It is not clear how to compute v in (2.7) or analyse it directly. Hence, in this section, we develop a way to approximate the value function v by a sequence of value functions, corresponding to simpler constant volatility optimal stopping problems.
Operator \(J_{i}\)
For the succinctness of notation, let \(\lambda _{i}:=\sum _{j \ne i} \lambda _{ij}\) denote the total intensity with which the volatility jumps from state \(\sigma _{i}\). Also, let us define
$$\begin{aligned} \eta ^{t}_{i}:= & {} \inf \{ s > 0 \,|\, \sigma (t+s) \ne \sigma (t) = \sigma _{i}\}, \end{aligned}$$
which is an Exp(\(\lambda _{i}\))-distributed random variable representing the duration up to the first volatility change if started from the volatility state \(\sigma _{i}\) at time t.
Furthermore, let us define an operator J acting on a bounded \(f : [0, T] \times (l,h) \rightarrow {\mathbb {R}}\) by
$$\begin{aligned}&(J f)(t, x, \sigma _{i})\nonumber \\&\quad := \sup _{\tau \in \mathcal {T}_{T-t}} {\tilde{{\mathbb {E}}}} \left[ e^{ \int _0^\tau {\hat{X}}^{t,x, \sigma _{i}}_{t+s} \,\mathrm {d}s } \mathbb {1}_{\{\tau < \eta ^{t}_{i} \}} + e^{ \int _0^{\eta ^{t}_{i}} {\hat{X}}^{t,x,\sigma _{i}}_{t+s} \,\mathrm {d}s }f\left( t+\eta ^{t}_{i}, {\hat{X}}^{t,x, \sigma _{i}}_{t+\eta ^{t}_{i}}\right) \mathbb {1}_{\{\tau \ge \eta ^{t}_{i}\}} \right] \nonumber \\ \end{aligned}$$
(3.1)
$$\begin{aligned}&\quad = \sup _{\tau \in \mathcal {T}_{T-t}} {\tilde{{\mathbb {E}}}} \left[ e^{ \int _0^\tau {\hat{X}}^{t,x, \sigma _{i}}_{t+s} -\lambda _{i} \,\mathrm {d}s } + \lambda _{i} \int _{0}^{\tau } e^{ \int _0^u {\hat{X}}^{t,x, \sigma _{i}}_{t+s} -\lambda _{i} \,\mathrm {d}s }f\left( t+u, {\hat{X}}^{t,x,\sigma _{i}}_{t+u}\right) \,\mathrm {d}u \right] , \nonumber \\ \end{aligned}$$
(3.2)
where \({\mathcal {T}}_{T-t}\) denotes the set of stopping times less or equal to \(T-t\) with respect to the usual augmentation of the filtration generated by \(\{ {\hat{X}}^{t,x,\sigma _{i}}_{t+s}\}_{s \ge 0}\) and \(\{ \sigma (t+s)\}_{s \ge 0}\). To simplify notation, we also define an operator \(J_{i}\) by
$$\begin{aligned} J_{i} f := (Jf)(\cdot , \cdot , \sigma _{i}). \end{aligned}$$
Intuitively, \((J_{i} f)\) represents a Markovian value function corresponding to optimal stopping before \(t+ \eta ^{t}_{i}\), i.e. before the first volatility change after t, when, at time \(t+\eta ^{t}_{i} < T\), the payoff \(f\left( t+\eta ^{t}_{i}, {\hat{X}}^{{t,x, \sigma _{i}}}_{t+\eta ^{t}_{i}} \right) \) is received provided stopping has not occurred yet.
Proposition 3.1
Let \(f : [0, T] \times (l,h) \rightarrow {\mathbb {R}}\) be bounded. Then
- (i)
Jf is bounded;
- (ii)
f increasing in the second variable x implies that Jf is increasing in the second variable x;
- (iii)
f decreasing in the first variable t implies that Jf is decreasing in the first variable t;
- (iv)
f increasing and convex in the second variable x implies that Jf is increasing and convex in the second variable x;
- (v)
J preserves order, i.e. \(f_{1} \le f_{2}\) implies \(J f_{1} \le J f_{2}\);
- (vi)
\(J f \ge 1\).
Proof
All except claim (iv) are straightforward consequences of the representation (3.2). To prove (iv), we will approximate the optimal stopping problem (3.2) by Bermudan options.
