When do creditors with heterogeneous beliefs agree to run?

Abstract

This paper explores, in a multiperiod setting, the funding liquidity of a borrower that finances its operations through short-term debt. The short-term debt is provided by a continuum of creditors with heterogeneous beliefs about the prospects of the borrower. In each period, creditors observe the borrower’s fundamentals and decide on the amount they invest in its short-term debt. We formalize this problem as a coordination game, and we show that there exists a unique reasonable Nash equilibrium. We show that the borrower is able to refinance if and only if the liquid net worth is above an illiquidity barrier, and we explicitly find this barrier in terms of the distribution of capital and beliefs across creditors.

Appendix

All proofs are presented in this appendix.

Before we come to the proof of Theorem 3.2, we first provide the

Proof of Lemma 3.3.

First, note that the function b _t (⋅) is decreasing for each t since β(⋅) and hence β ⁻¹(⋅) are decreasing. Hence, C(b _t (⋅)) is increasing, and hence m _t (⋅) is strictly increasing by (3.3) for each t.

By definition, b _T (⋅)=∞. Set $m_{ T+1}^{-1}(0):= \infty$. We show by backward induction on t≤T that b _t−1(x)≤b _t (x) and $m_{t}^{-1}(0) \leq m_{t+1}^{-1}(0)$. The initial step t=T is clear. Now let t<T and assume that b _t (x)≤b _t+1(x) and $m_{t+1}^{-1}(0) \leq m_{t+2}^{-1}(0)$. Then

$$m_{t} (x) = C\big(b_t( x)\big) + x \geq C\big(b_{t+1}( x)\big) + x = m_{t+1} (x) \mbox{ for all }x. $$

So $0 = m_{t} ( m_{t}^{-1}(0) ) \geq m_{t+1} ( m_{t}^{-1}(0) )$ and thus $m_{t+1}^{-1}(0) \geq m_{t}^{-1}(0)$ since $m_{t+1}^{-1}(\cdot)$ is increasing. This implies b _t (x)≥b _t−1(x) since β(⋅) is decreasing, finishing the induction step. □

We next show that under an allocation strategy $\bar{\pi}$ of the structure as in Theorem 3.2, no creditor increases her investment into the short-term debt at the time of default.

Lemma A.1

If $y_{t} \in\Delta_{t}(\bar{\pi})$, t>1, then for any creditor a, we have

$$\bar{\pi}_{t-1}(a, y_{t-1})\geq\bar{\pi}_t(a, y_t). $$

Proof

Since m _t (⋅) is strictly increasing, $y_{t} \in\Delta_{t}(\bar{\pi})$ and (3.3) imply

$$\begin{aligned} x_t^{\bar{\pi}}(y_t) < & m_t^{-1}(0),\\ x_{t-1}^{\bar{\pi}}(y_{t-1}) \geq& m_{t-1}^{-1}(0). \end{aligned}$$

Combined with Lemma 3.3, these equations imply

$$ x_t^{\bar{\pi}}(y_t) < m_t^{-1}(0)\leq m_{t-1}^{-1}(0)\leq x_{t-1}^{\bar{\pi}}(y_{t-1}). $$

Therefore, again from Lemma 3.3, we obtain

$$ b_{t-1} \big(x_{t-1}^{\bar{\pi}}(y_{t-1})\big)\leq b_t\big(x_{t-1}^{\bar{\pi}}(y_{t-1})\big)\leq b_t\big(x_{t}^{\bar{\pi}}(y_{t})\big), $$

which proves the statement. □

Proof of Theorem 3.2

We prove the claim by backward induction. The induction base at T holds by convention. Assume now that π ^∗ is an equilibrium at time t+1. Fix a∈A and take any allocation strategy π that satisfies $\pi _{s}(a',\cdot)= \bar{\pi}_{s}(a',\cdot)$ for all creditors a′ and all s=0,…,t−1, and also $\pi_{s}(a',\cdot) = \bar{\pi}_{s}(a',\cdot)$ for all creditors a′≠a and all s=t,t+1,…,T. The strategy π is obtained from $\bar{\pi}$ by modifying the strategy of player a alone, from time t on.

We have to show that

$$ g_t^{\bar{\pi}, a}(Y_t) \geq g_t^{\pi, a}(Y_t) $$

for each $Y_{t} \in\varGamma_{t-1}(\bar{\pi}) \times \mathbb {R}$. First, note that since the set {a} has measure zero, changing the strategy of a single player will have no impact on the default time, that is,

$$ T^{\pi}=T^{\bar{\pi}}. $$

(A.1)

By (A.1), and since consumption depends only on allocations up to t−1 by (2.5), we obtain $C_{t}^{\bar{\pi}, a} = C_{t}^{\pi, a}$; so it suffices to show

$$ \mathbb {E}^{a}[V_{t+1}^{\bar{\pi}, a} \mid Y_t] \geq \mathbb {E}^{a}[V_{t+1}^{\pi, a} \mid Y_t]. $$

(A.2)

We now distinguish two cases.

