Appendix A: Several extensions
In this section, we generalise our method for some more complicated situations to demonstrate the versatility of our approach.
1.1 A.1 Multi-sided Parisian options
Define a Parisian stopping time for an arbitrary set as
$$ \tau _{A}^{D} = \inf \{ t \ge 0 : 1_{\{ Y_{t} \in A \}} (t - g_{A, t}) \ge D \}, \qquad g_{A, t} = \sup \{s\le t : Y_{s} \notin A\}. $$
If \(A = (-\infty , L)\), then \(\tau _{A}^{D} = \tau _{L, D}^{-}\), and if \(A = (U, \infty )\), then \(\tau _{A}^{D} = \tau _{U, D}^{+}\). The multi-sided Parisian stopping time can be defined as
$$ \tau _{\mathcal{A}}^{D} = \min \{ \tau _{A}^{D}: A \in \mathcal{A} \}, $$
where \(\mathcal{A}\) is a collection of subsets of ℝ. The double-barrier case corresponds to \(\mathcal{A} = \{ (-\infty , L), (U, \infty ) \}\) with \(L< U\). We also let
$$ T^{-}_{A} = \inf \{ t \ge 0: Y_{t} \notin A \}, \qquad T^{+}_{A} = \inf \{ t \ge 0: Y_{t} \in A \}. $$
Consider a finite state CTMC \(Y\). For any \(x, y\) in its state space \(\mathbb{S}\), let
$$ h(q, x; y) = \mathbb{E}_{x} \big[ e^{-q \tau _{\mathcal{A}}^{D}} 1_{ \{ Y_{\tau _{\mathcal{A}}^{D}} = y \}} \big]. $$
Define \(B=\bigcup _{A\in \mathcal{A}}A\). Then by a conditioning argument, we obtain
$$\begin{aligned} h(q, x; y) &= \sum _{A \in \mathcal{A}} e^{-qD} \mathbb{E}_{x} \big[ 1_{ \{ T_{A}^{-} \ge D, Y_{D} = y \}} \big] 1_{\{ x \in A \}} \\ & \phantom{=:} + \sum _{A \in \mathcal{A}} \sum _{z \notin A} \mathbb{E}_{x}\big[ e^{-q T_{A}^{-}} 1_{\{ T_{A}^{-} < D, Y_{T_{A}^{-}} = z \}} \big] h(q, z; y) 1_{\{ x \in A \}} \\ & \phantom{=:} + \sum _{z \in B}\mathbb{E}_{x} \big[ e^{-q T_{B}^{+}} 1_{\{ Y_{T_{B}^{+}} = z \}} \big] h(q, z; y) 1_{\{ x \notin B \}}. \end{aligned}$$
Let \(v_{A}(D, x; y) = \mathbb{E}_{x} [ 1_{\{ T_{A}^{-} \ge D, Y_{D} = y \}} ]\). It solves
$$ \textstyle\begin{cases} \displaystyle \frac{\partial v_{A}}{\partial D}(D, x; y) = \mathbb{G} v_{A}(D, x; y) ,\qquad x \in A, D > 0, \\ v_{A}(D, x; y) = 0,\qquad x \notin A, D > 0, \\ v_{A}(0, x; y) = 1_{\{ x = y \}}. \end{cases} $$
Let \(u^{-}_{A}(q, D, x; z) = \mathbb{E}_{x}[ e^{-q T_{A}^{-}} 1_{\{ T_{A}^{-} < D, Y_{T_{A}^{-}} = z \}} ]\). It satisfies
$$ \textstyle\begin{cases} \displaystyle \frac{\partial u^{-}_{A}}{\partial D}(q, D, x; z) = \mathbb{G} u^{-}_{A}(q, D, x; z) - q u^{-}_{A}(q, D, x; z),\qquad x \in A, D > 0, \\ u^{-}_{A}(q, D, x; z) = 1_{\{ x = z \}},\qquad x \notin A, D > 0, \\ u^{-}_{A}(q, 0, x; z) = 0. \end{cases} $$
We can rewrite it as \(u^{-}_{A}(q, D, x; z) = u^{-}_{1,A}(q, x; z) - u^{-}_{2,A}(q, D, x; z)\), and these two parts satisfy
$$ \textstyle\begin{cases} \mathbb{G} u^{-}_{1,A}(q, x; z) - q u^{-}_{1,A}(q, x; z) = 0,\qquad x \in A, \\ u^{-}_{1,A}(q, x; z) = 1_{\{ x = z \}},\qquad x \notin A, \end{cases} $$
and
$$ \textstyle\begin{cases} \displaystyle \frac{\partial u^{-}_{2,A}}{\partial{D}}(q, D, x; z) = \mathbb{G} u^{-}_{2,A}(q, D, x; z) - q u^{-}_{2,A}(q, D, x; z), \qquad x \in A, \\ u^{-}_{2,A}(q, D, x; z) = 0,\qquad x \notin A, D > 0, \\ u^{-}_{2,A}(q, 0, x; z) = u^{-}_{1,A}(q, x; z). \end{cases} $$
Let \(u^{+}(q, x; z) = \mathbb{E}_{x} [ e^{-q T_{B}^{+}} 1_{\{ Y_{T_{B}^{+}} = z \}} ]\). It is the solution to
$$ \textstyle\begin{cases} \mathbb{G} u^{+}(q, x; z) - q u^{+}(q, x; z) = 0,\qquad x \notin B, \\ u^{+}(q, x; z) = 1_{\{ x = z \}},\qquad x \in B. \end{cases} $$
Let \({ I}_{A} = \operatorname{diag}((1_{\{x \in A\}})_{x \in \mathbb{S}})\). The solutions to the above equations are given as
$$\begin{aligned} V_{A} &= \big( v_{A}(D, x; y) \big)_{x,y \in \mathbb{S}}= \exp ( { I}_{A} { G}D ) { I}_{A}, \\ U_{1,A}^{-}(q) &= \big( u^{-}_{1,A}(q, x; z) \big)_{x, z\in \mathbb{S}} = \big( { I} - { I}_{A} - { I}_{A} ({ G} - q{ I} ) \big)^{-1} ({ I} - { I}_{A}), \\ U_{2,A}^{-}(q) &= \big( u^{-}_{2,A}(q,D, x; z) \big)_{x, z\in \mathbb{S}} = \exp ( { I}_{A} { G}D ) { I}_{A} { U}_{1,A}^{-}(q), \\ U^{-}_{A}(q) &= \big( u^{-}_{A}(q,D, x; z) \big)_{x, z\in \mathbb{S}} = { U}_{1,A}^{-}(q) - { U}_{2,A}^{-}(q), \\ U^{+}(q) &= \big( u^{+}(q, x; z) \big)_{x, z\in \mathbb{S}} = \big( { I}_{B} - ({ I} - { I}_{B}) ({ G } - q{ I}) \big)^{-1} { I}_{B}. \end{aligned}$$
Then we obtain
$$\begin{aligned} { H}(q) &= \big( h(q, x; y) \big)_{x, y \in \mathbb{S}} \\ &= e^{-qD} \bigg( { I} - \sum _{A \in \mathcal{A}} { I}_{A} { U}^{-}_{A}(q) - ({ I} - { I}_{B}) { U}^{+}(q) \bigg)^{-1} \sum _{A \in \mathcal{A}} { I}_{A} { V}_{A}. \end{aligned}$$
Now consider a multi-sided Parisian option with price
$$ u(t, x) = \mathbb{E}_{x}\big[ 1_{\{ \tau _{\mathcal{A}}^{D} \le t \}} f(Y_{t}) \big]. $$
Using the arguments from the single-sided case, we can derive that the Laplace transform \(\tilde{u}(q, x) = \int _{0}^{\infty} e^{-qt} u(t, x) dt\) is given by
$$ \widetilde{ u}(q) = \big( \tilde{u}(q, x) \big)_{x \in \mathbb{S}} = { H}(q) (q{ I} - { G})^{-1}{ f}. $$
1.2 A.2 Mixed barrier and Parisian options
The CTMC approximation can be generalised to derive the joint distribution of Parisian stopping times and first passage times. For example, Dassios and Zhang [24] introduce the so-called MinParisianHit option which is triggered either when the age of an excursion above \(L\) reaches time \(D\) or a barrier \(B > L\) is crossed by the underlying asset price process \(S\). The price of the MinParisianHit option can be approximated as
$$ \text{minPHC}_{i}^{u}(t, x; L, D, B) = e^{-r_{f}t} \mathbb{E}_{x} \big[ 1_{\{ \tau _{L, D}^{+} \wedge \tau _{B}^{+} \le t \}} f(Y_{t}) \big], $$
where \(Y\) is a CTMC with state space \(\mathbb{S}\) and transition rate matrix \({ G}\) to approximate the underlying price process. To price the option, it suffices to substitute \(\tau _{L, D}^{-}\) by \(\tau _{L, D}^{+} \wedge \tau _{B}^{+}\) in the proof of Theorem 2.7 and find the Laplace transform of \(\tau _{L, D}^{+} \wedge \tau _{B}^{+}\) under the model \(Y\). Let \(h_{1}(q, x; y) = \mathbb{E}_{x}[e^{-q\tau _{L, D}^{+} \wedge \tau _{B}^{+}} 1_{\{ \tau _{L, D}^{+} \wedge \tau _{B}^{+} = y \}}]\). Using a conditioning argument, we can show that \(h_{1}(q, x; y)\) satisfies the linear system
$$\begin{aligned} h_{1}(q, x; y) &= 1_{\{ x\ge B, x = y \}} + 1_{\{ x \le L \}} \sum _{z > L} \mathbb{E}_{x}\big[ e^{-q \overline{\tau}_{L}^{+}} 1_{\{ Y_{ \overline{\tau}_{L}^{+}} = z \}} \big] h_{1}(q, z; y) \end{aligned}$$
(A.1)
$$\begin{aligned} & \phantom{=:} + 1_{\{ L < x < B \}} \mathbb{E}_{x}\big[ e^{-q\tau _{B}^{+}} 1_{\{ \tau _{B}^{+} < D, Y_{\tau _{B}^{+}} = y \}} \big] \end{aligned}$$
(A.2)
$$\begin{aligned} & \phantom{=:} + 1_{\{ L < x < B \}} \sum _{z \le L} \mathbb{E}_{x}\big[ e^{-q \overline{\tau}_{L}^{-}} 1_{\{ \overline{\tau}_{L}^{-} < D \le \tau _{B}^{+}, Y_{\overline{\tau}_{L}^{-}} = z \}} \big] h_{1}(q, z; y) \end{aligned}$$
(A.3)
$$\begin{aligned} & \phantom{=:} + 1_{\{ L < x < B \}} e^{-qD} \mathbb{E}_{x}\big[ 1_{\{ \overline{\tau}_{L}^{-} \wedge \tau _{B}^{+} \ge D, Y_{D} = y \}} \big], \end{aligned}$$
(A.4)
where \(\overline{\tau}_{L}^{+} = \inf \{ t \ge 0: Y_{t} > L \}\) and \(\overline{\tau}_{L}^{-} = \inf \{ t \ge 0: Y_{t} \le L \}\) (they are slightly different from \(\tau _{L}^{+}\) and \(\tau _{L}^{-}\) defined in Sect. 2).