Let i and n be fixed. We will approximate the value function \(J_{i}f\) by a value function \(w^{(f)}_{i,n}\) of a corresponding Bermudan problem with stopping allowed only at times \(\left\{ \frac{kT}{2^{n}} \,:\, k\in \{0,1, \ldots , 2^{n}\}\right\} \). We define \(w^{(f)}_{i,n}\) recursively as follows. First,
$$\begin{aligned} w^{(f)}_{i,n}(T,x) := 1. \end{aligned}$$
Then, starting with \(k =2^{n}\) and continuing recursively down to \(k=1\), we define
$$\begin{aligned} w^{(f)}_{i,n}(t,x)= & {} \left\{ \begin{array}{ll} g\left( t,x, \frac{kT}{2^{n}}\right) , &{}\quad t \in \left( \frac{(k-1)T}{2^{n}}, \frac{kT}{2^{n}}\right) , \\ g\left( \frac{(k-1)T}{2^{n}},x, \frac{kT}{2^{n}}\right) \vee 1, &{}\quad t = \frac{(k-1)T}{2^{n}}, \\ \end{array} \right. \end{aligned}$$
(3.3)
where the function g is given by
$$\begin{aligned} g\left( t,x, \frac{kT}{2^{n}}\right):= & {} {\tilde{{\mathbb {E}}}} \bigg [ e^{ \int _t^{\frac{kT}{2^{n}}} {\hat{X}}^{t,x, \sigma _{i}}_{s} -\lambda _{i} \,\mathrm {d}s } w^{(f)}_{i,n}\left( \frac{kT}{n},{\hat{X}}^{t,x, \sigma _{i}}_{\frac{kT}{2^{n}}} \right) \nonumber \\&+ \int _{t}^{\frac{kT}{2^{n}}} e^{ \int _t^u {\hat{X}}^{t,x, \sigma _{i}}_{s} -\lambda _{i} \,\mathrm {d}s }f\Big (u, {\hat{X}}^{t,x,\sigma _{i}}_{u}\Big ) \,\mathrm {d}u \bigg ]. \end{aligned}$$
(3.4)
Next, we show by backward induction on k that \(w^{(f)}_{i,n}\) is increasing and convex in the second variable x. Suppose that for some \(k \in \{1, 2,\ldots , 2^{n}\}\), the function \(w^{(f)}_{i,n}\left( \frac{kT}{2^{n}}, \cdot \right) \) is increasing and convex (the assumption clearly holds for the base step \(k=2^{n}\)). Let \(t\in [ \frac{(k-1)T}{2^{n}}, \frac{kT}{2^{n}})\). Then, since f is also increasing and convex in the second variable x, we have that the function \(g(t,\cdot , \frac{kT}{2^{n}})\), and so \(w^{(f)}_{i,n}(t,\cdot )\), is convex by [14, Theorem 5.1]. Moreover, from (3.4) and [31, Theorem IX.3.7], it is clear that \(w^{(f)}_{i,n}(t,\cdot )\) is increasing. Consequently, by backward induction, we obtain that the Bermudan value function \(w^{(f)}_{i,n}\) is increasing and convex in the second variable.
Letting \(n \nearrow \infty \), the Bermudan value \(w^{(f)}_{i,n} \nearrow J_{i}f\) pointwise. As a result, \(J_{i}f\) is increasing and convex in the second argument, since convexity and monotonicity are preserved when taking pointwise limits. \(\square \)
The sets
$$\begin{aligned} \mathcal {C}^{f}_{i}:= & {} \{ (t,x) \in [0,T)\times (l, h) \,:\, (J_{i}f) (t,x) > 1 \}, \nonumber \\ \mathcal {D}^{f}_{i}:= & {} \{ (t,x) \in [0,T]\times (l, h) \,:\, (J_{i}f) (t,x) = 1 \} = [0,T]\times (l, h) \setminus \mathcal {C}^{f}_{i}, \end{aligned}$$
(3.5)
correspond to continuation and stopping sets for the stopping problem \(J_{i}f\) as the next proposition shows.
Proposition 3.2
(Optimal stopping time) The stopping time
$$\begin{aligned} \tau ^{f}_{\sigma _{i}}(t,x)= & {} \inf \{ u \in [0, T-t] \,:\, (t + u, {\hat{X}}^{t,x, \sigma _{i}}_{t+u}) \in \mathcal {D}^{f}_{i} \} \end{aligned}$$
(3.6)
is optimal for the problem (3.2).
Proof
A standard application of Theorem D.12 in [23]. \(\square \)
Proposition 3.3
If a bounded \(f :[0,T] \times (l,h) \rightarrow {\mathbb {R}}\) is decreasing in the first variable as well as increasing and convex in the second, then \(J_{i}f\) is continuous.
Proof
The argument is a trouble-free extension of the proof of the third part of Theorem 3.10 in [15]; still, we include it for completeness. Before we begin, in order to simplify notation, we will write \(u:= J_{i} f\).
Firstly, we let \(r\in (l, h)\) and will prove that there exists \(K>0\) such that, for every \(t\in [0,T]\), the map \(x\mapsto J_{i}f(t,x)\) is K-Lipschitz continuous on (l, r]. To obtain a contradiction, assume that there is no such K. Then, by convexity of u in the second variable, there is a sequence \(\{ t_{n}\}_{n\ge 0} \subset [0,T]\) such that the left-derivatives \(\partial ^{-}_{2} u(t_{n}, r) \nearrow \infty \). Hence, for \(r' \in (r, h)\), the sequence \(u(t_{n}, r') \rightarrow \infty \), which contradicts that \(u(t_{n}, r') \le u(0, r') < \infty \) for all \(n \in {\mathbb {N}}\).
Now, it remains to show that u is continuous in time. Assume for a contradiction that the map \(t \mapsto u(t, x_{0})\) is not continuous at \(t=t_{0}\) for some \(x_{0}\). Since u is decreasing in time, \(u(\cdot , x_{0})\) has a negative jump at \(t_{0}\). Next, we will investigate the cases \(u(t_{0}-, x_{0}) > u(t_{0}, x_{0})\) and \(u(t_{0}, x_{0}) > u(t_{0}+, x_{0})\) separately.