Case i) Suppose that $Y_{t} \in\Delta_{t}(\pi)=\Delta_{t}( \bar{\pi})$. By Lemma A.1 we have

$$\bar{\pi}_{t-1}(a, Y_{t-1}) \geq\bar{\pi}_t(a, Y_t), $$

and the allocations π and $\bar{\pi}$ are the same until time t−1. It follows that

$$\begin{aligned} \mathbb {E}^{a}[V_{t+1}^{\bar{\pi}, a} \mid Y_t] &= 1-\big(\bar{\pi}_{t-1}(a, Y_{t-1}) \vee\bar{\pi}_t(a, Y_t)\big) = 1 - \bar{\pi}_{t-1}(a, Y_{t-1}) \\ &\geq1 - \big( \bar{\pi}_{t-1}(a, Y_{t-1}) \vee \pi_t(a, Y_t) \big) = 1-\big(\pi_{t-1}(a, Y_{t-1}) \vee \pi_t(a, Y_t) \big)\\ &= \mathbb {E}^{a}[V_{t+1}^{\pi, a} \mid Y_t]. \end{aligned}$$

So we have (A.2).

Case ii) Suppose now that $Y_{t} \in\varGamma_{t}(\pi)=\varGamma_{t}( \bar{\pi})$. On the set $\{ T^{\bar{\pi}} = t+1 \}$, we have $0 > X_{t+1}^{\bar{\pi}} + C(b_{t+1}(X_{t+1}^{\bar{\pi}}) ) = X_{t+1}^{\bar{\pi}} + C(b_{t+1}(X_{t+1}^{\bar{\pi}}))$. So $b_{t+1}(X_{t+1}^{\bar{\pi}}) = \infty$ and therefore $\bar{\pi}_{t+1}(a, Y_{t+1}) = 0$ on $\{T^{\bar{\pi}} = t+1 \}$ by (3.6). Hence,

(A.3)

Since $T^{\pi}= T^{ \bar{\pi}}$, we have that $\mathbb {P}^{a}[T^{\pi}>t+1 \mid Y_{t}]$ and $\mathbb {P}^{a}[T^{\bar{\pi}}>t+1 \mid Y_{t}]$ coincide; so as in (A.3), we obtain

(A.4)

We also have the equivalence

$$\begin{aligned} T^{\bar{\pi}} > t+1 &\Longleftrightarrow X_{t+1}^{\bar{\pi}} + D_{t+1}^{\bar{\pi}} \geq0 \\ & \Longleftrightarrow X_{t+1}^{\bar{\pi}} + C\big(b_{t+1}(X_{t+1}^{\bar{\pi}}) \big) \geq0 \\ & \Longleftrightarrow X_{t+1}^{\bar{\pi}} \geq m_{t+1}^{-1}(0) \\ & \Longleftrightarrow X_t^{\bar{\pi}} + \theta_{t+1} - r C\big(b_t(X_t^{\bar{\pi}}) \big) \geq m_{t+1}^{-1}(0) \\ & \Longleftrightarrow- \theta_{t+1} + b(a) \leq X_t^{\bar{\pi}} - r C\big(b_t(X_t^{\bar{\pi}}) \big)- m_{t+1}^{-1}(0) + b(a), \end{aligned}$$

which yields

$$ \mathbb {P}^a[T^{\bar{\pi}}>t+1 \mid Y_t] = \varPhi\Big( X_t^{\bar{\pi}} - r C\big(b_t(X_t^{\bar{\pi}}) \big) - m_{t+1}^{-1}(0) + b(a) \Big). $$

(A.5)

Since $T^{\bar{\pi}} > t$, we have $X_{t}^{\bar{\pi}} + C(b_{t}(X_{t}^{\bar{\pi}})) = X_{t}^{\bar{\pi}} + C(b_{t}(X_{t}^{\bar{\pi}}) ) \geq0$ and thus $b_{t}(X_{t}^{\bar{\pi}}) = q(t,X_{t}^{\bar{\pi}})$. So it follows from (A.5) that we have the equivalence

$$\begin{aligned} & \mathbb {P}^a[T^{\bar{\pi}}>t+1|Y_t] \geq {\textstyle\frac{1}{1+r}} \\ &\Longleftrightarrow{} \varPhi^{-1} \big({\textstyle\frac{r}{1+r}} \big) \geq- X_t^{\bar{\pi}} + r C\big(q(t,X_t^{\bar{\pi}})\big) + m_{t+1}^{-1}(0) - b(a) \\ &\Longleftrightarrow{} b(a) \geq- X_t^{\bar{\pi}} + r C\big(q(t,X_t^{\bar{\pi}})\big) + m_{t+1}^{-1}(0) - \varPhi^{-1} \big( {\textstyle\frac{r}{1+r}} \big) = q(t,X_t^{\bar{\pi}}) = b_t(X_t^{\bar{\pi}}) \\ &\Longleftrightarrow{} \bar{\pi}_t(a, Y_t) = 1, \end{aligned}$$

(A.6)

where we used (3.1) and (3.4) in the first of the last two equalities, and (3.6) in the last equivalence. Combining (A.3) and (A.4) with (A.6) and π _t (a,Y _t )∈[0,1] yields (A.2). □

