3.1. Applications of Theorem 3.2

We now consider applications of Theorem 3.2. We first calculate the joint Laplace transform of $(U_{\mathbf{e} _q},X_{\mathbf{e} _q})$ , where $\mathbf{e} _q$ is an exponential time with parameter $q>0$ independent of X.

Corollary 3.2. Let X be a spectrally negative Lévy process. Let $q>0$ and $\alpha \in \mathbb{R}$ , $\beta \geq 0$ such that $q>\psi(\beta) \vee (\psi(\beta)-\alpha)$ . We have that for all $(u,x)\in E$ ,

(3.9) \begin{align}\mathbb{E}_{u,x}&\left( \mathrm{e}^{-\alpha U_{\mathbf{e} _q}+\beta X_{\mathbf{e} _q} } \right)\nonumber\\&=\frac{q\mathrm{e}^{\beta x}}{q-\psi(\beta)}+\mathrm{e}^{\Phi(q) x }\Phi'(q)\left[ \frac{q}{\Phi(q+\alpha)-\beta}-\frac{q}{\Phi(q)-\beta}\right]\nonumber\\&\qquad +\mathrm{e}^{\beta x}q \int_0^{x} \mathrm{e}^{-\beta y} [W^{(q)}(y)-\mathrm{e}^{-\alpha u} W^{(q+\alpha)}(y) ]\rm{d} y \nonumber\\&\qquad + \frac{q}{\Phi(q+\alpha)-\beta} \left[\mathrm{e}^{-\alpha u} W^{(q+\alpha)}(x)-W^{(q)}(x)\right].\end{align}

Proof. Consider the function $K(u,x)=\mathrm{e}^{-\alpha u +\beta x}$ for all $(u,x)\in E$ . We have that K is a continuous function and $K(u,x)\leq \mathrm{e}^{-(\alpha \wedge 0) u+\beta x}$ for all $(u,x)\in E$ . Taking $q>0$ such that $q> \psi(\beta)\vee (\psi(\beta)-\alpha)=\psi(\beta)-(\alpha \wedge 0)$ , we see that

\begin{align*}\mathbb{E}_x\left(\int_0^{\infty} \mathrm{e}^{-qr} \mathrm{e}^{-(\alpha \wedge 0) (u+r) +\beta X_r} \rm{d} r \right)&= \mathrm{e}^{\beta x-(\alpha \wedge 0)u}\int_0^{\infty}\mathrm{e}^{-(q+(\alpha \wedge 0)-\psi(\beta))r} \rm{d} r\\&=\frac{\mathrm{e}^{\beta x-(\alpha \wedge 0)u}}{ q+(\alpha \wedge 0)-\psi(\beta)}<\infty,\end{align*}

for all $u\geq 0$ and $x\in \mathbb{R}$ . Then for all $u> 0$ and $x> 0$ we have, by Fubini’s theorem and from equation (2.11), that

\begin{align*}K^+(u,x)&=\mathbb{E}_x\left( \int_0^{\tau_0^-}\mathrm{e}^{-qr}\mathrm{e}^{-\alpha (u+r)+\beta X_r}\rm{d} r \right)\\&=\mathrm{e}^{-\alpha u} \int_{(0,\infty)} \mathrm{e}^{\beta y}\int_0^{\infty} \mathrm{e}^{-(q+\alpha) r} \mathbb{P}_x(X_r\in \rm{d} y ,r<\tau_0^- )\rm{d} r \\&=\mathrm{e}^{-\alpha u} \int_{0}^{\infty} \mathrm{e}^{\beta y}\left[ \mathrm{e}^{-\Phi(q+\alpha) y}W^{(q+\alpha)}(x)-W^{(q+\alpha)}(x-y) \right] \rm{d} y \\&= \frac{\mathrm{e}^{-\alpha u} W^{(q+\alpha)}(x)}{\Phi(q+\alpha)-\beta}-\mathrm{e}^{-\alpha u}\mathrm{e}^{\beta x} \int_0^{x} \mathrm{e}^{-\beta y} W^{(q+\alpha)}(y) \rm{d} y.\end{align*}

Similarly, we calculate for any $x\in \mathbb{R}$ ,

\begin{align*}\int_{-\infty}^{0}\mathrm{e}^{\beta y}\left[\mathrm{e}^{\Phi(q) (x-y)}\Phi'(q) -W^{(q)}(x-y)\right] \rm{d} y& = \Phi'(q)\mathrm{e}^{\Phi(q)x} \int_0^{\infty}\mathrm{e}^{-(\beta -\Phi(q))y }\rm{d} y \nonumber\\&\quad-\mathrm{e}^{\beta x}\int_x^{\infty} \mathrm{e}^{-\beta y}W^{(q)}(y)\rm{d} y \\& = \frac{\Phi'(q)\mathrm{e}^{\Phi(q)x}}{\beta -\Phi(q)}-\frac{\mathrm{e}^{\beta x}}{\psi(\beta)-q} \nonumber\\&\quad+\mathrm{e}^{\beta x}\int_0^x \mathrm{e}^{-\beta y}W^{(q)}(y)\rm{d} y,\end{align*}

where the last equality follows from equation (2.3) and the last integral is understood as 0 when $x<0$ . Then from (3.7) we get that for all $(u,x)\in E$ ,

\begin{align*}&\mathbb{E}_{u,x}\left(\int_0^{\infty}\mathrm{e}^{-qr} \mathrm{e}^{-\alpha U_{r}+\beta X_{r} }\rm{d} r \right)\\&\quad=\frac{\mathrm{e}^{-\alpha u} W^{(q+\alpha)}(x)}{\Phi(q+\alpha)-\beta}-\mathrm{e}^{-\alpha u}\mathrm{e}^{\beta x} \int_0^{x} \mathrm{e}^{-\beta y} W^{(q+\alpha)}(y) \rm{d} y +\frac{\Phi'(q)\mathrm{e}^{\Phi(q)x}}{\beta -\Phi(q)}-\frac{\mathrm{e}^{\beta x}}{\psi(\beta)-q} \nonumber\\&\qquad +\mathrm{e}^{\beta x}\int_0^x \mathrm{e}^{-\beta y}W^{(q)}(y)\rm{d} y \\&\qquad +\mathrm{e}^{\Phi(q)x}\mathcal{I}^{(q,\Phi(q))}(x)\lim_{\varepsilon \downarrow 0 } \frac{ 1}{\psi'(\Phi(q)+) W^{(q)}(\varepsilon)} \left[\frac{ W^{(q+\alpha)}(\varepsilon)}{\Phi(q+\alpha)-\beta}-\mathrm{e}^{\beta \varepsilon} \int_0^{\varepsilon} \mathrm{e}^{-\beta y} W^{(q+\alpha)}(y) \rm{d} y \right]\\&\quad=\frac{\mathrm{e}^{-\alpha u} W^{(q+\alpha)}(x)}{\Phi(q+\alpha)-\beta}-\mathrm{e}^{-\alpha u}\mathrm{e}^{\beta x} \int_0^{x} \mathrm{e}^{-\beta y} W^{(q+\alpha)}(y) \rm{d} y +\frac{\Phi'(q)\mathrm{e}^{\Phi(q)x}}{\beta -\Phi(q)}-\frac{\mathrm{e}^{\beta x}}{\psi(\beta)-q}\nonumber\\&\qquad+\mathrm{e}^{\beta x}\int_0^x \mathrm{e}^{-\beta y}W^{(q)}(y)\rm{d} y +\mathrm{e}^{\Phi(q)x} \left[ 1-\psi'(\Phi(q)+) \mathrm{e}^{-\Phi(q) x}W^{(q)}(x) \right] \frac{\Phi'(q)}{\Phi(q+\alpha)-\beta},\end{align*}

where in the last equality we used the fact that $\Phi'(q)=1/\psi'(\Phi(q)+)$ , $W^{(q)}{(x)}$ is non-negative and strictly increasing on $[0,\infty)$ for all $q\geq 0$ , and that

\begin{align*}\lim_{\varepsilon \downarrow 0} \frac{W^{(q+\alpha)}(\varepsilon)}{W^{(q)}(\varepsilon)}=1.\end{align*}

The latter fact follows from the representation $W^{(q)}(x)=\sum_{k=0}^{\infty}q^k W^{*(k+1)}(x)$ and the estimate $W^{*(k+1)}(x)\leq x^k/k! W(x)^{k+1}$ (see equations (8.28) and (8.29) in [Reference Kyprianou29, pp.241--242]. Rearranging the terms and using that

\begin{align*}\mathbb{E}_{u,x}\left( \mathrm{e}^{-\alpha U_{\mathbf{e} _q}+\beta X_{\mathbf{e} _q} } \right)=q\mathbb{E}_{u,x}\left(\int_0^{\infty}\mathrm{e}^{-qr} \mathrm{e}^{-\alpha U_{r}+\beta X_{r} }\rm{d} r \right),\end{align*}

for all $(u,x)\in E$ , we obtain the desired result.

Remark 3.4. Note that from formula (3.9) we can recover some known expressions for spectrally negative Lévy processes. If we take $\alpha=0$ , we obtain, for all $\beta \geq 0$ , $q>\psi(\beta)\vee 0$ and $x\in \mathbb{R}$ , that

\begin{align*}\mathbb{E}_x(\mathrm{e}^{\beta X_{\mathbf{e} _q}})=\frac{q \mathrm{e}^{\beta x}}{q-\psi(\beta)}.\end{align*}

On the other hand, for any $\theta\geq 0$ , $q\geq 0$ and $x\in \mathbb{R}$ we have that

\begin{align*}\mathbb{E}_x(\mathrm{e}^{-\theta g_{\mathbf{e} _q}})=\int_0^{\infty}q\mathrm{e}^{-qt} \mathbb{E}_x(\mathrm{e}^{-\theta g_t})\rm{d} t=\int_0^{\infty}q\mathrm{e}^{-(q+\theta)t} \mathbb{E}_x(\mathrm{e}^{\theta U_t})\rm{d} t =\frac{q}{q+\theta} \mathbb{E}_x(\mathrm{e}^{\theta U_{\mathbf{e} _{q+\theta}}}),\end{align*}

where $\mathbf{e} _{q+\theta}$ is an exponential random variable with parameter $q+\theta$ . The result coincides with that found in [Reference Baurdoux5] (see Theorem 2).

Let $q>0$ , we consider the q-potential measure of (U, X) given by

\begin{align*}\int_0^{\infty} \mathrm{e}^{-qr }\mathbb{P}_{u,x}(U_r \in \rm{d} v, X_r \in \rm{d} y) \rm{d} r\end{align*}

for $(u,x),(v, y)\in E$ . From the fact that $U_t=0$ if and only if $X_t\leq 0$ for any $t>0$ , we have that for $(u,x)\in E$ and $y\leq 0$ ,

\begin{align*}\int_0^{\infty} \mathrm{e}^{-qr }\mathbb{P}_{u,x}(U_r =0, X_r \in \rm{d} y) \rm{d} r=\int_0^{\infty} \mathrm{e}^{-qr }\mathbb{P}_{x}( X_r \in \rm{d} y) \rm{d} r.\end{align*}

In the next corollary, we find an expression for a density when $v,x>0$ .

