Optimal Stopping for Dynamic Risk Measures with Jumps and Obstacle Problems

Abstract

We study the optimal stopping problem for a monotonous dynamic risk measure induced by a Backward Stochastic Differential Equation with jumps in the Markovian case. We show that the value function is a viscosity solution of an obstacle problem for a partial integro-differential variational inequality and we provide an uniqueness result for this obstacle problem.

Appendix

1.1 Some Useful Estimates

Let $T>0$ be a fixed terminal time. A map $f: [0,T] \times \Omega \times \mathbb {R}^2 \times L_{\nu }^2 \rightarrow \mathbb {R}; (t, \omega , y,z,k) \mapsto f(t, \omega , y,z,k)$ is said to be a Lipschitz driver if it is predictable, uniformly Lipchitz with respect to $y,z,k$ and such that $f(t,0,0,0) \in \mathbb {H}^2.$

Let $\xi _t^1, \xi _t^2 \in \fancyscript{S}^2$. Let $f^1 and f^2$ be two admissible Lipschitz drivers with Lipchitz constant $C$. For $i=1,2$, let $\fancyscript{E}^{i}$ be the $f^{i}$-conditional expectation associated with driver $f^{i}$, and let ($Y_t^{i}$) be the adapted process defined for each $t \in [0,T]$,

$$\begin{aligned} Y_t^{i}:=\mathrm{ess} \sup _{\tau \in \fancyscript{T}_t} \fancyscript{E}^{i}_{t, \tau }(\xi ^{i}_\tau ). \end{aligned}$$

(48)

Proposition 6.1

For $s \in [0,T]$, denote $\overline{Y}_s=Y_s^1-Y_s^2$, $\overline{\xi }_s=\xi _s^1-\xi _s^2$ and

$\overline{f}_s= \sup _{y,z,k}|f^1(s,y,z,k)-f^2(s,y,z,k)|$. Let $\eta , \beta >0$ be such that $\beta \ge \dfrac{3}{\eta }+2C$ and $\eta \le \dfrac{1}{C^2}$. Then for each $t$, we have

$$\begin{aligned} e^{\beta t} \overline{Y}_t^2 \le e^{\beta T}\left( \mathbb {E}\left[ \sup _{s \ge t} \overline{\xi _s}^2| \fancyscript{F}_t\right] + \eta \mathbb {E}\left[ \int _t^T{\overline{f}_s^2}\mathrm{d}s|\fancyscript{F}_t\right] \right) \textsc { } a.s. \end{aligned}$$

(49)

Proof

For $i=1,2$ and for each $\tau \in \fancyscript{T}_{0}$, let $(X^{i,\tau }$, $\pi _s^{i,\tau }, l_s^{i, \tau })$ be the solution of the BSDE associated with driver $f^i$, terminal time $\tau$, and terminal condition $\xi _{\tau }^i$. Set $\overline{X}_s^{\tau }=X_s^{1,\tau }-X_s^{2,\tau }$.

By a priori estimate on BSDEs (see Proposition $A.4$ in [5]), we have

$$\begin{aligned} e^{\beta t} (\overline{X}_t^{\tau })^2&\le e^{\beta T} \mathbb {E}\left[ \overline{\xi }_{\tau }^2|\fancyscript{F}_t\right] + \eta \mathbb {E}\left[ \int _t^Te^{\beta s}(f^1(s, X_s^{2,\tau },\pi _s^{2,\tau }, l_s^{2,\tau })\right. \nonumber \\ {}&\quad \left. -f^2(s, X_s^{2,\tau },\pi _s^{2,\tau }, l_s^{2,\tau }))^2 \mathrm{d}s| \fancyscript{F}_t\right] \quad \text { a.s. } \end{aligned}$$

(50)

from which, we derive that

$$\begin{aligned} e^{\beta t} (\overline{X}_t^{\tau })^2 \le e^{\beta T} \left( \mathbb {E}\left[ \sup _{s \ge t}\overline{\xi }_{s}^2|\fancyscript{F}_t\right] + \eta \mathbb {E}\left[ \int _t^T{\overline{f}_s^2 \mathrm{d}s}| \fancyscript{F}_t\right] \right) . \end{aligned}$$

(51)

Now, by definition of $Y^{i}$, we have $Y_t^i =\mathrm{ess} \sup _{\tau \ge t} X_t^{i, \tau }$ a.s. for $i=1,2$. We thus get $|\overline{Y}_t| \le \mathrm{ess} \sup _{\tau \ge t}|\overline{X}_t^{\tau }|$ a.s. The result follows. $\square$

Let $\xi _t \in \fancyscript{S}^2$. Let $f$ be a Lipschitz driver with Lipschitz constant $C>0$. Set

$$\begin{aligned} Y_t:=\mathrm{ess} \sup _{\tau \in \fancyscript{T}_t} \fancyscript{E}_{t, \tau }(\xi _\tau ), \end{aligned}$$

(52)

where $\fancyscript{E}$ is the $f$-conditional expectation associated with driver $f$.