Proof of Lemma 3.6

We prove the statement by contradiction. Let y _t ∈Δ _t (π). Assume that there exists a creditor a∈A such that π _t (a,y _t )>π _t−1(a,y ₋₁) and consider the strategy

$$\pi'_t(a, y_t)= \left\{ \begin{aligned} &\pi_s(a, y_s) \quad&&\mbox{ if}\ \ s \neq t, \\ &\pi_{s-1}(a, y_{s-1}) \quad&&\mbox{ if}\ \ s = t. \end{aligned} \right. $$

Then we have

$$ g_t^{\pi, a}(y_t)=1-\pi_t(a, y_t)< 1-\pi_{t-1}(a, y_{t-1})=g^{\langle\pi |\pi'(a) \rangle,a}_t(y_t), $$

in contradiction to the definition of a Nash equilibrium. □

Proof of Proposition 3.7

On the set {T ^π>t}, we have

Since π is an equilibrium at time t, Lemma 3.6 yields

and therefore we obtain

(A.7)

If $\mathbb {P}^{a}[T^{\pi}>t+1|Y_{t}] > \frac{1}{1+r}$, then π _t (a,Y _t )=1 maximizes the right-hand side of (A.7); hence, we have π _t (a,Y _t )=1. If $\mathbb {P}^{a}[T^{\pi}>t+1|Y_{t}] < \frac{1}{1+r}$, then π _t (a,Y _t )=0 maximizes the right-hand side of (A.7); hence, we have π _t (a,Y _t )=0. Finally, using (2.4), the fact that A is nonatomic, and the continuity of the distribution of θ _t+1 under $\mathbb {P}^{a}$, it follows easily that the set of creditors a∈A with $\mathbb {P}^{a}[T^{\pi }>t+1|Y_{t}] = \frac{1}{1+r}$ has measure zero. □

For the remainder of this Appendix, let b _t (⋅), m _t (⋅), and q(t,⋅) denote the functions defined by the recursion (3.2)–(3.4) and (3.8).

Proof of Theorem 3.8

We prove the assertion by backward induction on t=T,…,1. For t=T, the assertion is trivial. So let t<T and assume that the assertion is true for t+1. On {Y _t ∈Γ _t (π)}, we have the sequence of implications (using that $X_{t+1}^{\pi}= x_{t+1}^{\pi}(Y_{t+1})$)

$$\begin{aligned} Y_{t+1} \in\varGamma_{t+1}(\pi) &\Longleftrightarrow X_{t+1}^{\pi} + d_{t+1}^\pi(Y_{t+1}) \geq0 \\ & \Longrightarrow X_{t+1}^{\pi} + \hat{d}_{t+1}(X_{t+1}^{\pi}) \geq0 \\ & \Longleftrightarrow X_{t+1}^{\pi} \geq m_{t+1}^{-1}(0) \\ & \Longleftrightarrow X_t^{\pi} + \theta_{t+1} - r d_t^{\pi}(Y_t) \geq m_{t+1}^{-1}(0) \\ & \Longleftrightarrow- \theta_{t+1} + b(a) \leq X_t^{\pi} - r d_t^{\pi }(Y_t) - m_{t+1}^{-1}(0) + b(a), \end{aligned}$$

(A.8)

where the “⇒” in the second step follows from Y _t+1∈Γ _t+1(π) and the induction hypothesis. Now (A.8) yields

$$ \mathbb {P}^a[T^{\pi}>t+1|Y_t] \leq\varPhi\big( X_t^{\pi} - r d_t^{\pi}(Y_t) - m_{t+1}^{-1}(0) + b(a) \big). $$

Combining this with the equivalence

$$\begin{aligned} & \varPhi\big( X_t^{\pi} - r d_t^{\pi}(Y_t) - m_{t+1}^{-1}(0) + b(a) \big) < \frac{1}{1+r} \\ &\Longleftrightarrow b(a) < r d_t^{\pi}(Y_t) + m_{t+1}^{-1}(0) - \varPhi^{-1} \Big(\frac {r}{1+r}\Big) - X_t^{\pi} =: h_t(Y_t), \end{aligned}$$

we obtain the implications

$$ b(a) < h_t(Y_t) \ \Longrightarrow\ \mathbb {P}^a[T^{\pi }>t+1|Y_t] < \frac{1}{1+r} \ \Longrightarrow\ \pi _t(a, Y_t) = 0, $$

of which the last follows from Proposition 3.7. Therefore,

$$ d_t^{\pi}(Y_t) \leq C\big(h_t(Y_t)\big) = C\bigg(r d_t^{\pi}(Y_t) + m_{t+1}^{-1}(0) - \varPhi^{-1} \Big(\frac{r}{1+r}\Big) - X_t^{\pi}\bigg). $$

Setting $z = X_{t}^{\pi} - m_{t+1}^{-1}(0) + \varPhi^{-1} \big(\frac {r}{1+r}\big)$ and $s = d_{t}^{\pi}(Y_{t})$, Lemma A.2 below then implies

$$ d_t^{\pi}(Y_t) \leq C\Biggl(\beta^{-1}\bigg( X_t^{\pi} - m_{t+1}^{-1}(0) + \varPhi^{-1} \Big(\frac{r}{1+r}\Big) \bigg)\Biggr). $$

(A.9)

