }^2(w+1)}\frac{1-b}{1-a(1-q_\infty )^{w+1}} \le \root 4 \of {1-b} <1. \end{aligned}$$

Lastly, we show that there exists $M\ge 0$ such that $\partial h_2/\partial z_1 \le M$ for w large enough. It follows from Eq. (76) that for all $w\ge w_3$ and $(\beta ,q) \in \mathscr {Z}_1(w+1) \times \mathscr {Z}_2(w+1)$,

$$\begin{aligned} \begin{aligned} -\frac{\partial h_2(z)}{\partial z_1} =&f(\bar{F}^{-1}(h_2(z)))\alpha \left( r-\int _{-\infty }^{\bar{F}^{-1}(q)} F(x)\,\mathrm {d}x \right) \frac{\partial h_1(z)}{\partial z_1}\\ \le&f(\bar{F}^{-1}(h_2(z))\alpha r\frac{\partial h_1(z)}{\partial z_1} \le f(\bar{F}^{-1}(h_2(z))\alpha r. \end{aligned} \end{aligned}$$

(109)

The first inequality follows from Lemma 11. The second inequality follows from the first inequality in (103). It follows from the continuity of $f(\cdot )$ and $\bar{F}^{-1}(\cdot )$ that $f(\bar{F}^{-1}(x))$ is bounded on $[q_\infty , \bar{q}(w_3)]$. Recall that $h_2(z) \in \mathscr {Z}_2(w) =[q_\infty , \bar{q}(w)]$. Since $\bar{q}(w)$ is decreasing in w by Lemma 21, it follows that

$$\begin{aligned} q_\infty \le h_2(z) \le \bar{q}(w) \le \bar{q}(w_3),\ w\ge w_3\ \text{ and } \ (\beta ,q) \in \mathscr {Z}_1(w+1) \times \mathscr {Z}_2(w+1). \end{aligned}$$

Thus, the right-hand side of the third line in Eq. (109) is bounded. Letting M denote one such bound completes the proof.

In summary, letting $w_0 =w_3$, the four inequalities in (103) hold for all $w\ge w_0$ and $(\beta ,q) \in \mathscr {Z}_1(w+1) \times \mathscr {Z}_2(w+1)$

The next lemma is key to proving Lemma 5.

Lemma 24

The following holds:

$$\begin{aligned}&\lim _{w\rightarrow \infty } \sup \left\{ \left| \frac{\partial f_w(\beta ,q)}{\partial \beta }\right| : (\beta ,q)\in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)\right\} =0,\end{aligned}$$

(110)

$$\begin{aligned}&\lim _{w\rightarrow \infty } \sup \left\{ \left| \frac{\partial f_w(\beta ,q)}{\partial q}\right| : (\beta ,q)\in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)\right\} =0. \end{aligned}$$

(111)

Proof

We first show (110). To facilitate the analysis to follow, define a sequence $y_w$ for $w\ge 1$ as follows:

$$\begin{aligned} y_w =\sup \left\{ -\frac{\partial f_{w}(\beta ,q)}{\partial q}:(\beta ,q) \in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)\right\} . \end{aligned}$$

(112)

It follows from Eq. (94) that $y_w \ge 0$ for all $w\ge 1$. In addition, it follows from Eq. (90) that for any $(\beta ,q) \in \mathscr {Z}_1(w+1) \times \mathscr {Z}_2(w+1)$ and $w\ge 1$,

$$\begin{aligned} -\frac{\partial f_{w+1}(\beta ,q)}{\partial q}= & {} \frac{\displaystyle \frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_2}-\frac{\partial h_3(z(w+1))}{\partial z_2}}{\displaystyle \frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_3}+\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_3}-\frac{\partial h_3(z(w+1))}{\partial z_3}}\nonumber \\= & {} \frac{\displaystyle -\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_2}+\frac{\partial h_3(z(w+1))}{\partial z_2}}{\displaystyle -\frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_3}-\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_3}+\frac{\partial h_3(z(w+1))}{\partial z_3}}\nonumber \\\le & {} \left( -\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_2}+\frac{\partial h_3(z(w+1))}{\partial z_2}\right) \Bigg /\,\, \frac{\partial h_3(z(w+1))}{\partial z_3}\nonumber \\\le & {} \left( y_w\frac{\partial h_2(z(w+1))}{\partial z_2}+\frac{\partial h_3(z(w+1))}{\partial z_2}\right) \Bigg /\,\, \frac{\partial h_3(z(w+1))}{\partial z_3}, \end{aligned}$$

(113)

where $z(k) = z(k;w+1,\beta ,q)$ for $k = w, w+1$ and $e(w) = (z_1(w),z_2(w))$. We flip the signs of the terms in the second line of the right-hand side. Thus, it follows from Lemma 12 that every term in both the numerator and the denominator (of the right-hand side of the third line) is positive. This leads to the inequality in the fourth line. The last inequality follows from Eq. (112) because $(z_1(w),z_2(w)) \in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)$, which in turn follows from Lemma 22.

