We look for a solution in which the creditor always chooses to exercise the maturity extension option whenever
\(M\ge 1\). So assume that in (
14)–(
15)
\(\alpha \tilde{A}<F^{N,M-1}(\tilde{A})\) always holds whenever
\(\tilde{A}<\bar{A}^{R,N}\), which should be verified later. Also, let
\(H^{N,M}(A)\equiv F^{N,M}(A)+S^{N,M}(A)\). Equations (
14)–(
15) then reduce to
$$\begin{aligned} F^{N,M}(A)= & {} \left\{ \begin{array}{lll} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}^{R,N}\}}f+\mathbf{1}_{\{\tilde{A}<\bar{A}^{R,N}\}}\alpha \tilde{A}\right] &{}\quad \text {if}&{}M=0\\ e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}^{R,N}\}}f+\mathbf{1}_{\{\tilde{A}<\bar{A}^{R,N}\}}F^{N,M-1}(\tilde{A})\right] &{}\quad \text {if}&{}M\ge 1 \end{array}\right. \end{aligned}$$
(40)
$$\begin{aligned} H^{N,M}(A)= & {} \left\{ \begin{array}{lll} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}^{R,N}\}}H^{N,N}(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}^{R,N}\}}\alpha \tilde{A}\right] &{}\quad \text {if}&{}M=0\\ e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}^{R,N}\}}H^{N,N}(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}^{R,N}\}} H^{N,M-1}(\tilde{A}))\right] &{}\quad \text {if}&{}M\ge 1 \end{array}\right. .\nonumber \\ \end{aligned}$$
(41)
For a given
N, (
41) implies that
\(N+1\) equations,
\(H^{N,0}, \ldots , H^{N,N}\), are dependent on one another. Therefore, we must construct a contraction mapping that is applied to a vector of
\(N+1\) functions. Define a mapping
\(P^{\bar{A}}:V^{1+N}\rightarrow V^{1+N}\) by the r.h.s. of (
41) with
\(\bar{A}^{R,N}\) replaced by an arbitrary threshold
\(\bar{A}\) that is no greater than
\(\bar{A}^0\):
$$\begin{aligned} P^{\bar{A}}_0(g^0,\ldots , g^N)(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}\alpha \tilde{A}\right] ,\;\; M=0 \end{aligned}$$
(42)
$$\begin{aligned} P^{\bar{A}}_M(g^0,\ldots , g^N)(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}} g^{M-1}(\tilde{A}))\right] , \;\;M\ge 1.\nonumber \\ \end{aligned}$$
(43)
\(P^{\bar{A}}\) satisfies monotonicity and discounting. Monotonicity is straightforward. For discounting, note that for
\(M= 0\) and
\(a\in \mathbb {R}^{1+N}_{++}\),
$$\begin{aligned}&P^{\bar{A}}_0(g^0+a_1,\ldots , g^N+a_N)(A)\\&\quad =e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}(g^N(\tilde{A})+a_N)+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}\alpha \tilde{A}\right] \\&\quad = e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}\alpha \tilde{A}\right] +e^{-r\Delta }a_N E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}\right] \\&\quad \le P^{\bar{A}}(g^0,\ldots , g^N)(A)+\beta \max _j a_j, \end{aligned}$$
where
\(\beta =e^{-r\Delta } E_{\tilde{A}|A}[\mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}]<1\). For
\(M\ge 1\),
$$\begin{aligned}&P^{\bar{A}}_M(g^0+a_1,\ldots , g^N+a_N)(A)\\&\quad =e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}(g^N(\tilde{A})+a_N)+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}(g^{M-1}(\tilde{A})+a_{M-1})\right] \\&\quad = e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})\right. \\&\quad \quad \left. +\mathbf{1}_{\{\tilde{A}<\bar{A}\}}g^{M-1}(\tilde{A})\right] +e^{-r\Delta }\left( a_N E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}\right] +a_{M-1} E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}<\bar{A}\}}\right] \right) \\&\quad \le P^{\bar{A}}_M(g^0,\ldots , g^N)(A)+\beta \max _j a_j, \end{aligned}$$