Since T ^π>t, we obtain that

$$\begin{aligned} 0 \leq X_t^{\pi} + d_t^\pi(Y_t) &\leq X_t^{\pi} + C\Biggl(\beta ^{-1}\bigg( X_t^{\pi} - m_{t+1}^{-1}(0) + \varPhi^{-1} \Big(\frac {r}{1+r}\Big) \bigg)\Biggr)\\ &= X_t^{\pi} + C\big(q(t,X_t^{\pi})\big). \end{aligned}$$

This implies $\hat{b}_{t}(X_{t}^{\pi}) = \beta^{-1}( X_{t}^{\pi} - m_{t+1}^{-1}(0) + \varPhi^{-1} (\frac{r}{1+r}) )$, and hence (A.9) becomes $d_{t}^{\pi}(Y_{t}) \leq\hat{d}_{t}(X_{t}^{\pi})$, finishing the induction step. □

Lemma A.2

Fix $z \in \mathbb {R}$. The function

$$f(s) = s - C(rs - z) $$

is strictly increasing in $s \in \mathbb {R}$, and its zero is given by s=C(β ⁻¹(z)).

Proof

Clearly, f(⋅) is strictly increasing, and we have

$$\begin{aligned} f(s) = 0 & \Longleftrightarrow r C(rs - z) - rs = 0 \\ & \Longleftrightarrow\beta(rs-z) = z \\ & \Longleftrightarrow s = \frac{1}{r} \big( \beta^{-1}(z) + z \big) = C\big(\beta^{-1}(z)\big), \end{aligned}$$

where the last equation follows from (3.1). □

In the following, we use the notation

$$\hat{\pi}(a,y_t):= \hat{\pi}_t\big(a,x_t^{\hat{\pi}}(y_t)\big) $$

with the function $\hat{\pi}_{t}(a,x)$ introduced in (3.9). For the proof of Theorem 3.13, we use the following two immediate propositions.

Proposition A.3

Suppose that $y_{t} \in\varGamma_{t}(\hat{\pi})$. Then for all a∈A, we have

$$g_t^{\langle\hat{\pi}|\hat{\pi}(a) \rangle, a}(y_{t}) > g_t^{\langle\hat{\pi}|\tilde{\pi}(a) \rangle, a}(y_{t}) $$

for any $\tilde{\pi}_{t}(a,y_{t}) \neq\hat{\pi}_{t}(a,y_{t})$.

The following proposition states that given that strategy $\hat{\pi}$ is played from time t+1, and conditional on survival at time t, the strategy $\hat{\pi}$ is the best response at time t for any agent.

Proposition A.4

Let $\pi\in \mathcal {X}$ and t∈{1,…,T} such that for all s≥t+1 and $y_{s} \in\varGamma_{s-1}(\pi)\times \mathbb {R}$, we have $\pi_{s}(\cdot,y_{s}) = \hat{\pi}_{s}(\cdot, x_{s}^{\pi}(y_{s}))$. Suppose that y _t ∈Γ _t (π). Then for all a∈A, we have

$$g_t^{\langle\pi|\hat{\pi}(a) \rangle, a}(y_{t}) > g_t^{\langle\pi |\tilde{\pi}(a) \rangle, a}(y_{t}) $$

for any $\tilde{\pi}_{t}(a,y_{t}) \neq\hat{\pi}_{t}(a,x_{t}^{\pi}(y_{s}))$. Moreover, if π is an equilibrium at time t, then $\pi_{t}(\cdot, y_{t}) = \hat{\pi}_{t}(\cdot, x_{t}^{\pi}(y_{s}))$.

The last statement of Proposition A.4 follows by Theorem 3.8.

Proof of Theorem 3.13

For t=1,…,T, let Π _t be the set of strategies that are equilibria at time t and not weakly dominated for any a at times s=t,…,T. We first prove the following statement (H _t ) by backward induction on t=T,…,1.

(H _t ) We have that $\hat{\pi}\in\varPi_{t}$. Conversely, any π∈Π _t that is an equilibrium at time t−1 fulfils

$$\begin{aligned} \varGamma_t(\pi) = \{ y_{t-1} \in\varGamma_{t-1}(\pi), x_t^{\pi}(y_t)\geq m_{t}^{-1}(0) \},& \end{aligned}$$

(A.10)

$$\begin{aligned} \pi_t(a, y_t) \begin{cases} =\hat{\pi}_t(a, x_t^{\pi}(y_t)) &\quad \mbox{ if}\ \ y_{t-1} \in \varGamma_{t-1}(\pi) \text{ and}\ \ x_t^{\pi}(y_t) \geq m_{t}^{-1}(0),\\ \leq \pi_{t-1}(a,y_{t-1}) &\quad \mbox{ if}\ \ y_{t-1} \in \varGamma_{t-1}(\pi) \text{ and}\ \ x_t^{\pi}(y_t) < m_{t}^{-1}(0) \end{cases}& \end{aligned}$$

(A.11)

if t≥2, and $\pi_{1} \equiv\hat{\pi}_{1}$.

Note that (A.11) immediately implies that for any y _t−1∈Γ _t−1(π),

$$g_{t-1}^{\langle\pi|\hat{\pi}(a) \rangle, a}(y_{t-1}) = g_{t-1}^{\langle \hat{\pi}|\hat{\pi}(a) \rangle, a}(y_{t-1}). $$

Clearly, (H _T ) is true since any equilibrium π fulfils π _T ≡0 by definition. For the induction step, let 1≤t<T and assume that (H _t+1) is true. We now prove (H _t ) in the following steps.