Corollary 3.3. Let X be a spectrally negative Lévy process and $q>0$ . The q-potential measure of (U,X) has density given by

(3.10) \begin{align}\int_0^{\infty} \mathrm{e}^{-qr }\mathbb{P}_{u,x}(U_r \in \rm{d} v, X_r \in \rm{d} y) \rm{d} r&=\mathrm{e}^{-q(v-u)}\mathbb{P}_x(X_{v-u} \in \rm{d} y, v-u<\tau_0^-) \mathbb{I}_{\{v> u \}} \rm{d} v\nonumber\\&\qquad +\left[\mathrm{e}^{\Phi(q) x}\Phi'(q)-W^{(q)}(x) \right] \frac{y}{v} \mathrm{e}^{-qv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v\end{align}

for all $(u,x)\in E$ and $v,y>0$ . In particular, when $u=x=0$ we have that

\begin{align*}\int_0^{\infty} \mathrm{e}^{-qr }\mathbb{P}(U_r \in \rm{d} v, X_r \in \rm{d} y) \rm{d} r&=\Phi'(q) \frac{y}{v} \mathrm{e}^{-qv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v.\end{align*}

\begin{align*}\mathbb{E}_x\left(\int_0^{\infty}\mathrm{e}^{-qr} \mathbb{I}_{\{X_r \in Y \}} \rm{d} r \right)<\infty.\end{align*}

First, we calculate for all $u,x>0$ such that $u<u_1$ ,

\begin{align*}K^+(u,x)=\mathbb{E}_{u,x}\left(\int_0^{\tau_0^-} \mathrm{e}^{-qr}\mathbb{I}_{\{ U_r \in A, X_r \in Y\}} \rm{d} r \right)=\int_{A} \int_{Y} \mathrm{e}^{-q(r-u)}\mathbb{P}_x(X_{r-u} \in \rm{d} y, r-u<\tau_0^-) \rm{d} r.\end{align*}

For every $x\leq 0$ we have that

\begin{align*}K^-(x)= \mathbb{E}_{u,x}\left(\int_0^{\tau_0^+} \mathrm{e}^{-qr}\mathbb{I}_{\{ U_r \in A, X_r \in Y\}} \rm{d} r \right)=0.\end{align*}

Hence, for all $(u,x)\in E$ we obtain that

\begin{align*}\mathbb{E}_{u, x}&\left(\int_0^{\infty} \mathrm{e}^{-qr} \mathbb{I}_{\{U_r \in A, X_r \in Y \}} \rm{d} r \right)\\&=\int_{A} \int_Y \mathrm{e}^{-q(r-u)}\mathbb{P}_x(X_{r-u} \in \rm{d} y, r-u<\tau_0^-) \rm{d} r\\&\qquad + \mathrm{e}^{\Phi(q) x}\left[1-\psi'(\Phi(q)\!+\!) \mathrm{e}^{-\Phi(q) x}W^{(q)}(x) \right] \lim_{\varepsilon \downarrow 0 } \int_{A} \int_Y \frac{\mathrm{e}^{-qr}\mathbb{P}_{\varepsilon}(X_{r} \in \rm{d} y, r<\tau_0^-) }{ \psi'(\Phi(q)+) W^{(q)}(\varepsilon)} \rm{d} r.\end{align*}

We calculate the limit on the right-hand side of the equation above. Denote by $\mathbb{P}_{\varepsilon}^{\uparrow}$ the law of X starting from $\varepsilon$ conditioned to stay positive. We have, for all $x\in \mathbb{R}$ and $y>0$ , that

\begin{align*}\lim_{\varepsilon \downarrow 0 } \int_{A} \int_Y \frac{\mathrm{e}^{-qr}\mathbb{P}_{\varepsilon}(X_{r} \in \rm{d} y, r<\tau_0^-) }{ \psi'(\Phi(q)+) W^{(q)}(\varepsilon)} \rm{d} r&=\lim_{\varepsilon \downarrow 0 } \frac{W(\varepsilon)}{ \psi'(\Phi(q)+) W^{(q)}(\varepsilon)} \int_{A} \int_Y \frac{\mathrm{e}^{-qr}\mathbb{P}^{\uparrow}_{\varepsilon}(X_{r} \in \rm{d} y) }{ W(y)} \rm{d} r \\&=\frac{1}{ \psi'(\Phi(q)+) } \int_{A} \int_Y \frac{\mathrm{e}^{-qr}\mathbb{P}^{\uparrow}(X_{r} \in \rm{d} y) }{ W(y)} \rm{d} r,\end{align*}

where the first equality follows from the definition of $\mathbb{P}^{\uparrow}$ (see, for example, Section VII.3, equation (6) in [Reference Bertoin10]) and the last equality follows since $\lim_{\varepsilon \downarrow 0} W(\varepsilon)/W^{(q)}(\varepsilon)=1$ and $\mathbb{P}_{\varepsilon}^{\uparrow}$ converges to $\mathbb{P}^{\uparrow}$ in the sense of finite-dimensional distributions (see Proposition VII.3.14 in [Reference Bertoin10]). Moreover, we have for all $y,r>0$ that $\mathbb{P}^{\uparrow}(X_r\in \rm{d} y)= yW(y)\mathbb{P}(X_r \in \rm{d} y)/r$ (see Corollary VII.3.16 in [Reference Bertoin10]). Therefore, we obtain for all $(u,x)\in E$ that

\begin{align*}\mathbb{E}_{u,x}\left(\int_0^{\infty} \mathrm{e}^{-qr} \mathbb{I}_{\{U_r \in A, X_r \in Y \}} \rm{d} r \right)&=\int_{A} \int_Y \mathrm{e}^{-q(r-u)}\mathbb{P}_x(X_{r-u} \in \rm{d} y, r-u<\tau_0^-) \rm{d} r\\&\qquad + \left[\Phi'(q) \mathrm{e}^{\Phi(q) x}-W^{(q)}(x) \right] \int_{A} \int_Y \frac{y}{r}\mathrm{e}^{-qr} \mathbb{P}(X_{r} \in \rm{d} y) \rm{d} r,\end{align*}

where we also used the fact that $\Phi'(q)=1/\psi'(\Phi(q)\!+\!)$ . The proof is now complete.

Remark 3.5. Bingham [Reference Bingham12] showed that the q-potential measure of X has a density that is absolutely continuous with respect to the Lebesgue measure. This can be demonstrated by moving the killing barrier on the q-potential measure killed on entering the set $(\!-\!\infty,0]$ (see (2.11)) and taking limits. Alternatively, it can be deduced by taking limits on (3.7). Moreover, Corollary 3.3 provides an alternative method for finding the density mentioned above. For this, we use Kendall’s identity (see, for example, Corollary VII.3 in [Reference Bertoin10]) given by

(3.11) \begin{align}r\mathbb{P}(\tau_z^+ \in \rm{d} r) \rm{d} z=z\mathbb{P}(X_r \in \rm{d} z) \rm{d} r\end{align}

for all $r,z \geq 0$ . Indeed, let $u,y>0$ and $x\in \mathbb{R}$ ; integrating (3.10) with respect to the variable v, we obtain that

\begin{align*}\int_0^{\infty} \mathrm{e}^{-qr} &\mathbb{P}_x(X_r \in \rm{d} y) \rm{d} r\\&=\int_{(0,\infty)} \int_0^{\infty} \mathrm{e}^{-qr }\mathbb{P}_{u,x}(U_r \in \rm{d} v, X_r \in \rm{d} y) \rm{d} r \\&=\int_0^{\infty}\!\! \mathrm{e}^{-qv}\mathbb{P}_x(X_{v} \in \rm{d} y, v<\tau_0^-) \rm{d} v \!+ \int_0^{\infty}\!\left[\mathrm{e}^{\Phi(q) x}\Phi'(q)-W^{(q)}(x) \right] \frac{y}{v} \mathrm{e}^{-qv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v \\&=[\mathrm{e}^{-\Phi(q) y}W^{(q)}(x) \!-\!W^{(q)}(x-y)] \rm{d} y + \!\left[\mathrm{e}^{\Phi(q) x}\Phi'(q)-W^{(q)}(x) \right] \! \int_0^{\infty}\!\! \mathrm{e}^{-q v} \mathbb{P}(\tau_y^+ \in \rm{d}v)\rm{d} y ,\end{align*}

where the last equality follows from (2.11) and (3.11). Hence, using the formula for the Laplace transform of $\tau_y^+$ (see equation (2.2)) we have that

\begin{align*}\int_0^{\infty} \mathrm{e}^{-qr} \mathbb{P}_x(X_r \in \rm{d} y) \rm{d} r=\left( \mathrm{e}^{\Phi(q) (x-y)}\Phi'(q) -W^{(q)}(x-y)\right) \rm{d} y.\end{align*}

4.1. Optimal stopping problems

This section uses the results developed in the previous sections to solve a general optimal stopping problem. For the sake of simplicity, we will assume that X is a spectrally negative process with a Gaussian component. That is, we assume that $\sigma>0$ . We take $r\geq 0$ and let G be a continuous function on E such that

(4.1) \begin{align}\mathbb{E}_{u,x}\left(\int_0^{\infty} \mathrm{e}^{-rs}|G(U_s,X_s)|\rm{d} s \right)<\infty\end{align}

for all $(u,x)\in E$ . We further assume that there exists a value $x_G<0$ such that $G(0,x)<0$ for all $x<x_G$ with $\lim_{x\rightarrow -\infty}G(0,x)<0$ , and $G(u,x)\geq 0$ for all $(u,x)\in E$ such that $x\geq x_G$ . We also assume that the function

\begin{align*}K^+(u,x)\,:\!=\,\mathbb{E}_x\left(\int_0^{\tau_0^-} \mathrm{e}^{-rs}G(u+s,X_s)\rm{d} s \right)\end{align*}

is $C^{1,2}$ on $[0,\infty)\times[0,\infty)$ .

We consider the optimal stopping problem

(4.2) \begin{align}V(u,x)=\sup_{\tau \in \mathcal{T}} \mathbb{E}_{x,u}\left(\int_0^{\tau}\mathrm{e}^{-rs}G(U_s,X_s)\rm{d} s \right),\end{align}

where $\mathcal{T}$ is the set of all stopping times of X. Note that our assumptions suggest that it is never optimal to stop when X is taking positive values, and since G is negative for $x<x_G$ , we should stop as soon as X is below a value $z^*<x_G$ , for $|z^*|$ sufficiently large. The following theorem confirms this notion.