Rewriting Eq. (113) gives

$$\begin{aligned} -\frac{\partial f_{w+1}(\beta ,q)}{\partial q} \le \left( y_w\frac{\partial h_2(z(w+1))}{\partial z_2}+\frac{\partial h_3(z(w+1))}{\partial z_2}\right) \Bigg /\,\, \frac{\partial h_3(z(w+1))}{\partial z_3}. \end{aligned}$$

Taking the supremum of both sides over $(\beta ,q) \in \mathscr {Z}_1(w+1) \times \mathscr {Z}_2(w+1)$ gives the following:

$$\begin{aligned} y_{w+1}\le & {} \sup \nolimits _{(\beta ,q) \in \mathscr {Z}_1(w+1) \times \mathscr {Z}_2(w+1)} \left[ \left( y_w\frac{\partial h_2(z(w+1))}{\partial z_2}+\frac{\partial h_3(z(w+1))}{\partial z_2}\right) \Bigg /\,\, \frac{\partial h_3(z(w+1))}{\partial z_3}\right] \nonumber \\\le & {} \sup \nolimits _{(\beta ,q) \in \mathscr {Z}_1(w+1) \times \mathscr {Z}_2(w+1)} (1-q_\infty ) \left( y_w\frac{\partial h_2(z(w+1))}{\partial z_2}+\frac{\partial h_3(z(w+1))}{\partial z_2}\right) , \end{aligned}$$

(114)

where the last inequality follows from Eq. (81) and that $q\in [q_\infty , \bar{q}(w+1)]$. In particular,

$$\begin{aligned} \frac{\partial h_3(z(w+1))}{\partial z_3} = \frac{1}{1-z_2(w+1)} = \frac{1}{1-q}\ge \frac{1}{1-q_\infty }. \end{aligned}$$

(115)

Substituting the first two inequalities in Eqs. (103) into (114) yields the following:

$$\begin{aligned} y_{w+1} \le (1-q_\infty )\left( y_w + \frac{\epsilon }{2}q_\infty \right) \ \text{ for } \ w\ge w_0, \end{aligned}$$

where $w_0$ is as in Lemma 23. By induction, we obtain the following inequality: For all $w\ge w_0$ and $n\ge 1$,

$$\begin{aligned} \begin{aligned} y_{w+n} \le&(1-q_\infty ) y_{w+n-1} +\frac{\epsilon }{2}q_\infty (1-q_\infty ) \\ \le&(1-q_\infty )^2 y_{w+n-2} +\frac{\epsilon }{2}q_\infty (1-q_\infty )^2 +\frac{\epsilon }{2}q_\infty (1-q_\infty ) \\ \le&\cdots \le (1-q_\infty )^n y_{w} + \frac{\epsilon }{2}q_\infty \sum _{i=1}^n (1-q_\infty )^i\\ =&(1-q_\infty )^n y_{w} +\frac{\epsilon }{2}. \end{aligned} \end{aligned}$$

Now fix $w = w_0$. There exists $n_1$ such that for all $n \ge n_1$, $(1-q_\infty )^n y_{w_0} < \epsilon /2$. That is, for any $w \ge w_0+n_1$, $y_w < \epsilon$. Therefore, $y_w \rightarrow 0$, as $w\rightarrow \infty$. Since $y_w\ge 1$ for all $w\ge 1$, we deduce Eq. (110).