Step 1) Let us show that $\hat{\pi}\in\varPi_{T}$. By the induction hypothesis, $\hat{\pi}$ is not weakly dominated at times t+1,…,T. We now show that $\hat{\pi}$ is not weakly dominated at time t. Suppose, for a contradiction, that there exists a strategy $\tilde{\pi}$ with $\tilde{\pi}_{s} = \hat{\pi}_{s}$ for all s≤t−1 such that $\hat{\pi}$ is weakly dominated by $\tilde{\pi}$ for some creditor a at time t. Then we have the following two statements:

a) For all trajectories $y_{t} \in\varGamma_{t-1}(\hat{\pi}) \times \mathbb {R}$,

$$\hat{\pi}\in \mathcal {O}_t\Big(y_t, \big(d_1^{\hat{\pi}}(y_1), \dots, d_{t-1}^{\hat{\pi}}(y_{t-1})\big)\Big) $$

with $\hat{\pi}\in\varPi_{t+1}^{*}$, and we have

$$ g_{t}^{\langle\hat{\pi}|\hat{\pi}(a) \rangle, a}(y_{t}) \leq g_{t}^{ \langle\hat{\pi}|\tilde{\pi}(a) \rangle, a }(y_{t}). $$

(A.12)

b) There exist a trajectory $y_{t} \in\varGamma_{t-1}(\hat{\pi}) \times \mathbb {R}$ and a strategy

$$\pi^* \in \mathcal {O}_t\Big(y_t, \big(d_1^{\hat{\pi}}(y_1), \dots, d_{t-1}^{\hat{\pi}}(y_{t-1})\big)\Big) $$

with $\pi^{*} \in\varPi_{t+1}^{*}$ such that

$$ g_{t}^{\langle\pi^*|\hat{\pi}(a) \rangle, a}(y_{t}) < g_{t}^{ \langle\pi ^*|\tilde{\pi}(a) \rangle, a }(y_{t}). $$

(A.13)

For y _t and π ^∗ as in b), if y _t ∈Γ _t (π ^∗), then we have by Theorem 3.8 that $y_{t} \in\varGamma_{t}(\hat{\pi})$. Now the induction hypothesis implies that $g_{t}^{\langle\pi ^{*}|\hat{\pi}(a) \rangle, a}(y_{t})=g_{t}^{\langle\hat{\pi}|\hat{\pi}(a) \rangle, a}(y_{t})$ and $g_{t}^{\langle\pi^{*}|\tilde{\pi}(a) \rangle, a}(y_{t}) = g_{t}^{\langle\hat{\pi}|\tilde{\pi}(a) \rangle, a}(y_{t})$, and so (A.13) yields the inequality

$$g_{t}^{\langle\hat{\pi}|\hat{\pi}(a) \rangle, a}(y_{t}) < g_{t}^{ \langle \hat{\pi}|\tilde{\pi}(a) \rangle, a }(y_{t}), $$

in contradiction to the statement of Proposition A.3. So we must have y _t ∈Δ _t (π ^∗), and (A.13) becomes

$$ 1 - \big(\hat{\pi}_{t-1}(a,y_{t-1}) \vee\hat{\pi}_{t}(a,y_{t})\big) < 1 - \big( \tilde{\pi}_{t-1}(a,y_{t-1}) \vee\tilde{\pi}_{t}(a,y_{t}) \big). $$

Since $\tilde{\pi}_{t-1} = \hat{\pi}_{t-1}$, it follows that $\tilde{\pi}_{t}(a,y_{t}) < \hat{\pi}_{t}(a,y_{t})$, so in particular we have $\hat{\pi}_{t}(a,y_{t}) > 0$ and hence $y_{t} \in\varGamma_{t}(\hat{\pi})$ by definition of $\hat{\pi}$. Since $\hat{\pi}$ is an equilibrium, using Proposition A.3, this is a contradiction to (A.12).

Step 2) Now suppose t>1 and fix any π∈Π _t . Since by Theorem 3.8 we have

$$\varGamma_t(\pi) \subseteq\{y_{t-1} \in\varGamma_{t-1}(\pi),\ x_t^{\pi }(y_t) \geq m_{t}^{-1}(0) \}, $$

to show (A.10), it is sufficient to show that

$$ \{ y_{t-1} \in\varGamma_{t-1}(\pi), x_t^{\pi}(y_t) \geq m_{t}^{-1}(0) \} \subseteq\varGamma_t(\pi). $$

(A.14)

Step 2a) We first prove that for all $y_{t} \in\varGamma_{t-1}(\pi) \times \mathbb {R}$, we have

$$ \pi_{t}(a, y_t) \geq\mbox{min }\Big(\pi_{t-1}(a, y_{t-1}), \hat{\pi}_t \big(a, x_t^{\pi}(y_t)\big)\Big). $$

(A.15)