Theorem 4.1. Under the conditions stated above, we have that an optimal stopping time for (4.2) is given by

\begin{align*}\tau_{z^*}^-=\inf\{t>0\,:\, X_t\leq z^* \},\end{align*}

where $z^* \in (\!-\!\infty,0)$ is characterized as the unique solution for z in $(\!-\!\infty,0)$ to the equation

\begin{align*}\int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v+\int_{z}^0 G(0,y) \mathrm{e}^{-\Phi(r) y}\rm{d} y=0.\end{align*}

We have that $z^*<x_G$ , where we recall that $x_G=\inf\{x\in \mathbb{R}: G(0,x) \geq 0 \}<0$ . Moreover, the value function is given by

\begin{align*}V(u,x)= K^+(u,x) \!-\!W^{(r)}(x)\!\int_{(0,\infty)}\!\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v-\int_{z^*}^0 G(0,y)W^{(r)}(x-y) \rm{d} y,\end{align*}

for all $(u,x)\in E$ . Furthermore, there is smooth fit at $z^*$ , that is, $\frac{\partial}{\partial x} V(0,z^*+)=\frac{\partial}{\partial x} V(0,z^*-)$ .

We have the following remark regarding some of the assumptions in Theorem 4.1.

The proof of Theorem 4.1 relies on finding, with the help of the potential measure of (U, X) given in Corollary 3.3, a semi-explicit expression of the function $V_z(u,x)=\mathbb{E}_{u,x}\left(\int_0^{\tau_z^-} G(U_s,X_s)\rm{d} s \right)$ , for each $z\leq 0$ . Then, due to the properties of the scale functions $W^{(r)}$ and applying the version of the Itô formula derived in Theorem 3.1, we see that the two conditions given in Lemma 5.4 are satisfied. The reader should also note that formula (3.7) helped prove that condition (3.4) is satisfied for V. The proof is postponed to Section 5.3.

Motivated by Example 4.1 below, we have the following proposition as an application of Theorem 4.1. Its proof is postponed to Section 5.4.

Proposition 4.1. Let X be any spectrally negative Lévy process with $\sigma>0$ . Take $\beta \geq 0$ , $K\in (0,1)$ and $r>\psi(1)+\beta$ . We consider the optimal stopping problem

\begin{align*}V(u,x)&=\sup_{{\tau \in \mathcal{T}}}\mathbb{E}_{u,x}\left( \int_0^{\tau } \mathrm{e}^{-rs}[\mathrm{e}^{X_t +\beta U_s}-K]\rm{d} s \right),\quad (u,x)\in E.\end{align*}

Then the stopping time

\begin{align*}\tau_{z^*}^-=\inf\{t>0\colon X_t\leq z^* \}\end{align*}

is an optimal stopping time, where $z^*$ is the unique solution on $(\!-\!\infty,0)$ to the equation

(4.3) \begin{align}\frac{\mathrm{e}^{-(\Phi(r)-1)z}}{\Phi(r)-1}-\frac{K\mathrm{e}^{-\Phi(r)z}}{\Phi(r)}+\frac{1}{\Phi(r-\beta)-1}-\frac{1}{\Phi(r)-1}=0.\end{align}

Moreover, the value function takes the form

\begin{align*}V(u,x)&= \mathrm{e}^{\beta u} \left[ \frac{W^{(r-\beta)}(x)}{\Phi(r-\beta)-1} -\int_0^x \mathrm{e}^y W^{(r-\beta)}(x-y)\rm{d} y \right] +K\int_0^xW^{(r)}(y)\rm{d} y-\frac{K}{\Phi(r)}W^{(r)}(x)\\&\qquad-W^{(r)}(x)\left[\frac{1}{\Phi(r-\beta)-1}-\frac{K}{\Phi(r)} \right]-\int_{z^*}^0 [\mathrm{e}^{y}-K]W^{(r)}(x-y) \rm{d} y\end{align*}

for every $(u,x)\in E$ .

We have the following remark on the above proposition regarding some particular cases and how our results match the current literature.

We then present a setting where the result above becomes relevant.

Example 4.1. Following the model of corporate bankruptcy in [Reference Leland32] and [Reference Manso, Strulovici and Tchistyi33] (see also Section III.C in [Reference Quah and Strulovici40]), we consider that equity holders endogenously choose the bankruptcy time. Suppose that the performance of a firm Footnote ¹ at time $t\geq 0$ is given by $Y_t=\exp(X_t)$ , where X is a spectrally negative Lévy process such that $\sigma > 0$ . The performance measure $Y_t$ is normalized such that the values above the level 1 are considered good company performance, whereas values below 1 indicate a negative rating. Then we consider, for $t\geq 0$ , $V_t=t-\sup\{0\leq s\leq t\,:\, Y_s\leq 1 \}=t-U_t$ , the length of time since the last time the company performed poorly. Large values of $V_t$ can be interpreted as the firm’s financial stability.

Suppose that, until bankruptcy, the firm must pay a coupon rate c(v, y) to debt holders and receive a payout rate $\delta(v, y)$ in terms of the performance y and v, the current excursion above the level 1. Then the time of bankruptcy is determined by the optimal stopping problem

(4.4) \begin{align}V(u,x)&=\sup_{{\tau \in \mathcal{T}}}\mathbb{E}_{u,x}\left( \int_0^{\tau } \mathrm{e}^{-rs}[\delta(U_s,\mathrm{e}^{X_t})-c(U_s,\mathrm{e}^{X_t})]\rm{d} s \right),\end{align}

where $r\geq 0$ is the risk-free interest rate. Note that if $\delta(U_s,\mathrm{e}^{X_s})$ is lower than $c(U_s,\mathrm{e}^{X_s})$ , equity holders have a negative dividend rate. Then the firm will keep operating with a negative dividend rate if the firm’s prospects are good enough to compensate for the negative losses. Otherwise, the firm will stop operations, and bankruptcy will be declared.

We then see from Proposition 4.1 that, upon taking $\delta(v, y)=y\mathrm{e}^{\beta v}$ and $c(v, y)=K$ in (4.4), with K, $\beta$ and r as in Proposition 4.1, we conclude that the bankruptcy of the company occurs when $Y_t=\exp(X_t)$ crosses below the level $\mathrm{e}^{z^*}$ .

4.2. Optimal prediction problems

Let X be a stochastic process with state space in $\mathbb{R}$ and let $\theta$ be a last passage time of X, that is, $\theta =\sup\{t\geq 0: X_t\in A \}$ , where $A\subset \mathbb{R}$ . The recent literature has solved the problem of finding a stopping time approximating a specific last passage time. There are, for example, various papers in which the approximation is in the $L_1$ sense. That is, the following optimal prediction problem is solved:

(4.5) \begin{align}\inf_{\tau \in \mathcal{T}} \mathbb{E}(|\tau-\theta|).\end{align}

To mention a few: [Reference du Toit, Peskir and Shiryaev19] predicted the last zero of a Brownian motion with drift in a finite horizon setting; [Reference du Toit and Peskir18] predicted the time when the ultimate maximum at time $t=1$ is attained for a Brownian motion with drift; [Reference Shiryaev45] focused on the last time of the attainment of the ultimate maximum of a Brownian motion and proceeded to show that it is equivalent to predicting the last zero of the process in this setting; [Reference Glover, Hulley and Peskir21] predicted the time when a transient diffusion attains its ultimate minimum; [Reference Glover and Hulley20] predicted the last passage time of a level $z > 0$ for an arbitrary nonnegative time-homogeneous transient diffusion; [Reference Baurdoux and Van Schaik9] predicted the time at which a Lévy process attains its ultimate supremum; [Reference Baurdoux, Kyprianou and Ott6] predicted when a positive self-similar Markov process attains its path-wise global supremum or infimum before hitting zero for the first time; and [Reference Baurdoux and Pedraza7] predicted the last zero of a spectrally negative Lévy process.

From here onwards, consider X to be a spectrally negative Lévy process that drifts to infinity and let $g=\sup\{ t\geq 0\,:\, X_t\leq 0\}$ , the last zero of X. The problem (4.5) can be generalized to any convex function $d\,:\,\mathbb{R}_+ \times \mathbb{R}_+ \mapsto \mathbb{R}_+$ . That is, under the assumption that $\mathbb{E}(d(0,g))<\infty$ , consider the optimal prediction problem

(4.6) \begin{align}V_d=\inf_{\tau \in \mathcal{T}} \mathbb{E}(d(\tau,g)).\end{align}

As is in the case for problem (4.5), problem (4.6) cannot be solved using standard techniques of optimal stopping (cf. [Reference Peskir and Shiryaev38]) since the random variable g depends on the whole path of the process X and hence is only $\mathcal{F}$ measurable. However, the following lemma provides an equivalence between the optimal prediction problem above and an optimal stopping problem driven by the process $\{(g_t,t, X_t),t\geq 0\}$ .

Lemma 4.1. Let X be a spectrally negative Lévy process drifting to infinity and $d\,:\,\mathbb{R}_+ \times \mathbb{R}_+ \mapsto \mathbb{R}_+$ a convex function such that $\mathbb{E}(d(0,g))<\infty$ . We have for each $\tau \in \mathcal{T}$ ,

\begin{align*}\mathbb{E}(d(\tau,g))&=\mathbb{E}\left( \int_0^{\tau} G_d(g_s,s,X_s) \rm{d} s+d(0,g)\right),\end{align*}

where $G_d(\gamma,s,x)=\frac{\partial }{\partial x} d_+(s,\gamma)\psi'(0\!+\!)W(x)+\mathbb{E}_{x}\frac{\partial }{\partial x} d_+(s,g+s) \mathbb{I}_{\{g>0\}})$ and $\frac{\partial }{\partial x} d_+$ is the right-derivative with respect to the first argument of d.

\begin{align*}\mathbb{E}(d(\tau,g))&=\mathbb{E}\left( \int_0^{\tau} \frac{\partial }{\partial x} d_+(s,g)\rm{d} s+d(0,g)\right),\end{align*}

where $\frac{\partial }{\partial x} d_+$ is the right-derivative of d with respect to its first coordinate. Then, using Fubini’s theorem and the tower property for conditional expectation, we see that

\begin{align*}\mathbb{E}\left( \int_0^{\tau} \frac{\partial }{\partial x} d_+(s,g)\rm{d} s\right)&= \int_0^{\infty} \mathbb{E}\left[\mathbb{I}_{\{ s\leq \tau\}} \mathbb{E}\left( \frac{\partial }{\partial x} d_+(s,g) \bigg|\mathcal{F}_s\right) \rm{d} s\right] \\&=\mathbb{E}\left[\int_0^{\tau} \mathbb{E}\left( \frac{\partial }{\partial x} d_+(s,g) \bigg|\mathcal{F}_s\right) \rm{d} s \right]. \\\end{align*}

Hence, we proceed to find an expression for the conditional expectation inside the last integral. From the strong Markov property of the process $\{(g_t,t,X_t),t\geq 0\}$ we have that

\begin{align*}\mathbb{E}\left( \frac{\partial }{\partial x} d_+(s,g) \bigg|\mathcal{F}_s\right)&=\mathbb{E}_{g_s,s,X_s} \left( \frac{\partial }{\partial x} d_+(s,g)\right).\end{align*}

From (3.2), we have that for $(\gamma,s,x)\in E_g$ ,

\begin{align*}\mathbb{E}_{\gamma,s,x} \left( \frac{\partial }{\partial x} d_+(s,g)\right)&=\mathbb{E}_{x} \left( \frac{\partial }{\partial x} d_+(s,\gamma)\mathbb{I}_{\{\tau_0^-=\infty \}}\right)+\mathbb{E}_{x} \left( \frac{\partial }{\partial x} d_+(s,g+s)\mathbb{I}_{\{\tau_0^-<\infty \}}\right)\\&=\frac{\partial }{\partial x} d_+(s,\gamma) \psi'(0\!+\!)W(x)+\mathbb{E}_x\left(\frac{\partial }{\partial x} d_+(s,g+s)\mathbb{I}_{\{ g>0\}}\right)\\&=G_d(\gamma,s,x).\end{align*}

Thus, for any $s\geq 0$ ,

\begin{align*}\mathbb{E}\left( \frac{\partial }{\partial x} d_+(s,g) \bigg|\mathcal{F}_s\right)=G_d(g_s,s,X_s).\end{align*}

The result then follows.