where
\(\beta =e^{-r\Delta }<1\). By Lemma
4.1,
\(P^{\bar{A}}\) is a contraction. Define sets
\(Z^{'}_1\subset Z_1\subset V^{1+N}\) by
$$\begin{aligned} Z_1\equiv & {} \{(g^0,\ldots , g^N): H^0(A)\le g^0(A)\\\le & {} \cdots \le g^N(A)\le A, \forall A, \,\text {and } g'\text {s are nondecreasing.}\} \end{aligned}$$
and
$$\begin{aligned} Z^{'}_1\equiv & {} \{(g^0,\ldots , g^N): H^0(A)< g^0(A)< \cdots< g^N(A)\\< & {} A, \forall A, \,\text {and }g'\hbox {s are strictly increasing.}\}. \end{aligned}$$
Then since
\(\bar{A}\le \bar{A}^0\), for any
\(G=(g^0,\ldots , g^N)\in Z_1\), we have
$$\begin{aligned} P^{\bar{A}}_0(g^0,\ldots , g^N)(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}\alpha \tilde{A}\right] \\\ge & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}^0\}}H^0(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}^0\}}\alpha \tilde{A}\right] =H^0(A), \end{aligned}$$
[see Eq. (
22)], and
$$\begin{aligned} P^{\bar{A}}_0(g^0,\ldots , g^N)(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}\alpha \tilde{A}\right] \\< & {} e^{-r\Delta }E_{\tilde{A}|A}[\tilde{A}]=A. \end{aligned}$$
Next, we have
$$\begin{aligned} P^{\bar{A}}_1(g^0,\ldots , g^N)(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}g^0(\tilde{A})\right] \\> & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}\alpha \tilde{A}\right] \\= & {} P^{\bar{A}}_0(g^0,\ldots , g^N)(A), \end{aligned}$$
and
$$\begin{aligned} P^{\bar{A}}_1(g^0,\ldots , g^N)(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}g^0(\tilde{A})\right] \\\le & {} e^{-r\Delta }E_{\tilde{A}|A}[\tilde{A}]=A, \end{aligned}$$
where the inequality is strict if
\(g^0<A\). Furthermore, for
\(M\ge 2\),
$$\begin{aligned} P^{\bar{A}}_M(g^0,\ldots , g^N)(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}g^{M-1}(\tilde{A})\right] \\\ge & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}g^{M-2}(\tilde{A})\right] \\= & {} P^{\bar{A}}_{M-1}(g^0,\ldots , g^N)(A), \end{aligned}$$
where the inequality is strict if
\(g^{M-2}<g^{M-1}\), and
$$\begin{aligned} P^{\bar{A}}_M(g^0,\ldots , g^N)(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}g^{M-1}(\tilde{A})\right] \\\le & {} e^{-r\Delta }E_{\tilde{A}|A}[\tilde{A}]=A, \end{aligned}$$
where the inequality is strict if
\(g^{M-1}<A\). Lastly, we have
$$\begin{aligned} P^{\bar{A}}_0(g^0,\ldots , g^N)(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}\alpha \tilde{A}\right] \\<&e^{-r\Delta }E_{\tilde{A}|A^{'}}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}\alpha \tilde{A}\right] \\= & {} P^{\bar{A}}_0(g^0,\ldots , g^N)(A^{'}) \end{aligned}$$
if
\(A<A^{'}\), and
$$\begin{aligned} P^{\bar{A}}_M(g^0,\ldots , g^N)(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}g^{M-1}(\tilde{A})\right] \\<&e^{-r\Delta }E_{\tilde{A}|A^{'}}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}\}}g^N(\tilde{A})+\mathbf{1}_{\{\tilde{A}<\bar{A}\}}g^{M-1}(\tilde{A})\right] \\= & {} P^{\bar{A}}_M(g^0,\ldots , g^N)(A^{'}) \end{aligned}$$