Next, we prove (111) in a similar fashion. Define a sequence $x_w$ as follows:

$$\begin{aligned} x_w =\sup \left\{ \frac{\partial f_{w}(\beta ,q)}{\partial \beta }:(\beta ,q) \in \mathscr {Z}_1(w) \times \mathscr {Z}_2(w)\right\} , \ w\ge 1. \end{aligned}$$

It follows from Eq. (94) that $x_w \ge 0$ for all $w\ge 1$. In addition, it follows from Eq. (89) that the following holds: For any $w \ge 1$ and $(\beta ,q) \in \mathscr {Z}_1(w+1)\times \mathscr {Z}_2(w+1)$

$$\begin{aligned} \frac{\partial f_{w+1}(\beta ,q)}{\partial \beta }= & {} -\frac{\displaystyle \frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_1}+\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_1}}{\displaystyle \frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_3}+\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_3}-\frac{\partial h_3(z(w+1))}{\partial z_3}}\\= & {} \frac{\displaystyle \frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_1}+\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_1}}{\displaystyle -\frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_3}-\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_3}+\frac{\partial h_3(z(w+1))}{\partial z_3}}\\\le & {} \left( \frac{\partial f_w(e(w))}{\partial \beta }\frac{\partial h_1(z(w+1))}{\partial z_1}+\frac{\partial f_w(e(w))}{\partial q}\frac{\partial h_2(z(w+1))}{\partial z_1}\right) \Bigg /\,\, \frac{\partial h_3(z(w+1))}{\partial z_3}\\\le & {} \left( x_w\frac{\partial h_1(z(w+1))}{\partial z_1}+y_w\left( -\frac{\partial h_2(z(w+1))}{\partial z_1}\right) \right) \Bigg /\,\,{\frac{\partial h_3(z(w+1))}{\partial z_3}}. \end{aligned}$$

The first inequality holds because every term in both the numerator and denominator of the right-hand side of the second line is positive. The last inequality holds because $(z_1(w),z_2(w))\in \mathscr {Z}_1(w)\times \mathscr {Z}_2(w)$. Taking the supremum of both sides over $(\beta ,q) \in \mathscr {Z}_1(w+1) \times \mathscr {Z}_2(w+1)$ yields the following: For $w\ge 1$,

$$\begin{aligned} x_{w+1} \le \sup \nolimits _{(\beta ,q) \in \mathscr {Z}_1(w+1) \times \mathscr {Z}_2(w+1)}\left( x_w\frac{\partial h_1(z(w+1))}{\partial z_1}-y_w\frac{\partial h_2(z(w+1))}{\partial z_1}\right) \Bigg /\,\,{\frac{\partial h_3(z(w+1))}{\partial z_3}}. \end{aligned}$$

Substituting the last two inequalities in Eqs. (103) and  (115) into this inequality yields the following:

$$\begin{aligned} x_{w+1} \le (1-q_\infty ) (x_w + My_w)\ \text{ for } \ w\ge w_0. \end{aligned}$$

We have just shown that $y_w\rightarrow 0$ as $w\rightarrow \infty$. Therefore, for $\epsilon >0$, there exists $w_1 \ge w_0$ such that $y_w< q_\infty \epsilon /2M$ for $w\ge w_1$. Thus, the following holds:

$$\begin{aligned} x_{w+1} \le (1-q_\infty ) \left( x_w + \frac{\epsilon }{2}q_\infty \right) \ \text{ for } \ w\ge w_1. \end{aligned}$$

Fixing $w =w_1$, we have that

$$\begin{aligned} \begin{aligned} x_{w_1+n} \le&(1-q_\infty ) x_{w_1+n-1} +\frac{\epsilon }{2}q_\infty (1-q_\infty ) \\ \le&(1-q_\infty )^2 x_{w_1+n-2} +\frac{\epsilon }{2}q_\infty (1-q_\infty )^2+\frac{\epsilon }{2}q_\infty (1-q_\infty )\\ \le&\cdots \le (1-q_\infty )^n x_{w_1} + \frac{\epsilon }{2}q_\infty \sum _{i=1}^n (1-q_\infty )^i\\ \le&(1-q_\infty )^n x_{w_1} + \frac{\epsilon }{2}. \end{aligned} \end{aligned}$$

There exists $n_2$ such that for $n \ge n_2$, $(1-q_\infty )^n x_{w_1} <\epsilon /2$. Thus, for $w \ge w_1+n_2$, $x_w < \epsilon$. Since $x_w\ge 1$, we have that $\lim _{w\rightarrow \infty } x_w = 0$, which gives (111).

1.4 Characterizing the equilibrium quantities with $f_w(\cdot )$

This subsection relates the equilibrium quantities to $f_w(\cdot )$. The following lemma shows that the equilibrium $e^*$ can be characterized by $h(\cdot )$.

Lemma 25

For any equilibrium $e^*$, the following holds:

$$\begin{aligned} (\beta ^*(w),q^*(w),\bar{G}^*(w)) = h(\beta ^*(w+1),q^*(w+1),\bar{G}^*(w+1)), \ w\ge 1. \end{aligned}$$

Proof

It follows from (4) to (70) that $\beta ^*(w) = h_1(\beta ^*(w+1),q^*(w+1),\bar{G}^*(w+1))$ for $w\ge 1$.