The lemma above directly implies that solving the optimal prediction problem (4.6) is equivalent to solving the optimal stopping problem

(4.7) \begin{align}\inf_{\tau \in \mathcal{T}} \mathbb{E}_{\gamma,t,x}\left[ \int_0^{\tau} G_d(g_{s+t},s+t,X_{s+t}) \rm{d} s \right],\end{align}

for each $(\gamma,t,x)\in E_g$ . In [Reference Baurdoux and Pedraza8], the case when $d(x,y)=|x-y|^p$ with $p>1$ is solved, that is, g is approximated by a stopping time using an $L_p$ distance. In this case, problem (4.7) reads

(4.8) \begin{align}V(u,x)=\inf_{\tau \in \mathcal{T}}\mathbb{E}_{u,x}\left( \int_0^{\tau} G(U_s,X_s)\rm{d} s \right),\end{align}

for $(u,x)\in E$ , where $G(u,x)=u^{p-1}\psi'(0\!+\!)W(x)-\mathbb{E}_x(g^{p-1})$ . Although in [Reference Baurdoux and Pedraza8] a rather general spectrally negative Lévy process is considered (only integrability conditions on the Lévy measure are imposed), we include here the main results (see Theorem 3.3 in [Reference Baurdoux and Pedraza8]) when X is a Brownian motion with positive drift (with Gaussian coefficient $\sigma>0$ ). It is shown that an optimal stopping time for (4.8) is given by

\begin{align*}\tau_D=\inf\{t\geq 0\,:\, X_t\geq b(U_t) \},\end{align*}

where b is a strictly positive, non-increasing and continuous function such that $\lim_{u\rightarrow\infty } b(u)=0$ and $\lim_{u\downarrow 0}b(u)=\infty$ . Moreover, the function b and the value V(0, 0) are characterized as the only solution to the non-linear equations

\begin{align*}0&=V(0, 0)\frac{\sigma^2}{2}W'(b(u))+ \mathbb{E}_{b(u)}\left( \int_0^{\tau_0^-} G(u+s,X_s) \mathbb{I}_{\{X_s < b(u+s) \}}\rm{d} s \right),\\0&=V(0, 0)\frac{\sigma^2}{2}W''(0\!+\!)+\frac{\partial}{\partial x} \mathbb{E}_{x}\left( \int_0^{\tau_0^-} G(u+s,X_s) \mathbb{I}_{\{X_s < b(u+s) \}}\rm{d} s \right)\bigg|_{x=0,u=0}\nonumber\\&\qquad+\int_{[0,\infty)} \mathbb{E}_{-u}(g^{p-1}) W(\rm{d} u),\end{align*}

where b is considered in the class of continuous functions bounded by below by $h(u)\,:\!=\,\inf\{x\geq 0\,:\, G(u,x)\geq 0 \}$ and $V(0, 0)<0$ .

Note that properties of the stochastic process (U, X) were needed to derive the above result. For instance, the Markov property of (U, X) is crucial to solving the optimal stopping problem (4.8) using the standard theory of optimal stopping. Moreover, the explicit version of the infinitesimal generator (3.6) and formula (3.7) played a crucial role in deriving the non-linear equations presented above. In particular, given the unusual shape of the set E, (3.7) gives us a method to show that there is smooth pasting at the point (0, 0) for the function V, which allowed us to propose a characterization of the value V(0, 0).

5.1. Proof of Proposition 3.1

From the definition of $g_t$ , it is easy to note that for all $t\geq 0$ , we have $X_t \leq 0$ if and only if $g_t=t$ , from which we obtain that $(g_t,t, X_t)$ can take only values in $E_g$ . We now proceed to show the strong Markov property holds. Consider a measurable positive function $h: E_g \mapsto \mathbb{R}$ . Then we have for any stopping time $\tau$ and $s\geq 0$ ,

\begin{align*}\mathbb{E}( h(g_{\tau+s},\tau+s,X_{\tau+s})|\mathcal{F}_{\tau})&=\mathbb{E}(h(g_{\tau} \vee \sup\{ r\in [\tau,s+\tau]\,:\, X_{r} \leq 0 \},\tau+s,X_{\tau+s} )|\mathcal{F}_{\tau})\\&=\mathbb{E}( h(g_{\tau} \vee \sup\{ r\in [\tau,s+\tau]\,:\, \widetilde{X}_{r-\tau}+X_{\tau} \leq 0 \},\tau+s,\widetilde{X}_s+X_{\tau} )|\mathcal{F}_{\tau}),\end{align*}

where $\widetilde{X}_r=X_{r+\tau}-X_{\tau}$ and $a\vee b\,:\!=\,\max\{ a,b \}$ for any $a,b\in \mathbb{R}$ . Using the strong Markov property for Lévy processes and the fact that $g_{\tau}$ and $X_{\tau}$ are $\mathcal{F}_{\tau}$ measurable, we obtain that

\begin{align*}\mathbb{E}( h(g_{\tau+s},\tau+s,X_{\tau+s})|\mathcal{F}_{\tau})=f_s(g_{\tau},\tau,X_{\tau}),\end{align*}

where for any $x\in \mathbb{R}$ and $0\leq \gamma \leq t$ , the function $f_s$ is given by

\begin{align*}f_s(\gamma,t ,x)&=\mathbb{E}_x( h(\gamma \vee \sup\{ r\in [t,s+t]: X_{r-t} \leq 0 \},t+s,X_s )).\end{align*}

Note that, on the event $\{\sigma_{0}^->s\}$ , the set $\{ r\in [t,s+t]: X_{r-t} \leq 0 \}=\emptyset$ . Then $\gamma \vee \sup\{ r\in [t,s+t]\,:\, X_{r-t} \leq 0 \}=\gamma$ , where we used the convention that $\sup \emptyset =0$ . Otherwise, in the event $\{\sigma_{0}^-\leq s\}$ , we have that $\{ r\in [t,s+t]\,:\, X_{r-t} \leq 0 \}\neq \emptyset$ and then $\sup\{ r\in [t,s+t]\,:\, X_{r-t} \leq 0 \} \geq t\geq \gamma$ . Hence, we have that, in the event $\{\sigma_{0}^-\leq s\}$ ,

\begin{align*}\gamma \vee \sup\{ r\in [t,s+t]\,:\, X_{r-t} \leq 0 \}&=\sup\{ r\in [t,s+t]\,:\, X_{r-t} \leq 0 \}\\&=t+\sup\{ r\in [0,s]\,:\, X_{r} \leq 0 \}\\&=t+g_s.\end{align*}

Therefore, for any $x\in \mathbb{R}$ and $0\leq \gamma \leq t$ , the function $f_s$ takes the form

\begin{align*}f_s(\gamma,t ,x)&=\mathbb{E}_x(h(\gamma,t+s, X_s)\mathbb{I}_{\{ \sigma_0^->s\}})+\mathbb{E}_x(h(g_s+t,t+s, X_s)\mathbb{I}_{\{ \sigma_0^-\leq s\}}).\end{align*}

On the other hand, similar calculations lead us to

\begin{align*}\mathbb{E}( h(g_{\tau+s},\tau+s,X_{\tau+s})|\sigma(g_{\tau},\tau,X_{\tau}))=f_s(g_{\tau},\tau,X_{\tau}).\end{align*}

Hence, for any measurable positive function h, we obtain

\begin{align*}\mathbb{E}( h(g_{\tau+s},\tau+s,X_{\tau+s})|\mathcal{F}_{ \tau})&=\mathbb{E}( h(g_{\tau+s},\tau+s,X_{\tau+s})|\sigma(g_{\tau},\tau,X_{\tau})).\end{align*}

Therefore, we conclude that the process $\{ (g_t,t,X_t),t\geq 0 \}$ is a strong Markov process.

5.2. Perturbed Lévy process

Suppose that X is a spectrally negative Lévy process of finite variation. Then, with probability 1, it takes a positive amount of time to cross below 0, that is, $\tau_0^->0$ $\mathbb{P}$ -almost surely. Hence, stopping at the consecutive times at which X is below zero and together with the ideas mentioned in Remark 3.1, we can fully describe the behaviour of $g_t$ and then derive the results in Theorems 3.1 and 3.2. However, when X is of infinite variation, it is well known that the closure of the set of zeros of X is perfect and nowhere dense, and the approach mentioned is no longer useful (since we have that $\tau_0^-=0$ $\mathbb{P}$ -almost surely). Therefore, we use a perturbation method to exploit the idea applicable to finite variation processes. This method, which is mainly based on the work of [Reference Dassios and Wu17] and [Reference Revuz and Yor41] (see Theorem VI.1.10), consists of constructing a new ‘perturbed’ process $X^{(\varepsilon)}$ (for $\varepsilon$ sufficiently small) that approximates X, with the property that $X^{(\varepsilon)}$ visits the level zero a finite number of times before any time $t\geq 0$ . Then we approximate $g_t$ by the corresponding last zero process of $X^{(\varepsilon)}$ .

We formally describe the construction of the ‘perturbed’ process $X^{(\varepsilon)}$ . Let $\varepsilon>0$ , and define the stopping times $\rho_{1,\varepsilon}^-=0$ and, for any $k \geq 1$ ,

\begin{align*}\rho_{k,\varepsilon}^+&\,:\!=\,\inf\{ t>\rho_{k,\varepsilon}^-\,:\, X_t \geq \varepsilon \},\\\rho_{k+1,\varepsilon}^-&\,:\!=\,\inf\{ t>\rho_{k,\varepsilon}^+\,:\, X_t < 0\},\end{align*}

where we use the usual convention that $\inf \emptyset=\infty$ . We define the auxiliary process $X^{(\varepsilon)}=\{X_t^{(\varepsilon)} ,t\geq 0 \}$ , where for $t\geq 0$ ,

(5.1) \begin{align}X_t^{(\varepsilon)}=\left\{\begin{array}{l@{\quad}l}X_t-\varepsilon, & \rho_{k,\varepsilon}^-\leq t< \rho_{k,\varepsilon }^+, \\X_t, & \rho_{k,\varepsilon}^+\leq t < \rho_{k+1,\varepsilon }^- . \\\end{array}\right.\end{align}

In Figure 1 we include a sample path of the process $X^{(\varepsilon)}$ compared with the original process X.