so the result of the mapping is strictly increasing. All the above together imply that applying
\(P^{\bar{A}}\) on
\(Z_1\) repeatedly
N times guarantees
\((P^{\bar{A}})^N(Z_1)\subset Z^{'}_1\). Therefore, the fixed point of
\(P^{\bar{A}}\), say
\((h^{N,0}(\cdot ;\bar{A}), \ldots , h^{N,N}(\cdot ;\bar{A}))\), is in
\(Z^{'}_1\). That is, for all
\(A>0\),
$$\begin{aligned} H^0(A)<h^{N,0}(A;\bar{A})< \cdots< h^{N,N}(A;\bar{A})<A \end{aligned}$$
(44)
and in particular
$$\begin{aligned} H^0(\bar{A})<h^{N,0}(\bar{A};\bar{A})< \cdots< h^{N,N}(\bar{A};\bar{A})<\bar{A}. \end{aligned}$$
(45)
Letting
\(\bar{A}^{R,N}\) be such that
\(h^{N,N}(\bar{A}^{R,N};\bar{A}^{R,N})=f\), we have
$$\begin{aligned} (\alpha \bar{A}^{R,N}<\alpha \bar{A}^0<)f<\bar{A}^{R,N}<\bar{A}^0. \end{aligned}$$
(46)
Also, (
44) holds for this particular threshold and we have
\(H^0(A)<h^{N,0}(A;\bar{A}^{R,N})< \cdots< h^{N,N}(A;\bar{A}^{R,N})<A\), or
$$\begin{aligned} H^0(A)<H^{N,0}(A)< \cdots< H^{N,N}(A)<A. \end{aligned}$$
(47)
Now we turn to debt values. We have
$$\begin{aligned} F^{N,0}(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}^{R,N}\}}f+\mathbf{1}_{\{\tilde{A}<\bar{A}^{R,N}\}}\alpha \tilde{A}\right] \end{aligned}$$
(48)
$$\begin{aligned}> & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}^{0}\}}f+\mathbf{1}_{\{\tilde{A}<\bar{A}^{0}\}}\alpha \tilde{A}\right] =F^0(A) \end{aligned}$$
(49)
where the inequality comes from
\(\bar{A}^{R,N}<\bar{A}^0<f/\alpha \). Also
$$\begin{aligned} F^{N,1}(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}^{R,N}\}}f+\mathbf{1}_{\{\tilde{A}<\bar{A}^{R,N}\}}F^{N,0}(\tilde{A})\right] \end{aligned}$$
(50)
$$\begin{aligned}> & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}^{R,N}\}}f+\mathbf{1}_{\{\tilde{A}<\bar{A}^{R,N}\}}\alpha \tilde{A}\right] =F^{N,0}(A), \end{aligned}$$
(51)
where the inequality comes from
\((\tilde{A}<\bar{A}^{R,N}<\bar{A}^0<\underline{A}^1)\Rightarrow (F^{N,0}(\tilde{A})>F^0(\tilde{A})>\alpha \tilde{A})\). By induction, we have
\(F^{N,M}(\cdot )>F^{N,M-1}(\cdot ), M=1,\ldots ,N\). By letting
\(\underline{A}^{N,M}\equiv \min \{\tilde{A}>0: F^{N,M}(\tilde{A})-\alpha \tilde{A}=0\}\), we have
\(\bar{A}^{R,N}<\bar{A}^0<\underline{A}^{N,1}<\underline{A}^{N,2}<\cdots \). Therefore the initial assumption that the creditor always chooses to extend the maturity has been confirmed. Since
\(S^{N,M}(\cdot )=H^{N,M}(\cdot )-F^{N,M}(\cdot )\), we have
$$\begin{aligned} S^{N,0}(A)=e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}^{R,N}\}}(H^{N,N}(\tilde{A})-f)\right] , \end{aligned}$$
(52)
so
\(S^{N,0}\) is nonnegative-valued and strictly increasing. Also
$$\begin{aligned} S^{N,1}(A)= & {} e^{-r\Delta }E_{\tilde{A}|A}\left[ \mathbf{1}_{\{\tilde{A}\ge \bar{A}^{R,N}\}}(H^{N,N}(\tilde{A})-f)+\mathbf{1}_{\{\tilde{A}<\bar{A}^{R,N}\}}S^{N,0}(\tilde{A})\right] \nonumber \\> & {} S^{N,0}(A). \end{aligned}$$
(53)
where the inequality follows (
52). Then by induction, we have
\(S^{N,M}>S^{N,M-1}\) for
\(M=1,\ldots ,N\).
\(\square \)