Optimal Stopping of a Hilbert Space Valued Diffusion: An Infinite Dimensional Variational Inequality

Abstract

A finite horizon optimal stopping problem for an infinite dimensional diffusion X is analyzed by means of variational techniques. The diffusion is driven by a SDE on a Hilbert space $\mathcal {H}$ with a non-linear diffusion coefficient $\sigma (X)$ and a generic unbounded operator A in the drift term. When the gain function $\Theta$ is time-dependent and fulfils mild regularity assumptions, the value function $\mathcal {U}$ of the optimal stopping problem is shown to solve an infinite-dimensional, parabolic, degenerate variational inequality on an unbounded domain. Once the coefficient $\sigma (X)$ is specified, the solution of the variational problem is found in a suitable Banach space $\mathcal {V}$ fully characterized in terms of a Gaussian measure $\mu$. This work provides the infinite-dimensional counterpart, in the spirit of Bensoussan and Lions (Application of variational inequalities in stochastic control, 1982), of well-known results on optimal stopping theory and variational inequalities in $\mathbb {R}^n$. These results may be useful in several fields, as in mathematical finance when pricing American options in the HJM model.

Appendix

Proof of Proposition 4.1

Fix $(t,x^{(n)})\in [0,T]\times \mathbb {R}^n$ and take $\overline{R}>0$ such that $x^{(n)}\in \mathcal {O}_{\overline{R}}$. Now for all $R\ge \overline{R}$ we have

$$\begin{aligned} 0&\le \mathcal {U}^{(n)}_{\alpha }(t,x^{(n)})-\mathcal {U}^{(n)}_{\alpha ,R}(t,x^{(n)})\\&\le \sup _{t\le \sigma \le T}\mathbb {E}\left\{ \left( \Theta ^{(n)}(\sigma ,X^{(\alpha )t,x;n}_{\sigma })-\Theta ^{(n)} (\tau _{R}, X^{(\alpha )t,x;n}_{\tau _{R}})\right) I_{\{\sigma >\tau _{R}\}}\right\} \!\le \! 2\,\overline{\Theta }\,\mathbb {P}\big (\tau _R\!<\!T\big ), \end{aligned}$$

by (2.11) and with $I_{\{\sigma >\tau _{R}\}}$ the indicator function of the set $\{\sigma >\tau _{R}\}$. By Markov inequality and standard estimates for strong solutions of SDEs in $\mathbb {R}^n$ (cf. for instance [18] Chapter 2, Section 5, Corollary 12), it follows

$$\begin{aligned} \mathbb {P}\big (\tau _R<T\big )&\le \mathbb {P}\Big (\sup _{t\le s\le T}\big \Vert X^{(\alpha )t,x;n}_s-x^{(n)}\big \Vert _{\mathbb {R}^n}>R-\overline{R}\Big )\\&\le \frac{\mathbb {E}\Big \{\sup _{t\le s\le T}\big \Vert X^{(\alpha )t,x;n}_s\!-\!x^{(n)}\big \Vert _{\mathbb {R}^n}\Big \}}{(R-\overline{R})}\!\le \! C_{n,\alpha ,T}\Big (1\!+\!\big \Vert x^{(n)}\big \Vert _{\mathbb {R}^n}\Big )\frac{(T\!-\!t)^{\frac{1}{2}}}{(R\!-\!\overline{R})} \end{aligned}$$

with $C_{n,\alpha ,T}>0$, only depending on $(\alpha ,n,T)$ and bounds on $\sigma$.

Therefore

$$\begin{aligned} \lim _{R\rightarrow \infty }\sup _{(t,x^{(n)})\in [0,T]\times \mathcal {K}}\big |\mathcal {U}^{(n)}_{\alpha ,R}(t,x^{(n)})-\mathcal {U}^{(n)}_{\alpha }(t,x^{(n)})\big |=0 \end{aligned}$$

for every compact subset $\mathcal {K}\subset \mathbb {R}^n$. If all $\mathcal {U}^{(n)}_{\alpha ,R}$, are continuous, then $\mathcal {U}^{(n)}_{\alpha }$ is continuous on every compact subset $[0,T]\times \mathcal {K}$ and this is enough for global continuity in $\mathbb {R}^n$. $\square$

Proof of Corollary 4.3

By the regularity of $\bar{u}$ in Corollary 4.2, it is well known that the expression (4.11) is equivalent to

$$\begin{aligned} \max \left\{ \frac{\partial \bar{u}}{\partial t}+\mathcal {L}_{\alpha ,n}\bar{u}+f_{\alpha ,n},\,-\bar{u}\right\} =0, \qquad \text {a.e.}\,\in [0,T]\times \overline{\mathcal {O}}_R. \end{aligned}$$

(see for instance [4], Chapter 3, Section 1, p. 191).