Suppose \(u(t_{0}-, x_{0}) > u(t_{0}, x_{0})\). By Lipschitz continuity in the second variable, there exists \(\delta >0\) such that, writing \(\mathcal {R} = (t_{0}-\delta , t_{0}) \times (x_{0} - \delta , x_{0}+\delta )\),
$$\begin{aligned} \inf _{(t,x) \in \mathcal {R}} u(t, x) > u(t_{0}, x_{0} + \delta ). \end{aligned}$$
(3.7)
Thus \(\mathcal {R} \subseteq \mathcal {C}^{f}_{i}\). Let \(t \in (t_{0}-\delta , t_{0})\) and \(\tau _{\mathcal {R}} := \inf \{ s \ge 0\,:\, (t+s, {\hat{X}}^{t,x,\sigma _{i}}_{t+\tau _{\mathcal {R}}}) \notin \mathcal {R} \}\). Then, by the martingality in the continuation region,
$$\begin{aligned} u(t,x_0)= & {} {\tilde{{\mathbb {E}}}} \bigg [ e^{ \int _0^{\tau _{{\mathcal {R}}}} {\hat{X}}^{t,x_0, \sigma _{i}}_{t+u} - \lambda _{i} \,\mathrm {d}u }u\Big (t+\tau _{\mathcal {R}}, {\hat{X}}^{t,x_0, \sigma _{i}}_{t+\tau _{{\mathcal {R}}}}\Big ) \\&+ \int _{0}^{\tau _{\mathcal {R}}} e^{ \int _0^u {\hat{X}}^{t,x_{0}, \sigma _{i}}_{t+s} -\lambda _{i} \,\mathrm {d}s }f\Big (t+u, {\hat{X}}^{t,x_{0},\sigma _{i}}_{t+u}\Big ) \,\mathrm {d}u \bigg ]\\\le & {} {\tilde{{\mathbb {E}}}} \bigg [ e^{ (t_0-t)(x_{0}+\delta )^{+} }u(t,x_0+\delta )\mathbb {1}_{\{t+\tau _{\mathcal {R}}<t_0\}} \\&+\,e^{ (t_0-t)(x_0+\delta )^{+}}u(t_0,x_0+\delta )\mathbb {1}_{\{t+\tau _{\mathcal {R}} = t_0\}}\\&+ \int _{0}^{t_{0}-t} e^{ \int _0^u {\hat{X}}^{t,x_{0}, \sigma _{i}}_{t+s} -\lambda _{i} \,\mathrm {d}s }\Big |f\Big (t+u, {\hat{X}}^{t,x_{0},\sigma _{i}}_{t+u}\Big )\Big | \,\mathrm {d}u \bigg ]\\\le & {} e^{(t_0-t)(x_0+\delta )^+}u(t,x_0+\delta ) {\tilde{{{\mathbb {P}}}}}(t+\tau _{\mathcal {R}} <t_0) + e^{(t_0-t)(x_0+\delta )^+}u(t_0,x_0+\delta )\\&+\int _{0}^{t_{0}-t} {\tilde{{\mathbb {E}}}} \left[ e^{ \int _0^u {\hat{X}}^{t,x_{0}, \sigma _{i}}_{t+s} -\lambda _{i} \,\mathrm {d}s } \Big |f\Big (t+u, {\hat{X}}^{t,x_{0},\sigma _{i}}_{t+u}\Big )\Big |\right] \,\mathrm {d}u \\\rightarrow & {} u(t_0,x_0+\delta ) \end{aligned}$$
as \(t \rightarrow t_{0}\), contradicting (3.7).
The other case to consider is \(u(t_{0}, x_{0}) > u(t_{0}+, x_{0})\); we look into the situation \(u(t_{0}, x_{0})> u(t_{0}+, x_{0})>1\) first. The local Lipschitz continuity in the second variable and the decay in the first variable imply that there exist \(\epsilon >0\) and \(\delta >0\) such that, writing \(\mathcal {R} = (t_{0}, t_{0}+\epsilon ] \times [x_{0}-\delta , x_{0}+\delta ]\),
$$\begin{aligned} u(t_{0}, x_{0})> \sup _{(t,x) \in \mathcal {R}} u(t,x) \ge \inf _{(t,x) \in \mathcal {R}} u(t,x) > 1. \end{aligned}$$
(3.8)
Hence, \({\mathcal {R}} \subseteq \mathcal {C}^{f}_{i}\) and writing \(\tau _{\mathcal {R}}:=\inf \{s\ge 0: (t_{0}+s, {\hat{X}}^{t_{0},x_0, \sigma _{i}}_{t_{0}+s})\notin {\mathcal {R}}\}\) we have