Figure 1.

Left: Sample path of X. Right: Sample path of the perturbed process $X^{(\varepsilon)}$ . The vertical lines correspond to the sequence of stopping times $\{(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^+),k\geq 1 \}$ .

It is straightforward from the definition of $X^{(\varepsilon)}$ that $X_t -\varepsilon \leq X_t^{(\varepsilon)} \leq X_t$ , and that $X^{(\varepsilon)} \uparrow X $ uniformly when $\varepsilon \downarrow 0$ , that is,

\begin{align*}\lim_{\varepsilon \downarrow 0} \sup_{t\geq 0} |X_t^{(\varepsilon)}-X_t|=0.\end{align*}

In addition, we define the last zero process $g_{\varepsilon,t}$ associated to the process $X^{(\varepsilon)}$ , that is,

\begin{align*}g_{\varepsilon,t}=\sup\{0\leq s\leq t\,:\, X_s^{(\varepsilon)}\leq 0 \}\end{align*}

for $\varepsilon>0$ and $t\geq 0$ . The inequality $g_t\leq g_{\varepsilon,t} \leq g_t^{(\varepsilon)}$ holds for all $t\geq 0$ . Taking $\varepsilon\downarrow 0$ , and by the right-continuity of $x\mapsto g_t^{(x)}$ , we obtain that $ g_{\varepsilon,t} \downarrow g_t$ when $\varepsilon \downarrow 0$ for all $t\geq 0$ . Moreover, we have that $t-g_{\varepsilon,t}\,=\!:\,U_{\varepsilon,t} \uparrow U_t$ when $\varepsilon \downarrow 0$ for all $t\geq 0$ .

Recall that the local time at the level $a\in \mathbb{R}$ , $L^a=\{L_t^a,t\geq 0 \}$ , is a continuous process defined in terms of the Itô–Tanaka formula (see, for example, Theorem IV.68 on p. 216 in [Reference Protter39] and (2.13)) and its measure $\rm{d} L_t^a$ is carried by the set $\{s\geq 0: X_{s-}=X_s=a \}$ . For ease of notation, we denote $L_t=L_t^0$ for every $t\geq 0$ , the local time at the level zero. In this case (see, for example, Corollary IV.1 on p. 219 in [Reference Protter39]), we have that for any bounded and measurable function g,

\begin{align*}\int_{-\infty}^{\infty} L_t^a g(a)\rm{d} a=\sigma \int_0^t g(X_s) \rm{d} s, \quad \text{almost surely}.\end{align*}

For each $\varepsilon>0$ and $t\geq 0$ , we define

\begin{align*}M_t^{(\varepsilon)}=\sum_{k=1}^{\infty} \mathbb{I}_{\{\rho_{k,\varepsilon}^- < t \}}.\end{align*}

Note that $M_t^{(\varepsilon)}-1$ is the number of downcrossings of the level zero at time $t> 0$ of the process $X_t^{(\varepsilon)}$ . It turns out that $M_t^{(\varepsilon)}$ works as an approximation of the local time at zero in some sense. We have the following lemma. The proof follows an argument similar to that in [Reference Revuz and Yor41] (see Theorem VI.1.10 on p. 227).

Lemma 5.1. Suppose that X is a spectrally negative Lévy process. Then for all $t\geq 0$ ,

\begin{align*}\lim_{\varepsilon \downarrow 0}\varepsilon M_t^{(\varepsilon)} =\frac{1}{2} L_t \quad \textit{in probability}.\end{align*}

Proof. From the Meyer–Itô formula (see, for example, Theorems IV.68 and IV.70 in pp. 216 and 218 in [Reference Protter39]) we know that

\begin{align*}X_t^+=X_0^++\int_{(0,t]} \mathbb{I}_{\{X_{s-}>0\}}\rm{d} X_s +\int_{(0,t]} \int_{(\!-\!\infty,0)} (X_{s-}+y)^- \mathbb{I}_{\{X_{s-}>0\}} N(\rm{d} s \times \rm{d} y)+\frac{1}{2} L_t,\end{align*}

where $x^+$ and $x^-$ are the positive and negative part, respectively, of x defined by $x^+=\max\{ x,0\}$ and $x^-=-\min\{x,0\}$ . Hence, for $t\geq 0$ and $1 \leq k\leq M_t^{(\varepsilon)}$ we get that

\begin{align*}X_{\rho_{k,\varepsilon}^+ \wedge t}^+ -X_{\rho_{k,\varepsilon}^-}^+&=\int_{(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^+\wedge t]} \mathbb{I}_{\{X_{s-}>0\}}\rm{d} X_s +\int_{(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^+ \wedge t]} \int_{(\!-\!\infty,0)} (X_{s-}+y)^- \mathbb{I}_{\{X_{s-}>0\}} N(\rm{d} s \times \rm{d} y)\\&\qquad +\frac{1}{2} (L_{\rho_{k,\varepsilon}^+ \wedge t}-L_{\rho_{k,\varepsilon}^-}).\end{align*}

From the definition of the stopping times $\rho_{k,\varepsilon}^-$ , we have that $X_r>0$ when $r\in [\rho_{k,\varepsilon}^+, \rho_{k+1,\varepsilon}^-)$ for some $k \geq 1$ , and, since L is continuous and only charge points in the set of zeros of X, we have that $L_{\rho_{k,\varepsilon}^+}=L_{\rho_{k+1,\varepsilon}^-}$ and $L_{t \wedge \rho_{M_t^{(\varepsilon)},\varepsilon}^+}=L_t$ . Hence, using a telescopic sum and the fact that $g_{\varepsilon,r-}=r$ if and only if $r \in (\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^+]$ , for some $k\geq 1$ , we have that

\begin{align*}X_{t \wedge \rho_{M_t^{(\varepsilon)},\varepsilon}^+}^+ &-X_{\rho_{M_t^{(\varepsilon)},\varepsilon}^-}^+ +\sum_{k=1}^{M_t^{(\varepsilon)}-1} (X_{\rho_{k,\varepsilon}^+}^+ -X_{\rho_{k,\varepsilon}^-}^+ ) \\& =\int_{(0,t]} \mathbb{I}_{\{ g_{\varepsilon,s-}=s\}}\mathbb{I}_{\{X_{s-}\gt 0\}}\rm{d} X_s +\int_{(0,t]} \int_{(\!-\!\infty,0)} (X_{s-}+y)^- \mathbb{I}_{\{g_{\varepsilon,s-}=s\}} \mathbb{I}_{\{X_{s-}\gt 0\}} N(\rm{d} s \times \rm{d} y)\nonumber\\&\qquad+\frac{1}{2} L_t.\end{align*}

Thus, since $X_{\rho_{k,\varepsilon}^-}\leq 0$ and $X_{\rho_{k,\varepsilon}^+}=\varepsilon$ on the event $\{\rho_{k,\varepsilon}^+<\infty \}$ for all $k\geq 1$ , we obtain that for any $t\geq 0$ and $\varepsilon>0$ ,

\begin{align*}X_{t \wedge \rho_{M_t^{(\varepsilon)},\varepsilon}^+}^+ & +\varepsilon (M_t^{(\varepsilon)}-1) \\& =\int_{(0,t]} \mathbb{I}_{\{ g_{\varepsilon,s-}=s\}}\mathbb{I}_{\{X_{s-}\gt 0\}}\rm{d} X_s +\int_{(0,t]} \int_{(\!-\!\infty,0)} (X_{s-}+y)^- \mathbb{I}_{\{ g_{\varepsilon,s-}=s\}} \mathbb{I}_{\{X_{s-}\gt 0\}} N(\rm{d} s \times \rm{d} y)\nonumber\\ &\qquad+\frac{1}{2} L_t.\end{align*}

Note that $0\leq X_{t \wedge \rho_{M_t^{(\varepsilon)},\varepsilon}^+}^+ \leq \varepsilon$ and then $\lim_{\varepsilon \downarrow 0} X_{t \wedge \rho_{M_t^{(\varepsilon)},\varepsilon}^+}^+=0$ . Moreover, from the inequality $g_t\leq g_{\varepsilon,t} \leq g_t^{(\varepsilon)}$ we see that

\begin{align*}\mathbb{I}_{\{g_{t-}^{(\varepsilon)} < t \}}\leq \mathbb{I}_{\{g_{\varepsilon,t-} < t \}}\leq \mathbb{I}_{\{g_{t-} < t \}}\end{align*}

for any $t > 0$ and $\varepsilon>0$ . Thus, since $g_{t-}^{(\varepsilon)} \downarrow g_{t-}$ when $\varepsilon \downarrow 0$ , and the mapping $x\mapsto \mathbb{I}_{\{ x < t \}}$ is right-continuous, for fixed $t>0$ , we conclude that $\lim_{\varepsilon \downarrow 0} \mathbb{I}_{\{g_{\varepsilon,t-} < t \}}=\mathbb{I}_{\{g_{t-} < t \}}$ for all $t>0$ . Therefore, from the dominated convergence theorem for stochastic integrals (see, for example, Theorems IV.15 and IV.32 on pp. 166 and 176 in [Reference Protter39]), we have that the first term on the right-hand side of the equation above converges to 0 uniformly on compacts in probability, that is, for all $t> 0$ ,

\begin{align*}\sup_{{0\leq s\leq t}} \left|\int_{(0,s]} \mathbb{I}_{\{ g_{\varepsilon,r-}=r\}}\mathbb{I}_{\{X_{r-} > 0\}}\rm{d} X_r \right|\end{align*}

converges to 0 in probability when $\varepsilon \downarrow 0$ . Note that, for all $s\geq 0$ , we have that $(X_{s-}+y)^- \mathbb{I}_{\{ g_{\varepsilon,s-}=s\}} \mathbb{I}_{\{X_{s-}>0\}} \leq (X_{s-}+y)^- \mathbb{I}_{\{X_{s-}>0\}}$ and that

\begin{align*} \int_{(0,t]} \int_{(\!-\!\infty,0)} (X_{s-}+y)^- \mathbb{I}_{\{X_{s-} > 0\}} N(\rm{d} s \times \rm{d} y) < \infty,\end{align*}

for all $t\geq 0$ . Then, by the dominated convergence theorem,

\begin{align*}\lim_{\varepsilon \downarrow 0}\int_{(0,t]} \int_{(\!-\!\infty,0)} (X_{s-}+y)^- \mathbb{I}_{\{g_{\varepsilon,s-}=s\}} \mathbb{I}_{\{X_{s-}>0\}} N(\rm{d} s \times \rm{d} y)=0,\end{align*}

for any $t\geq 0$ . Thus, for fixed $t\geq 0$ , we have that $\varepsilon M_t^{(\varepsilon)}$ converges to $L_t/2$ in probability when $\varepsilon \downarrow 0$ .