It follows from Eq. (11) that

$$\begin{aligned} \begin{aligned} J^*(w) =&\mathbb {E}_{\varepsilon }[\bar{F}^{-1}(q^*(w))-(\varepsilon (1)-\varepsilon (0))]^+ =\int _{-\infty }^{\bar{F}^{-1}(q^*(w))} F(x)\,\mathrm {d}x. \end{aligned} \end{aligned}$$

Substituting this equation and $\beta ^*(w) = h_1(\beta ^*(w+1),q^*(w+1),\bar{G}^*(w+1))$ into (10) and comparing it with (71), we have that

$$\begin{aligned} \begin{aligned} q^*(w) =&\bar{F}\left( -c+\alpha \left\{ \beta ^*(w)r+(1-\beta ^*(w))\int _{-\infty }^{\bar{F}^{-1}(q^*(w))} F(x)\,\mathrm {d}x\right\} \right) \\ =&h_2(\beta ^*(w+1),q^*(w+1),\bar{G}^*(w+1)). \end{aligned} \end{aligned}$$

In addition, it follows from Eqs. (1) to (72) that

$$\begin{aligned} \bar{G}^*(w)= & {} \frac{\bar{G}^*(w+1)}{1-q^*(w+1)} = \min \left( 1, \frac{\bar{G}^*(w+1)}{1-q^*(w+1)} \right) \\= & {} h_3(\beta ^*(w+1),q^*(w+1),\bar{G}^*(w+1)), \end{aligned}$$

where the second equality holds because $\bar{G}^*(w)= \prod _{i=1}^w (1-q^*(i)) \le 1$.

Thus, it is immediate that $f_w(\cdot )$ characterizes $\bar{G}^*(w)$ in terms of $\beta ^*(w)$ and $q^*(w)$, which is formalized in the following corollary.

Corollary 5

We have that $\bar{G}^*(w) = f_w(e^*(w))$ for all $w\ge 1$.

Proof

Fixing a $w\ge 1$ and substituting $z(w) = (\beta ^*(w),q^*(w),\bar{G}^*(w))$ into Eq. (83), by applying Lemma 25 inductively we have that $z(1) = (\beta ^*(1),q^*(1),\bar{G}^*(1))$. Note that the resulting $z(1) = (\beta ^*(1),q^*(1),\bar{G}^*(1))$ satisfies condition (84). In particular, $\bar{G}^*(1) = 1-q^*(1)$. Thus, it follows from the definition of the function $f_w(\cdot )$ that $\bar{G}^*(w) = f_w(e^*(w))$ for all $w\ge 1$.

The following lemma shows that the equilibrium quantities live in the set $\mathscr {Z}_1(w)\times \mathscr {Z}_2(w)$.

Lemma 26

For any equilibrium $e^* = (\beta ^*,q^*)$, we have that

$$\begin{aligned} e^*(w) = (\beta ^*(w),q^*(w)) \in \mathscr {Z}_1(w)\times \mathscr {Z}_2(w), \ w\ge 1. \end{aligned}$$

Proof

By Lemma 4 and Corollaries 3–4, $\beta ^*(w) \le b$ and $q^*(w) \ge q_\infty$. Thus, it suffices to show that $\beta ^*(w) \ge \underline{\beta }(w)$ and $q^*(w) \le \bar{q}(w)$ for all $w\ge 1$.

We first show that $\beta ^*(w) \ge \underline{\beta }(w)$. It follows from Proposition 1 that

$$\begin{aligned} \begin{aligned} \beta ^*(w) =&\left( 1+\sum _{t=w}^\infty \prod _{i=w}^t \frac{1-b}{1-a\bar{G}^*(i+1)}\right) ^{-1}\\ \ge&\left( 1+\sum _{t=w}^\infty \prod _{i=w}^t \frac{1-b}{1-a\bar{G}^*(w)}\right) ^{-1}\\ =&\left( 1+\sum _{i=1}^\infty \left( \frac{1-b}{1-a\bar{G}^*(w)}\right) ^i\right) ^{-1} \\ =&1-\frac{1-b}{1-a\bar{G}^*(w)} \\ \ge&1- \frac{1-b}{1-a(1-q_\infty )^{w}} = \frac{b-a(1-q_\infty )^{w}}{1-a(1-q_\infty )^{w}} =\underline{\beta }(w), \end{aligned} \end{aligned}$$

where the first inequality follows from

$$\begin{aligned} \bar{G}^*(i+1) = \bar{G}^*(w)\prod _{j=w+1}^{i+1} (1-q^*(j)) \le \bar{G}^*(w),\ i \ge w, \end{aligned}$$

and the last inequality follows from Lemma 16. In particular, it follows from

$$\begin{aligned} \bar{G}^*(w) = z_3(w;w,\beta ^*(w),q^*(w))\le (1-q_\infty )^w. \end{aligned}$$

Therefore, we have that $\beta ^*(w) \in \mathscr {Z}_1(w)$.