To this end, take any a∈A and assume, for a contradiction, that there exists a trajectory $\tilde{y}_{t} \in\varGamma_{t-1}(\pi) \times \mathbb {R}$ such that

$$ \pi_{t}(a, \tilde{y}_{t}) < \mbox{min }\Big(\pi_{t-1}(a, \tilde{y}_{t-1}), \hat{\pi}_t \big(a, x_t^{\pi}(\tilde{y}_{t})\big)\Big). $$

(A.16)

Since $\hat{\pi}_{t}$ is {0,1}-valued, and π _t−1 is {0,1}-valued for almost all creditors by Proposition 3.7, we have $\mbox{min}(\pi_{t-1}(a, \tilde{y}_{t-1}), \hat{\pi}_{t} (a, x_{t}^{\pi }(\tilde{y}_{t})))=1$ for almost any creditor a that satisfies (A.16). This yields

$$\begin{aligned} \pi_{t-1}(a, \tilde{y}_{t-1}) = \hat{\pi}_t \big(a, x_t^{\pi}(\tilde{y}_{t})\big) &= 1, \end{aligned}$$

(A.17)

$$\begin{aligned} \pi_t(a,\tilde{y}_t) &< 1. \end{aligned}$$

(A.18)

Define the allocation strategy $\tilde{\pi}$ by

$$ \tilde{\pi}_s(a, y_s)= \begin{cases} 1 &\quad\mbox{ if}\ \ y = \tilde{y} \text{ and}\ \ s = t, \\ \pi_s(a, y_s) &\quad\mbox{ otherwise.} \end{cases} $$

(A.19)

We claim that π is weakly dominated by $\tilde{\pi}$ for creditor a at time t. To prove this, let $y_{t} \in\varGamma_{t-1}(\pi) \times \mathbb {R}$ and consider a strategy

$$ \pi^* \in \mathcal {O}_t\Big(y_t, \big(d_1^{\pi}(y_1), \dots, d_{t-1}^{\pi }(y_{t-1})\big)\Big) $$

(A.20)

that is an equilibrium on A∖{a} and is not weakly dominated for any creditor in A∖{a} at times t=t+1,…,T. Since {a} has measure zero, we can extend π ^∗ to an equilibrium on A, again denoted by π ^∗, which is not weakly dominated for any creditor in A at times t=t+1,…,T. We first have to show that

$$ g_{t}^{ \langle\pi^*|\pi(a) \rangle, a }(y_{t}) \leq g_{t}^{\langle\pi ^*|\tilde{\pi}(a) \rangle, a}(y_{t}). $$

(A.21)

If $y \neq\tilde{y}$, then $\tilde{\pi}= \pi$, and we have equality in (A.21). So we now assume that $y = \tilde{y}$. First, observe that $T^{ \langle \pi^{*}|\pi(a) \rangle} = T^{ \langle\pi^{*}|\tilde{\pi}(a) \rangle} = T^{\pi^{*}}$. Since $\tilde{\pi}_{t-1}(a, \tilde{y}_{t-1}) = \pi_{t-1}(a, \tilde{y}_{t-1})$, we have $C_{t}^{\langle\pi^{*} | \pi(a) \rangle, a} = C_{t}^{\langle\pi^{*} |\tilde{\pi}(a) \rangle, a}$. We now distinguish two cases.

i) Suppose that $\tilde{y}_{t} \in\Delta_{t}(\pi^{*})$: Since $\tilde{\pi}_{t-1}(a,\tilde{y}_{t-1}) = \pi_{t-1}(a,\tilde{y}_{t-1}) = 1$ by (A.17) and (A.19), it follows from (2.5) and (2.6) that $g_{t}^{\langle\pi^{*} |\tilde{\pi}(a) \rangle, a}(\tilde{y}_{t}) = g_{t}^{\langle\pi^{*}|\pi(a) \rangle,a}(\tilde{y}_{t}) = 0$.

ii) Suppose that $\tilde{y}_{t} \in\varGamma_{t}(\pi^{*})$: By the induction hypothesis, whenever y _t ∈Γ _t (π ^∗) and $x_{t+1}^{\pi^{*}}(y_{t+1}) \geq m_{t+1}^{-1}(0)$, it holds that $\pi^{*}_{t +1}(a,y_{t+1}) = \hat{\pi}_{t +1}(a,x_{t+1}^{\pi^{*}}(y_{t+1}))$. Moreover, $\pi^{*}_{t +1}(a,y_{t+1}) \leq\pi^{*}_{t}(a,y_{t})$ whenever y _t ∈Γ _t (π ^∗) and $x_{t+1}^{\pi^{*}}(y_{t+1}) < m_{t+1}^{-1}(0)$. By Proposition A.4, $\hat{\pi}_{t} (a, x_{t}^{\pi^{*}}(\tilde{y}_{t}))$ is a strict maximizer over π(a) of $g_{t}^{ \langle\pi ^{*}|\pi(a) \rangle, a }(\tilde{y}_{t})$. By (A.17)–(A.19) and the consistency condition (A.20) (see Remark 3.10), we have

$$\begin{aligned} \tilde{\pi}_t(a,\tilde{y}_t) &= 1 = \hat{\pi}_t \big(a, x_t^{\pi}(\tilde{y}_{t})\big) = \hat{\pi}_t \big(a, x_t^{\pi^*}(\tilde{y}_{t})\big),\\ \pi_t(a,\tilde{y}_t) &< 1 = \hat{\pi}_t \big(a, x_t^{\pi}(\tilde{y}_{t})\big) = \hat{\pi}_t \big(a, x_t^{\pi^*}(\tilde{y}_{t})\big). \end{aligned}$$

Therefore,

$$ g_{t}^{ \langle\pi^*|\pi(a) \rangle, a }(\tilde{y}_{t}) < g_{t}^{\langle \pi^*|\tilde{\pi}(a) \rangle, a}(\tilde{y}_{t}). $$

So we have shown (A.21).