$$\begin{aligned} u(t_{0},x_0)= & {} {\tilde{{\mathbb {E}}}} \bigg [ e^{ \int _{0}^{\tau _{{\mathcal {R}}}} {\hat{X}}^{t_{0},x_0 , \sigma _{i}}_{t_{0}+u} - \lambda _{i} \,\mathrm {d}u } u\Big (t_0+\tau _{\mathcal {R}}, {\hat{X}}^{t_{0},x_0, \sigma _{i}}_{t_0+\tau _{{\mathcal {R}}}}\Big ) \\&+ \int _{0}^{\tau _{\mathcal {R}}} e^{ \int _0^u {\hat{X}}^{t_0,x_0, \sigma _{i}}_{t_0+s} -\lambda _{i} \,\mathrm {d}s }f\Big (t_0+u, {\hat{X}}^{t_0,x_0,\sigma _{i}}_{t_0+u}\Big ) \,\mathrm {d}u \bigg ] \\\le & {} {\tilde{{\mathbb {E}}}} \left[ e^{\epsilon (x_0+\delta )^+}u(t_{0},x_0+\delta )\mathbb {1}_{\{\tau _{\mathcal {R}}<\epsilon \}}\right] \\&+\,\,{\tilde{{\mathbb {E}}}} \bigg [e^{\epsilon (x_0+\delta )^+}u(t_0+\epsilon ,x_0+\delta )\mathbb {1}_{\{\tau _{\mathcal {R}} = \epsilon \}} \\&+\int _{0}^{\epsilon } e^{ \int _0^u {\hat{X}}^{t_0,x, \sigma _{i}}_{t_0+s} -\lambda _{i} \,\mathrm {d}s }\Big |f\Big (t_0+u, {\hat{X}}^{t_0,x_0,\sigma _{i}}_{t_0+u}\Big )\Big | \,\mathrm {d}u\bigg ]\\\le & {} e^{\epsilon (x_0+\delta )^+}u(t_{0},x_0+\delta ) {\tilde{{{\mathbb {P}}}}}(\tau _{\mathcal {R}} <\epsilon ) + e^{\epsilon (x_0+\delta )^+}u(t_0+\epsilon ,x_0+\delta )\\&+\int _{0}^{\epsilon } {\tilde{{\mathbb {E}}}} \left[ e^{ \int _0^u {\hat{X}}^{t_0,x_0, \sigma _{i}}_{t_0+s} -\lambda _{i} \,\mathrm {d}s }\Big |f\Big (t_0+u, {\hat{X}}^{t_0,x_0,\sigma _{i}}_{t_0+u}\Big )\Big | \right] \,\mathrm {d}u \\\rightarrow & {} u(t_0+,x_0+\delta ) \end{aligned}$$
as \(\epsilon \searrow 0\), which contradicts (3.8).
Lastly, suppose that \(u(t_{0}, x_{0}) > u(t_{0}+, x_{0}) = 1\). By Lipschitz continuity in the second variable, there exists \(\delta >0\) such that
$$\begin{aligned} \inf _{x\in (x_{0}-\delta , x_{0})}u(t_{0},x)>u(t_0+,x_0)=1. \end{aligned}$$
(3.9)
Consequently, \((t_{0}, T]\times (x_{0}-\delta , x_{0}) \subseteq \mathcal {D}^{f}_{i}\). Hence the process \({\hat{X}}^{t_{0}, x_{0}-\delta /2, \sigma _{i}}\) hits the stopping region immediately and so \((t_{0}, x_{0}-\delta /2) \in \mathcal {D}^{f}_{i}\), which contradicts (3.9). \(\square \)
Proposition 3.4
(Optimal stopping boundary) Let \(f: [0,T] \times (l, h) \rightarrow {\mathbb {R}}\) be bounded, decreasing in the first variable as well as increasing and convex in the second variable. Then the following hold.
- (i)
There exists a function \(b^{f}_{\sigma _{i}}: [0,T) \rightarrow [l,h]\) that is both increasing, right-continuous with left limits, and satisfies
$$\begin{aligned} \mathcal {C}^{f}_{i} = \{ (t,x) \in [0,T) \times (l, h) \,:\, x > b^{f}_{\sigma _{i}}(t) \}. \end{aligned}$$
(3.10)
- (ii)
The pair \((J_{i}f, b^{f}_{\sigma _{i}})\) satisfies the free-boundary problem
$$\begin{aligned} \left\{ \begin{array}{ll} \partial _{t}u(t,x) + {\sigma _{i}} \phi (x, \sigma _{i}) \partial _{x} u(t,x) + \frac{1}{2} \phi (x, \sigma _{i})^{2} \partial _{xx} u(t,x)\\ \quad +\,(x-\lambda _{i})u(t,x)+\lambda _{i} f(t,x) = 0, &{}\quad \text { if } x > b^{f}_{\sigma _{i}}(t), \\ u(t,x) = 1, &{}\quad \text { if } x \le b^{f}_{\sigma _{i}}(t) \text { or } t=T. \end{array} \right. \end{aligned}$$
(3.11)
Proof
-
(i)
By Proposition 3.1 (iv), there exists a unique function \(b^{f}_{\sigma _{i}}\) satisfying (3.10). Moreover, by Proposition 3.1 (iii), this boundary \(b^{f}_{\sigma _{i}}\) is increasing. Hence, using Proposition 3.3, we also obtain that \(b^{f}_{\sigma _{i}}\) is right-continuous with left limits.
-
(ii)
The proof follows a well-known standard argument (e.g. see [23, Theorem 7.7 in Chapter 2]), thus we omit it.