We now state the following proposition for stopped processes that will be of use in the proof of Theorem 3.1. We introduce some additional notation: for any $-\infty\leq b < a\leq \infty$ and any stopping time $\tau$ , we define the stopping time

\begin{align*}T_{a,b,\tau}=\inf\{t\geq \tau\,:\, X_t\notin [b,a) \}.\end{align*}

We also recall that $i=1$ if X is of finite variation and $i=2$ when X is of infinite variation.

Proposition 5.1. Suppose that X is any spectrally negative process. Take $-\infty\leq b < a\leq \infty$ and define $C=[0,\infty)\times (b,a)$ . Let $F\,:\,\mathbb{R}_+\times \mathbb{R} \mapsto \mathbb{R}$ be a function such that F is $C^{1,i}$ on $\bar{C}$ and, if $a<\infty$ , we assume that F satisfies $\lim_{h \downarrow 0} F(t,a-h)=F(t,a)$ for all $t\geq 0$ . Further, if $\sigma>0$ and $b>-\infty$ , assume that $\lim_{h\downarrow 0}F(t,b+h)=F(t,b)$ for all $t\geq 0$ . Then, for any stopping time $\tau$ and $t\geq 0$ , we have

(5.2) \begin{align}F(t\wedge T_{a,b,\tau}&,X_{t\wedge T_{a,b,\tau}})\nonumber\\&= F(t\wedge \tau ,X_{t\wedge \tau})+\int_{t\wedge \tau}^{t\wedge T_{a,b,\tau}} \frac{\partial F}{\partial t}(s,X_{s-})\rm{d} s+\int_{(t\wedge \tau, t\wedge T_{a,b,\tau}]} \frac{\partial F}{\partial x} (s,X_{s-})\rm{d} X_s \nonumber\\&\qquad +\!\frac{1}{2}\sigma^2 \int_{t\wedge \tau}^{t\wedge T_{a,b,\tau}} \frac{\partial^2 F}{\partial x^2} (s,X_{s-})\rm{d} s \nonumber\\&\qquad+\!\int_{(t\wedge \tau,t\wedge T_{a,b,\tau}]} \!\int_{(\!-\!\infty,0)}\![F(s,X_{s-}+y)-F(s,X_{s-})-y \frac{\partial F}{\partial x} (s,X_{s-}) ]N(\rm{d} s\times \rm{d} y).\end{align}

Proof. We first note that $t\wedge \tau \leq t\wedge T_{a,b,\tau} $ and that the statement trivially holds when $t\leq \tau$ or when $X_{\tau} \notin [b,a) $ , as $T_{a,b,\tau}=\tau$ in the latter case. Moreover, if X is of infinite variation and $X_{\tau}=b$ , it can also be easily seen that $T_{a,b,\tau}=\tau$ . Further, from the Markov property we see that if $a<\infty$ ,

\begin{align*}\mathbb{P}(X_{ T_{a,b,\tau} }>a,X_{\tau}\in [b,a), T_{a,b,\tau}<\infty)=\mathbb{E}\left[\mathbb{I}_{\{X_{\tau}\in [b,a), T_{a,b,\tau}<\infty \}} \mathbb{P}_{X_{\tau} }( X_{ \tau_a^+ \wedge \tau_b^-} >a )\right]=0,\end{align*}

where we used that $X_{\tau_a^+}=a$ with probability 1 on the event $\{ 0<\tau_a^+ < \infty\}$ and $X_{\tau_b^-}\leq b$ on the event $\{0<\tau_b^-<\infty \}$ , whenever $b>-\infty$ . Moreover, if $\sigma=0$ and $b>-\infty$ , we deduce by the strong Markov property that

\begin{align*}\mathbb{P}(X_{ T_{a,b,\tau} }=b,X_{\tau}\in (b,a), T_{a,b,\tau} < \infty)&=\mathbb{E}\left[ \mathbb{I}_{\{X_{\tau}\in (b,a) \}} \mathbb{P}_{X_{\tau} }( X_{ \tau_b^-} =b, \tau_b^- < \tau_a^+ )\right]\\&\leq \mathbb{E}\left[ \mathbb{I}_{\{X_{\tau}\in (b,a) \}} \mathbb{P}_{X_{\tau} }( X_{ \tau_b^-} =b, \tau_b^- < \infty )\right]\\&=0,\end{align*}

where we used that $X_{\tau_b^-}=b$ with positive probability, on the event $\{0< \tau_b^- < \infty\}$ , only when $\sigma>0$ (see (2.8)). Similarly, when X is of finite variation, we can see that $\mathbb{P}(X_{ T_{a,b,\tau} }=b, T_{a,b,\tau}<\infty,X_{\tau}=b)=0$ . We define the auxiliary function

\begin{align*}\widehat{F}(t,x)=\left\{\begin{array}{l@{\quad}l}F(t,a\!-\!),\, & t\geq 0 \text{ and } x\geq a,\\F(t,x),\, & t\geq 0 \text{ and } b < x < a,\\F(t,b\!+\!), \,& t\geq 0 \text{ and } x\leq b.\end{array}\right.\end{align*}

Hence, we have that $\widehat{F}$ is continuous on $\mathbb{R}_+\times \mathbb{R}$ and is such that $\widehat{F}$ is $C^{1,i}$ on $\bar{C}$ , $\widehat{F}$ is $C^{1,i}$ on $\bar{D}_1$ and $\widehat{F}$ is $C^{1,i}$ on $\bar{D}_2$ , where $D_1=[0,\infty)\times (a,\infty)$ and $D_2=[0,\infty)\times (\!-\!\infty,b)$ . Using that $F(t,a)=F(t,a-)$ and $F(t,b+)=F(t,b)$ , when $\sigma>0$ , we see that

\begin{align*}F(t\wedge T_{a,b,\tau},X_{t\wedge T_{a,b,\tau}})&=F(t,X_{t})\mathbb{I}_{\{t < T_{a,b,\tau} \}} +F(T_{a,b,\tau},a)\mathbb{I}_{\{T_{a,b,\tau} \leq t \}}\mathbb{I}_{\{X_{T_{a,b,\tau}} =a \}}\\&\quad +F( T_{a,b,\tau},b)\mathbb{I}_{\{T_{a,b,\tau}\leq t \}}\mathbb{I}_{\{X_{T_{a,b,\tau}}=b \}}\nonumber\\&\quad +F( T_{a,b,\tau},X_{ T_{a,b,\tau}})\mathbb{I}_{\{T_{a,b,\tau}\leq t \}}\mathbb{I}_{\{X_{T_{a,b,\tau}} < b \}}\\&=\widehat{F}(t\wedge T_{a,b,\tau},X_{t\wedge T_{a,b,\tau}})+[F( T_{a,b,\tau},X_{T_{a,b,\tau}})\nonumber\\&\quad -F(T_{a,b,\tau},b+)]\mathbb{I}_{\{T_{a,b,\tau}\leq t \}}\mathbb{I}_{\{X_{T_{a,b,\tau}} < b \}}\\&=\widehat{F}(t\wedge T_{a,b,\tau},X_{t\wedge T_{a,b,\tau}})\\&\quad +\int_{(t\wedge \tau ,t\wedge T_{a,b,\tau}]}\int_{(\!-\!\infty,0)}[F( s,X_{ s-}+y)\nonumber\\&\quad -F(s,b+)]\mathbb{I}_{\{X_{s-}+y \leq b\}} N(\rm{d} s \times \rm{d} y).\end{align*}

Thus, applying the version for multiple curves of formula (2.12) (see Remarks 2.2 and 3.3 in [Reference Peskir37]) at the times $t\wedge T_{a,b,\tau}$ and $t\wedge \tau$ respectively, upon using that $L_{t\wedge T_{a,b,\tau}}^a=L_{t\wedge \tau}^a$ and $L_{t\wedge T_{a,b,\tau}}^b=L_{t\wedge \tau}^b$ as the local time is continuous and $X_{s-}\in (b,a)$ for $s\in ( \tau,t\wedge T_{a,b,\tau})$ when $\tau < t$ and $\tau < T_{a,b,\tau}$ , we obtain that

\begin{align*}\widehat{F}&(t\wedge T_{a,b,\tau},X_{t\wedge T_{a,b,\tau}})\\&= \widehat{F}(t\wedge \tau,X_{t\wedge \tau})+\int_{t\wedge \tau}^{t\wedge T_{a,b,\tau}} \frac{\partial \widehat{F}}{\partial t}(s,X_{s-})\rm{d} s+\int_{(t\wedge \tau, t\wedge T_{a,b,\tau}]} \frac{\partial \widehat{F}}{\partial x} (s,X_{s-})\rm{d} X_s\nonumber\\&\quad+\frac{1}{2}\sigma^2 \int_{t\wedge \tau}^{t\wedge T_{a,b,\tau}} \frac{\partial^2 \widehat{F}}{\partial x^2} (s, X_{s-}) \rm{d} s\\&\quad+\int_{(t\wedge \tau,t\wedge T_{a,b,\tau}]} \int_{(\!-\!\infty,0)}[\widehat{F}(s,X_{s-}+y)-\widehat{F}(s,X_{s-})-y \frac{\partial \widehat{F}}{\partial x} (s,X_{s-}) ]N(\rm{d} s\times \rm{d} y)\\&= F(t\wedge \tau,X_{t\wedge\tau})+\int_{t\wedge \tau}^{t\wedge T_{a,b,\tau}} \frac{\partial F}{\partial t}(s,X_{s-})\rm{d} s+\int_{(t\wedge \tau, t\wedge T_{a,b,\tau}]} \frac{\partial F}{\partial x} (s,X_{s-})\rm{d} X_s\nonumber\\&\quad+\frac{1}{2}\sigma^2 \int_{t\wedge \tau}^{t\wedge T_{a,b,\tau}} \frac{\partial^2 F}{\partial x^2} (s, X_{s-}) \rm{d} s\\&\quad+\int_{(t\wedge\tau,t\wedge T_{a,b,\tau}]} \int_{(\!-\!\infty,0)}[F(s,X_{s-}+y)-F(s,X_{s-})-y \frac{\partial F}{\partial x} (s,X_{s-}) ]\mathbb{I}_{\{X_{s-}+y> b \}}N(\rm{d} s\times \rm{d} y)\\&\quad+\int_{(t\wedge\tau,t\wedge T_{a,b,\tau}]} \int_{(\!-\!\infty,0)}[F(s,b+)-F(s,X_{s-})-y \frac{\partial F}{\partial x} (s,X_{s-}) ]\mathbb{I}_{\{ X_{s-}+y\leq b\}}N(\rm{d} s\times \rm{d} y).\end{align*}

Therefore, we obtain the desired result by combining the two equations above.