We then prove that $q^*(w)\in \mathscr {Z}_2(w)$. We first show that $J^*(w)\ge \underline{J}(w)$ for all $w\ge 1$ by contradiction, where $J^*(w)$ is the expected discounted utility of waiting. Suppose this is not true and there exists $w_0$ such that $J^*(w_0)< \underline{J}(w_0)$.

We first show by induction that $J^*(w) < \underline{J}(w_0)$ for all $w \ge w_0$. It is true for $w_0$ by assumption. By the induction hypothesis, suppose it is true for $k = w$. In particular, $J^*(w)< \underline{J}(w_0)$. We then show that $J^*(w+1)< \underline{J}(w_0)$. Substituting Eq. (8) and $\underline{J}(w_0) = \kappa (\underline{\beta }(w_0), \underline{J}(w_0))$ into $J^*(w)< \underline{J}(w_0)$, we obtain that

$$\begin{aligned} \begin{aligned}&\mathbb {E}[-c+\alpha [{\beta }^*(w)r+(1-\beta ^*(w))J^*(w+1)]-(\varepsilon (1)-\varepsilon (0))]^+\\&\quad =\mathbb {E}[\max \{\varepsilon (1),-c+\alpha [{\beta }^*(w)r+(1-{\beta }^*(w))J^*(w+1)]+\varepsilon (0)\}]\\&\quad = J^*(w) <\underline{J}(w_0) \\&\quad = \mathbb {E}[-c+\alpha [\underline{\beta }(w_0)r+(1-\underline{\beta }(w_0))\underline{J}(w_0)]-(\varepsilon (1)-\varepsilon (0))]^+, \end{aligned} \end{aligned}$$

where the first equality holds because $\mathbb {E}[\varepsilon (1)] = 0$. Comparing the first line and right-hand side of the last line, we conclude that the following inequality holds:

$$\begin{aligned}\begin{aligned}&{\beta }^*(w)r+(1-{\beta }^*(w))J^*(w+1) \\&\quad < \underline{\beta }(w_0)r+(1-\underline{\beta }(w_0))\underline{J}(w_0)\\&\quad ={\beta }^*(w)r+(1-{\beta }^*(w))\underline{J}(w_0) +(\underline{\beta }(w_0)-{\beta }^*(w))(r-\underline{J}(w_0)). \end{aligned} \end{aligned}$$

Rearranging the terms, we have that

$$\begin{aligned} J^*(w+1) <\underline{J}(w_0) +\frac{(\underline{\beta }(w_0)-{\beta }^*(w))(r-\underline{J}(w_0))}{1-\beta ^*(w)}. \end{aligned}$$

(116)

Note that the last term on the right-hand side of Eq. (116) is non-positive. To see this, recall that we have shown $\beta ^*(w) \ge \underline{\beta }(w)$ at the beginning of this proof. It follows from property (i) of Lemma 21 that $\underline{\beta }(w) \ge \underline{\beta }(w_0)$ for $w\ge w_0$. Combining the two inequalities, we have that $\underline{\beta }(w_0) - \beta ^*(w) \le 0$. In addition, it follows from Lemma 21 that $r - \underline{J}(w_0)\ge 0$. Since $1-\hat{\beta }^*(w) >0$, we have that the last term on the right-hand side of Eq. (116) is non-positive.

By dropping the non-positive term on the right-hand side of (116), we have that $J^*(w+1) < \underline{J}(w_0)$, which completes the induction argument. In summary, $J^*(w) < \underline{J}(w_0)$ for all $w \ge w_0$.

On the one hand, by the induction argument, we prove that $J^*(w)< \underline{J}(w_0)$ for all $w\ge w_0$. Thus, it follows from Corollary 4 that $J_\infty = \lim _{w\rightarrow \infty } J^*(w) \le \underline{J}(w_0)$. On the other hand, it follows from properties (i)–(ii) in Lemma 21 that $\underline{J}(w_0) < J_\infty$. This leads to a contradiction. Therefore, there exists no $w_0$ such that $J^*(w_0) < \underline{J}(w_0)$. In other words, $J^*(w)\ge \underline{J}(w)$ for all $w\ge 1$.