The regularity of $\partial \mathcal {O}_R$ and [1], Theorem 3.22 enable us to find a sequence $(u_j)_{j\in \mathbb {N}}$, such that $u_j\in C^\infty _c(\mathbb {R}^{n+1})$ and

$$\begin{aligned} \Vert u_j-\bar{u}\Vert _{W^{1\,2,p}((0,T)\times \mathcal {O}_R)}\rightarrow 0\quad \text {as}\quad \,j\rightarrow \infty . \end{aligned}$$

(6.1)

In fact it suffices to take a partition of the domain and use the standard mollification on each element of the partition. Then (6.1) follows from the usual properties of the mollifiers and the fact that the operators $\partial _t$, D and $D^2$ are closed in $L^p$. Moreover, the continuity of $\bar{u}$ and that of a suitable extension to $\mathbb {R}^{n+1}$ imply that the convergence is also uniform on any compact set $\mathcal {O}^\prime$ such that $[0,T]\times \overline{\mathcal {O}}_R\subset \mathcal {O}^\prime$; that is

$$\begin{aligned} \Vert u_j-\bar{u}\Vert _{L^\infty }\rightarrow 0,\qquad \,\text {as}\,j\rightarrow \infty ,\quad \text {on}\quad \,\mathcal {O}^{\prime }. \end{aligned}$$

(6.2)

Now we fix an arbitrary $t\in [0,T]$ and a stopping time $\tau \in [t,T]$. An application of Dynkin’s formula from t to $\tau \wedge \tau _R$ gives

$$\begin{aligned}&{\mathbb {E}}\left\{ u_j(\tau \wedge \tau _{R},X^{(\alpha )t,x;n}_{\tau \wedge \tau _{R}})\right\} =u_j(t,x^{(n)})\nonumber \\&\quad +\mathbb {E} \left\{ \int _t^{\tau \wedge \tau _{R}}{\left( \frac{\partial u_j}{\partial s}+\mathcal {L}_{\alpha ,n}u_j\right) (s,X^{(\alpha )t,x;n}_s)ds}\right\} . \end{aligned}$$

(6.3)

On the other hand by [4], Chapter 2, Lemma 8.1 there exists a constant $C_{T,R}>0$ such that

$$\begin{aligned}&\left| \mathbb {E}\left\{ \int _t^{\tau \wedge \tau _{R}}{\left( \frac{\partial }{\partial s}+\mathcal {L}_{\alpha ,n}\right) \left( u_j-\bar{u}\right) (s,X^{(\alpha )t,x;n}_s)ds}\right\} \right| \nonumber \\&\quad \le C_{T,R}\left\| \left( \frac{\partial }{\partial s}+\mathcal {L}_{\alpha ,n}\right) \left( u_j-\bar{u}\right) \right\| _{L^2((0,T)\times \mathcal {O}_R)}, \end{aligned}$$

(6.4)

hence by taking the limit as $j\rightarrow \infty$ and by using (6.1) and (6.2) we obtain

$$\begin{aligned}&\mathbb {E}\left\{ \bar{u}(\tau \wedge \tau _{R},X^{(\alpha )t,x;n}_{\tau \wedge \tau _{R}})\right\} =\bar{u}(t,x^{(n)})\nonumber \\&\quad +\mathbb {E} \left\{ \int _t^{\tau \wedge \tau _{R}}{\left( \frac{\partial \bar{u}}{\partial s}+\mathcal {L}_{\alpha ,n}\bar{u}\right) (s,X^{(\alpha )t,x;n}_s)ds}\right\} \,\quad \text {for all}\quad \,\tau \in [t,T]. \end{aligned}$$

(6.5)

Recall that (4.12) holds almost everywhere in $(0,T)\times \mathcal {O}_R$ and, being the diffusion uniformly non degenerate, the law of $X^{(\alpha )t,x;n}$ is absolutely continuous with respect to the Lebesgue measure on $(0,T)\times \mathcal {O}_R$. Then

$$\begin{aligned} \bar{u}(t,x^{(n)})\ge \mathbb {E}\left\{ \int _t^{\tau \wedge \tau _R}{f_{\alpha ,n}(s,X^{(\alpha )t,x;n}_s)ds}\right\} \qquad \text {for all}\,\tau \in [t,T]; \end{aligned}$$