Proposition 6.2

Let $\eta , \beta >0$ be such that $\beta \ge \dfrac{3}{\eta }+2C$ and $\eta \le \dfrac{1}{C^2}$. Then for each $t$, we have

$$\begin{aligned} e^{\beta t} Y_t^2 \le e^{\beta T}\left( \mathbb {E}\left[ \sup _{s \ge t} {\xi _s}^2| \fancyscript{F}_t\right] + \eta \mathbb {E}\left[ \int _t^T{f(s,0,0,0)^2}\mathrm{d}s|\fancyscript{F}_t\right] \right) \textsc { } a.s. \end{aligned}$$

(53)

Proof

Let $X_t^{\tau }$ be the solution of the BSDE associated with driver $f$, terminal time $\tau$, and terminal condition $\xi _{\tau }$. By applying inequality (50) with $f^1=f$, $\xi _1=\xi$, $f^2=0$ and $\xi ^2=0$, we get

$$\begin{aligned} e^{\beta t} (X_t^{\tau })^2 \le e^{\beta T} \mathbb {E} [\xi _{\tau }^2|\fancyscript{F}_t]+ \eta \mathbb {E} [\int _t^T{e^{\beta s}(f(s,0,0,0))^2}|\fancyscript{F}_t]. \end{aligned}$$

(54)

The result follows. $\square$

Remark 6.1

If the drivers satisfy Assumption 3.1 in [5], then $Y$ (resp. $Y^{i}$) is the solution of the RBSDE associated with driver $f$ (resp.$f^{i}$) and obstacle $\xi$ (resp. $\xi ^{i}$). Hence the above estimates provide some new estimates on RBSDEs. Note that $\eta$ and $\beta$ are universal constants, i.e., they do not depend on $T$, $\xi , \xi ^1, \xi ^2, f, f^1, f^2$. This was not the case for the estimates given in the previous literature (see e.g. [6]).

1.2 Some Properties of the Value Function $u$

We prove below the continuity and polynomial growth of the function $u$ defined by (8).

Lemma 6.1

The function $u$ is continuous in $(t,x)$.

Proof

It is sufficient to show that when $(t_n, x_n) \rightarrow (t,x)$, $|u(t_n, x_n)-u(t,x)| \rightarrow 0$.

Let $\bar{h}$ be the map defined by $\bar{h} (t,x) =h(t,x)$ for $t<T$ and $\bar{h} (T,x) =g(x)$, so that, for each $(t,x)$, we have $\xi ^{t,x}_s = \bar{h}(s, X^{t,x}_s)$, $0 \le s \le T$ a.s.

By applying Proposition 6.1 with $X_s^1=X_s^{t_n,x_n}$, $X^2_s= X_s^{t,x}$, $f^1(s, \omega , y,z,q):= \mathbf 1 _{[t,T]}(s) f(s, X_s^{t,x}(\omega ),y,z,q)$ and

$f^2(s, \omega , y,z,q):= \mathbf 1 _{[t_n,T]}(s) f(s, X_s^{t_n,x_n}(\omega ),y,z,q)$, we obtain

$$\begin{aligned} |u(t_n, x_n)-u(t,x)|^2 \le K_{C,T} \mathbb {E}\left[ \sup _{0\le s\le T}|\overline{h}(s, X_{s}^{t_n, x_n}) - \overline{h}(s, X_{s}^{t,x})|^2+ \int _{0}^{T}(\overline{f}_s^n)^2\right] , \end{aligned}$$

where

$$\begin{aligned}&K_{C,T}:=e^{(3C^2+2C)T}\max \left( 1, \dfrac{1}{C^2}\right) \\&\overline{f}_s^n(\omega ):=\sup _{y,z,q}|\mathbf 1 _{[t,T]}f(s, X_s^{t,x}(\omega ),y,z,q)-\mathbf 1 _{[t_n,T]}f(s, X_s^{t_n,x_n}(\omega ),y,z,q)|. \end{aligned}$$

The continuity of $u$ is then a consequence of the following convergences as $n\rightarrow \infty$:

$$\begin{aligned}&\mathbb {E}\left( \sup _{0\le s\le T}|\overline{h}(s, X_{s}^{t,x}\right) - \overline{h}(s, X_{s}^{t_n}( x_n))|^{2})\rightarrow 0\\&\mathbb {E}\left[ \int _{0}^{T}(\overline{f}_s^n)^{2}\mathrm{d}s\right] \rightarrow 0, \end{aligned}$$

which follow from the Lebesgue’s theorem, using the continuity assumptions and polynomial growth of $f$ and $h$. $\square$

Lemma 6.2

The function $u$ has at most polynomial growth at infinity.

Proof

By applying Prop. 6.2 , we obtain the following estimate:

$$\begin{aligned} u(t,x)^2 \le K_{C,T}(\mathbb {E}\left( \int _{0}^{T} f(s, X_{s}^{t,x},0,0,0)^{2}\mathrm{d}s + \sup _{0\le s\le T}\overline{h}(s,X_{s}^{t,x})^{2}\right) . \end{aligned}$$

(55)

Using now the hypothesis of polynomial growth on $f,h,g$, and the standard estimate

$$\begin{aligned} \mathbb {E}\left[ \sup _{0\le s\le T}|X_{s}^{t,x}|^{2}\right] \le C'(1+x^{2}), \end{aligned}$$

we derive that there exist $\bar{C} \in \mathbb {R}$ and $p\in \mathbb {N}$ such that $|u(t,x)| \le \bar{C}(1+x^{p}),\ \forall t\in [0,T]$, $\forall x\in \mathbb {R}.$ $\square$

Remark 6.2

By (55), if $(t,x) \mapsto f(t,x, 0, 0), \; h$ and $g$ are bounded, then $u$ is bounded.