5.2.1. Proof of Theorem 3.1

We first assume that F and its derivatives are bounded. It is easy to see that for each $\varepsilon>0$ we have $\sup_{n\geq 0} \rho_{k,\varepsilon}^-=\infty$ . Then, for each $t\geq 0$ and $\varepsilon>0$ , there exists a value $n\geq 1$ such that $\rho_{n,\varepsilon}^-\leq t< \rho_{n+1,\varepsilon}^-$ , so that $M_{t}^{(\varepsilon)}=n$ . We consider the case when $X_t>0$ (we omit the proof of the case when $X_t\leq 0$ as it is similar). Then, for any $\varepsilon>0$ sufficiently small, there exists a value $n\geq 1$ (depending on the value of $\varepsilon$ ) such that $\rho_{n,\varepsilon}^+ \leq t <\rho_{n+1,\varepsilon}^-$ . Note that in this case we have that $X_{t}^{(\varepsilon)}>0$ and $g_{\varepsilon,t}=\rho_{n,\varepsilon}^+$ . Using a telescopic sum, we obtain that

\begin{align*}F(g_{\varepsilon,t},t,X_t^{(\varepsilon)})&=F(g_{\varepsilon,0},0,X^{(\varepsilon)}_0+\varepsilon)+\sum_{k=1}^{M_t^{(\varepsilon)}}[ F(g_{\varepsilon,\rho_{k,\varepsilon}^-},\rho_{k,\varepsilon}^-,X^{(\varepsilon)}_{\rho_{k,\varepsilon}^-})-F(g_{\varepsilon,\rho_{k,\varepsilon}^-},\rho_{k,\varepsilon}^-,X^{(\varepsilon)}_{\rho_{k,\varepsilon}^-}+\varepsilon)]\\&\qquad +\sum_{k=1}^{M_t^{(\varepsilon)}}[ F(g_{\varepsilon,\rho_{k,\varepsilon}^+},\rho_{k,\varepsilon}^+,X^{(\varepsilon)}_{\rho_{k,\varepsilon}^+}-\varepsilon)-F(g_{\varepsilon,\rho_{k,\varepsilon}^-},\rho_{k,\varepsilon}^-,X^{(\varepsilon)}_{\rho_{k,\varepsilon}^-})]\\&\qquad +\sum_{k=1}^{M_t^{(\varepsilon)}}[ F(g_{\varepsilon,\rho_{k,\varepsilon}^+},\rho_{k,\varepsilon}^+,X^{(\varepsilon)}_{\rho_{k,\varepsilon}^+})-F(g_{\varepsilon,\rho_{k,\varepsilon}^+},\rho_{k,\varepsilon}^+,X^{(\varepsilon)}_{\rho_{k,\varepsilon}^+}-\varepsilon)]\\&\qquad +\sum_{k=1}^{M_t^{(\varepsilon)}-1}[ F(g_{\varepsilon,\rho_{k+1,\varepsilon}^-},\rho_{k+1,\varepsilon}^-,X^{(\varepsilon)}_{\rho_{k+1,\varepsilon}^-}+\varepsilon)-F(g_{\varepsilon,\rho_{k,\varepsilon}^+},\rho_{k,\varepsilon}^+,X^{(\varepsilon)}_{\rho_{k,\varepsilon}^+})]\\&\qquad+[F(g_{\varepsilon,t},t,X^{(\varepsilon)}_t)-F(g_{\varepsilon,\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+},\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+,X^{(\varepsilon)}_{\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+})].\end{align*}

From Remark 5.1 we know that $g_{\varepsilon,\rho_{k,\varepsilon}^-}=\rho_{k,\varepsilon}^- $ and $g_{\varepsilon,\rho_{k,\varepsilon}^+}=\rho_{k,\varepsilon}^+$ for all $k\geq 1$ . Thus, from the definition of $X^{(\varepsilon)}$ (see (5.1)) we see that

(5.3) \begin{align}F\big(g_{\varepsilon,t},t,X_t^{(\varepsilon)}\big)&=F\big(g_{\varepsilon,0},0,X^{(\varepsilon)}_0+\varepsilon\big)+\sum_{k=1}^{M_t^{(\varepsilon)}} \left[ F\big(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,X_{\rho_{k,\varepsilon}^-}-\varepsilon \big)-F\big(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,X_{\rho_{k,\varepsilon}^-}\big)\right]\nonumber\\&\qquad +\sum_{k=1}^{M_t^{(\varepsilon)}}[ F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+}-\varepsilon)-F(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,X_{\rho_{k,\varepsilon}^-}-\varepsilon)]\nonumber\\&\qquad +\sum_{k=1}^{M_t^{(\varepsilon)}}[ F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+})-F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+}-\varepsilon)]\nonumber\\&\qquad +\sum_{k=1}^{M_t^{(\varepsilon)}-1}[ F(\rho_{k+1,\varepsilon}^-,\rho_{k+1,\varepsilon}^-,X_{\rho_{k+1,\varepsilon}^-})-F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+})]\nonumber\\&\qquad+[F(\rho^+_{M_{t}^{(\varepsilon)},\varepsilon},t,X_t)-F(\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+,\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+,X_{\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+})].\end{align}

We first show that the sum of the terms on the second, fourth and last lines on the right-hand side of the equation above converges in probability to the terms on the right-hand side of (3.5).

Lemma 5.2. Under the assumptions of Theorem 3.1 and further assuming that F and its derivatives are bounded, we have that

(5.4) \begin{align}\lim_{\varepsilon \downarrow 0}&\bigg[\sum_{k=1}^{M_t^{(\varepsilon)}}[ F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+}-\varepsilon)-F(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,X_{\rho_{k,\varepsilon}^-}-\varepsilon)]\nonumber\\&\qquad+\sum_{k=1}^{M_t^{(\varepsilon)}-1}[ F(\rho_{k+1,\varepsilon}^-,\rho_{k+1,\varepsilon}^-,X_{\rho_{k+1,\varepsilon}^-})-F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+})]\nonumber\\&\qquad+[F(\rho^+_{M_{t}^{(\varepsilon)},\varepsilon},t,X_t)-F(\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+,\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+,X_{\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+})]\bigg]\nonumber\\&= \int_{0}^{t} \frac{\partial F_g }{\partial t } (s,X_{s-})\mathbb{I}_{\{g_{s-}=s \}}\rm{d} s+ \int_{0}^{t} \frac{\partial F }{\partial t} (g_{s-},s,X_{s-})\mathbb{I}_{\{ g_{s-} < s\}} \rm{d} s+\int_{0}^{t}\frac{\partial F }{\partial x} (g_{s-},s,X_{s-})\rm{d} X_s \nonumber\\&\qquad+\frac{1}{2} \sigma^2 \int_{0}^{t} \frac{\partial^2 F}{\partial x^2} (g_{s-},s,X_{s-}) \rm{d} s\nonumber\\& \qquad +\!\int_{[0,t] } \int_{(\!-\!\infty,0)} \!\left[ F(s,s,X_{s-}+y) -F(s,s,X_{s-})-y\frac{\partial F }{\partial x} (s,s,X_{s-}) \right]\mathbb{I}_{\{g_{s-}=s \}}N(\rm{d} s\times \rm{d} y)\nonumber\\&\qquad+\int_{[0,t]}\int_{(\!-\!\infty,0)} \left[F(g_{s-},s,X_{s-}+y)-F(g_{s-},s,X_{s-})-y\frac{\partial F }{\partial x} (g_{s-},s,X_{s-})\right]\nonumber\\&\qquad\qquad\qquad \times \mathbb{I}_{\{X_{s-}+y >0 \}}\mathbb{I}_{\{ g_{s-} < s\}} N(\rm{d} s\times \rm{d} y) \nonumber\\&\qquad+ \int_{[0,t]}\int_{(\!-\!\infty,0)} \left[F(s,s,X_{s-}+y)-F(g_{s-},s,X_{s-})-y\frac{\partial F }{\partial x} (g_{s-},s,X_{s-}) \right]\nonumber\\&\qquad\qquad\qquad \times \mathbb{I}_{\{X_{s-}+y \leq 0 \}} \mathbb{I}_{\{ g_{s-} < s\}}N(\rm{d} s\times \rm{d} y)\end{align}

in probability.

Proof. We recall that $E_g=\{ (\gamma,t,x) \, :\, 0\leq \gamma < t \text{ and } x>0\}\cup \{ (\gamma,t,x) \,:\, 0 \leq \gamma=t \text{ and } x\leq 0\}$ . We define the auxiliary function $\widetilde{F}\,:\,\mathbb{R}_2^+\times \mathbb{R}\mapsto \mathbb{R}$ via

\begin{align*}\widetilde{F}(\gamma,t,x)=\left\{\begin{array}{l@{\quad}l}F(\gamma,t,x), & (\gamma,t,x)\in \bar{E}_g,\\F(t,t,x), & (\gamma,t,x)\in (\mathbb{R}_2^+\times \mathbb{R}) \setminus \bar{E}_g.\end{array} \right.\end{align*}

Note that, in particular, we have that $\widetilde{F}(\gamma,t,x)=F(t,t,x)$ for any $\gamma,t\geq 0$ and $x\leq 0$ . Hence, since $X_{\rho_{k+1,\varepsilon}^-}\leq 0$ for all $k\leq M_t^{(\varepsilon)}-1$ , we can write

\begin{align*}\sum_{k=1}^{M_t^{(\varepsilon)}-1}&\left[ F\left(\rho_{k+1,\varepsilon}^-,\rho_{k+1,\varepsilon}^-,X_{\rho_{k+1,\varepsilon}^-}\right)-F\left(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+}\right)\right] \\&\qquad+\left[F\left(\rho^+_{M_{t}^{(\varepsilon)},\varepsilon},t,X_t\right)-F\left(\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+,\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+,X_{\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+}\right)\right]\\&=\sum_{k=1}^{M_t^{(\varepsilon)}}\left[ \widetilde{F}\left(\rho_{k,\varepsilon}^+,\rho_{k+1,\varepsilon}^- \wedge t,X_{\rho_{k+1,\varepsilon}^- \wedge t}\right)-\widetilde{F}\left(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+}\right)\right].\end{align*}