We next analyse each term. For (A.1), let \(\overline{u}(q, x; z) = \mathbb{E}_{x}[ e^{-q \overline{\tau}_{L}^{+}} 1_{\{ Y_{\overline{\tau}_{L}^{+}} = z \}} ]\). It satisfies
$$ \textstyle\begin{cases} \mathbb{G} \overline{u}(q, x; z) - q \overline{u}(q, x; z) = 0, \qquad x \in (-\infty , L] \cap \mathbb{S}, \\ \overline{u}(q, x; z) = 1_{\{ x = z \}},\qquad x \in (L, \infty ) \cap \mathbb{S}. \end{cases} $$
For (A.2), let \(v(q, D, x; y) = \mathbb{E}_{x}[ e^{-q\tau _{B}^{+}} 1_{\{ \tau _{B}^{+} < D, Y_{\tau _{B}^{+}} = y \}} ]\). It is the solution to
$$ \textstyle\begin{cases} \displaystyle \frac{\partial v}{\partial D}(q, D, x; y) = \mathbb{G} v(q, D, x; y) - q v(q, D, x; y), \\ \qquad \quad \ \, \qquad \qquad D >0, x \in (-\infty , B) \cap \mathbb{S}, \\ v(q, D, x; y) = 1_{\{ x = y \}},\qquad D>0, x \in [B, \infty ) \cap \mathbb{S}, \\ v(q, 0, x; y) = 0,\qquad x \in \mathbb{S}. \end{cases} $$
\(v(q, D, x; y)\) can be split as \(v_{1}(q, x; y) - v_{2}(q, D, x; y)\) with \(v_{1}(q, x ; y)\) satisfying
$$ \textstyle\begin{cases} \mathbb{G} v_{1}(q,x; y) - qv_{1}(q, x; y) = 0,\qquad x \in (-\infty , B) \cap \mathbb{S}, \\ v_{1}(q, x; y) = 1_{\{ x = y \}},\qquad x \in [B, \infty ) \cap \mathbb{S}, \end{cases} $$
and \(v_{2}(q, D, x; y)\) satisfying
$$ \textstyle\begin{cases} \displaystyle \frac{\partial v_{2}}{\partial D}(q, D, x; y) = \mathbb{G} v_{2}(q, D, x; y) - q v_{2}(q, D, x; y), \\ \qquad \quad \ \, \qquad \qquad D >0, x \in (-\infty , B) \cap \mathbb{S}, \\ v_{2}(q, D, x; y) = 0,\qquad D>0, x \in [B, \infty ) \cap \mathbb{S}, \\ v_{2}(q, 0, x; y) = v_{1}(q, x; y),\qquad x \in \mathbb{S}. \end{cases} $$
For (A.3), let \(u(q, D, x; z) = \mathbb{E}_{x}[ e^{-q\overline{\tau}_{L}^{-}} 1_{\{ \overline{\tau}_{L}^{-} < D \le \tau _{B}^{+}, Y_{\overline{\tau}_{L}^{-}} = z \}} ]\). It solves
$$ \textstyle\begin{cases} \displaystyle \frac{\partial u}{\partial D}(q, D, x; z) = \mathbb{G} u(q, D, x; z) - q u(q, D, x; z), D > 0, x \in (L, B) \cap \mathbb{S}, \\ u(q, D, x; z) = 1_{\{ x = z \}} \mathbb{E}_{x} [ 1_{\{ \tau _{B}^{+} \ge D \}} ],\qquad D>0, x \in (-\infty , L] \cap \mathbb{S}, \\ u(q, D, x; z) = 0,\qquad D > 0, x \in [B, \infty ) \cap \mathbb{S}, \\ u(q, 0, x; z) = 0, \qquad x \in \mathbb{S}. \end{cases} $$
The term \(u(q, D, x; z)\) can be split as \(u_{1}(q, x; z) - u_{2}(q, D, x; z)\) with \(u_{1}(q, x; z)\) satisfying
$$ \textstyle\begin{cases} \mathbb{G} u_{1}(q, x; z) - q u_{1}(q, x; z) = 0, \qquad x \in (L, B) \cap \mathbb{S}, \\ u_{1}(q, x; z) = 1_{\{ x = z \}} \mathbb{E}_{x} [ 1_{\{ \tau _{B}^{+} \ge D \}} ], \qquad x \in (-\infty , L] \cap \mathbb{S}, \\ u_{1}(q, x; z) = 0, \qquad x \in [B, \infty ) \cap \mathbb{S}, \end{cases} $$
and \(u_{2}(q, D, x; z)\) satisfying
$$ \textstyle\begin{cases} \displaystyle \frac{\partial u_{2}}{\partial D}(q, D, x; z) = \mathbb{G} u_{2}(q, D, x; z) - q u_{2}(q, D, x; z), \\ \qquad \quad \ \, \qquad \qquad D > 0, x \in (L, B) \cap \mathbb{S}, \\ u_{2}(q, D, x; z) = 0, \qquad D>0, x \in (-\infty , L] \cap \mathbb{S}, \\ u_{2}(q, D, x; z) = 0, \qquad D > 0, x \in [B, \infty ) \cap \mathbb{S}, \\ u_{2}(q, 0, x; z) = u_{1}(q, x; z),\qquad x \in \mathbb{S}. \end{cases} $$
For (A.4), let \(w(D, x; y) = \mathbb{E}_{x}[ 1_{\{ \overline{\tau}_{L}^{-} \wedge \tau _{B}^{+} \ge D, Y_{D} = y \}} ]\). It solves
$$ \textstyle\begin{cases} \displaystyle \frac{\partial w}{\partial D}(D, x; y) = \mathbb{G} w(D, x; y), \qquad D>0, x \in (L, B) \cap \mathbb{S}, \\ w(D, x; y) = 0, \qquad D>0, x \in \mathbb{S} \backslash (L, B), \\ w(0, x; y) = 1_{\{ x = y \}}, \qquad x \in \mathbb{S}. \end{cases} $$
Let \(\overline{ I}_{L}^{+} = \operatorname{diag}((1_{\{ x > L \}})_{x \in \mathbb{S}})\), \(\overline{ I}_{L}^{-} = \operatorname{diag}((1_{\{ x \le L \}})_{x \in \mathbb{S}})\), \({ I}_{L, B} = \overline{ I}_{L}^{+} { I}_{B}^{-}\) (they are slightly different from \({ I}_{L}^{+}\) and \({ I}_{L}^{-}\) defined in Sect. 2). The solutions to the above equations are given by
$$\begin{aligned} \overline{U}(q) &= \big( \overline{u}(q, x; z) \big)_{x, z \in \mathbb{S}} = ( q\overline{ I}_{L}^{-} - \overline{ I}_{L}^{-}{G} + \overline{ I}_{L}^{+})^{-1} \overline{ I}_{L}^{+}, \\ {V}_{1}(q) &= \big( v_{1}(q, x; y) \big)_{x, y \in \mathbb{S}}= ( q { I}_{B}^{-} - { I}_{B}^{-} { G} + { I}_{B}^{+} )^{-1} { I}_{B}^{+}, \\ {V}_{2}(q) &= \big( v_{2}(q, D, x; y) \big)_{x, y \in \mathbb{S}} = \exp ( { I}_{B}^{-} { G} - q{ I}_{B}^{-} ) { I}_{B}^{-} { V}_{1}(q), \\ {U}_{1}(q) &= \big( u_{1}(q, x; z) \big)_{x, z \in \mathbb{S}} \\ &= ( q{ I}_{L, B} - { I}_{L, B} { G} + { I} - { I}_{L, B} )^{-1} \overline{ I}_{L}^{-} \operatorname{diag} \big(\exp ({I}_{B}^{-} { G} D ) { 1}_{B}^{-}\big) , \\ {U}_{2}(q) &= \big( u_{2}(q, D, x; z) \big)_{x, z \in \mathbb{S}} = \exp ( { I}_{L, B} { G} - q { I}_{L, B} ) { I}_{L, B} { U}_{1}(q), \\ {W} &= \big( w(D, x; y) \big)_{x, y \in \mathbb{S}} = \exp ({I}_{L, B} { G} D ) { I}_{L, B}, \end{aligned}$$
where \({ 1}_{B}^{-} = (1_{\{ x < B \}})_{x \in \mathbb{S}}\). Let \({ H}_{1}(q) = ( h_{1}(q,x; y) )_{x, y \in \mathbb{S}}\), which satisfies
$$\begin{aligned} { H}_{1}(q) &= { I}_{B}^{+} + { I}_{L, B} \big({ V}_{1}(q) - { V}_{2}(q) \big) + { I}_{L, B} \big({ U}_{1}(q) - { U}_{2}(q) \big) { H}_{1}(q) \\ & \phantom{=:} + \overline{ I}_{L}^{-} \overline{ U}(q) { H}_{1}(q) + e^{-qD} { I}_{L, B} { W}. \end{aligned}$$
The solution is given by
$$ { H}_{1}(q) = \big( { I} - { U}(q) \big)^{-1} { V}(q), $$
where
$$\begin{aligned} &{ U}(q) = { I}_{L, B} \big({ U}_{1}(q) - { U}_{2}(q) \big) + \overline{ I}_{L}^{-} \overline{ U}(q), \\ &{ V}(q) = { I}_{B}^{+} + { I}_{L, B} \big({ V}_{1}(q) - { V}_{2}(q) \big) + e^{-qD} { I}_{L, B} { W}. \end{aligned}$$
Consider the Laplace transform
$$ \widetilde{u}_{i}(q, x) = \int _{0}^{\infty} e^{-qt} \text{minPHC}_{i}^{u}(t, x; L, D, B) dt,\qquad \Re (q) > 0. $$
It can be obtained as
$$ \widetilde{ u}_{i}(q) = \big( \widetilde{u}_{i}(q, x) \big)_{x \in \mathbb{S}} = { H}_{1}(q + r_{f}) \big( (q + r_{f}) { I} - { G} \big)^{-1} { f}. $$
1.3 A.3 Pricing Parisian bonds
Recently, Dassios et al. [20] proposed a Parisian type of bonds whose payoff depends on whether the excursion of the interest rate above some level \(L\) exceeds a given duration before maturity, i.e., \(h(R_{\tau}) 1_{\{\tau < T \}}\), where \(T\) is the bond maturity, \(R\) is the short rate process, \(h(\cdot )\) is the payoff function and
$$ \tau = \inf \{ t > 0: U_{t} = D \}, \qquad U_{t} = t - \sup \{ s < t: R_{s} \le L \}. $$
The bond price can be written as
$$ P(T, x) = \mathbb{E}_{x}\big[ e^{-\int _{0}^{\tau} R_{t} dt} f(R_{ \tau}) 1_{\{ \tau < T \}} \big], $$
where \(\mathbb{E}_{x}[\cdot ] = \mathbb{E}[\cdot |R_{0} = x]\). We can calculate its Laplace transform with respect to \(T\) as
$$ \widetilde{P}(q, x) = \int _{0}^{\infty} e^{-qT} P(T, x) dT = \frac{1}{q} \mathbb{E}_{x}\big[ e^{-\int _{0}^{\tau} (q + R_{t}) dt} f(R_{ \tau}) \big]. $$
Suppose that \(R\) is a CTMC with state space \(\mathbb{S}_{R}\) to approximate the original short rate model (e.g. the CIR model considered in Dassios et al. [20]). Let
$$ h(q, x) = \mathbb{E}_{x}\big[ e^{-\int _{0}^{\tau} (q + R_{t}) dt} f(R_{ \tau}) \big] $$
and define \(\overline{\tau}_{L}^{+}\) and \(\overline{\tau}_{L}^{-}\) as in Sect. A.2. Then using a conditioning argument, we obtain
$$\begin{aligned} h(q, x) &= \mathbb{E}_{x}\big[ e^{-\int _{0}^{D} (q + R_{t}) dt} f(R_{D}) 1_{\{ \overline{\tau}_{L}^{-} \ge D \}} \big] 1_{\{ x > L \}} \\ & \phantom{=:} + \sum _{z \le L} \mathbb{E}_{x}\big[ e^{-\int _{0}^{\overline{\tau}_{L}^{-}} (q + R_{t}) dt} 1_{\{ \overline{\tau}_{L}^{-} < D, R_{\overline{\tau}_{L}^{-}} = z \}} \big] 1_{\{ x > L \}} h(q, z) \\ & \phantom{=:} + \sum _{z > L} \mathbb{E}_{x}\big[ e^{-\int _{0}^{\overline{\tau}_{L}^{+}} (q + R_{t}) dt} 1_{\{ R_{\overline{\tau}_{L}^{+}} = z \}} \big] 1_{\{ x \le L \}} h(q, z). \end{aligned}$$
Let \({ G}\) be the generator matrix of \(R\). Consider
$$ v(q, D, x) = \mathbb{E}_{x}\big[ e^{-\int _{0}^{D} (q + R_{t}) dt} f(R_{D}) 1_{\{ \overline{\tau}_{L}^{-} \ge D \}} \big]. $$
It satisfies
$$ \textstyle\begin{cases} \displaystyle \frac{\partial v}{\partial D}(q, D, x) = (\mathbb{G} - q-x) v(q, D, x), \qquad D > 0, x \in (L, \infty ) \cap \mathbb{S}_{R}, \\ v(q, D, x) = 0,\qquad D > 0, x \in (-\infty , L] \cap \mathbb{S}_{R}, \\ v(q, 0, x) = f(x), \qquad x \in \mathbb{S}_{R}. \end{cases} $$
Let \(u^{-}(q, D, x; z) = \mathbb{E}_{x}[ e^{-\int _{0}^{\overline{\tau}_{L}^{-}} (q + R_{t}) dt} 1_{\{ \overline{\tau}_{L}^{-} < D, R_{\overline{\tau}_{L}^{-}} = z \}} ]\). It solves
$$ \textstyle\begin{cases} \displaystyle \frac{\partial u^{-}}{\partial D}(q, D, x; z) = ( \mathbb{G} - q-x) u^{-}(q, D, x; z), \qquad D > 0, x \in (L, \infty ) \cap \mathbb{S}_{R}, \\ u^{-}(q, D, x; z) = 1_{\{ x = z \}}, \qquad D>0, x \in (-\infty , L] \cap \mathbb{S}_{R}, \\ u^{-}(q, 0, x; z) = 0. \end{cases} $$
We can decompose \(u^{-}(q, D, x; z)\) as \(u^{-}(q, D, x; z) = u_{1}^{-}(q, x; z) - u_{2}^{-}(q, D, x; z)\) with them satisfying
$$ \textstyle\begin{cases} (\mathbb{G} - q-x) u_{1}^{-}(q, x; z) = 0, \qquad x \in (L, \infty ) \cap \mathbb{S}_{R}, \\ u_{1}^{-}(q, x; z) = 1_{\{ x = z \}}, \qquad x \in (-\infty , L] \cap \mathbb{S}_{R}, \end{cases} $$
and
$$ \textstyle\begin{cases} \displaystyle \frac{\partial u_{2}^{-}}{\partial D}(q, D, x; z) = ( \mathbb{G} - q-x) u_{2}^{-}(q, D, x; z), \qquad D > 0, x \in (L, \infty ) \cap \mathbb{S}_{R}, \\ u_{2}^{-}(q, D, x; z) = 0, \qquad D>0, x \in (-\infty , L] \cap \mathbb{S}_{R}, \\ u_{2}^{-}(q, 0, x; z) = u_{1}^{-}(q, x; z), \qquad x \in \mathbb{S}_{R}. \end{cases} $$
Let \(u^{+}(q, x; z) = \mathbb{E}_{x}[ e^{-\int _{0}^{\overline{\tau}_{L}^{+}} (q + R_{t}) dt} 1_{\{ R_{\overline{\tau}_{L}^{+}} = z \}} ]\). It satisfies
$$ \textstyle\begin{cases} (\mathbb{G} - q-x) u^{+}(q, x; z) = 0, \qquad x \in (-\infty , L] \cap \mathbb{S}_{R}, \\ u^{+}(q, x; z) = 1_{\{ x = z \}}, \qquad x \in (L, \infty ) \cap \mathbb{S}_{R}. \end{cases} $$
Let \({ f} = (f(x))_{x \in \mathbb{S}_{R}}\) and
$$\begin{aligned} h(q) &= \big( h(q, x) \big)_{x \in \mathbb{S}_{R}},\qquad { v}(q) = \big( v(q, D, x) \big)_{x \in \mathbb{S}_{R}}, \\ U^{-}(q) &= \big( u^{-}(q, D, x; z) \big)_{x, z \in \mathbb{S}_{R}}, \\ U_{1}^{-}(q) &= \big( u_{1}^{-}(q, x; z) \big)_{x, z \in \mathbb{S}_{R}}, \qquad { U}_{2}^{-}(q) = \big( u_{2}^{-}(q, D, x; z) \big)_{x, z \in \mathbb{S}_{R}}, \\ U^{+}(q) &= \big( u^{+}(q, x; z) \big)_{x, z \in \mathbb{S}_{R}}, \qquad { R}_{q} = \text{diag}\big( (q + x)_{x \in \mathbb{S}_{R}} \big). \end{aligned}$$
They can be calculated as
$$\begin{aligned} v(q) &= \exp \big( \overline{ I}^{+}_{L} ({ G} - { R}_{q}) D \big) \overline{ I}^{+}_{L} { f}, \\ U_{1}^{-}(q) &= \big( \overline{{ I}}^{-}_{L} - \overline{ I}^{+}_{L} ({ G} - { R}_{q}) \big)^{-1} \overline{{ I}}^{-}_{L}, \\ U_{2}^{-}(q) &= \exp \big( \overline{ I}^{+}_{L} ({ G} - { R}_{q}) D \big) \overline{ I}^{+}_{L} { U}_{1}^{-}(q), \\ U^{-}(q) &= { U}_{1}^{-}(q) - { U}_{2}^{-}(q), \\ U^{+}(q) &= \big( \overline{ I}^{+}_{L} - \overline{ I}^{-}_{L} ({ G} - { R}_{q}) \big)^{-1} \overline{ I}^{+}_{L}, \\ h(q) &= \big( { I} - \overline{ I}^{+}_{L} { U}^{-}(q) \overline{ I}^{-} - \overline{ I}^{-}_{L} { U}^{+}(q) \overline{ I}^{+}_{L} \big)^{-1} { v}(q). \end{aligned}$$
We then calculate \(\widetilde{P}(q, x)\) by dividing \(h(q,x)\) by \(q\) and obtain the bond price \(P(T, x)\) by Laplace inversion.