(6.6)

in particular, with $\tau ^\star$ defined by

$$\begin{aligned} \tau ^\star :=\inf \{s\ge t\,:\,\bar{u}(s,X^{(\alpha )t,x;n}_s)=0\}\wedge {{\tau _R}}\wedge T, \end{aligned}$$

(6.7)

(4.12) implies

$$\begin{aligned} \bar{u}(t,x^{(n)})=\mathbb {E}\left\{ \int _t^{{{\tau ^\star }}}{f_{\alpha ,n}(s,X^{(\alpha )t,x;n}_s)ds}\right\} . \end{aligned}$$

(6.8)

Therefore, by using (4.10) and by recalling (4.2) we have

$$\begin{aligned} \bar{u}(t,x^{(n)})&=\sup _{t\le \tau \le T}\mathbb {E}\left\{ \int _t^{\tau \wedge \tau _R}{\left( \frac{\partial \Theta ^{(n)}}{\partial s}+\mathcal {L}_{\alpha ,n}\Theta ^{(n)}\right) (s,X^{(\alpha )t,x;n}_s)ds}\right\} \nonumber \\&=\sup _{t\le \tau \le T}\mathbb {E}\left\{ \Theta ^{(n)}(\tau \wedge \tau _R,X^{(\alpha )t,x;n}_{\tau \wedge \tau _R})\right\} -\Theta ^{(n)}(t,x)\nonumber \\&=\mathcal {U}^{(n)}_{\alpha ,R}(t,x)-\Theta ^{(n)}(t,x). \end{aligned}$$

(6.9)

It now follows that $\bar{u}=u^{(n)}_{\alpha ,R}$ and $\tau ^\star =\tau ^\star _{\alpha ,n,R}$.

Notice that for any stopping time $\tau \le \tau ^\star _{\alpha ,n,R}$, combining (6.9) and (6.5) gives

$$\begin{aligned} \mathcal {U}^{(n)}_{\alpha ,R}(t,x^{(n)})=\mathbb {E}\left\{ \mathcal {U}^{(n)}_{\alpha ,R}({{\tau }},X^{(\alpha )t,x;n} _{{{\tau }}})\right\} , \end{aligned}$$

(6.10)

i.e. the dynamic programming principle for $\mathcal {U}^{(n)}_{\alpha ,R}$ holds. $\square$

Proof of Lemma 4.6

Set $u^R:=u^{(n)}_{\alpha ,R}$ and recall Corollary 4.3. An application of Dynkin’s formula based on the same arguments as those that lead to (6.5) gives

$$\begin{aligned} \mathbb {E}&\left\{ e^{-\int ^{\tau ^x_R}_t{\,\frac{1}{\varepsilon }\nu (s)ds}}\,u^R\big (\tau ^x_R,X^{(\alpha )t,x;n}_{\tau ^x_R}\big )- e^{-\int ^{\tau ^x_R\wedge \tau ^y_R}_t{\,\frac{1}{\varepsilon }\nu (s)ds}}\,u^R\big (\tau ^x_R\wedge \tau ^y_R,X^{(\alpha )t,x;n}_{\tau ^x_R\wedge \tau ^y_R}\big )\right\} \nonumber \\ \le&-\mathbb {E}\left\{ \int ^{\tau ^x_R}_{\tau ^x_R\wedge \tau ^y_R}{ e^{-\int ^s_t{\frac{1}{\varepsilon }\nu (u)du}}f_{\alpha ,n}\big (s,X^{(\alpha )t,x;n}_s\big )ds}\right\} . \end{aligned}$$

(6.11)

For the left-hand side of (6.11) we observe that on the set $\big \{\tau ^x_R\le \tau ^y_R\big \}$ the difference inside the expectation is zero, whereas on the set $\big \{\tau ^x_R>\tau ^y_R\big \}$ one has

$$\begin{aligned} u^R\big (\tau ^x_R,X^{(\alpha )t,x;n}_{\tau ^x_R}\big )=0=u^R\big (\tau ^x_R\wedge \tau ^y_R,X^{(\alpha )t,y;n}_{\tau ^x_R\wedge \tau ^y_R}\big )\qquad \mathbb {P}-\text { a.s.} \end{aligned}$$

(6.12)