\(\square \)
A Sequence of Approximating Problems
Let us define a sequence of stopping times \(\{\xi ^{t}_{n}\}_{n \ge 0}\) recursively by
$$\begin{aligned} \xi ^{t}_{0}:= & {} 0, \\ \xi ^{t}_{n}:= & {} \inf \big \{ s> \xi ^{t}_{n-1}\,:\, \sigma ( t + s) \ne \sigma (t + \xi ^{t}_{n-1}) \big \}, \quad n > 0. \end{aligned}$$
Here \(\xi ^{t}_{n}\) represents the duration until the n-th volatility jump since time t. Furthermore, let us define a sequence of operators \(\{ J^{(n)} \}_{n\ge 0}\) by
$$\begin{aligned}&(J^{(n)} f)(t,x, \sigma _{i}) \nonumber \\&\quad := \sup _{\tau \in \mathcal {T}_{T-t}} {\tilde{{\mathbb {E}}}} \left[ e^{\int _0^\tau {\hat{X}}^{t,x, \sigma _{i}}_{t+s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau < \xi ^{t}_{n} \}} + e^{\int _0^{\xi ^{t}_{n}} {\hat{X}}^{t,x, \sigma _{i}}_{t+s} \,\mathrm {d}s }f\Big (t+\xi ^{t}_{n}, {\hat{X}}^{t,x, \sigma _{i}}_{t+\xi ^{t}_{n}}\Big ) \mathbb {1}_{\{\tau \ge \xi ^{t}_{n}\}} \right] ,\nonumber \\ \end{aligned}$$
(3.12)
where \(f : [0, T] \times (l,r) \rightarrow {\mathbb {R}}\) is bounded. In particular, note that \(J^{(0)} f = f\) and \(J^{(1)}f=Jf\). Similarly as for the operator J, we define \(J^{(n)}_{i}\) by
$$\begin{aligned} J^{(n)}_{i} f := (J^{(n)}f)(\cdot , \cdot , \sigma _{i}). \end{aligned}$$
Proposition 3.5
Let \(n \ge 0\) and \(i \in \{0,\ldots , m\}\). Then
$$\begin{aligned} J^{(n+1)}_{i}= J_{i} \left( \sum _{j\ne i} \frac{\lambda _{ij}}{\lambda _{i}} J^{(n)}_{j} \right) . \end{aligned}$$
(3.13)
Proof
The proof is by induction. In order to present the argument of the proof while keeping intricate notation at bay, we will only prove that, for a bounded \(f:[0,T]\times (l,h) \rightarrow {\mathbb {R}}\) and \(x\in (l,h)\), the identity \((J^{(2)}_{i}f)(t,x)= (J_{i}(\sum _{j\ne i} \frac{\lambda _{ij}}{\lambda _{i}} J_{j}f))(t,x)\) holds. The induction step \(J^{(n+1)}_{i}= J_{i} \left( \sum _{j\ne i} \frac{\lambda _{ij}}{\lambda _{i}} J^{(n)}_{j} \right) \) follows a similar argument, though with more abstract notation. Note that without loss of generality, we can assume \(t=0\), which we do.
Firstly, we will show \((J_{i}^{(2)}f)(0,x) \le J_{i}\bigg ( \sum _{j \ne i}\frac{\lambda _{ij}}{\lambda _{i}} (J_{j}f)\bigg )(0,x)\) and then the opposite inequality. For \(j \in {\mathbb {N}}\), we will write \(\xi _{j}\) instead of \(\xi ^{0}_{j}\) as well as will use the notation \(\eta _{j}:=\xi _{j}-\xi _{j-1}\). Let \(\tau \in \mathcal {T}_{T}\) and consider
$$\begin{aligned}&A(\tau )\nonumber \\&\quad := {\tilde{{\mathbb {E}}}} \left[ e^{\int _0^\tau {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau< \eta _{1} \}} + e^{\int _0^\tau {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \eta _{1} \le \tau< \xi _{2} \}} +e^{\int _0^{\xi _{2}} {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s }f(\xi _{2}, {\hat{X}}^{0,x, \sigma _i}_{\xi _{2}}) \mathbb {1}_{\{\tau \ge \xi _{2}\}} \right] \nonumber \\&\quad = {\tilde{{\mathbb {E}}}} \bigg [ e^{\int _0^\tau {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau< \eta _{1} \}} + {\tilde{{\mathbb {E}}}} \Big [ e^{\int _0^\tau {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \eta _{1} \le \tau < \xi _{2} \}} \nonumber \\&\qquad +\,e^{\int _0^{\xi _{2}} {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s }f(\xi _{2}, {\hat{X}}^{0,x, \sigma _i}_{\xi _{2}}) \mathbb {1}_{\{\tau \ge \xi _{2}\}} \, | \, {\mathcal {F}}^{{\hat{X}}^{0,x,\sigma _{i}}, N}_{\eta _{1}} \Big ] \bigg ], \end{aligned}$$
(3.14)