1.4 A.4 Regime-switching and stochastic volatility models
Suppose the asset price satisfies \(S_{t} = \zeta (X_{t}, \widetilde{v}_{t})\) for some function \(\zeta (\cdot , \cdot )\) and a regime-switching process \(X\). The regime process \(\widetilde{v}\) has values in \(\mathbb{S}_{v} = \{ v_{1}, v_{2}, \dots , v_{n_{v}} \}\), and in each regime, \(X\) is a general jump-diffusion. We approximate the dynamics of \(X\) in each regime by a CTMC \(\widetilde{X}\) with state space \(\mathbb{S}_{X} = \{ x_{1}, x_{2}, \dots , x_{n} \}\). Hence \(S_{t}\) can be approximated by \(\widetilde{S}_{t} = \zeta (\widetilde{X}_{t}, \widetilde{v}_{t})\). The analysis of single-barrier Parisian stopping times for this type of models can be done similarly as in Sect. 2. Let
$$ \widetilde{ x} = \big((x_{1}, v_{1}), (x_{1}, v_{2}), \dots , (x_{1}, v_{n_{v}}), \dots , (x_{n}, v_{1}), (x_{n}, v_{2}), \dots , (x_{n}, v_{n_{v}}) \big) \in \mathbb{R}^{nn_{v}}. $$
Let \(\widetilde{{ G}}\) be the generator matrix of \((\widetilde{X}, \widetilde{v})\), which can be constructed as
$$\begin{aligned} \widetilde{ G} = { I} \otimes \Lambda + \sum _{i = 1}^{n_{v}} { G}_{i} \otimes { I}_{i} \in \mathbb{R}^{nn_{v} \times nn_{v}}, \end{aligned}$$
(A.5)
where \(I\) is the identity matrix in \(\mathbb{R}^{n \times n}\), \(\Lambda \in \mathbb{R}^{n_{v}\times n_{v}}\) is the generator matrix of \(\widetilde{v}\), ⊗ stands for the Kronecker product, \(G_{i}\) is the generator matrix of \(\widetilde{X}\) in regime \(v_{i}\), and \({ I}_{i}\) is a matrix in \(\mathbb{R}^{n_{v}\times n_{v}}\) with the \((i, i)\)-element being 1 and all other elements being zero. Let
$$ { H}(q) = \big(h(q, x, v; y, u)\big)_{x, y \in \mathbb{S}_{X}, u, v \in \mathbb{S}_{v}} $$
with
$$ h(q, x, v; y, u) = \mathbb{E}_{x, v}\big[ e^{-q \widetilde{\tau}_{L, D}^{-}} 1_{\{ \widetilde{X}_{\widetilde{\tau}_{L, D}^{-}} = y, \widetilde{v}_{ \widetilde{\tau}_{L, D}^{-}} = u \}} \big], $$
where \(\widetilde{\tau}_{L, D}^{-} = \inf \{ t \ge 0: 1_{\{ \widetilde{S}_{t} < L \}} (t - \widetilde{g}^{-}_{L, t}) \ge D \}\) with
$$ \widetilde{g}^{-}_{L, t} = \sup \{ s \le t: \widetilde{S}_{s} \ge L \}. $$
We can solve the Parisian problem in the same way as for 1D CTMCs. Let
$$\begin{aligned} \widetilde{{ V}} &= \exp \big( \widetilde{ I}_{L}^{-} \widetilde{{ G}} D \big) \widetilde{ I}_{L}^{-}, \\ \widetilde{ U}^{+}_{1}(q) &= \widetilde{ I}_{L}^{-} \big(q \widetilde{ I}_{L}^{-} - \widetilde{ I}_{L}^{-} \widetilde{ G} + \widetilde{ I}_{L}^{+} \big)^{-1} \widetilde{ I}_{L}^{+}, \\ \widetilde{ U}^{+}_{2}(q) & = \widetilde{ I}_{L}^{-} \exp \big( \widetilde{ I}_{L}^{-} \widetilde{ G} D \big) \widetilde{ I}_{L}^{-} \widetilde{ U}_{1}(q), \\ \widetilde{ U}^{-}(q) &= \widetilde{ I}_{L}^{+}\big(q\widetilde{ I}_{L}^{+} - \widetilde{ I}_{L}^{+} \widetilde{ G} + \widetilde{ I}_{L}^{-} \big)^{-1} \widetilde{ I}_{L}^{-}, \\ \widetilde{ U}(q) &= \widetilde{ U}_{1}^{+}(q) - \widetilde{ U}_{2}^{+}(q) + \widetilde{ U}^{-}(q), \end{aligned}$$
where \(\widetilde{ I}_{L}^{+} = \operatorname{diag}(1_{\{ \zeta (\widetilde{ x}) \ge L \}})\) and \(\widetilde{ I}_{L}^{-} = \operatorname{diag}(1_{\{ \zeta (\widetilde{ x}) < L \}})\). Then we have
$$ \widetilde{ H}(q) = e^{-qD} \widetilde{ V} + \widetilde{ U}(q) \widetilde{ H}(q). $$
We solve for \(\widetilde{ H}(q)\) and obtain
$$ \widetilde{ H}(q) = e^{-qD} \big(\widetilde{ I} - \widetilde{ U}(q) \big)^{-1} \widetilde{ V}, $$
where \(\widetilde{ I}\) is the identity matrix in \(\mathbb{R}^{nn_{v} \times n n_{v}}\). For option pricing, let
$$ \widetilde{{ u}}(q) = \big( \widetilde{u}(q, x, v) \big)_{x\in \mathbb{S}_{X}, v \in \mathbb{S}_{v}}, $$
where \(\widetilde{u}(q, x, v) \) is the Laplace transform of the option price given by
$$ \widetilde{u}(q, x, v) = \int _{0}^{\infty} e^{-qt} \mathbb{E}_{x, v} \big[ f(\widetilde{X}_{t}, \widetilde{v}_{t}) 1_{\{ \widetilde{\tau}_{L, D}^{-} \le t \}} \big] dt. $$
We obtain \(\widetilde{{ u}}(q)\) as
$$ \widetilde{{ u}}(q) = e^{-qD} \big(\widetilde{ I} - \widetilde{ U}(q) \big)^{-1} \widetilde{ V} (q \widetilde{ I} - \widetilde{ G} )^{-1} \widetilde{ f}, $$
where \(\widetilde{{ f}} = { {\mathbf{1}}}_{n_{v}} \otimes (f(x_{1}), f(x_{2}), \dots , f(x_{n}))\), and \({ {\mathbf{1}}}_{n_{v}}\) is an all-one vector in \(\mathbb{R}^{n_{v}}\).
Remark A.1
Cui et al. [13] show that general stochastic volatility models can be approximated by a regime-switching CTMC. Consider
$$ \textstyle\begin{cases} dS_{t}=\omega (S_{t}, v_{t}) d t+m (v_{t} ) \Gamma (S_{t}) d W_{t}^{(1)}, \\ d v_{t}=\mu (v_{t}) d t+\sigma (v_{t}) d W_{t}^{(2)}, \end{cases} $$
where \([W^{(1)}, W^{(2)} ]_{t} = \rho t\) with \(\rho \in [-1, 1]\). As in Cui et al. [13], consider
$$ X_{t} = g(S_{t}) - \rho f(v_{t}) $$
with \(g(x):=\int _{0}^{x} \frac{1}{\Gamma (u)} d u\) and \(f(x):=\int _{0}^{x} \frac{m(u)}{\sigma (u)} d u\). It follows that
$$ dX_{t} =\theta (X_{t}, v_{t}) d t+\sqrt{1-\rho ^{2}} m (v_{t} ) d W_{t}^{*}, $$
where \(W^{\ast}\) is a standard Brownian motion independent of \(W^{(2)}\) and
$$ \theta (x, v) = \frac{\omega (\zeta (x, v), v)}{\Gamma (\zeta (x, v))}- \frac{\Gamma ^{\prime}(\zeta (x, v))}{2} m^{2} (v )-\rho h (v ) $$
with
$$\begin{aligned} \zeta (x, v)& := g^{-1}\big(x + \rho f(v)\big), \\ h(x)&: =\mu (x) \frac{m(x)}{\sigma (x)}+\frac{1}{2}\big(\sigma (x) m^{ \prime}(x)-\sigma ^{\prime}(x) m(x)\big). \end{aligned}$$
Then we can use a two-layer approximation for \((X, v)\). First, construct a CTMC \(\widetilde{v}\) with state space \(\mathbb{S}_{v} = \{ v_{1}, \dots , v_{n_{v}} \}\) and generator matrix \(\Lambda \) in \(\mathbb{R}^{n_{v} \times n_{v}}\) to approximate \(v\). Second, for each \(v_{\ell }\in \mathbb{S}_{v}\), construct a CTMC with state space \(\mathbb{S}_{X} = \{ x_{1}, \dots , x_{n} \}\) and generator matrix \(\mathcal{G}_{v}\) to approximate the dynamics of \(X\) conditionally on \(\widetilde{v}=v_{\ell}\). Then \((X, v)\) is approximated by a regime-switching CTMC \((\widetilde{X}, \widetilde{v})\), where \(\widetilde{X}\) transitions on \(\mathbb{S}_{X}\) following \(\mathcal{G}_{v}\) when \(\widetilde{v} = v\) for each \(v \in \mathbb{S}_{v}\), and \(\widetilde{v}\) evolves according to its transition rate matrix \(\Lambda \).
Remark A.2
We can significantly reduce the complexity of our algorithm when \(X\) is a regime-switching diffusion. In this case, it is approximated by a regime-switching birth-and-death process \(\widetilde{X}\). As \(\widetilde{X}\) can only move to the neighbouring states at each transition, we have
$$\begin{aligned} h(q, x, v; y, u) &= \mathbb{E}_{x, v}\big[ e^{-q \widetilde{\tau}_{L, D}^{-}} 1_{\{ \widetilde{X}_{\widetilde{\tau}_{L, D}^{-}} = y, \widetilde{v}_{ \widetilde{\tau}_{L, D}^{-}} = u \}} \big] \\ &= 1_{\{ x < L \}} e^{-qD} \widetilde{v}(D, x, v; y, u) \\ & \phantom{=:} + \sum _{w \in \mathbb{S}_{v}} 1_{\{ x < L \}} \widetilde{u}^{+}(q, D, x, v; w) h(q, L^{+}, w; y, u) \\ & \phantom{=:} + \sum _{w \in \mathbb{S}_{v}} 1_{\{ x \ge L \}} \widetilde{u}^{-}(q, x, v; w) h(q, L^{-}, w; y, u), \end{aligned}$$
where
$$\begin{aligned} &\widetilde{v}(D, x, v; y, u) = \mathbb{E}_{x, v}\big[ 1_{\{ \widetilde{T}_{L}^{+} \ge D, \widetilde{X}_{D} = y, \widetilde{v}_{D} = u \}} \big], \\ &\widetilde{u}^{+}(q, D, x, v; w) = \mathbb{E}_{x, v} \big[ e^{-q \widetilde{T}_{L}^{+}} 1_{\{ \widetilde{v}_{\widetilde{T}_{L}^{+}} = w, \widetilde{T}_{L}^{+} < D \}} \big], \\ &\widetilde{u}^{-}(q, x, v; w) = \mathbb{E}_{x, v} \big[ e^{-q \widetilde{T}_{L}^{-}} 1_{\{ \widetilde{v}_{\widetilde{T}_{L}^{-}} = w \}} \big]. \end{aligned}$$
Setting \(x = L^{+},L^{-}\) yields
$$\begin{aligned} h(q, L^{+}, v; y, u) &= \sum _{w \in \mathbb{S}_{v}} \widetilde{u}^{-}(q, L^{+}, v; w) h(q, L^{-}, w; y, u), \\ h(q, L^{-}, v; y, u) &= e^{-qD} \widetilde{v}(D, L^{-}, v; y, u) \\ & \phantom{=:} + \sum _{w \in \mathbb{S}_{v}} \widetilde{u}^{+}(q, D, L^{-}, v; w) h(q, L^{+}, w; y, u). \end{aligned}$$
As each \({ G}_{i}\), \(1 \le i \le n_{v}\), is a tridiagonal matrix, \(\widetilde{ G}\) defined in (A.5) is a block tridiagonal matrix. We can solve a block tridiagonal linear system associated with \(\widetilde{ G}\) using the block LU decomposition (see Quarteroni et al. [46, Sect. 3.8.3]), and the cost is only \(\mathcal{O}(n_{v}^{3} n)\) which is linear in \(n\). Using the analysis in Remark 2.6, we can conclude that the cost of computing \(\widetilde{{ u}}(q)\) is \(\mathcal{O}(m n_{v}^{3} n)\) if the approximation (2.6) is applied.