Therefore from (6.11), (4.9), (2.12), (2.20) and Lemma 2.9 we obtain

$$\begin{aligned} \mathbb {E}\bigg \{\int ^{\tau ^x_R}_{\tau ^x_R\wedge \tau ^y_R}&{ e^{-\int ^s_t{\frac{1}{\varepsilon }\nu (u)du}}f_{\alpha ,n}\big (s,X^{(\alpha )t,x;n}_s\big )ds}\bigg \}\nonumber \\ \le&\mathbb {E}\left\{ \big |u^R\big (\tau ^x_R\wedge \tau ^y_R,X^{(\alpha )t,y;n}_{\tau ^x_R\wedge \tau ^y_R}\big )- u^R\big (\tau ^x_R\wedge \tau ^y_R,X^{(\alpha )t,x;n}_{\tau ^x_R\wedge \tau ^y_R}\big )\big |\right\} \\ \le&\big (L_\Theta +L_\mathcal {U}\big )C_{1,T}\big \Vert x^{(n)}-y^{(n)}\big \Vert _{\mathcal {H}}.\nonumber \end{aligned}$$

(6.13)

To obtain (4.17) we need to find a similar bound for the first member of (6.13) but from below. For that we introduce the auxiliary problem

$$\begin{aligned} v^R(t,x^{(n)}):=\inf _{t\le \tau \le T}\mathbb {E}\left\{ \int _t^{\tau \wedge \tau _R}{f_{\alpha ,n}(s,X^{(\alpha )t,x;n}_s)ds}\right\} \qquad \text {for } (t,x^{(n)})\in [0,T]\times \mathbb {R}^n \end{aligned}$$

(6.14)

and we observe that same arguments as those used to obtain Proposition 4.2 and Corollary 4.3 give $v^R\in L^p(0,T;W^{1,p}_0(\mathcal {O}_R))\cap L^p(0,T;W^{2,p}(\mathcal {O}_R))$ and $\frac{\partial \,v^R}{\partial \,t}\in L^p(0,T;L^p(\mathcal {O}_R))$, for all $1\le p<+\infty$. Moreover $v^R$ uniquely solves, in the almost everywhere sense, the obstacle problem

$$\begin{aligned} \left\{ \begin{array}{ll} \max \left\{ -\displaystyle {\frac{\partial v}{\partial t}}-\mathcal {L}_{\alpha ,n}v-f_{\alpha ,n},\,v\right\} (t,x^{(n)})= 0,\qquad (t,x^{(n)})\in (0,T)\!\times \!\mathcal {O}_R, &{}\\ v(t,x^{(n)})\le 0\quad \text {on}\quad [0,T]\times \overline{\mathcal {O}}_R;\quad v(T,x^{(n)})\!=\!0,\qquad x^{(n)}\!\in \!\overline{\mathcal {O}}_R.&{}\\ \end{array} \right. \nonumber \\ \end{aligned}$$

(6.15)

Again, by arguing as above for (6.11) and by replacing $u^R$ by $v^R$, the reversed inequality is obtained. Hence, the analogous for $v^R$ of (6.12) gives

$$\begin{aligned} \mathbb {E}\bigg \{\int ^{\tau ^x_R}_{\tau ^x_R\wedge \tau ^y_R}{e^{-\int ^s_t{\frac{1}{\varepsilon }\nu (u)du}}f_{\alpha ,n}\big (s,X^{(\alpha )t,x;n}_s\big )ds}\bigg \}\ge -\big (L_\Theta +L_\mathcal {U}\big )C_{1,T}\big \Vert x^{(n)}-y^{(n)}\big \Vert _{\mathcal {H}}. \end{aligned}$$

(6.16)

Now (4.17) follows by (6.13) and (6.16). $\square$

Proof of Lemma 4.7

It is enough to show that $\Vert u^R_\varepsilon (t,x^{(n)})-u^R_\varepsilon (t,y^{(n)})\Vert \le L_P\Vert x^{(n)}-y^{(n)}\Vert _{\mathcal {H}}$ for all $t\in [0,T]$ and $x,y\in \mathcal {H}$. Recalling (4.10) and (4.19), we find

(6.17)