Consider now a strategy ρ defined recursively by ρ _s (a,⋅)=π _s (a,⋅) for s<t and $\rho_{s}(a,y_{s}) = \hat{\pi}_{s}(a, x_{s}^{\rho}(y_{s}))$ for s≥t and y _s ∈Γ _s (ρ). We have by Step 1 that ρ∈Π _t . We have $\hat{\pi}_{t} (a, x_{t}^{\rho}(\tilde{y}_{t})) = \hat{\pi}_{t} (a, x_{t}^{\pi}(\tilde{y}_{t})) = 1$. This implies that $\tilde{y}_{t} \in\varGamma (\rho)$. So for π ^∗=ρ, we are in case ii), and the inequality in (A.21) is strict for $y_{t} = \tilde{y}_{t}$. This shows that $\tilde{\pi}$ weakly dominates π for creditor a at time t, in contradiction to our assumption. Hence, we have (A.15).

Step 2b) Let Y _t−1∈Γ _t−1(π). It follows immediately from Step 2a) that

$$D_{t}^{\pi} \geq\min\big(\hat{d}_t(X_t^{\pi}), D_{t-1}^{\pi}\big). $$

Set

$$\mu_{t-1}(y_{t-1}):= \max\big( m_{t}^{-1}(0), -d_{t-1}^\pi(y_{t-1}) \big). $$

We then have

$$\begin{aligned} T^{\pi} > t & \Longleftrightarrow X_{t}^{\pi} + D_{t}^\pi\geq0 \\ & \Longleftrightarrow X_{t} + d_{t}^\pi(Y_{t-1},X_{t}^{\pi}) \geq0 \\ & \Longleftarrow X_{t}^{\pi} + C\big(b_t(X_{t}^{\pi}) \big) \geq0 \quad\text{and} \quad X_{t}^{\pi} + d_{t-1}^\pi(Y_{t-1}) \geq0 \\ & \Longleftrightarrow X_{t}^{\pi} \geq m_{t}^{-1}(0) \quad\text {and} \quad X_{t}^{\pi} \geq-d_{t-1}^\pi(Y_{t-1}) \\ & \Longleftrightarrow X_{t}^{\pi} \geq\mu_{t-1}(Y_{t-1}) \end{aligned}$$

(A.22)

$$\begin{aligned} & \Longleftrightarrow X_{t-1}^{\pi} + \theta_{t} - r d_{t-1}^{\pi }(Y_{t-1}) \geq\mu_{t-1}(Y_{t-1}) \\ & \Longleftrightarrow- \theta_{t} + b(a) \leq X_{t-1}^{\pi} - r d_{t-1}^{\pi}(Y_{t-1}) - \mu_{t-1}(Y_{t-1}) + b(a), \end{aligned}$$

(A.23)

where the “⇐” in the third line follows from $D_{t}^{\pi} \geq\min(\hat{d}_{t}(X_{t}^{\pi}), D_{t-1}^{\pi})$. Now (A.23), using that $- \theta_{t} + b(a) \sim \mathcal{N}(0,1)$ under $\mathbb {P}^{a}$, yields

$$ \mathbb {P}^a[T^{\pi}>t|Y_{t-1}] \geq\varPhi\big( X_{t-1}^{\pi} - r d_{t-1}^{\pi}(Y_{t-1}) - \mu_{t-1}(Y_{t-1}) + b(a) \big). $$

Combining this with the equivalence

$$\begin{aligned} & \varPhi\big( X_{t-1}^{\pi} - r d_{t-1}^{\pi}(Y_{t-1}) - \mu _{t-1}(Y_{t-1}) + b(a) \big) > \frac{1}{1+r} \\ & \Longleftrightarrow b(a) > r d_{t-1}^{\pi}(Y_{t-1}) + \mu_{t-1}(Y_{t-1}) - \varPhi^{-1} \Big(\frac{r}{1+r}\Big) - X_{t-1}^{\pi} =: h_{t-1}(Y_{t-1}), \end{aligned}$$

we obtain the implications

$$b(a) > h_{t-1}(Y_{t-1}) \ \Longrightarrow\ \mathbb {P}^a[T^{\pi}>t|Y_{t-1}] > \frac{1}{1+r} \ \Longrightarrow \ \pi_{t-1}(a, Y_{t-1}) = 1, $$

of which the last follows from Proposition 3.7. Therefore,

$$\begin{aligned} d_{t-1}^{\pi}(Y_{t-1}) &\geq C\big(h_{t-1}(Y_{t-1})\big)\\ &= C\bigg(r d_{t-1}^{\pi}(Y_{t-1}) + \mu_{t-1}(Y_{t-1}) - \varPhi^{-1} \Big(\frac{r}{1+r}\Big) - X_{t-1}^{\pi}\bigg). \end{aligned}$$