Appendix B: Proofs
Proof of Proposition 2.3
We derive the equation for \(v_{n}(D, x; y)\); the others can be obtained in a similar manner. Let \(\mathbb{T}_{\delta }= \{ i\delta : i = 0, 1, 2, \dots \}\) and correspondingly \(\tau _{L}^{\delta ,+} = \inf \{ t \in \mathbb{T}_{\delta}: Y_{t} \ge L \}\). As \(Y\) is a piecewise constant process, we have \(\tau _{L}^{\delta ,+} \downarrow \tau _{L}^{+}\) as \(\delta \to 0\). Using the monotone convergence theorem, we have
$$ v_{n,\delta}(D, x; y) := \mathbb{E}_{x}\big[ 1_{\{ \tau _{L}^{\delta ,+} \ge D \}} 1_{\{ Y_{D} = y \}} \big] \longrightarrow v_{n}(D, x; y). $$
Denote the transition probability of \(Y\) by \(p_{n}(\delta , x, z)\). We have that
$$ p_{n}(\delta , x, z) = { G}(x, z) \delta + o(\delta ) $$
for \(z \ne x\) and \(p_{n}(\delta , x, x) = 1 + { G}(x, x) \delta \). Then for \(x < L\),
$$\begin{aligned} v_{n,\delta}(D, x; y) &= \sum _{z \in \mathbb{S}_{n}} p_{n}(\delta , x, z) v_{n,\delta}(D-\delta , z; y) \\ &= \big(1 + G\delta + o(\delta ) \big) v_{n,\delta}(D-\delta , x; y). \end{aligned}$$
Therefore,
$$ \frac{v_{n,\delta}(D, x; y) - v_{n,\delta}(D-\delta , x; y)}{\delta} = Gv_{n,\delta}(D-\delta , x; y) + o(1). $$
Taking \(\delta \) to 0 shows the ODE. The boundary and initial conditions are obvious. □
Proof of Proposition 2.4
It is easy to see that the solution matrix for \({ V}\) is unique. Suppose there are two solutions \({ V'}\) and \({ V''}\). Their difference \(V'- V''\) satisfies a homogeneous linear ODE system with zero as the initial and boundary conditions. Thus it must equal the zero matrix, and hence \({V}'= {V}''\). Similarly, we obtain the uniqueness of the solution matrix for \({ U}^{+}_{2}(q)\). The solution matrix for \({ U}^{+}_{1}(q)\) is also unique. Note that for \(D>0\) and \(\Re (q)>0\), we have
$$ (q{ I}_{L}^{-} - { I}_{L}^{-} { G} + { I}_{L}^{+} ){ U}^{+}_{1}(q) = { I}_{L}^{+}. $$
Let \({A}=(a_{ij})=q{ I}_{L}^{-} - { I}_{L}^{-} { G} + { I}_{L}^{+}\). Observe that
$$ \Re (a_{ii}) > \sum _{j \ne i} |a_{ij} |, \qquad i = 0, 1, \ldots , n. $$
By Gershgorin’s circle theorem (see Geršgorin [30]), all the eigenvalues of \(A\) have a strictly positive real part, which implies the invertibility of \(A\). Similarly, we can show the uniqueness of the solution matrix for \({ U}^{-}(q)\). □
Proof of Theorem 3.3
The assumptions imply that for any \(\varepsilon > 0\), there exists \(M > 0\) such that for any \(n\), we have
$$\begin{aligned} \big| \mathbb{E}\big[ |f(Y_{t}^{n}) |1_{\{ |f(Y_{t}^{n})| \ge M \}} \big] \big| < \varepsilon , \\ \big| \mathbb{E}\big[ |f(X_{t}) | 1_{\{ |f(X_{t})| \ge M \}}\big] \big| < \varepsilon . \end{aligned}$$
It follows that
$$\begin{aligned} & \big|\mathbb{E}\big[ 1_{\{ \tau _{L, D}^{(n), -} \le t \}} f(Y^{(n)}_{t}) \big] - \mathbb{E}\big[ 1_{\{ \tau _{L, D}^{-} \le t \}} f(X_{t}) \big] \big| \\ &\le \big|\mathbb{E}\big[ 1_{\{ \tau _{L, D}^{(n), -} \le t \}} f_{M}(Y^{(n)}_{t}) \big] - \mathbb{E}\big[ 1_{\{ \tau _{L, D}^{-} \le t \}} f_{M}(X_{t}) \big] \big| + 2\varepsilon , \end{aligned}$$
where
$$ f_{M}(x) = M 1_{\{ f(x) > M \}} + f(x) 1_{\{ |f(x)| \le M \}} - M 1_{ \{ f(x) < - M \}}. $$
Noting that \(\varepsilon \) can be taken arbitrarily small, it suffices to show that
$$ \mathbb{E}\big[ 1_{\{ \tau _{L, D}^{(n), -} \le t \}} f_{M}(Y^{(n)}_{t}) \big] \longrightarrow \mathbb{E}\big[ 1_{\{ \tau _{L, D}^{-} \le t \}} f_{M}(X_{t}) \big] \qquad \text{as } n \to \infty . $$
But \(Y^{(n)} \Rightarrow X\) implies that if \(g: D(\mathbb{R}) \to \mathbb{R}\) is bounded and continuous on some subset \(C\) of \(D(\mathbb{R})\) such that \(\mathbb{P}[X \in C] = 1\), then
$$ \mathbb{E} [g(Y^{(n)}) ] \longrightarrow \mathbb{E} [g(X) ] \qquad \text{as } n \to \infty . $$
With the assumptions that
$$\begin{aligned} \mathbb{P}[X_{t} \in \mathcal{D}] &=\mathbb{P}[\tau _{L, D}^{-} = t] = 0, \\ \mathbb{P}[X \in V] &= \mathbb{P}[X \in W] = \mathbb{P}[X \in U^{+}] = \mathbb{P}[X \in U^{-}] = 1, \end{aligned}$$
it suffices to establish the continuity of \(\tau _{L, D}^{-}(\omega )\) on \(V \cap W \cap U^{+} \cap U^{-}\).
For \(\omega \in V \cap W \cap U^{+} \cap U^{-}\), suppose \(\omega ^{(n)} \to \omega \) as \(n \to \infty \) on \(D(\mathbb{R})\). Note that for \(\omega \in U^{+} \cap U^{-}\), we have \(\sigma _{k+1}^{-}(\omega )>\sigma _{k}^{+}(\omega )>\sigma _{k}^{-}( \omega )\) for \(k\ge 1\).
– Suppose \(\tau _{L, D}^{-}(\omega ) < s\). There exists \(k \ge 1\) such that \(\sigma _{k}^{+}(\omega ) \wedge s - \sigma ^{-}_{k}(\omega ) > D\). Let \(J(\omega )=\{t: \omega (t)\ne \omega (t-)\}\). Since \(\mathbb{R}_{+}\backslash J(\omega )\) is dense, we can find an \(\varepsilon > 0\) that is small enough such that \(\sigma _{k}^{+}(\omega ) \wedge s - \sigma ^{-}_{k}(\omega ) - 2 \varepsilon > D\), \(\omega \) is continuous at \(\sigma _{k}^{+}(\omega ) \wedge s - \varepsilon \) and \(\sigma ^{-}_{k}(\omega )+\varepsilon \), and
$$ \sup \{ \omega _{u}: \sigma ^{-}_{k}(\omega ) + \varepsilon \le u \le \sigma _{k}^{+}(\omega ) \wedge s - \varepsilon \} < L. $$
By Jacod and Shiryaev [33, Proposition 2.4] which shows the continuity of the supremum process, \(\sup \{ \omega _{u}^{(n)}: \sigma ^{-}_{k}(\omega ) + \varepsilon \le u \le \sigma _{k}^{+}(\omega ) \wedge s - \varepsilon \} < L\). Since \(\sigma _{k}^{+}(\omega ) \wedge s - \sigma ^{-}_{k}(\omega ) - 2 \varepsilon > D\), we conclude that \(\tau _{L, D}^{ -}(\omega ^{(n)}) < s\) for sufficiently large \(n\). Therefore, we have
$$ \limsup _{n \to \infty} \tau _{L, D}^{ -}(\omega ^{(n)}) \le \tau _{L, D}^{-}(\omega ). $$
– Suppose \(\tau _{L, D}^{-}(\omega ) > s\). Since \(\omega \in V\), there is \(\overline{k} \ge 1\) such that \(\sigma _{\overline{k}}^{+}(\omega ) \ge s \) and \(\sigma _{\overline{k}}^{-}(\omega ) < s\). As \(\omega \in W\), \(\max \{ \sigma _{k}^{+}(\omega ) \wedge s - \sigma _{k}^{-}(\omega ) : 1\le k\le \overline{k}\} < D\). Due to the denseness of \(\mathbb{R}_{+}\backslash J(\omega )\), we can find a small enough \(\varepsilon >0\) such that \(\sigma _{k}^{+}(\omega ) \wedge s - \sigma _{k}^{-}(\omega ) + 2\varepsilon < D\), \(\omega \) is continuous at \(\sigma _{k}^{+} \wedge s + \varepsilon \) and \(\sigma _{k}^{-}-\varepsilon \) for any \(1 \le k \le \overline{k}\), and \(\inf \{ \omega _{u}: \sigma _{k}^{+}(\omega ) + \varepsilon \le u \le \sigma _{k+1}^{-}(\omega ) -\varepsilon \} > L\) for any \(1 \le k < \overline{k}\). In the same spirit as [33, Proposition 2.4], we can show the continuity of the infimum process. For sufficiently large \(n\), we thus have \(\inf \{ \omega _{u}^{(n)}: \sigma _{k}^{+}(\omega ) + \varepsilon \le u \le \sigma _{k+1}^{-}(\omega ) - \varepsilon \} > L\) for any \(1 \le k < \overline{k}\). It follows that the age of the excursion below \(L\) of \(\omega ^{(n)}\) up to time \(s\) cannot exceed \(\max \{ \sigma _{k}^{+}(\omega ) \wedge s - \sigma _{k}^{-}(\omega ) + 2\varepsilon : 1\le k\le \overline{k}\} < D\). This implies \(\tau _{L, D}^{-}(\omega ^{(n)}) > s\) for sufficiently large \(n\). Therefore,
$$ \liminf _{n \to \infty} \tau _{L, D}^{ -}(\omega ^{(n)}) \ge \tau _{L, D}^{-}(\omega ). $$
Combining the arguments above, we obtain the continuity of the Parisian stopping time \(\tau _{L,D}^{-}(\omega )\) on \(V \cap W \cap U^{+} \cap U^{-}\). This concludes the proof. □
Proof of Proposition 3.4
As
$$ Y_{t}^{(n)} - Y_{0}^{(n)} - \int _{0}^{t} \mathbb{G}g(Y_{s}^{(n)}) ds, \qquad t \geq 0, $$
is a martingale with \(g(y) = y\), we have
$$ \mathbb{E}_{x}[Y_{t}^{(n)}] = x + \mathbb{E}_{x} \bigg[ \int _{0}^{t} \mathbb{G}g(Y_{s}^{(n)}) ds \bigg]. $$
We next bound \(\mathbb{G}g(Y_{s}^{(n)})\) by writing
$$\begin{aligned} \mathbb{G}g(Y_{s}^{(n)}) &= \sum _{y \in \mathbb{S}_{n}} { G}(Y_{s}^{(n)}, y) y = \sum _{y \in \mathbb{S}_{n}} { G}(Y_{s}^{(n)}, y) (y - Y_{s}^{n}) \\ &= \widetilde{\mu}(Y_{s}^{(n)}) + \sum _{y \in \mathbb{S}_{n} \backslash \{ Y_{s}^{(n)} \}} (y - Y_{s}^{(n)}) \int _{I_{y}- Y_{s}^{(n)}} \nu (Y_{s}^{(n)}, dz) \\ &= \mu (Y_{s}^{(n)}) - \sum _{y \in \mathbb{S}_{n} \backslash \{ Y_{s}^{(n)} \}} (y - Y_{s}^{(n)}) \int _{I_{y} - Y_{s}^{(n)}} 1_{\{ |z| \le 1 \}} \nu (Y_{s}^{(n)}, dz) \\ & \phantom{=:} + \sum _{y \in \mathbb{S}_{n}\backslash \{ Y_{s}^{(n)} \}} (y - Y_{s}^{(n)}) \int _{I_{y}- Y_{s}^{(n)}} \nu (Y_{s}^{(n)}, dz) \\ &= \mu (Y_{s}^{(n)}) + \sum _{y \in \mathbb{S}_{n} \backslash \{ Y_{s}^{(n)} \}} (y - Y_{s}^{(n)}) \int _{I_{y} - Y_{s}^{(n)}} 1_{\{ |z| > 1 \}} \nu (Y_{s}^{(n)}, dz) \\ &\le c_{1} Y_{s}^{(n)} + c_{2}. \end{aligned}$$
It follows that
$$ \mathbb{E}_{x}[Y_{t}^{(n)}] \le x + c_{1} \int _{0}^{t} \mathbb{E}_{x}[Y_{s}^{(n)}] ds + c_{2}t. $$
Using Gronwall’s inequality, we obtain
$$ \mathbb{E}_{x}[Y_{t}^{(n)}] \le x + c_{2}t + c_{1}\int _{0}^{t} (x+c_{2}s)e^{c_{1}(t-s)}ds, $$
which shows the claim. □
Proof of Lemma 4.4
(1) This result can be found in Zhang and Li [48, Proposition 1 and Corollary 1].