We complete the proof by showing $q^*(w) \le \bar{q}(w)$ for $w\ge 1$. It follows from Eq. (11) that

$$\begin{aligned} \begin{aligned} J^*(w) = \mathbb {E}_{\varepsilon }[\bar{F}^{-1}(q^*(w))-(\varepsilon (1)-\varepsilon (0))]^+ =\int _{-\infty }^{\bar{F}^{-1}(q^*(w))} F(x)\,\mathrm {d}x. \end{aligned} \end{aligned}$$

In addition, recall from Lemma 21 that

$$\begin{aligned}\begin{aligned} \underline{J}(w) = \int _{-\infty }^{\bar{F}^{-1}(\bar{q}(w))} F(x)\,\mathrm {d}x. \end{aligned} \end{aligned}$$

By comparing these two equations, we can conclude that $q^*(w) \le \bar{q}(w)$ because $J^*(w)\ge \underline{J}(w)$ for $w\ge 1$. $\square$

1.5 Proof of Lemma 5

It follows from Corollary 5 that

$$\begin{aligned} \bar{G}^*_i(w) = f_w(\beta ^*_i(w),q^*_i(w)),\ i=1,2\ \text{ and } \ w\ge 1. \end{aligned}$$

Applying the mean value theorem for multivariable functions to $f_w(\cdot )$, the following holds: For $w\ge 1$,

$$\begin{aligned} \begin{aligned} \delta _{\bar{G}}(w) =&f_w((\beta ^*_1(w),q^*_1(w))) - f_w((\beta ^*_2(w),q^*_2(w)))\\ =&\frac{\partial f_w(\tilde{e}(w))}{\partial q} \delta _q(w) + \frac{\partial f_w(\tilde{e}(w))}{\partial \beta } \delta _{\beta }(w), \end{aligned} \end{aligned}$$

(117)

where $\tilde{e}(w) = C(w)e^*_1(w)+(1-C(w))e^*_2(w)$ for some $C(w)\in (0,1)$. It follows from Lemma 26 that $e^*_1(w), e^*_2(w) \in \mathscr {Z}_1(w)\times \mathscr {Z}_2(w)$. Since $\mathscr {Z}_1(w)\times \mathscr {Z}_2(w)$ is convex, $\tilde{e}(w) \in \mathscr {Z}_1(w)\times \mathscr {Z}_2(w)$ for all $w\ge 1$. It follows from Lemma 24 that

$$\begin{aligned} \lim _{w\rightarrow \infty } \frac{\partial f_w(\tilde{e}(w))}{\partial q} = 0\ \text{ and } \ \lim _{w\rightarrow \infty } \frac{\partial f_w(\tilde{e}(w))}{\partial \beta } = 0. \end{aligned}$$

Thus, we conclude that for any $\epsilon >0$, there exists a nonnegative constant $w_1$ such that the following inequalities are satisfied:

$$\begin{aligned} \left| \frac{\partial f_w(\tilde{e}(w))}{\partial q}\right| \le \epsilon \ \text{ and } \ \left| \frac{\partial f_w(\tilde{e}(w))}{\partial \beta }\right| \le \epsilon ,\ w\ge w_1. \end{aligned}$$

Substituting the two inequalities into Eq. (117), we obtain that

$$\begin{aligned} | \delta _{\bar{G}}(w)|\le \epsilon (|\delta _{\beta }(w)|+|\delta _{q}(w)|),\ w\ge w_1. \end{aligned}$$

$\square$

Appendix 3: The road map for the proof of uniqueness

This appendix provides a detailed road map of the uniqueness proof (Proposition 4). The proof is done by contradiction. In what follows, we first provide an overview of the key steps that lead to the contradiction using various auxiliary lemmas (see Fig. 8). We then summarize the key steps to proving Lemma 5 provided in Appendix 2, which is an important technical lemma for the uniqueness proof. Two auxiliary functions, denoted by $h(\cdot )$ and $f_w(\cdot )$, and several lemmas in Appendix 2 facilitate the proof of Lemma 5. Figure 9 provides a diagram to show how the lemmas in Appendix 2 are used to prove Lemma 5.