Setting $z = X_{t-1}^{\pi} - \mu_{t-1}(Y_{t-1}) + \varPhi^{-1} (\frac {r}{1+r})$ and $s = d_{t-1}^{\pi}(Y_{t-1})$, Lemma A.2 then implies

$$ d_{t-1}^{\pi}(Y_{t-1}) \geq C\Biggl( \beta^{-1} \bigg( X_{t-1}^{\pi} - \mu_{t-1}(Y_{t-1}) + \varPhi^{-1} \Big(\frac{r}{1+r}\Big) \bigg) \Biggr). $$

(A.24)

Since T ^π>t−1, we have $X_{t-1}^{\pi} + d_{t-1}^{\pi}(Y_{t-1}) \geq0$. Suppose, for a contraction, that $-d_{t-1}^{\pi}(Y_{t-1}) > m_{t}^{-1}(0)$. It follows that

$$X_{t-1}^{\pi} \geq- d_{t-1}^\pi(Y_{t-1}) = \max\big( m_{t}^{-1}(0), -d_{t-1}^\pi(Y_{t-1}) \big) = \mu_{t-1}(Y_{t-1}), $$

and so we obtain, since β ⁻¹ is decreasing,

$$\beta^{-1} \bigg( X_{t-1}^{\pi} - \mu_{t-1}(Y_{t-1}) + \varPhi^{-1} \Big(\frac{r}{1+r}\Big) \bigg) \leq\beta^{-1} \bigg( \varPhi^{-1} \Big(\frac {r}{1+r}\Big) \bigg). $$

With (A.24), we get

$$ d_{t-1}^{\pi}(Y_{t-1}) \geq C\Biggl( \beta^{-1} \bigg( \varPhi^{-1} \Big(\frac{r}{1+r}\Big) \bigg) \Biggr) \geq C\Big( \hat{b}_t\big( m_{t}^{-1}(0)\big) \Big) = -m_{t}^{-1}(0), $$

where the last inequality follows from

$$\begin{aligned} \beta^{-1} \bigg( \varPhi^{-1} \Big(\frac{r}{1+r}\Big) \bigg) &\leq\beta ^{-1} \bigg( m_{t}^{-1}(0) - m_{t+1}^{-1}(0) + \varPhi^{-1} \Big(\frac {r}{1+r}\Big) \bigg) \\&= q\big(t,m_{t}^{-1}(0)\big) \leq\hat{b}_t\big(m_{t}^{-1}(0)\big). \end{aligned}$$

This contradicts our assumption. Hence, we have

$$ -d_{t-1}^\pi(Y_{t-1}) \leq m_{t}^{-1}(0) $$

and thus $\mu_{t-1}(Y_{t-1}) = m_{t}^{-1}(0)$. Now (A.22) reads as (A.14), which implies (A.10).

Step 3) Let us now complete the induction step by showing (A.11). First, suppose that t>1. If $Y_{t} \in\varGamma_{t-1}(\pi) \times \mathbb {R}$ and $X_{t}^{\pi} < m_{t}^{-1}(0)$, then the inequality in (A.11) follows from Lemma 3.6 and Theorem 3.8. If $Y_{t} \in\varGamma_{t-1}(\pi) \times \mathbb {R}$ and $X_{t}^{\pi} \geq m_{t}^{-1}(0)$, then Y _t ∈Γ _t (π) by (A.10), so as in (A.7), we have

$$\begin{aligned} \mathbb {E}^{a}[ V_{t+1}^{\pi, a} \mid Y_t] &= 1 + \pi_t(a, Y_t) \big( \mathbb {P}^a[ T^{\pi} > t+1 \mid Y_t](1+r) - 1 \big) \\ &= 1 + \pi_t(a, Y_t) \big( \mathbb {P}^a[ X_{t+1}^{\pi} < m_{t+1}^{-1}(0) \mid Y_t](1+r) - 1 \big) , \end{aligned}$$

(A.25)

where the second equation follows from the induction hypothesis. Since π is an equilibrium at time t, maximizing the above expectation with respect to π _t (a,Y _t ) yields $\pi_{t}(a, Y_{t}) = \hat{\pi}_{t}(a, X_{t}^{\pi})$ for almost all a, and we obtain (A.11).

Next suppose that t=1. Since Y ₁=0∈Γ ₁(π), we have (A.25) as in the case t>1, and again we obtain $\pi _{1}(a,y_{1}) = \hat{\pi}_{1}(a,y_{1})$ for almost all a. This concludes the induction step, and so we have proved (H _t ) for all t=1,…,T.

To complete the proof of the theorem, first note that $\hat{\pi}$ is a reasonable Nash equilibrium by (H ₁). Conversely, let π be any reasonable Nash equilibrium. Forward induction and (A.10) immediately yield $\varGamma_{t}(\pi) = \varGamma_{t}(\hat{\pi})$. So we have condition 1 of Definition 2.2. Equation (A.11) yields condition 2 of Definition 2.2, and condition 3 of Definition 2.2 follows from (A.11) and Lemma 3.6. Hence, π and $\hat{\pi}$ are identical. □