1.3 An Extension of the Comparison Result for BSDEs with Jumps

We provide here an extension of the comparison theorem for BSDEs given in [4] which formally states that if two drivers $f_1, f_2$ satisfy $f_1 \ge f_2 + \varepsilon$ then the associated solutions $X^1$ and $X^2$ satisfy $X_0^1 > X_0^2$.

Proposition 6.3

Let $t_0 \in [0,T]$ and let $\theta$ be a stopping time such that $\theta > t_0$ a.s.

Let $\xi _1$ and $\xi _2$ $\in$ $L^2(\mathcal{F}_{\theta })$. Let $f_1$ be a driver. Let $f_2$ be a Lipschitz driver. For $i=1,2$, let $(X^i_t, \pi ^i_t, l^i_t)$ be a solution in $S^{2} \times \mathbb {H}^{2} \times \mathbb {H}_{\nu }^{2}$ of the BSDE

$$\begin{aligned} -dX^i_t = \displaystyle f_i (t, X^i_t, \pi ^i_t, l^i_t) \mathrm{d}t - \pi ^i_t \mathrm{d}W_t - \int _{\mathbb {R}^*} l^i_t(u) \tilde{N} (\mathrm{d}t,\mathrm{d}u); \quad X^i_{\theta } = \xi _i. \end{aligned}$$

(56)

Assume that there exists a bounded predictable process $(\gamma _t)$ such that $\mathrm{d}t\otimes dP \otimes \nu (\mathrm{d}e)$-a.s. $\gamma _t(e) \ge -1$ and $|\gamma _t(e) | \le C(1 \wedge |e|)$ and such that

$$\begin{aligned} f_2(t, X^2_t, \pi ^2_t, l^1_t) - f_2(t,X^2_t,\pi ^2_t,l^2_t) \ge \langle \gamma _t\,,\, l^1_t- l^2_t \rangle _\nu , \quad t_0 \le t \le \theta ,\; \; \mathrm{d}t\otimes dP \quad \text { a.s.} \end{aligned}$$

(57)

Suppose also that

$$\begin{aligned}&\xi _1 \ge \xi _2 \quad \text { a.s. } \\&f_1 (t, X^1_t, \pi ^1_t, l^1_t) \ge f_2 (t, X^1_t, \pi ^1_t, l^1_t) +\varepsilon , \quad t_0 \le t \le \theta ,\; \; \mathrm{d}t\otimes dP \quad \text { a.s.}, \end{aligned}$$

where $\varepsilon$ is a real constant. Then

$$\begin{aligned} X_{t_0}^1-X_{t_0}^2 \ge \varepsilon \alpha \quad \mathrm{a.s.}, \end{aligned}$$

where $\alpha$ is a non-negative $\mathcal{F}_{t_0}$-measurable r.v. which does not depend on $\varepsilon$, with $P(\alpha >0) >0$.

Proof

From inequality (4.22) in the proof of the Comparison Theorem in [4], we derive that

$$\begin{aligned} X_{t_0}^1-X_{t_0}^2 \ge e^{-CT}\mathbb {E} \left[ \int _{t_0}^{\theta } H_{t_0,s}\, \varepsilon \,\mathrm{d}s | \fancyscript{F}_{t_0} \right] \; \quad \mathrm{a.s.}\, \end{aligned}$$

where $C$ is the Lipschitz constant of $f_2$, and $(H_{t_0,s})_{s \in [t_0,T]}$ is the square integrable non-negative martingale satisfying

$$\begin{aligned} d H_{t_0,s} = \displaystyle H_{t_0,s^-} \left[ \beta _s d W_s + \int _{\mathbb {R}^*} \gamma _s(u) \tilde{N}(\mathrm{d}s,\mathrm{d}u)\right] ; \quad H_{t_0,t_0} = 1, \end{aligned}$$

$(\beta _s)$ being a predictable process bounded by $C$. We get

$$\begin{aligned} X_{t_0}^1-X_{t_0}^2 \ge e^{-CT}\varepsilon \, \mathbb {E} \left[ H_{t_0,\theta }\, (\theta - t_0) | \fancyscript{F}_{t_0} \right] \; \quad \mathrm{a.s.}\, \end{aligned}$$

Since $\theta > t_0$ a.s. , we have $H_{t_0,\theta }\, (\theta - t_0) \ge 0$ a.s. and $P( H_{t_0,\theta }\, (\theta - t_0) >0) >0$. Setting $\alpha := e^{-CT} \, \mathbb {E} \left[ H_{t_0,\theta }\, (\theta - t_0) | \fancyscript{F}_{t_0} \right]$, the result follows. $\square$