The proof (of uniqueness) by contradiction proceeds as follows: Suppose that there are two different equilibria and define their difference as $(\delta _{\beta },\delta _{q})$. The contradiction is built on the limiting properties of the difference $(\delta _{\beta }(w),\delta _{q}(w))$ as w tends to infinity. Figure 8 shows how the contradiction is constructed.

Fig. 8

The logic flow for proving uniqueness of the equilibrium

On the one hand, Corollaries 3–4 provide the limits of equilibrium quantities (in any potential equilibrium). It is immediate from these two corollaries that the difference of the equilibrium quantities (of two different equilibria) vanishes as w goes to infinity, i.e.,

$$\begin{aligned} (\delta _{\beta }(w),\delta _{q}(w)) \rightarrow 0\ \text{ as } \ w\rightarrow \infty . \end{aligned}$$

On the other hand, Lemmas 7 and 8 show that this convergence cannot hold. Lemma 7 shows that the difference of the equilibrium quantities $(\delta _{\beta }(w),\delta _{q}(w))$, $w\ge 0$, is characterized by a dynamical system. To be more specific, the function characterizing the evolution of this dynamical system has two parts: a constant matrix A with a special structure and a matrix function $B(\cdot )$. In addition, the perturbation function B(w) vanishes as w goes to infinity, i.e., $|||B(w)|||_\infty \rightarrow 0$ as $w\rightarrow \infty$. This property is proved with the help of the technical Lemma 5. Then Lemma 8 shows that the dynamical system given in Lemma 7 cannot converge to zero. Combining Lemmas 7 and 8, we conclude that the difference $(\delta _{\beta }(w),\delta _{q}(w)))$ cannot converge to zero, which leads to the contradiction.

The rest of this section summarizes the critical steps in Appendix 2 to prove the technical Lemma 5, which characterizes $\delta _{\bar{G}}(w)$ in terms of $\delta _{\beta }(w)$ and $\delta _{q}(w)$ for $w\ge 0$ using the functions $g_1(\cdot )$ and $g_2(\cdot )$. Figure 9 illustrates how various lemmas are used (and relate to one another) to prove Lemma 5. To be specific, “Definition of the auxiliary function $f_w(\cdot )$,” “Partial derivatives of the auxiliary function $f_w(\cdot )$” and “Properties of $f_w(\cdot )$ on a restricted set” in Appendix 2 construct two auxiliary functions $h(\cdot )$ and $f_w(\cdot )$ and provide various properties of these two functions. “Characterizing the equilibrium quantities with $f_w(\cdot )$” in Appendix 2 shows the characterization of the equilibrium quantities using the auxiliary functions $h(\cdot )$ and $f_w(\cdot )$. Thus, the properties of the auxiliary functions provided in “Definition of the auxiliary function $f_w(\cdot )$,” “Partial derivatives of the auxiliary function $f_w(\cdot )$,” and “Properties of $f_w(\cdot )$ on a restricted set” in Appendix 2 are applicable to the equilibrium quantities. Appendix 2 proves Lemma 5.

Fig. 9

The logic flow for proving Lemma 5

The proof of Lemma 5 in Appendix 2 includes two parts. The first part constructs the functions $g_1(\cdot )$ and $g_2(\cdot )$ in two steps. In the first step, we use the auxiliary function $f_w(\cdot )$ defined in “Definition of the auxiliary function $f_w(\cdot )$” in Appendix 2 to characterize $\bar{G}(w)$ in terms of $\beta (w)$ and q(w), i.e., $\bar{G} = f_w(\beta (w),q(w))$ for $w\ge 0$; see Corollary 5 in “Characterizing the equilibrium quantities with $f_w(\cdot )$” in Appendix 2. In the second step, we apply the mean value theorem and construct the functions $g_1$ and $g_2$ using the partial derivatives of the function $f_w(\cdot )$; see Eq. (117).

The second part of the proof of Lemma 5 shows that the two functions $g_1(w)$ and $g_2(w)$ converge to zero as $w\rightarrow \infty$. To show this, it is sufficient to show that the supremum norm of the partial derivatives of the function $f_w(\cdot )$ converges to zero as $w\rightarrow \infty$; see Eq. (117). However, this statement is not true in general, but is valid if we restrict the arguments of the function $f_w(\cdot )$ to be in the set $\mathscr {L}_1(w)\times \mathscr {L}_2(w)$ defined in “Properties of $f_w(\cdot )$ on a restricted set” in Appendix 2. The convergence result of the partial derivatives of the function $f_w(\cdot )$ restricted in the set $\mathscr {L}_1(w)\times \mathscr {L}_2(w)$ is given by Lemma 24. In addition, Lemma 26 ensures that the equilibrium quantities lie in the $\mathscr {L}_1(w)\times \mathscr {L}_2(w)$. Thus, applying Lemma 24 completes the second part of the proof of Lemma 5.