(2) We first apply a Liouville transform to the eigenvalue problem. Let
$$\begin{aligned} &y = \int _{\ell}^{x} \frac{1}{\sigma (z)} dz, \qquad B = \int _{ \ell}^{b} \frac{1}{\sigma (z)} dz, \\ & Q(y) = \frac{ ( (m (x(y) )/s (x(y) ) )^{1/4} )''}{ (m (x(y) )/s (x(y) ) )^{1/4}}, \qquad \mu _{k}^{+}(B) = \lambda ^{+}_{k}(b). \end{aligned}$$
Then the eigenvalue problem is cast in the Liouville normal form as (see Fulton and Pruess [29, Eqs. (2.1)–(2.5)])
$$ \textstyle\begin{cases} -\partial _{yy} \phi _{k}^{+}(y, B) + Q(y) \phi _{k}^{+}(y, B) = - \mu _{k}^{+}(B) \phi _{k}^{+}(y, B), \qquad y \in (0, B), \\ \phi _{k}^{+}( 0, B) = \phi _{k}^{+}(B, B) = 0. \end{cases} $$
As shown in [29, Eq. (2.13)],
$$ \| \varphi _{k}^{+}( \cdot , b) \|_{2}^{2} = \int _{\ell}^{b} \varphi _{k}^{+}( x, b)^{2} m(x) dx = \int _{0}^{B} \phi _{k}^{+}(y, B)^{2} dy = \| \phi _{k}^{+}(\cdot , B) \|_{2}^{2}. $$
Hence the normalised eigenfunction can be recovered as
$$ \varphi _{k}^{+}( x, b) = \bigg(\frac{s (x(y) )}{m (x(y) )}\bigg)^{1/4} \frac{\phi _{k}^{+}(y, B)}{ \| \phi _{k}^{+}(\cdot , B) \|}. $$
We study the sensitivities of \(\phi _{k}^{+}( y, B)\) and \(\mu _{k}^{+}( B)\) with respect to \(y\) and \(B\) from which we can obtain the sensitivities of \(\varphi _{k}^{+}( x, b)\) and \(\lambda _{k}^{+}( b)\) with respect to \(x\) and \(b\). Let
$$ s_{k}(B) = \sqrt{\mu _{k}^{+}( B)}. $$
By [29, Eq. (3.7)], we have
$$\begin{aligned} \phi _{k}^{+}( y, B) &= \frac{\sin (s_{k}(B) y) }{s_{k}(B) } \\ & \phantom{=:} + \frac{1}{s_{k}(B) } \int _{0}^{y} \sin \big(s_{k}(B) (y - z)\big) Q(z) \phi _{k}^{+}( z, B) dz. \end{aligned}$$
(B.1)
By Gronwall’s inequality, \(\phi _{k}^{+}( y, B) = \mathcal{O}(1/k)\). Differentiating on both sides yields
$$\begin{aligned} \partial _{y} \phi _{k}^{+}( y, B) &= \sin \big(s_{k}(B) y\big)+\int _{0}^{y} \cos \big(s_{k}(B) (y - z)\big) Q(z) \phi _{k}^{+}( z, B) dz, \\ \end{aligned}$$
(B.2)
$$\begin{aligned} \partial _{y}^{2} \phi _{k}^{+}( y, B) &= s_{k}(B) \cos \big(s_{k}(B) y \big) + Q(y) \phi _{k}^{+}( y, B) \\ & \phantom{=:} -s_{k}(B) \int _{0}^{y} \sin \big(s_{k}(B) (y - z)\big) Q(z) \phi _{k}^{+}( z, B) dz, \\ \end{aligned}$$
(B.3)
$$\begin{aligned} \partial _{y}^{3} \phi _{k}^{+}( y, B) &= -s_{k}^{2}(B) \sin \big(s_{k}(B) y\big) + Q'(y) \phi _{k}^{+}( y, B) + Q(y) \partial _{y} \phi _{k}^{+}( y, B) \\ & \phantom{=:} -s_{k}^{2}(B) \int _{0}^{y} \cos \big(s_{k}(B) (y - z)\big) Q(z) \phi _{k}^{+}( z, B) dz. \end{aligned}$$
(B.4)
Thus we get
$$ | \partial _{y} \phi _{k}^{+}( y, B) | = \mathcal{O}(1), \qquad | \partial _{y}^{2} \phi _{k}^{+}( y, B) | = \mathcal{O}(k), \qquad | \partial _{y}^{3} \phi _{k}^{+}( y, B) | = \mathcal{O}(k^{2}). $$
By Kong and Zettl [35, Theorem 3.2], we have
$$ s_{k}'(B) = - \frac{ (\partial _{y} \phi _{k}^{+}(B, B) )^{2}}{ \| \phi _{k}^{+}(\cdot , B) \|^{2} }. $$
(B.5)
As in [29, Eq. (6.11)], we have \(1/ \| \phi _{k}^{+}(\cdot , B) \| = O(k)\). Moreover, (B.2) implies that \(\partial _{y} \phi _{k}^{+}( B, B) = \mathcal{O}(1)\). Therefore \(s_{k}'(B) = \mathcal{O}(k^{2})\).
Differentiating on both sides of (B.1), we obtain
$$\begin{aligned} \partial _{B} \phi ^{+}_{k}( y, B) &= \frac{y \cos (s_{k}(B) y) s_{k}^{\prime}(B) s_{k}(B)-\sin (s_{k}(B) y) s_{k}^{\prime}(B)}{s_{k}(B)^{2}} \\ & \phantom{=:} +\int _{0}^{y} \frac{(x-z) \cos (s_{k}(B)(y-z)) s_{k}^{\prime}(B) s_{k}(B)}{s_{k}(B)^{2}} Q(z) \phi ^{+}_{k}( z, B) d z \\ & \phantom{=:} -\int _{0}^{y} \frac{\sin (s_{k}(B)(y-z)) s_{k}^{\prime}(B)}{s_{k}(B)^{2}} Q(z) \phi ^{+}_{k}( z, B) d z \\ & \phantom{=:} +\int _{0}^{y} \sin \big(s_{k}(B)(y-z)\big) Q(z) \partial _{B} \phi ^{+}_{k}( z, B) d z. \end{aligned}$$
Thus there exist constants \(c_{2}, c_{3} > 0\) such that
$$ |\partial _{B} \phi ^{+}_{k}( y, B) | \le c_{2} k + c_{3} \int _{0}^{y} |\partial _{B} \phi ^{+}_{k}( z, B) | dz. $$
Applying Gronwall’s inequality again shows that
$$ |\partial _{B} \phi ^{+}_{k}( y, B) | \le c_{2}k \exp ( c_{3} B ) = \mathcal{O}(k). $$
Differentiating with respect to \(B\) on both sides of (B.2)–(B.4) and applying the above estimate, we obtain
$$\begin{aligned} | \partial _{B} \partial _{y} \phi _{k}^{+}( y, B) | &= \mathcal{O}(k^{2}), \\ | \partial _{B} \partial _{y}^{2} \phi _{k}^{+}( y, B) | &= \mathcal{O}(k^{3}), \\ | \partial _{B} \partial _{y}^{3} \phi _{k}^{+}( y, B) | &= \mathcal{O}(k^{4}). \end{aligned}$$
We can also derive that
$$ \partial _{B} \| \phi _{k}^{+}( \cdot , B) \|^{2} = \big(\phi _{k}^{+}( \cdot , B)\big)^{2} + 2\int _{0}^{B} \partial _{B} \phi _{k}^{+}( z, B) \phi _{k}^{+}( z, B) dz = O(1/k). $$
Differentiating on both sides of (B.5), we obtain
$$ s_{k}''(B) = \frac{\partial _{B} \| \phi _{k}^{+}(\cdot , B) \|^{2} (\partial _{y} \phi _{k}^{+}(B, B))^{2} - \| \phi _{k}^{+}(\cdot , B) \|^{2} \partial _{B} (\partial _{y} \phi _{k}^{+}(B, B))^{2}}{ \| \phi _{k}^{+}(\cdot , B) \|^{4}} = \mathcal{O}(k^{4}). $$
Subsequently, we obtain
$$\begin{aligned} \partial _{B} \mu _{k}( B) &= 2s_{k}(B) s_{k}'(B) = \mathcal{O}(k^{3}), \\ \partial _{BB} \mu _{k}( B) &= 2s_{k}(B) s_{k}''(B) + 2 \big(s_{k}'(B) \big)^{2} = \mathcal{O}(k^{5}). \end{aligned}$$
After some calculations based on the relationship between \((\lambda _{k}^{+}( b) , \varphi _{k}^{+} ( x, b))\) and \((\mu _{k}^{+}( B), \phi _{k}^{+}(y, B))\), we obtain
$$\begin{aligned} \partial _{b} \lambda ^{+}_{k}( b) &= \mathcal{O}(k^{3}), \qquad \partial _{bb} \lambda ^{+}_{k}( b) = \mathcal{O}(k^{5}), \\ | \partial _{b} \varphi _{k}^{+} ( x, b) | &= \mathcal{O}(k^{2}), \qquad | \partial _{b} \partial _{x} \varphi _{k}^{+} ( x, b) | = \mathcal{O}(k^{3}), \\ | \partial _{b} \partial _{x}^{2} \varphi _{k}^{+} ( x, b) | & = O(k^{4}), \qquad | \partial _{b} \partial _{x}^{3} \varphi _{k}^{+} ( x, b) | = O(k^{5}). \end{aligned}$$
By Zhang and Li [48, Proposition 2], we have \(\lambda _{n, k}^{+} = \lambda _{k}( L^{+}) + \mathcal{O}(k^{4} \delta _{n}^{2})\) and hence
$$ \lambda _{n, k}^{+} = \lambda _{k}( L) + k^{3}\mathcal{O}(L^{+} - L) + \mathcal{O}(k^{4} \delta _{n}^{2}). $$
For (4.5), by [48, Proposition 3], we have
$$ \varphi _{n, k}^{+}( x) = \varphi _{k}^{+}( x, L^{+}) + \mathcal{O}(k^{4} \delta _{n}^{2}) = \varphi _{k}^{+}( x, L) + k \mathcal{O}(L^{+} - L) + \mathcal{O}(k^{4} \delta _{n}^{2}). $$
The same proposition also shows that \(\nabla ^{+} \varphi ^{+}_{n, k}( L^{-}) = \nabla ^{+} \varphi ^{+}_{k}( L^{-}; L^{+}) + \mathcal{O}(k^{6} \delta _{n}^{2})\). Thus we obtain
$$\begin{aligned} \varphi ^{+}_{n, k}( L^{-}) &= \varphi ^{+}_{k}( L^{-}, L^{+}) + \mathcal{O}(k^{6} \delta _{n}^{3}) \\ &= \varphi ^{+}_{k}( L^{-}, L^{+}) - \varphi ^{+}_{k}( L, L^{+}) + \varphi ^{+}_{k}( L, L^{+})- \varphi ^{+}_{k}( L^{+}, L^{+}) + \mathcal{O}(k^{6} \delta _{n}^{3}) \\ &= \partial _{x} \varphi ^{+}_{k}( L, L^{+}) (L^{-} - L) + \frac{1}{2} \partial _{xx} \varphi ^{+}_{k}( L, L^{+}) (L^{-} - L)^{2} \\ & \phantom{=:} - \partial _{x} \varphi ^{+}_{k}( L, L^{+}) (L^{+} - L) - \frac{1}{2} \partial _{xx} \varphi ^{+}_{k}( L, L^{+}) (L^{+} - L)^{2} + \mathcal{O}(k^{6} \delta _{n}^{3}) \\ &= -\partial _{x} \varphi ^{+}_{k}( L, L^{+}) \delta ^{+} L^{-} + \frac{1}{2} \partial _{xx} \varphi ^{+}_{k}( L, L^{+}) (\delta ^{+} L^{-})^{2} + \mathcal{O}(k^{6} \delta _{n}^{3}) \\ & \phantom{=:} + \partial _{xx} \varphi ^{+}_{k}( L, L^{+}) \delta ^{+} L^{-}(L^{+} - L) + \mathcal{O}(k^{6} \delta _{n}^{3}) \\ &= -\partial _{x} \varphi ^{+}_{k}( L, L) \delta ^{+} L^{-} + \frac{1}{2} \partial _{xx} \varphi ^{+}_{k}( L, L) (\delta ^{+} L^{-})^{2} + k^{4} \delta ^{+} L^{-} \mathcal{O}(L^{+} - L) \\ & \phantom{=:} + \mathcal{O}(k^{6} \delta _{n}^{3}). \end{aligned}$$
(B.6)
The rest of the results in the second part follows from [48, Lemmas 3, 5 and 6].
(3) These results can be proved using arguments similar to those for Lemma 4.6 and Proposition 4.7. □
Proof of Lemma 4.6
Let \(\psi ^{+}(\cdot )\) and \(\psi ^{-}(\cdot )\) be two independent solutions to the Sturm–Liouville problem \(\mu (x) \psi '(x) + \frac{1}{2} \sigma ^{2}(x) \psi ''(x) - q\psi (x) = 0\), where \(\psi ^{+}(\cdot )\) is strictly increasing and \(\psi ^{-}(\cdot )\) is strictly decreasing and they are \(C^{4}\) by Gilbarg and Trudinger [31, Theorem 6.19]. Then we can construct \(u_{1}^{+}(q, x, b)\) as
$$ u_{1}^{+}(q, x, b) = \frac{\psi ^{+}(\ell ) \psi ^{-}(x) - \psi ^{+}(x) \psi ^{-}(\ell )}{\psi ^{+}(\ell ) \psi ^{-}(b) - \psi ^{+}(b) \psi ^{-}(\ell )}, $$
from which it is easy to see that \(b\mapsto u_{1}^{+}(q, x, b)\in C^{3}\).
Similarly to Li and Zhang [41, Theorem 3.22], we can show that
$$ u_{1, n}^{+}(q, x; L^{+}) = u_{1}^{+}(q, x, L^{+}) + \mathcal{O}( \delta _{n}^{2}). $$
So the first equation in the lemma follows by the smoothness of \(b \mapsto u_{1}^{+}(q, x; b)\). For the second, let \(e_{n}(x) = u_{1, n}^{+}(q, x; L^{+}) - u^{+}_{1}(q, x; L^{+})\). For \(x \in \mathbb{S}_{n} \cap (\ell , L^{+})\), we have
$$\begin{aligned} &\mathbb{G}_{n} e_{n}(x) \\ &= \mathbb{G}_{n} u_{1, n}^{+}(q, x; L^{+}) - \big(\mathbb{G}_{n} u^{+}_{1}(q, x; L^{+}) - \mathcal{G} u^{+}_{1}(q, x, L^{+}) \big) + \mathcal{G} u^{+}_{1}(q, x, L^{+}) \\ &= q \big( u_{1, n}^{+}(q, x; L^{+}) - u_{1}^{+}(q, x; L^{+}) \big) - \big(\mathbb{G}_{n} u_{1}^{+}(q, x, L^{+}) - \mathcal{G} u_{1}^{+}(q, x, L^{+}) \big) \\ &= \mathcal{O}(\delta _{n}^{2}). \end{aligned}$$
Therefore, for any \(x, y \in \mathbb{S}_{n} \cap [\ell , L^{-}]\), we obtain
$$ \frac{1}{s_{n}(y)} \nabla ^{+} e_{n}(y) - \frac{1}{s_{n}(x)} \nabla ^{+} e_{n}(x) = \sum _{z \in \mathbb{S}_{n} \cap [x^{+}, y]} m_{n}(z) \delta z \mathbb{G}_{n} e_{n}(z) = \mathcal{O}(\delta _{n}^{2}). $$
We also have \(\sum _{z \in \mathbb{S}_{n} \cap [\ell , L^{-}]} \delta ^{+} z \nabla ^{+} e_{n}(z) = e_{n}(L^{+}) - e_{n}(\ell ) = 0\), from which we conclude that there must exist \(x, y \in \mathbb{S}_{n} \cap [\ell , L^{-}]\) such that
$$ \frac{1}{s_{n}(y)} \nabla ^{+} e_{n}(y) \frac{1}{s_{n}(x)} \nabla ^{+} e_{n}(x) \le 0. $$
Therefore we obtain
$$ \bigg| \frac{1}{s_{n}(x)} \nabla ^{+} e_{n}(x) \bigg| \le \bigg| \frac{1}{s_{n}(y)} \nabla ^{+} e_{n}(y) - \frac{1}{s_{n}(x)} \nabla ^{+} e_{n}(x) \bigg| = \mathcal{O}(\delta _{n}^{2}). $$
It follows that
$$ \bigg| \frac{1}{s_{n}(L^{-})} \nabla ^{+} e_{n}(L^{-}) \bigg| \le \bigg| \frac{1}{s_{n}(x)} \nabla ^{+} e_{n}(x) \bigg| + \mathcal{O}( \delta _{n}^{2}) = \mathcal{O}(\delta _{n}^{2}), $$
and hence \(\nabla ^{+} e_{n}(L^{-}) = \mathcal{O}(\delta _{n}^{2})\) holds. Therefore we get
$$ u_{1, n}^{+}(q, L^{-}; L^{+}) - u_{1}^{+}(q, L^{-}, L^{+}) = \delta ^{+} L^{-} \nabla ^{+} e_{n}(L^{-}) = \mathcal{O}(\delta _{n}^{3}). $$
Using the arguments for obtaining (B.6), we arrive at the second equation. □
Proof of Lemma 4.8
We can construct \(u^{-}(q, x, b)\) as
$$ u^{-}(q, x, b) = \frac{\psi ^{+}(r) \psi ^{-}(x) - \psi ^{+}(x) \psi ^{-}(r)}{\psi ^{+}(r) \psi ^{-}(b) - \psi ^{+}(b) \psi ^{-}(r)}, $$
where \(\psi ^{+}\) and \(\psi ^{-}\) are given in the proof of Lemma 4.6. From this expression, it is easy to deduce that \(b \mapsto u^{-}(q, x, b)\) is \(C^{3}\).