where \(\{N_{t}\}_{t\ge 0}\) denotes the process counting the volatility jumps. The inner conditional expectation in (3.14) satisfies
$$\begin{aligned}&{\tilde{{\mathbb {E}}}} \Big [ e^{\int _0^\tau {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \eta _{1} \le \tau< \xi _{2} \}} +e^{\int _0^{\xi _{2}} {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s }f(\xi _{2}, {\hat{X}}^{0,x, \sigma _i}_{\xi _{2}}) \mathbb {1}_{\{\tau \ge \xi _{2}\}} \, | \, {\mathcal {F}}^{{\hat{X}}^{0,x,\sigma _{i}}, N}_{\eta _{1}} \Big ] \nonumber \\&\quad = e^{\int _0^{\eta _{1}} {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \eta _{1} \le \tau \}}{\tilde{{\mathbb {E}}}} \Big [ e^{\int _{\eta _{1}}^{\tau } {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau< \xi _{2} \}} \nonumber \\&\qquad +\,e^{\int _{\eta _{1}}^{\xi _{2}} {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s }f(\xi _{2}, {\hat{X}}^{0,x, \sigma _i}_{\xi _{2}}) \mathbb {1}_{\{\tau \ge \xi _{2}\}} \, | \, {\mathcal {F}}^{{\hat{X}}^{0,x,\sigma _{i}}, N}_{\eta _{1}} \Big ] \nonumber \\&\quad = e^{\int _0^{\eta _{1}} {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \eta _{1} \le \tau \}} \sum _{j \ne i} \frac{\lambda _{ij}}{\lambda _{i}} {\tilde{{\mathbb {E}}}}^{\eta _{1},{\hat{X}}^{0,x,\sigma _i}_{\eta _{1}}, \sigma _{j}} \bigg [ e^{\int _{0}^{{\tilde{\tau }}} {\hat{X}}_{\eta _{1}+s} \,\mathrm {d}s} \mathbb {1}_{\{{\tilde{\tau }} < \eta _{2}\}}\nonumber \\&\qquad +\, e^{\int _{0}^{\eta _{2}} {\hat{X}}_{\eta _{1}+s} \,\mathrm {d}s }f(\eta _{1}+\eta _{2}, {\hat{X}}_{\eta _{1}+\eta _{2}}) \mathbb {1}_{\{{\tilde{\tau }} \ge \eta _{2}\}}\bigg ], \end{aligned}$$
(3.15)
where \({\tilde{\tau }} = \tau - \eta _{1}\) in the case \(\eta _{1} \le \tau \le T\). Therefore, substituting (3.15) into (3.14) and then taking a supremum over \({\tilde{\tau }}\), we get
$$\begin{aligned} A(\tau )\le & {} {\tilde{{\mathbb {E}}}} \bigg [ e^{\int _0^\tau {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau< \eta _{1} \}} \nonumber \\&+\,e^{\int _0^{\eta _{1}} {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau \ge \eta _{1} \}} \sum _{j \ne i} \frac{\lambda _{ij}}{\lambda _{i}} \sup _{{\tilde{\tau }} \in \mathcal {T}_{T-T\wedge \eta _{1}}} {\tilde{{\mathbb {E}}}}^{\eta _{1},{\hat{X}}^{0,x,\sigma _i}_{\eta _{1}}, \sigma _j} \Big [ e^{\int _{0}^{{\tilde{\tau }}} {\hat{X}}_{\eta _{1}+s} \,\mathrm {d}s} \mathbb {1}_{\{{\tilde{\tau }}< \eta _{2}\}} \nonumber \\&+\,e^{\int _{0}^{\eta _{2}} {\hat{X}}_{\eta _{1}+s} \,\mathrm {d}s }f(\eta _{1}+\eta _{2}, {\hat{X}}_{\eta _{1}+\eta _{2}}) \mathbb {1}_{\{{\tilde{\tau }} \ge \eta _{2}\}}\Big ]\bigg ] \nonumber \\= & {} {\tilde{{\mathbb {E}}}} \bigg [ e^{\int _0^\tau {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau < \eta _{1} \}} + e^{\int _0^{\eta _{1}} {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau \ge \eta _{1} \}} \sum _{j \ne i} \frac{\lambda _{ij}}{\lambda _{i}} (J_{j}f)(\eta _{1}, X^{0,x,\sigma _i}_{\eta _{1}}) \bigg ] \nonumber \\ \end{aligned}$$
(3.16)
Taking a supremum over \(\tau \) in (3.16), we obtain
$$\begin{aligned} (J_{i}^{(2)}f)(0,x) = \sup _{\tau \in \mathcal {T}_{T}} A(\tau ) \le J_{i}\bigg ( \sum _{j \ne i}\frac{\lambda _{ij}}{\lambda _{i}} (J_{j}f)\bigg )(0,x). \end{aligned}$$
(3.17)
It remains to establish the opposite inequality. Let \(\tau \in \mathcal {T}_{T}\) and define
$$\begin{aligned} \check{\tau }:= & {} \tau \mathbb {1}_{\{ \tau \le \eta _{1} \}} + (\eta _{1} \wedge T +\tau _{\sigma (\eta _{1})}) \mathbb {1}_{\{ \tau > \eta _{1} \}}, \end{aligned}$$
(3.18)