From Itô’s formula, (4.10) and Lemma 4.6 one finds

$$\begin{aligned} \mathbb {E}\bigg \{e^{-\int ^{\tau ^x_R}_t{\frac{1}{\varepsilon }\nu (s)}ds}\,&\Theta ^{(n)}\big (\tau ^x_R,X^{(\alpha )t,x;n}_{\tau ^x_R}\big )\bigg \}\\ \le&L_f\big \Vert x^{(n)}-y^{(n)}\big \Vert +\mathbb {E}\bigg \{e^{-\int ^{\tau ^x_R\wedge \tau ^y_R}_t{\frac{1}{\varepsilon } \nu (s)}ds}\, \Theta ^{(n)}\big (\tau ^x_R\wedge \tau ^y_R,X^{(\alpha )t,x;n}_{\tau ^x_R\wedge \tau ^y_R}\big )\bigg \}\nonumber \\&-\mathbb {E}\bigg \{\int ^{\tau ^x_R}_{\tau ^x_R\wedge \tau ^y_R}{e^{-\int ^s_t{\frac{1}{\varepsilon }\nu (u)du}} \frac{1}{\varepsilon }\nu (s)\Theta ^{(n)}\big (s,X^{(\alpha )t,x;n}_s\big )ds}\bigg \}\nonumber \end{aligned}$$

(6.18)

and similarly,

$$\begin{aligned} \mathbb {E}\bigg \{e^{-\int ^{\tau ^y_R}_t{\frac{1}{\varepsilon }\nu '(s)}ds}\,&\Theta ^{(n)}\big (\tau ^y_R,X^{(\alpha )t,y;n}_{\tau ^y_R}\big )\bigg \}\\ \ge&-L_f\big \Vert x^{(n)}-y^{(n)}\big \Vert +\mathbb {E}\bigg \{e^{-\int ^{\tau ^x_R\wedge \tau ^y_R}_t{\frac{1}{\varepsilon } \nu '(s)}ds} \,\Theta ^{(n)}\big (\tau ^x_R\wedge \tau ^y_R,X^{(\alpha )t,y;n}_{\tau ^x_R\wedge \tau ^y_R}\big )\bigg \}\nonumber \\&-\mathbb {E}\bigg \{\int ^{\tau ^y_R}_{\tau ^x_R\wedge \tau ^y_R}{e^{-\int ^s_t{\frac{1}{\varepsilon }\nu '(u)du}} \frac{1}{\varepsilon }\nu '(s)\Theta ^{(n)}\big (s,X^{(\alpha )t,y;n}_s\big )ds}\bigg \}.\nonumber \end{aligned}$$

(6.19)

Take now

$$\begin{aligned} \nu '(s)=\nu (s) \text { for } s\in (t,\tau ^x_R\wedge \tau ^y_R]\, \text { and }\, \nu '(s)=0 \text { for } s>\tau ^x_R\wedge \tau ^y_R, \end{aligned}$$

(6.20)

then from (6.17), (6.18), (6.19), (6.20) and recalling (2.12) and Lemma 2.9 we obtain

$$\begin{aligned} u^R_\varepsilon (t,x^{(n)})&-u^R_\varepsilon (t,y^{(n)}) \nonumber \\&\le \big (2L_f+L_\Theta \big )\big \Vert x^{(n)}-y^{(n)}\big \Vert _{\mathcal {H}}\nonumber \\&\quad +\mathbb {E}\left\{ \Big |\Theta ^{(n)}\big (\tau ^x_R\wedge \tau ^y_R,X^{(\alpha )t,x;n}_{\tau ^x_R\wedge \tau ^x_R}\big )- \Theta ^{(n)}\big (\tau ^y_R\wedge \tau ^x_R,X^{(\alpha )t,y;n}_{\tau ^y_R\wedge \tau ^x_R}\big )\Big |\right\} \\&\quad + \sup _\nu \mathbb {E}\bigg \{\int ^{\tau ^x_R}_{\tau ^x_R\wedge \tau ^y_R} e^{-\int ^s_t{\frac{1}{\varepsilon }\nu (u)du}}\frac{1}{\varepsilon }\nu (s)\big (\Theta ^{(n)}\big (s,X^{(\alpha )t,x;n}_s\big )\nonumber \\&\quad -\Theta ^{(n)}\big (s,X^{(\alpha )t,y;n}_s\big )\big )ds \bigg \}\nonumber \\ \le&\big (2L_f+L_\Theta +2L_\Theta C_{1,T}\big )\big \Vert x^{(n)}-y^{(n)}\big \Vert _{\mathcal {H}}.\nonumber \end{aligned}$$

(6.21)

One can argue in a similar way to bound $u^R_\varepsilon (t,y^{(n)})-u^R_\varepsilon (t,x^{(n)})$. $\square$