Direct calculations show that
$$\begin{aligned} \partial _{b} u^{-}(q, b, b) &= - \frac{\psi ^{+}(r) (\psi ^{-})'(b) - (\psi ^{+})'(b) \psi ^{-}(r)}{\psi ^{+}(r) \psi ^{-}(b) - \psi ^{+}(b) \psi ^{-}(r)}, \\ \partial _{bb} u^{-}(q, x, b) &= - \frac{\psi ^{+}(r) (\psi ^{-})''(b) - (\psi ^{+})''(b) \psi ^{-}(r)}{\psi ^{+}(r) \psi ^{-}(b) - \psi ^{+}(b) \psi ^{-}(r)} \\ & \phantom{=:} + 2 \frac{(\psi ^{+}(r) (\psi ^{-})'(b) - (\psi ^{+})'(b) \psi ^{-}(r))^{2}}{(\psi ^{+}(r) \psi ^{-}(b) - \psi ^{+}(b) \psi ^{-}(r))^{2}}. \end{aligned}$$
Using these equations, we can verify that
$$\begin{aligned} &\mu (L) \partial _{b} u^{-}(q, L, L) - \sigma ^{2}(L) \big(\partial _{b} u^{-}(q, L, L)\big)^{2} + \frac{1}{2} \sigma ^{2}(L) \partial _{bb} u^{-}(q, L, L) + q \\ &= - \frac{\psi ^{+}(r) ( \mu (b) (\psi ^{-})'(b) + \frac{1}{2} \sigma ^{2}(b) (\psi ^{-})''(b) - q\psi ^{-}(b) ) }{\psi ^{+}(r) \psi ^{-}(b) - \psi ^{+}(b) \psi ^{-}(r)} \\ & \phantom{=:} + \frac{\psi ^{-}(r) ( \mu (b) (\psi ^{+})'(b) + \frac{1}{2} \sigma ^{2}(b) (\psi ^{+})''(b) - q\psi ^{+}(b) ) }{\psi ^{+}(r) \psi ^{-}(b) - \psi ^{+}(b) \psi ^{-}(r)} = 0. \end{aligned}$$
We can factorise \(u_{n}^{-}(q, x; L^{-})\) as
$$\begin{aligned} u_{n}^{-}(q, x; L^{-}) &= \mathbb{E}_{x}\big[ e^{-q\tau _{L}^{-}} 1_{ \{ Y_{\tau _{L}^{-}} = L^{-} \}} \big] \\ &= \mathbb{E}_{x}\big[ e^{-q\tau _{L^{++}}^{-}} 1_{\{ Y_{\tau _{L^{++}}^{-}} = L^{+} \}} \big] \mathbb{E}_{L^{+}}\big[ e^{-q\tau _{L}^{-}} 1_{\{ Y_{ \tau _{L}^{-}} = L^{-} \}} \big] \\ &= \widetilde{u}_{n}^{-}(q, x; L^{+}) u_{n}^{-}(q, L^{+}; L^{-}). \end{aligned}$$
To derive this, we use the fact that for \(Y^{(n)}\) to arrive at \(L^{-}\) from the above, it should first touch \(L^{+}\) because it is a birth-and-death process. Using the smoothness of \(u^{-}(q, x; b)\) in \(b\) and
$$ u_{n}^{-}(q, x; L^{+}) = u^{-}(q, x; L^{+}) + \mathcal{O}(\delta _{n}^{2}), $$
we obtain
$$ u_{n}^{-}(q, x; L^{+}) = u^{-}(q, x, L) + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}). $$
Similarly to the proof of Lemma 4.6, we can show that
$$ u_{n}^{-}(q, L^{+}; L^{-}) = u^{-}(q, L^{+}, L^{-}) + \mathcal{O}( \delta _{n}^{3}). $$
Thus we obtain
$$\begin{aligned} &u^{-}(q, L^{+}, L^{-}) - 1 \\ &= u^{-}(q, L^{+}, L^{-}) - u^{-}(q, L^{+}, L) + u^{-}(q, L^{+}, L) - u^{-}(q, L^{+}, L^{+}) \\ &= \partial _{b} u^{-}(q, L^{+}, L)(L^{-} - L) + \frac{1}{2} \partial _{bb} u^{-}(q, L^{+}, L)(L^{-} - L)^{2} \\ &\quad - \partial _{b} u^{-}(q, L^{+}, L)(L^{+} - L) - \frac{1}{2} \partial _{bb} u^{-}(q, L^{+}, L)(L^{+} - L)^{2} + \mathcal{O}( \delta _{n}^{3}) \\ &= -\partial _{b} u^{-}(q, L^{+}, L) \delta ^{+} L^{-} + \frac{1}{2} \partial _{bb} u^{-}(q, L^{+}, L)(\delta ^{+} L^{-})^{2} \\ & \phantom{=:} - \frac{1}{2} \partial _{bb} u^{-}(q, L^{+}, L) \delta ^{+} L^{-} (L^{+} - L) + \mathcal{O}(\delta _{n}^{3}) \\ &= -\partial _{b} u^{-}(q, L^{+}, L) \delta ^{+} L^{-} + \frac{1}{2} \partial _{bb} w^{-}(q, L^{+}, L)(\delta ^{+} L^{-})^{2} + \mathcal{O}(\delta ^{+} L^{-}) (L^{+} - L) \\ & \phantom{=:} + \mathcal{O}(\delta _{n}^{3}). \end{aligned}$$
This concludes the proof. □
Proof of Proposition 4.5
By Lemma 4.4, there exist constants \(c_{1}, c_{2} > 0\) such that
$$\begin{aligned} &v_{n}(D, x; y) / \big(m_{n}(y) \delta y\big) \\ &= \sum _{k = 1}^{n_{e}} e^{-\lambda _{n, k}^{+} D} \varphi _{n, k}^{+}(q, x) \varphi _{n, k}^{+}(y) \\ &= \sum _{k = 1}^{n_{e}} e^{-\lambda _{k}^{+}(L)D} \varphi _{k}^{+}(x, L) \varphi _{k}^{+}(y, L) + \big( \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}) \big)\sum _{k = 1}^{n_{e}} k^{11} e^{-c_{1}k^{2} D} \\ &= \sum _{k = 1}^{\infty} e^{-\lambda _{k}^{+}(L)D} \varphi _{k}^{+}(x, L) \varphi _{k}^{+}(y, L) + \mathcal{O}\bigg(\sum _{k = n_{e} + 1}^{ \infty} e^{-c_{2} k^{2} D}\bigg) \\ & \phantom{=:} + \big( \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}) \big) \sum _{k = 1}^{\infty} k^{11} e^{-c_{1}k^{2} D} \\ &= \bar{v}(D, x, y) + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}). \end{aligned}$$
In the last equality, we use the fact that \(\sum _{k = n_{e} + 1}^{\infty} e^{-c_{2} k^{2} D} \le c_{3} e^{-c_{4}/ \delta _{n}^{2}} \le c_{5} \delta _{n}^{2}\) for some constants \(c_{3}, c_{4}, c_{5} > 0\). We use this result directly hereafter. Using (4.5), we also obtain
$$\begin{aligned} v_{n}(D, L^{-}; y) / \big(m_{n}(y) \delta y\big) &= \sum _{k = 1}^{n_{e}} e^{-\lambda _{n, k}^{+} D} \varphi _{n, k}^{+}(L^{-}) \varphi _{n, k}^{+}(y) \\ &= \sum _{k = 1}^{n_{e}} e^{-\lambda _{k}^{+}(L) D} \varphi _{n, k}^{+}(L^{-}) \varphi _{k}^{+}(y, L) \\ & \phantom{=:} + \mathcal{O}\big((L^{+} - L)\delta ^{+}L^{-}\big) + \mathcal{O}( \delta _{n}^{3}) \\ &= -\delta ^{+} L^{-}\sum _{k = 1}^{n_{e}} e^{-\lambda _{k}^{+}(L) D} \partial _{x} \varphi ^{+}_{k}(L, L) \varphi _{k}^{+}(y, L) \\ & \phantom{=:} + \frac{1}{2} (\delta ^{+} L^{-})^{2} \sum _{k = 1}^{n_{e}} e^{- \lambda _{k}^{+}(L) D} \partial _{xx} \varphi ^{+}_{k}(L, L) \varphi _{k}^{+}(y, L) \\ & \phantom{=:} + \mathcal{O}\big((L^{+} - L)\delta ^{+}L^{-}\big) + \mathcal{O}( \delta _{n}^{3}) \\ &= -\delta ^{+} L^{-}\sum _{k = 1}^{\infty} e^{-\lambda _{k}^{+}(L) D} \partial _{x} \varphi ^{+}_{k}(L, L) \varphi _{k}^{+}(q, y; L) \\ & \phantom{=:} + \frac{1}{2} (\delta ^{+} L^{-})^{2} \sum _{k = 1}^{\infty} e^{- \lambda _{k}^{+}(L) D} \partial _{xx} \varphi ^{+}_{k}(L, L) \varphi _{k}^{+}(y, L) \\ & \phantom{=:} + \mathcal{O}\big((L^{+} - L)\delta ^{+}L^{-}\big) + \mathcal{O}( \delta _{n}^{3}) \\ &= - \bar{v}_{x}(D, L; y) \delta ^{+} L^{-} + \frac{1}{2} \partial _{xx} \bar{v}(D, L; y) ( \delta ^{+} L^{-})^{2} \\ & \phantom{=:} + \mathcal{O}\big((L^{+} - L)\delta ^{+}L^{-}\big) + \mathcal{O}( \delta _{n}^{3}). \end{aligned}$$
For (4.8), we have
$$ \mu (L)\partial _{x} \varphi _{k}^{+}(L, L) + \frac{1}{2} \sigma ^{2}(L) \partial _{xx} \varphi _{k}^{+}(L, L) - q \varphi _{k}^{+}(L, L) = - \lambda _{k}(L) \varphi _{k}^{+}(L, L). $$
As \(\varphi _{k}^{+}(L,L) = 0\), we get \(\mu (L)\partial _{x} \varphi _{k}^{+}(L, L) + \frac{1}{2} \sigma ^{2}(L) \partial _{xx} \varphi _{k}^{+}(L, L) = 0\). It follows that
$$\begin{aligned} &\mu (L) \partial _{x} \bar{v}(D, L; y) + \frac{1}{2} \sigma ^{2}(L) \partial _{xx} \bar{v}(D, L; y) \\ &= \sum _{k = 1}^{\infty }e^{-\lambda _{k}^{+}(L) D} \bigg( \mu (L) \partial _{x} \varphi _{k}^{+}(L, L) + \frac{1}{2} \sigma ^{2}(L) \partial _{xx} \varphi _{k}^{+}(L, L) \bigg) = 0, \end{aligned}$$
where the interchange of summation and differentiation can be verified with the estimates of \(\lambda _{k}^{+}(L)\) and \(\varphi _{k}^{+}(L, L)\) in Lemma 4.4.