“Definition of the auxiliary function $f_w(\cdot )$,” “Partial derivatives of the auxiliary function $f_w(\cdot )$,” and “Properties of $f_w(\cdot )$ on a restricted set” in Appendix 2 are dedicated to proving Lemma 24 using the auxiliary functions $h(\cdot )$ and $f_w(\cdot )$. To be specific, “Definition of the auxiliary function $f_w(\cdot )$” in Appendix 2 defines the auxiliary functions $h(\cdot )$ and $f_w(\cdot )$. “Partial derivatives of the auxiliary function $f_w(\cdot )$” in Appendix 2 provides the recursive equations to characterize the partial derivatives of the function $f_w(\cdot )$ and the signs of the partial derivatives. “Properties of $f_w(\cdot )$ on a restricted set” in Appendix 2 constructs the restricted set $\mathscr {L}_1(w)\times \mathscr {L}_2(w)$ and proves the convergence of the partial derivatives of the function $f_w(\cdot )$ in the restricted set.

The auxiliary function $f_w(\beta ,q)$ is defined implicitly through Eqs. (82)–(84), which are rewritten for convenience as follows:

$$\begin{aligned}&z(w) = (\beta , q, f_w(\beta ,q)),\\&z(k-1) = h(z(k))\ \ \text{ for }\ \ k=w,\ldots , 2,\\&z_2(1) = 1-z_3(1). \end{aligned}$$

Lemma 14 ensures that the function $f_w(\beta ,q)$ is well-defined. In order to make sense of this definition of the implicit function $f_w(\cdot )$, the auxiliary function $h(\cdot )$ needs to be introduced. The function $h(\cdot )$ is constructed such that it characterizes the time-reversed evolution of the equilibrium quantities; see Lemma 25 in “Characterizing the equilibrium quantities with $f_w(\cdot )$” in Appendix 2. This immediately leads to the observation that if we substitute the equilibrium quantities at time w into z(w) in Eq. (82), i.e., $z(w) = (\beta ^*(w), q^*(w), \bar{G}^*(w))$, then the values of z(k) (for $k=w-1,\ldots , 1$) in Eq. (83) equal the equilibrium quantities as well, i.e.,

$$\begin{aligned} z(k) = (\beta ^*(k), q^*(k), \bar{G}^*(k))\ \ \text{ for }\ \ k=w-1,\ldots , 1. \end{aligned}$$

In addition, the condition in Eq. (84) is automatically satisfied by the definition of $\bar{G}$ in Eq. (1). Therefore, the function $f_w(\cdot )$ is the implicit function that characterizes the equilibrium quantity $\bar{G}^*(w)$ in terms of $\beta ^*(w), q^*(w)$, i.e., $\bar{G}^*(w) = f_w(\beta ^*(w), q^*(w))$, $w\ge 0$; see Corollary 5 in “Characterizing the equilibrium quantities with $f_w(\cdot )$” in Appendix 2.

Lemma 24 provides the convergence property of the partial derivatives of the implicit function $f_w(\cdot )$. In order to prove this lemma, we first provide a recursive characterization of the partial derivatives of the implicit function $f_w(\cdot )$. Since the function $f_w(\cdot )$ is defined implicitly by using the function $h(\cdot )$ recursively, the partial derivatives of the implicit function $f_w(\cdot )$ are characterized using the partial derivatives of the function $h(\cdot )$ (provided in Lemma 12) recursively; see Lemma 17. By analyzing the partial derivatives of the functions $f_w(\cdot )$ and $h(\cdot )$ (provided in Lemmas 12 and 17), we provide useful properties of the partial derivatives. These properties eventually lead to the convergence property in Lemma 24; see Fig. 9.

We end this section by providing a comment on Lemma 23. Lemma 23 provides critical bounds of the partial derivatives of the function $h(\cdot )$ to prove Lemma 24. However, these bounds only hold after we restrict the arguments of the function $h(\cdot )$ to the set $\mathscr {L}_1(w)\times \mathscr {L}_2(w)$. The set $\mathscr {L}_1(w)\times \mathscr {L}_2(w)$ is carefully constructed to satisfy two conditions. First, the set is narrow enough such that the bounds in Lemma 23 hold. Second, the set is wide enough to ensure that the equilibrium quantities lie in the set; see Lemma 26.