where \(\tau _{\sigma (\eta _{1})} := \tau ^{f}_{\sigma (\eta _{1})} (\eta _{1} \wedge T, {\hat{X}}^{0,x,\sigma _{i}}_{\eta _{1} \wedge T})\). Clearly, \(\check{\tau } \in \mathcal {T}_{T}\). Then
$$\begin{aligned}&(J^{(2)}_{1}f)(0,x) \\&\quad \ge A(\check{\tau }) \\&\quad = {\tilde{{\mathbb {E}}}} \bigg [ e^{\int _0^{\tau } {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau< \eta _{1} \}} + e^{\int _0^{\eta _{1}} {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau \ge \eta _{1}\}} \sum _{j \ne i} \frac{\lambda _{ij}}{\lambda _{i}} {\tilde{{\mathbb {E}}}}^{\eta _{1},{\hat{X}}^{0,x,\sigma _i}_{\eta _{1}}, \sigma _j} \Big [ e^{\int _{0}^{\tau _{\sigma _j}} {\hat{X}}_{\eta _{1}+s} \,\mathrm {d}s} \mathbb {1}_{\{\tau _{\sigma _j}< \eta _{2}\}} \\&\qquad +\,\, e^{\int _{0}^{\eta _{2}} {\hat{X}}_{\eta _{1}+s} \,\mathrm {d}s }f(\eta _{1}+\eta _{2}, {\hat{X}}_{\eta _{1}+\eta _{2}}) \mathbb {1}_{\{ \tau _{\sigma _j} \ge \eta _{2}\}}\Big ] \bigg ] \\&\quad = {\tilde{{\mathbb {E}}}} \bigg [ e^{\int _0^{\tau } {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau < \eta _{1} \}} +\,e^{\int _0^{\eta _{1}} {\hat{X}}^{0,x,\sigma _i}_{s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau \ge \eta _{1}\}} \sum _{j \ne i} \frac{\lambda _{ij}}{\lambda _{i}} (J_{j}f)(\eta _{1}, {\hat{X}}^{0,x,\sigma _i}_{\eta _{1}}) \bigg ], \end{aligned}$$
where Proposition 3.2 was used to obtain the last equality. Hence, by taking supremum over stopping times \(\tau \in \mathcal {T}_{T}\), we get
$$\begin{aligned} (J^{(2)}_{i}f)(0,x) \ge J_{i} \bigg ( \sum _{j \ne i}\frac{\lambda _{ij}}{\lambda _{i}} (J_{j}f) \bigg )(0,x). \end{aligned}$$
(3.19)
Finally, (3.17) and (3.19) taken together imply
$$\begin{aligned} (J^{(2)}_{i}f)(0,x) = J_{i} \bigg ( \sum _{j \ne i}\frac{\lambda _{ij}}{\lambda _{i}} (J_{j}f) \bigg )(0,x). \end{aligned}$$
\(\square \)
Remark 3.6
In [24], the authors use the same approximation procedure for an optimal stopping problem with regime switching volatility as in this article. Unfortunately, a mistake is made in equation (18) of [24], which wrecks the subsequent approximation procedure when the number of volatility states is greater than 2. The identity (18) therein should be replaced by (3.13).
Convergence to the Value Function
Proposition 3.7
(Properties of the approximating sequence)
- (i)
The sequence of functions \(\{ J^{(n)} 1 \}_{n \ge 0}\) is increasing, bounded from below by 1 and from above by \(e^{hT}\).
- (ii)
Every \(J^{(n)} 1 \) is decreasing in the first variable t as well as increasing and convex in the second variable x.
- (iii)
The sequence of functions
$$\begin{aligned} J^{(n)} 1 \nearrow v \quad \text { pointwise as } n \nearrow \infty . \end{aligned}$$
Moreover, the approximation error
$$\begin{aligned} \Vert v - J^{(n)} 1 \Vert _{\infty } \le e^{hT} \lambda T \frac{(\lambda T)^{n-1}}{(n-1)!} \text { as } n \rightarrow \infty , \end{aligned}$$
(3.20)
where \(\lambda := \max \{ \lambda _{i}\,:\, 1 \le i \le m \}\).
- (iv)
For every \(n \in {\mathbb {N}}\cup \left\{ 0\right\} \),
$$\begin{aligned} J_{m}^n 1 \le J^{(n)} 1 \le J_{1}^n 1. \end{aligned}$$
(3.21)
Proof
-
(i)
The statement that \(\{ J^{(n)}_{i} 1 \}_{n \ge 0}\) is increasing, bounded from below by 1 and from above by \(e^{hT}\) is a direct consequence of the definition (3.12).
-
(ii)
The claim that every \(J^{(n)}_{i} 1 \) is decreasing in the first variable t as well as increasing and convex in the second variable x follows by a straightforward induction on n, using Proposition 3.1 (iii),(iv) and Proposition 3.5 at the induction step.