The results for \(v_{n}(D, x;\ell )\) and \(v(D, L, L)\) can be proved by arguments similar to those for the third part of Lemma 4.6 and the equation
$$ v(D, x, L) = v_{1}(x, L) - v_{2}(D, x, L) $$
along with the differential equation for \(v_{1}(x, L)\) and the eigenfunction expansion for \(v_{2}(D, x, L)\). □
Proof of Proposition 4.7
For the first and second claims, we only prove that
$$ c_{n, k}(q) = c_{k}(q, L) + O(k^{4} \delta _{n}^{2}). $$
The other steps are almost identical to those in the proof of Proposition 4.5. But
$$\begin{aligned} &c_{n, k}(q) - c_{k}(q, L) \\ &= \sum _{y \in \mathbb{S}_{n}^{o} \cap (-\infty , L^{+})}\!\!\! \varphi _{n, k}^{+}(y) u_{1, n}^{+}(q, y; L^{+}) m_{n}(y) \delta y - \int _{\ell}^{L} \varphi _{k}^{+}(y, L) u_{1}^{+}(q, y, L) m(y) dy \\ &= \sum _{y \in \mathbb{S}_{n}^{o} \cap (-\infty , L^{+})} \!\!\! \varphi _{k}^{+}(y, L) u_{1}^{+}(q, y, L) m_{n}(y) \delta y + \mathcal{O}\big(k^{4} \delta _{n}^{2}\big) + \mathcal{O}\big(k^{2} (L^{+} - L)\big) \\ & \phantom{=:} \qquad - \int _{\ell}^{L} \varphi _{k}^{+}(y, L) u_{1}^{+}(q, y, L) m(y) dy \\ &= \sum _{y \in \mathbb{S}_{n}^{o} \cap (-\infty , L^{+})} \!\!\! \varphi _{k}^{+}(y, L) u_{1}^{+}(q, y, L) m(y) \delta y + \mathcal{O} \big(k^{4} \delta _{n}^{2}\big) + \mathcal{O}\big(k^{2} (L^{+} - L) \big) \\ & \phantom{=:} \qquad - \int _{\ell}^{L} \varphi _{k}^{+}(y, L) u_{1}^{+}(q, y, L) m(y) dy \\ &= \sum _{z \in \mathbb{S}_{n}^{-} , z< L^{-}} \frac{1}{2} \big( \varphi _{k}^{+}(z, L) u_{1}^{+}(q, z, L) m(z) + \varphi _{k}^{+}(z^{+}, L) u_{1}^{+}(q, z^{+}, L) m(z^{+}) \big) \delta ^{+} z \\ & \phantom{=:} \quad - \int _{z}^{z^{+}} \varphi _{k}^{+}(y, L) u_{1}^{+}(q, y, L) m(y) dy ) + \int _{L^{-}}^{L} \varphi _{k}^{+}(y, L) u_{1}^{+}(q, y, L) m(y) dy \\ & \phantom{=:} \quad + \mathcal{O} (k^{4} \delta _{n}^{2} ) + \mathcal{O}\big(k^{2} (L^{+} - L)\big) \\ &= \mathcal{O} (k^{4} \delta _{n}^{2} ) + \mathcal{O}\big(k^{2} (L^{+} - L)\big). \end{aligned}$$
For the last claim, using the arguments in the proof of Proposition 4.5, we obtain
$$ \mu (L) \partial _{x} u_{2}^{+}(q, D, L; L) + \frac{1}{2} \sigma ^{2}(L) \partial _{xx} u^{+}_{2}(q, D, L; L) = 0. $$
It follows that
$$\begin{aligned} &\mu (L) \partial _{x} u^{+}(q, D, L, L) + \frac{1}{2} \sigma ^{2}(L) \partial _{xx} u^{+}(q, D, L, L) - q \\ &= \mu (L) \partial _{x} u_{1}^{+}(q, L, L) + \frac{1}{2} \sigma ^{2}(L) \partial _{xx} u_{1}^{+}(q, L, L) - qu_{1}^{+}(q, L, L) = 0, \end{aligned}$$
where we use the differential equation for \(u_{1}^{+}(q, x, L)\) at \(x = L\) and the boundary condition \(u_{1}^{+}(q, L, L) = 1\). □
Proof of Proposition 4.10
By Athanasiadis and Stratis [3, Theorem 1.4], (4.17) admits a unique solution \(w(\cdot )\) that belongs to \(C^{1}([\ell , r]) \cap C^{2}([\ell , r] \backslash \mathcal{D})\). The equation for \(\widetilde{w}(q, z)\) can be written in a self-adjoint form as
$$ \frac{1}{m(x)} \partial _{x}\bigg( \frac{1}{s(x)} \partial _{x} \widetilde{w}(q, x)\bigg) - q \widetilde{w}(q, x) = f(x). $$
Multiplying both sides with \(m(x)\) and integrating from \(\ell ^{+}_{1/2} = \ell + \delta ^{+} \ell /2\) to \(z \in \mathbb{S}_{n}\) yields
$$\begin{aligned} &\frac{1}{s(z)} \partial _{x} \widetilde{w}(q, z) - \frac{1}{s(\ell ^{+}_{1/2})} \partial _{x} \widetilde{w}(q, l^{+}_{1/2}) - q \int _{\ell ^{+}_{1/2}}^{z} m(y) \widetilde{w}(q, y) dy dz \\ &= \int _{\ell ^{+}_{1/2}}^{z} m(y) f(y) dy. \end{aligned}$$
Further multiplying both sides with \(s(z)\) and integrating from \(x\) to \(x^{+}\) gives
$$\begin{aligned} &\widetilde{w}(q, x^{+}) - \widetilde{w}(q, x) - \frac{\int _{x}^{x^{+}} s(z) dz}{s(\ell ^{+}_{1/2})} \partial _{x} \widetilde{w}(q, \ell ^{+}_{1/2}) \\ & \phantom{=::} \qquad \, \ \ - q \int _{x}^{x^{+}} s(z) \int _{\ell ^{+}_{1/2}}^{z} m(y) \widetilde{w}(q, y) dy dz \\ &= \int _{x}^{x^{+}} s(z) \int _{\ell ^{+}_{1/2}}^{z} m(y) f(y) dy dz. \end{aligned}$$
Moreover, it is clear that \(\widetilde{w}_{n}(q, z)\) satisfies
$$ \frac{\delta ^{-} x}{m_{n}(x) \delta x} \nabla ^{-}\bigg( \frac{1}{s_{n}(x)} \nabla ^{+} \widetilde{w}_{n}(q, x) \bigg) - q \widetilde{w}_{n}(q, x) = f(x). $$
Multiplying both sides with \(m_{n}(x) \delta x\) and summing from \(\ell ^{+}\) to \(x\), we obtain
$$\begin{aligned} &\frac{1}{s_{n}(x)} \nabla ^{+} \widetilde{w}_{n}(q, x) - \frac{1}{s_{n}(\ell )} \nabla ^{+} \widetilde{w}_{n}(q,\ell ) - q \sum _{\ell < y \le x} \widetilde{w}_{n}(q, y) m_{n}(y) \delta y \\ &= \sum _{\ell < y \le x} f(y) m_{n}(y) \delta y. \end{aligned}$$
It follows that
$$\begin{aligned} s_{n}(x) \delta ^{+} x \sum _{\ell < y \le x} f(y) m_{n}(y) \delta y &= \widetilde{w}_{n}(q, x^{+}) - \widetilde{w}_{n}(q, x) \\ & \phantom{=:} - s_{n}(x) \frac{\delta ^{+} x}{s_{n}(\ell )} \nabla ^{+} \widetilde{w}_{n}(q,\ell ) \\ & \phantom{=:} - q s_{n}(x) \delta ^{+} x\sum _{\ell < y \le x} \widetilde{w}_{n}(q, y) m_{n}(y) \delta y. \end{aligned}$$
Let \(e(x) = \widetilde{w}_{n}(q, x) - \widetilde{w}(q, x)\). We have
$$\begin{aligned} &e(x^{+}) - e(x) \\ &= s_{n}(x) \delta ^{+} x \bigg( \frac{1}{s_{n}(\ell ^{+}_{1/2})} \nabla ^{+} \widetilde{w}_{n}(q,\ell ) - \frac{1}{s(\ell ^{+}_{1/2})} \partial _{x} \widetilde{w}(q,\ell ) \bigg) \\ & \phantom{=:} + \bigg( s_{n}(x) \delta ^{+} x - \int _{x}^{x^{+}} s(z) dz \bigg) \frac{1}{s(\ell ^{+}_{1/2})} \partial _{x} \widetilde{w}(q,\ell ) \end{aligned}$$
(B.7)
$$\begin{aligned} & \phantom{=:} + q s_{n}(x) \delta ^{+} x \sum _{\ell < y \le x} e(y) m_{n}(y) \delta y \\ & \phantom{=:} + q \bigg(s_{n}(x) \delta ^{+} x - \int _{x}^{x^{+}} s(z) dz\bigg) \sum _{\ell < y \le x} \widetilde{w}(q, y) m_{n}(y) \delta y \end{aligned}$$
(B.8)
$$\begin{aligned} & \phantom{=:} + q \int _{x}^{x^{+}} s(z) \bigg( \sum _{\ell < y \le x} \widetilde{w}(q, y) m_{n}(y) \delta y - \int _{\ell ^{+}_{1/2}}^{z} \widetilde{w}(q, y) m(y) dy \bigg) dz \end{aligned}$$
(B.9)
$$\begin{aligned} & \phantom{=:} + \bigg(s_{n}(x) \delta ^{+} x - \int _{x}^{x^{+}} s(z) dz\bigg) \sum _{\ell < y \le x} f(y) m_{n}(y) \delta y \end{aligned}$$
(B.10)
$$\begin{aligned} & \phantom{=:} + \int _{x}^{x^{+}} s(z) \bigg( \sum _{\ell < y \le x} f(y) m_{n}(y) \delta y - \int _{\ell ^{+}_{1/2}}^{z} f(y) m(y) dy \bigg) dz. \end{aligned}$$
(B.11)
The quantities in (B.7), (B.8) and (B.10) are all \(\mathcal{O}(\delta _{n}^{3})\). For the quantity in (B.11), note that
$$\begin{aligned} s(z) &= s(x) + s'(x)(z - x) + \mathcal{O}(\delta _{n}^{2}), \\ \int _{\ell ^{+}_{1/2}}^{z} f(y) m(y) dy &= \int _{\ell ^{+}_{1/2}}^{x^{+}_{1/2}} f(y) m(y) dy + f(x) m(x) (z - x^{+}_{1/2}) \\ & \phantom{=:} + \mathcal{O}(\delta _{n}^{\gamma }1_{\{ x \in \mathcal{D}_{N} \}} + \delta _{n}^{2}), \end{aligned}$$
where \(x^{+}_{1/2} = x + \delta ^{+}x/2\) for \(z \in [x, x^{+}]\),
$$ \int _{\ell ^{+}_{1/2}}^{x^{+}_{1/2}} f(y) m(y) dy = \sum _{\ell < y \le x} f(y) m_{n}(y) dy + \mathcal{O}(\delta _{n}^{\gamma}), $$
and \(\mathcal{D}_{N} = \{ y \in \mathbb{S}_{n}^{o}: [y^{-}, y^{+}] \cap \mathcal{D} \ne \emptyset \}\). Therefore,
$$\begin{aligned} \text{(B.11)} &= s(x) \delta ^{+} x \bigg( \sum _{\ell < y \le x} f(y) m_{n}(y) dy - \int _{\ell ^{+}_{1/2}}^{x^{+}_{1/2}} f(y) m(y) dy \bigg) \\ & \phantom{=:} + s(x) f(x) m(x) \int _{x}^{x^{+}} (x^{+}_{1/2} - z) dz + \mathcal{O} ( \delta _{n}^{1 + \gamma} 1_{\{ x\in \mathcal{D}_{N} \}} + \delta _{n}^{3} + \delta _{n}^{1 + \gamma} ) \\ &= \mathcal{O} (\delta _{n}^{1 + \gamma} 1_{\{ x\in \mathcal{D}_{N} \}} + \delta _{n}^{3} ). \end{aligned}$$
By the same argument, we can show that \(\text{(B.9)} = \mathcal{O}(\delta _{n}^{3})\). Putting these estimates back and letting \(e^{\ast}(x) = -e(x)\), we deduce that there exists a constant \(c_{1} > 0\) independent of \(\delta _{n}\) such that
$$\begin{aligned} e(x^{+}) &\le s_{n}(x) \delta ^{+} x \bigg( \frac{1}{s_{n}(\ell ^{+}_{1/2})} \nabla ^{+} \widetilde{w}_{n}(q, \ell ) - \frac{1}{s(\ell ^{+}_{1/2})} \partial _{x} \widetilde{w}(q, \ell ) \bigg) \\ & \phantom{=:} + e(x) + q s_{n}(x) \delta ^{+} x \sum _{\ell < y \le x} e(y) m_{n}(y) \delta y + c_{1} (\delta _{n}^{1 + \gamma} 1_{\{ x\in \mathcal{D}_{N} \}} + \delta _{n}^{3} ), \\ e^{\ast}(x^{+}) &\le -s_{n}(x) \delta ^{+} x \bigg( \frac{1}{s_{n}(\ell ^{+}_{1/2})} \nabla ^{+} \widetilde{w}_{n}(q, \ell ) - \frac{1}{s(\ell ^{+}_{1/2})} \partial _{x} \widetilde{w}(q, \ell ) \bigg) \\ & \phantom{=:} + e^{\ast}(x) + q s_{n}(x) \delta ^{+} x \sum _{\ell < y \le x} e^{ \ast}(y) m_{n}(y) \delta y + c_{1} (\delta _{n}^{1 + \gamma} 1_{\{ x \in \mathcal{D}_{N} \}} + \delta _{n}^{3} ). \end{aligned}$$
Using the positive lower and upper bounds for \(s_{n}(x)\), \(m_{n}(x)\) which are independent of \(\delta _{n}\) and the discrete Gronwall inequality, we conclude that there exist constants \(c_{2}, c_{3}, c_{4}, c_{5} > 0\) independent of \(\delta _{n}\) such that
$$\begin{aligned} e(x) &\le c_{2}\bigg( c_{1} h_{n}^{\gamma }+ c_{3} \Big( \frac{1}{s_{n}(\ell ^{+}_{1/2})} \nabla ^{+} \widetilde{w}_{n}(q, \ell ) - \frac{1}{s(\ell ^{+}_{1/2})} \partial _{x} \widetilde{w}(q, \ell ) \Big) \bigg), \end{aligned}$$
(B.12)
$$\begin{aligned} e^{\ast}(x) &\le c_{4}\bigg( c_{1} h_{n}^{\gamma }- c_{5} \Big( \frac{1}{s_{n}(\ell ^{+}_{1/2})} \nabla ^{+} \widetilde{w}_{n}(q, \ell ) - \frac{1}{s(\ell ^{+}_{1/2})} \partial _{x} \widetilde{w}(q, \ell ) \Big) \bigg). \end{aligned}$$
(B.13)
Note that \(e(r) = e^{\ast}(r) = 0\). Then there exist constants \(c_{6}, c_{7} > 0\) independent of \(\delta _{n}\) such that