-
(iii)
First, let \(i \in \{ 1, \ldots , m\}\) and note that, for any \(n \in {\mathbb {N}}\),
$$\begin{aligned} J^{(n)}_{i}1 \le v_{i}. \end{aligned}$$
Here the inequality holds by suboptimality, since \(J^{(n)}_{i}1\) corresponds to an expected payoff of a particular stopping time in the problem (2.4). Next, define
$$\begin{aligned} U^{(i)}_{n}(t,x):= & {} \sup _{\tau \in \mathcal {T}_{T-t}} {\tilde{{\mathbb {E}}}} \left[ e^{\int _0^\tau {\hat{X}}^{t,x, \sigma _{i}}_{t+s} \,\mathrm {d}s } \mathbb {1}_{\{ \tau < \xi ^{t}_{n} \}}\right] . \end{aligned}$$
Then
$$\begin{aligned} U^{(i)}_{n}(t,x) \le (J^{(n)}_{i}1) (t,x) \le v_{i}(t,x) \le U^{(i)}_{n}(t,x) + e^{h(T-t)} {{\mathbb {P}}}(\xi ^{t}_{n} \le T-t). \end{aligned}$$
(3.22)
Since it is a standard fact that the \(n^{\text {th}}\) jump time, call it \(\zeta _{n}\), of a Poisson process with jump intensity \(\lambda := \max \{ \lambda _{i}\,:\, 1 \le i \le m \}\) follows the Erlang distribution, we have
$$\begin{aligned} {{\mathbb {P}}}(\xi ^{t}_{n} \le T-t)\le & {} {{\mathbb {P}}}(\zeta _{n} \le T-t)\\= & {} \frac{1}{(n-1)!} \int _{0}^{\lambda (T-t)} u^{n-1} e^{-u} \,\mathrm {d}u \\\le & {} \lambda T \frac{(\lambda T)^{n-1}}{(n-1)!}. \end{aligned}$$
Therefore, by (3.22),
$$\begin{aligned} \Vert v - J^{(n)} 1 \Vert _{\infty } \le e^{hT} \lambda T \frac{(\lambda T)^{n-1}}{(n-1)!} \text { as } n \rightarrow \infty . \end{aligned}$$
-
(iv)
The string of inequalities (3.21) will be proved by induction. First, the base step is obvious. Now, suppose (3.21) holds for some \(n \ge 0\). Hence, for any \(i \in \{1, \ldots , m\}\),
$$\begin{aligned} J^{n}_{m} 1 \le \sum _{j\ne i} \frac{\lambda _{ij}}{\lambda _{i}} J^{(n)}_{j} 1 \le J^n_1 1. \end{aligned}$$
(3.23)
Let us fix \(i \in \left\{ 1, \ldots , m \right\} \). By Proposition 3.1 (iv), every function in (3.23) is convex in the spatial variable x, thus [14, Theorem 6.1] yields
$$\begin{aligned} J^{n+1}_{m} 1 \le J_i \left( \sum _{j\ne i} \frac{\lambda _{ij}}{\lambda _{i}} J^{(n)}_{j} 1 \right) \le J^{n+1}_1 1. \end{aligned}$$
As i was arbitrary, we also have
$$\begin{aligned} J_{\sigma _{m}}^{n+1} 1 \le J^{(n+1)} 1 \le J_{\sigma _{1}}^{n+1} 1. \end{aligned}$$
(3.24)
\(\square \)
Remark 3.8
If instead of 1 we choose the constant function \(e^{hT}\) to apply the operators \(J^{(n)}_{i}\) to, then, following the same strategy as above, \(\{ J^{(n)}_{i} e^{hT} \}_{n \ge 0}\) is a decreasing sequence of functions with the limit \( J^{(n)}_{i} e^{hT} \searrow v_{i}\) pointwise as \(n \nearrow \infty \).
Let \(\mathcal {B}_{b}([0,T]\times (l,h); {\mathbb {R}})\) denote the set of bounded functions from \([0,T]\times (l,h)\) to \({\mathbb {R}}\) and define an operator \( {\tilde{J}} : \mathcal {B}_{b}([0,T]\times (l,h); {\mathbb {R}})^{m} \rightarrow \mathcal {B}_{b}([0,T]\times (l,h); {\mathbb {R}})^{m}\) by
$$\begin{aligned} {\tilde{J}} \left( \begin{array}{c} f_{1} \\ \vdots \\ f_{m} \end{array}\right):= & {} \left( \begin{array}{c} J_{1} ( \sum _{j \ne 1} \frac{\lambda _{1j}}{\lambda _{1}} f_{j} ) \\ \vdots \\ J_{m} (\sum _{j \ne m} \frac{\lambda _{mj}}{\lambda _{m}} f_{j} )\end{array} \right) . \end{aligned}$$
Proposition 3.9
-
(i)
Let \(f \in \mathcal {B}_{b}([0,T]\times (l,h); {\mathbb {R}})^{m}\). Then
$$\begin{aligned} \lim _{n\rightarrow \infty } {\tilde{J}}^{n} f= & {} \left( \begin{array}{c} v_{1}\\ \vdots \\ v_{m} \end{array} \right) . \end{aligned}$$
-
(ii)
The vector \((v_{1}, \ldots , v_{m})^{tr}\) of value functions is a fixed point of the operator \({\tilde{J}}\), i.e.
$$\begin{aligned} {\tilde{J}} \left( \begin{array}{c} v_{1} \\ \vdots \\ v_{m} \end{array} \right)= & {} \left( \begin{array}{c} v_{1} \\ \vdots \\ v_{m} \end{array}\right) . \end{aligned}$$
(3.25)
Proof
-
(i)
Observe that the argument in the proof of part (iii) of Proposition 3.7 also gives that \(J^{(n)}_{i} g \rightarrow v_{i}\) as \(n \rightarrow \infty \) for any bounded g. Hence to finish the proof it is enough to recall the relation (3.13) in Proposition 3.5.
-
(ii)
Let \(i \in \{1, \ldots , m\}\). By Proposition 3.5,
$$\begin{aligned} J^{(n+1)}_{i}1= J_{i} \left( \sum _{j\ne i} \frac{\lambda _{ij}}{\lambda _{i}} J^{(n)}_{j} 1\right) . \end{aligned}$$
(3.26)
By Proposition 3.7 (iii), for every \(j \in \{1, \ldots , m\}\), the sequence \(J^{(n)}_{j} 1 \nearrow v_{j}\) as \(n \nearrow \infty \), so, letting \(n \nearrow \infty \) in (3.26), the monotone convergence theorem tells us that
$$\begin{aligned} v_{i}= J_{i} \left( \sum _{j\ne i} \frac{\lambda _{ij}}{\lambda _{i}} v_{j} \right) . \end{aligned}$$
(3.27)
\(\square \)