$$\begin{aligned} \bigg( \frac{1}{s_{n}(\ell ^{+}_{1/2})} \nabla ^{+} \widetilde{w}_{n}(q, \ell ) - \frac{1}{s(\ell ^{+}_{1/2})} \partial _{x} \widetilde{w}(q, \ell ) \bigg) &\le c_{6} \delta _{n}^{\gamma}, \\ -\bigg( \frac{1}{s_{n}(\ell ^{+}_{1/2})} \nabla ^{+} \widetilde{w}_{n}(q, \ell ) - \frac{1}{s(\ell ^{+}_{1/2})} \partial _{x} \widetilde{w}(q, \ell ) \bigg) &\le c_{7} \delta _{n}^{\gamma}. \end{aligned}$$
Hence \(| \frac{1}{s_{n}(\ell ^{+}_{1/2})} \nabla ^{+} \widetilde{w}_{n}(q, \ell ) - \frac{1}{s(\ell ^{+}_{1/2})} \partial _{x} \widetilde{w}(q, \ell ) | = \mathcal{O}(\delta _{n}^{\gamma})\). Substituting these estimates back into (B.12) and (B.13), we obtain \(e(x) = \mathcal{O}(\delta _{n}^{\gamma})\) and \(-e(x) = \mathcal{O}(\delta _{n}^{\gamma})\). Thus we have \(e(x) = \mathcal{O}(\delta _{n}^{\gamma})\) and the proof is complete. □
Proof of Theorem 4.9
The smoothness of \(h(q, x; y)\) and its limit at \(y =\ell \) and value at \(y = L\) are direct consequences of the properties of \(v(D, x; y)\). By (4.7), (4.11) and (4.15), we have
$$\begin{aligned} &u_{n}^{-}(q, L^{+}; L^{-}) v_{n}(D, L^{-}; y) / \big(m_{n}(y) \delta y\big) \\ &= \bigg(- \bar{v}_{x} \delta ^{+} L^{-} + \frac{1}{2} \bar{v}_{xx} ( \delta ^{+} L^{-})^{2} + \mathcal{O}(\delta ^{+} L^{-})(L^{+} - L) \bigg) \\ & \phantom{=:} \times \bigg(1 - u^{-}_{b} \delta ^{+} L^{-} + \frac{1}{2} u^{-}_{bb} ( \delta ^{+} L^{-})^{2} + \mathcal{O}(\delta ^{+} L^{-})(L^{+} - L) \bigg) + \mathcal{O}(\delta _{n}^{3}) \\ &= -\bar{v}_{x} \delta ^{+}L^{-} + (\delta ^{+}L^{-})^{2} \bigg( \bar{v}_{x} u_{b}^{-} + \frac{1}{2} \bar{v}_{xx} \bigg) + \mathcal{O}( \delta ^{+} L^{-}) (L^{+} - L) + \mathcal{O}(\delta _{n}^{3}). \end{aligned}$$
Here, we neglect the arguments \((D, L; y)\) for \(\bar{v}\) and \((q, L, L)\) for \(u^{-}\) to make the equations shorter. Moreover, using (4.11) and (4.15), we obtain
$$\begin{aligned} &1 - u_{n}^{+}(q,D, L^{-}; L^{+}) u_{n}^{-}(q, L^{+}; L^{-}) \\ &= 1 - \bigg(1 - u^{+}_{x} \delta ^{+} L^{-} + \frac{1}{2} u^{+}_{xx} ( \delta ^{+} L^{-})^{2} + \mathcal{O}(\delta ^{+} L^{-})(L^{+} - L) \bigg) \\ & \phantom{::} \qquad \times \bigg(1 - u^{-}_{b} \delta ^{+} L^{-} + \frac{1}{2} u^{-}_{bb} (\delta ^{+} L^{-})^{2} + \mathcal{O}(\delta ^{+} L^{-})(L^{+} - L) \bigg) + \mathcal{O}(\delta _{n}^{3}) \\ &= \delta ^{+} L^{-} \big( u_{b}^{-} + u_{x}^{+} \big) - (\delta ^{+} L^{-})^{2} \bigg( \frac{1}{2} u_{bb}^{-} + u_{x}^{+} u_{b}^{-} + \frac{1}{2} u_{xx}^{+} \bigg) \\ & \phantom{=:} + \mathcal{O}(\delta ^{+} L^{-}) (L^{+} - L) + \mathcal{O}(\delta _{n}^{3}). \end{aligned}$$
Here, we neglect the arguments \((q, D, x, L)\) for \(u^{+}\) for the same reason. It follows that
$$\begin{aligned} &h_{n}(q, L^{+}; y)/\big(m_{n}(y)\delta y\big) \\ &= e^{-qD} \frac{-v_{x} + \delta ^{+}L^{-} ( \bar{v}_{x} u_{b}^{-} + \frac{1}{2} \bar{v}_{xx} ) + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2})}{ u_{b}^{-} + u_{x}^{+} - \delta ^{+} L^{-} ( \frac{1}{2} u_{bb}^{-} + u_{x}^{+} u_{b}^{-} + \frac{1}{2} u_{xx}^{+} ) + \mathcal{O} (L^{+} - L) + \mathcal{O}(\delta _{n}^{2})} \\ &= h(q, L; y) / m(y) \\ & \phantom{=:} + e^{-qD} \delta ^{+} L^{-} \frac{-\frac{1}{2} \bar{v}_{x} u_{bb}^{-} - \frac{1}{2} \bar{v}_{x} u_{xx}^{+} + v_{x}^{-} (u_{b}^{-} )^{2} + \frac{1}{2} \bar{v}_{xx} u_{b}^{-} + \frac{1}{2} u_{x}^{+} \bar{v}_{xx} }{ ( u_{b}^{-} + u_{x}^{+} )^{2} + \mathcal{O}(\delta _{n})} \\ & \phantom{=:} + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}). \end{aligned}$$
By (4.12) and (4.8), we have \(\bar{v}_{xx} = -\frac{2\mu (L)}{\sigma ^{2}(L)} \bar{v}_{x}\) and \(u^{+}_{xx} = -\frac{2\mu (L)}{\sigma ^{2}(L)} u^{+}_{x} + \frac{2q}{\sigma ^{2}(L)}\). Subsequently, we obtain
$$\begin{aligned} &-\frac{1}{2} \bar{v}_{x} u_{bb}^{-} - \frac{1}{2} \bar{v}_{x} u_{xx}^{+} + v_{x}^{-} (u_{b}^{-} )^{2} + \frac{1}{2} \bar{v}_{xx} u_{b}^{-} + \frac{1}{2} u_{x}^{+} \bar{v}_{xx} \\ &= -\frac{1}{2} \bar{v}_{x} u_{bb}^{-}- \frac{1}{2} \bar{v}_{x} \bigg( -\frac{2\mu (L)}{\sigma ^{2}(L)} u^{+}_{x} + \frac{2q}{\sigma ^{2}(L)} \bigg) \\ & \phantom{=:} + v_{x}^{-} (u_{b}^{-} )^{2} - \frac{1}{2} \frac{2\mu (L)}{\sigma ^{2}(L)} \bar{v}_{x} u_{b}^{-} - \frac{1}{2} \frac{2\mu (L)}{\sigma ^{2}(L)} \bar{v}_{x} u_{x}^{+} \\ &= -\frac{1}{\sigma ^{2}(L) }\bigg( \frac{1}{2} \sigma ^{2}(L) u_{bb}^{-} + q - \sigma ^{2}(L) (u_{b}^{-})^{2} + \mu (L) u_{b}^{-} \bigg) = 0. \end{aligned}$$
The last equality follows from (4.13). Therefore, we have that
$$ h_{n}(q, L^{+}; y)/\big(m_{n}(y)\delta y\big) = h(q, L; y) / m(y) + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}). $$
Then by (4.14), we obtain
$$\begin{aligned} &h_{n}(q, L^{-}; y) u_{n}^{-}(q, x; L^{-}) / \big(m_{n}(y) \delta y \big) \\ &= h_{n}(q, L^{-}; y) u_{n}^{-}(q, x; L^{+}) u_{n}^{-}(q, L^{+}; L^{-}) / \big(m_{n}(y) \delta y\big) \\ &= h_{n}(q, L^{+}; y) u_{n}^{-}(q, x; L^{+}) / \big(m_{n}(y) \delta y \big) \\ &= h(q, L; y) u_{n}^{-}(q, x; L^{+}) / m(y) + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}) \\ &= h(q, L; y) u^{-}(q, x, L) / m(y) + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}). \end{aligned}$$
Based on the previous estimates, we deduce that
$$\begin{aligned} &h_{n}(q, x; y) / \big(m_{n}(y) \delta y\big) \\ &= 1_{\{ x< L \}} e^{-qD} v_{n}(D, x; y) / \big(m_{n}(y) \delta y \big) \\ & \phantom{=:} + 1_{\{ x < L \}} u_{n}^{+}(q, D, x; L^{+}) h_{n}(q, L^{+}; y) / \big(m_{n}(y) \delta y\big) \\ & \phantom{=:} + 1_{\{ x \ge L \}} u^{-}_{n}(q, x; L^{-}) h_{n}(q, L^{-}; y) / \big(m_{n}(y) \delta y\big) \\ &= 1_{\{ x< L \}} e^{-qD} \bar{v}(D, x; y) + 1_{\{ x < L \}} u^{+}(q, D, x, L) h(q, L; y) / m(y) \\ & \phantom{=:} + 1_{\{ x \ge L \}} h(q, L; y) u^{-}(q, x, L) / m(y) + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}). \end{aligned}$$
□
Proof of Theorem 4.11
Recall that
$$\begin{aligned} h_{n}(q, x;\ell ) &= h(q, x;\ell ) + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}), \\ \widetilde{w}(q,\ell ) &= \widetilde{w}_{n}(q,\ell ) = f(\ell )/q, \end{aligned}$$
which will be used below. We have
$$\begin{aligned} &\widetilde{u}_{n}(q, x) - \widetilde{u}(q, x) \\ &= \sum _{z \in \mathbb{S}_{n} \cap (\ell , L)} \!\!\! h_{n}(q, x; z) \widetilde{w}_{n}(q, z) - \int _{\ell}^{L} h(q, x; z) \widetilde{w}(q, z) dz \\ & \phantom{=:} \quad \ \ + h_{n}(q, x;\ell ) \widetilde{w}_{n}(q,\ell ) - h(q, x; \ell ) \widetilde{w}(q,\ell ) \\ &= \sum _{z \in \mathbb{S}_{n} \cap (\ell , L)}\!\!\! h_{n}(q, x; z) \widetilde{w}_{n}(q, z) - \int _{\ell}^{L} h(q, x; z) \widetilde{w}(q, z) dz + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}) \\ &= \sum _{z \in \mathbb{S}_{n} \cap (\ell , L)} \!\!\! h_{n}(q, x; z) / \big(m_{n}(z) \delta z\big) \widetilde{w}_{n}(q, z) m_{n}(z) \delta z - \int _{\ell}^{L} h(q, x; z) \widetilde{w}(q, z) dz \\ & \phantom{=:} \quad \ \ + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{2}) \\ &= \sum _{z \in \mathbb{S}_{n} \cap (\ell , L)}\!\!\! h(q, x; z) \widetilde{w}(q, z) m_{n}(z) / m(z) \delta z - \int _{\ell}^{L} h(q, x; z) \widetilde{w}(q, z) dz \\ & \phantom{=:} \quad \ \ + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{\gamma}) \\ &= \sum _{z \in \mathbb{S}_{n} \cap (\ell , L)} \!\!\! h(q, x; z) \widetilde{w}(q, z) \delta z - \int _{\ell}^{L} h(q, x; z) \widetilde{w}(q, z) dz \\ & \phantom{=:} \quad \ \ + \frac{1}{2}\sum _{z \in \mathbb{S}_{n} \cap (\ell , L)} \! \!\! h(q, x; z) \widetilde{w}(q, z) \frac{\mu (z)}{\sigma ^{2}(z) } \big( (\delta ^{+} z)^{2} - (\delta ^{-} z)^{2} \big) \\ & \phantom{=:} \quad \ \ + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{\gamma}) \\ &= \sum _{z \in \mathbb{S}_{n}^{-} \cap (\ell , L^{-})}\!\! \bigg( \frac{1}{2} (h(q, x; z) + h(q, x; z^{+})) \delta ^{+} z - \int _{z}^{z^{+}} h(q, x; y) \widetilde{w}(q, y) dy \bigg) \\ & \phantom{=:} \quad \ \ + h(q, x; \ell ^{+}) \widetilde{w}(q, \ell ^{+}) \delta ^{-} \ell ^{+}/2 + h(q, x; L^{-}) \widetilde{w}(q, L^{-}) \delta ^{+} L^{-}/2 \\ & \phantom{=:} \quad \ \ - \int _{y \in (\ell , \ell ^{+}) \cup (L^{-}, L)} h(q, x; y) \widetilde{w}(q, y) dy \\ & \phantom{=:} \quad \ \ + \frac{1}{2} \sum _{z \in \mathbb{S}_{n}^{-} \cap ( \ell , L^{-})} \!\! \bigg( h(q, x; z) \widetilde{w}(q, z) \frac{\mu (z)}{\sigma ^{2}(z) } \\ & \phantom{=:} \qquad \qquad \qquad \qquad \ \ - h(q, x; z^{+}) \widetilde{w}(q, z^{+}) \frac{\mu (z^{+})}{\sigma ^{2}(z^{+}) } \bigg) (\delta ^{+} z)^{2} \\ & \phantom{=:} \quad \ \ - \frac{1}{2} h(q, x; \ell ^{+}) \widetilde{w}(q, \ell ^{+}) \frac{\mu (\ell ^{+})}{\sigma ^{2}(\ell ^{+}) } (\delta ^{-}\ell ^{+})^{2} \\ & \phantom{=:} \quad \ \ + \frac{1}{2} h(q, x; L^{-}) \widetilde{w}(q, L^{-}) \frac{\mu (L^{-})}{\sigma ^{2}(L^{-}) } (\delta ^{+}L^{-})^{2} \\ & \phantom{=:} \quad \ \ + \mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{ \gamma}) \\ &= \sum _{z \in \mathbb{S}_{n} \cap (\ell , L^{-}), [z, z^{+}] \mathcal{D} = \emptyset} \max _{y \in [z, z^{+}]} \partial _{yy} \big( h(q, x; \cdot ) \widetilde{w}(q, \cdot ) \big) (y) \mathcal{O}( \delta _{n}^{3} ) \\ & \phantom{=:} \qquad \qquad \ + \sum _{z \in \mathbb{S}_{n} \cap (\ell , L^{-}), [z, z^{+}] \mathcal{D} \ne \emptyset} \max _{y \in [z, z^{+}]} \partial _{y} \big( h(q, x; \cdot ) \widetilde{w}(q, \cdot ) \big) (y) \mathcal{O}( \delta _{n}^{2} ) \\ & \phantom{=:} \qquad \qquad \ + \mathcal{O}(L^{+} - L) + \mathcal{O}( \delta _{n}^{\gamma}) \\ &=\mathcal{O}(L^{+} - L) + \mathcal{O}(\delta _{n}^{\gamma}). \end{aligned}$$
In the second-to-last second equality, we use the error estimate of the trapezoidal rule and the smoothness of \(h(q, x; z)\) and \(\widetilde{w}(q, z)\). □