From the assumptions on F, we see that for fixed $\gamma\geq 0$ , the mapping $(t,x)\mapsto \widetilde{F}(\gamma,t,x)$ is $C^{1,i}$ on $[\gamma,\infty)\times [0,\infty)$ . Moreover, if $\sigma>0$ , since $\lim_{h\downarrow 0} F(\gamma,t,h)=F(t,t,0)$ for all $0\leq \gamma \leq t$ , we see that $ \lim_{h\downarrow 0} \widetilde{F}(\gamma,t,h)=\widetilde{F}(\gamma,t,0)$ for all $\gamma\geq 0$ and $t\geq 0$ . Applying Itô's formula (see Proposition 5.1) to $s\mapsto \widetilde{F}(\rho_{k,\varepsilon}^+,s , X_{s}) $ on time intervals of the form $(\rho_{k,\varepsilon}^+ \wedge t,\rho_{k+1,\varepsilon}^-\wedge t]$ , for each $k\leq M_t^{(\varepsilon)}$ , we deduce that

\begin{align*}&\sum_{k=1}^{M_t^{(\varepsilon)}}[ \widetilde{F}(\rho_{k,\varepsilon}^+,\rho_{k+1,\varepsilon}^- \wedge t,X_{\rho_{k+1,\varepsilon}^- \wedge t})-\widetilde{F}(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+})]\\&=\sum_{k=1}^{M_t^{(\varepsilon)}} \left[ \int_{\rho_{k,\varepsilon}^+}^{\rho_{k+1,\varepsilon}^- \wedge t} \frac{\partial \widetilde{F} }{\partial t} (\rho_{k,\varepsilon}^+,s,X_{s-}) \rm{d} s +\int_{(\rho_{k,\varepsilon}^+,\rho_{k+1,\varepsilon}^- \wedge t]} \frac{\partial \widetilde{F} }{\partial x} (\rho_{k,\varepsilon}^+,s,X_{s-}) \rm{d} X_s \right]\\&\qquad +\sum_{k=1}^{M_t^{(\varepsilon)}} \frac{1}{2}\sigma^2\int_{\rho_{k,\varepsilon}^+}^{\rho_{k+1,\varepsilon}^- \wedge t} \frac{\partial^2 \widetilde{F} }{\partial x^2} (\rho_{k,\varepsilon}^+,s,X_{s-}) \rm{d} s \\&\qquad+\sum_{k=1}^{M_t^{(\varepsilon)}} \int_{(\rho_{k,\varepsilon}^+,\rho_{k+1,\varepsilon}^- \wedge t]} \int_{(\!-\!\infty,0)} \left[\widetilde{F}(\rho_{k,\varepsilon}^+,s,X_{s-}+y)-\widetilde{F}(\rho_{k,\varepsilon}^+,s,X_{s-}) -y\frac{\partial \widetilde{F}}{\partial x} ( \rho_{k,\varepsilon}^+,s,X_{s-}) \right]\\&\qquad\qquad\qquad\times N(\rm{d} s\times \rm{d} y).\end{align*}

From the definition of $\widetilde{F}$ and since $g_{\varepsilon,s-} < s$ if and only if $s\in (\rho_{k,\varepsilon}^+, \rho_{k+1,\varepsilon}^-]$ for some $k\geq 1$ (in this case $g_{\varepsilon,s-}=\rho_{k,\varepsilon}^+$ and $X_{s-}^{(\varepsilon)}=X_{s-}$ ), we have that

\begin{align*}&\sum_{k=1}^{M_t^{(\varepsilon)}}[ \widetilde{F}(\rho_{k,\varepsilon}^+,\rho_{k+1,\varepsilon}^- \wedge t,X_{\rho_{k+1,\varepsilon}^- \wedge t})-\widetilde{F}(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+})]\\&= \int_{0}^{t} \frac{\partial F }{\partial t} (g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)})\mathbb{I}_{\{ g_{\varepsilon,s-} < s\}} \rm{d} s +\int_{0}^{t} \frac{\partial F }{\partial x} (g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)})\mathbb{I}_{\{ g_{\varepsilon,s-} < s \}} \rm{d} X_s \\&\qquad + \frac{1}{2}\sigma^2\int_{0}^{t} \frac{\partial^2 F }{\partial x^2} (g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)})\mathbb{I}_{\{ g_{\varepsilon,s-} < s\}} \rm{d} s \\&\qquad+\int_{[0,t]}\int_{(-\infty,0)} \left[F(g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)}+y)-F(g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)})-y\frac{\partial F }{\partial x} (g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)})\right]\\&\qquad\qquad\qquad \times \mathbb{I}_{\{X_{s-}^{(\varepsilon)}+y >0 \}}\mathbb{I}_{\{ g_{\varepsilon,s-} < s\}} N(\rm{d} s\times \rm{d} y) \\&\qquad+ \int_{[0,t]}\int_{(-\infty,0)} \left[F(s,s,X_{s-}^{(\varepsilon)}+y)-F(g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)})-y\frac{\partial F }{\partial x} (g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)}) \right]\\&\qquad\qquad\qquad \times \mathbb{I}_{\{X_{s-}^{(\varepsilon)}+y \leq 0 \}} \mathbb{I}_{\{ g_{\varepsilon,s-}< s\}}N(\rm{d} s\times \rm{d} y).\end{align*}

Similarly, since $(t,x)\mapsto F(t,t,x)$ is $C^{1,i}$ on $[0,\infty)\times (\!-\!\infty,0]$ and F is continuous on $E_g$ , we can apply Proposition 5.1 on intervals of the form $(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^+\wedge t]=(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^+]$ , upon noticing that $X_s-\varepsilon\leq 0$ for all $s\in (\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^+]$ , to see that

\begin{align*}&\sum_{k=1}^{M_t^{(\varepsilon)}}[ F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+}-\varepsilon)-F(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,X_{\rho_{k,\varepsilon}^-}-\varepsilon)]\\&=\sum_{k=1}^{M_t^{(\varepsilon)}}\left[ \int_{\rho_{k,\varepsilon}^-}^{\rho_{k,\varepsilon}^+ } \frac{\partial F_g }{\partial t } (s,X_{s-}-\varepsilon)\rm{d} s+\int_{(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^+] }\frac{\partial F }{\partial x} (s,s,X_{s-}-\varepsilon)\rm{d} X_s \right.\end{align*}\begin{align*}&\quad \left.+\frac{1}{2} \sigma^2 \int_{\rho_{k,\varepsilon}^-}^{\rho_{k,\varepsilon}^+ } \frac{\partial^2 F}{\partial x^2} (s,s,X_{s-}-\varepsilon)\rm{d} s\right]\\& \quad +\sum_{k=1}^{M_t^{(\varepsilon)}}\int_{(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^+] } \int_{(\!-\!\infty,0)} \left[ F(s,s,X_{s-}+y-\varepsilon) \vphantom{\frac{\partial F }{\partial x}}\right.\nonumber\\&\quad\left.-F(s,s,X_{s-}-\varepsilon)-y\frac{\partial F }{\partial x} (s,s,X_{s-}-\varepsilon) \right]N(\rm{d} s\times \rm{d} y)\\&= \int_{0}^{t} \frac{\partial F_g }{\partial t } (s,X_{s-}^{(\varepsilon)})\mathbb{I}_{\{g_{\varepsilon,s-}=s \}}\rm{d} s+\int_{0}^{t}\frac{\partial F }{\partial x} (s,s,X_{s-}^{(\varepsilon)})\mathbb{I}_{\{g_{\varepsilon,s-}=s \}}\rm{d} X_s \\&\quad+\frac{1}{2} \sigma^2 \int_{0}^{t} \frac{\partial^2 F}{\partial x^2} (s,s,X_{s-}^{(\varepsilon)}) \mathbb{I}_{\{g_{\varepsilon,s-}=s \}}\rm{d} s\\& \quad +\int_{[0,t] } \int_{(\!-\!\infty,0)} \left[ F(s,s,X_{s-}^{(\varepsilon)}+y) -F(s,s,X_{s-}^{(\varepsilon)})-y\frac{\partial F }{\partial x} (s,s,X_{s-}^{(\varepsilon)}) \right] \mathbb{I}_{\{g_{\varepsilon,s-}=s \}}N(\rm{d} s\times \rm{d} y),\end{align*}

where we recall that $F_g(t,x)=F(t,t,x)$ for $t\geq 0$ and $x\leq 0$ . Therefore, we conclude that

\begin{align*}&\sum_{k=1}^{M_t^{(\varepsilon)}}[ F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+}-\varepsilon)-F(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,X_{\rho_{k,\varepsilon}^-}-\varepsilon)]\\&\qquad+\sum_{k=1}^{M_t^{(\varepsilon)}-1}[ F(\rho_{k+1,\varepsilon}^-,\rho_{k+1,\varepsilon}^-,X_{\rho_{k+1,\varepsilon}^-})-F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+})]\\&\qquad+[F(\rho_{M_{t}^{(\varepsilon)}},t,X_t)-F(\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+,\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+,X_{\rho_{M_{t}^{(\varepsilon)},\varepsilon}^+})]\\&= \int_{0}^{t} \frac{\partial F_g }{\partial t } (s,X_{s-}^{(\varepsilon)})\mathbb{I}_{\{g_{\varepsilon,s-}=s \}}\rm{d} s+ \int_{0}^{t} \frac{\partial F }{\partial t} (g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)})\mathbb{I}_{\{ g_{\varepsilon,s-} < s\}} \rm{d} s+\int_{0}^{t}\frac{\partial F }{\partial x} (g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)})\rm{d} X_s \\&\qquad+\frac{1}{2} \sigma^2 \int_{0}^{t} \frac{\partial^2 F}{\partial x^2} (g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)}) \rm{d} s\\& \qquad +\int_{[0,t] } \int_{(\!-\!\infty,0)} \left[ F(s,s,X_{s-}^{(\varepsilon)}+y) -F(s,s,X_{s-}^{(\varepsilon)})-y\frac{\partial F }{\partial x} (s,s,X_{s-}^{(\varepsilon)}) \right]\mathbb{I}_{\{g_{\varepsilon,s-}=s \}}N(\rm{d} s\times \rm{d} y)\\&\qquad+\int_{[0,t]}\int_{(\!-\!\infty,0)} \left[F(g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)}+y)-F(g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)})-y\frac{\partial F }{\partial x} (g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)})\right]\\&\qquad\qquad\qquad \times \mathbb{I}_{\{X_{s-}^{(\varepsilon)}+y >0 \}}\mathbb{I}_{\{ g_{\varepsilon,s-} < s\}} N(\rm{d} s\times \rm{d} y) \\&\qquad+ \int_{[0,t]}\int_{(\!-\!\infty,0)} \left[F(s,s,X_{s-}^{(\varepsilon)}+y)-F(g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)})-y\frac{\partial F }{\partial x} (g_{\varepsilon,s-},s,X_{s-}^{(\varepsilon)}) \right]\\&\qquad\qquad\qquad \times \mathbb{I}_{\{X_{s-}^{(\varepsilon)}+y \leq 0 \}} \mathbb{I}_{\{ g_{\varepsilon,s-} < s\}}N(\rm{d} s\times \rm{d} y).\end{align*}

Hence, since we are assuming that F and its derivatives are continuous and bounded, by using the mean value theorem when X is of finite variation, Taylor’s theorem when X is of infinite variation, and the dominated convergence theorem for stochastic integrals (see, for example, Theorems IV.15 and IV.32 on pp. 166 and 176 in [Reference Protter39]) we conclude that (5.4) indeed holds.

Next, to complete the proof of Theorem 3.1, we show the existence of a sequence such that the remaining terms on the right-hand side of (5.3) converge to zero almost surely.

Lemma 5.3 Under the assumptions of Theorem 3.1 and further assuming that F and its derivatives are bounded and that $\Pi(\!-\!\infty,0)<\infty$ , we have that there exists a sequence $\{ \varepsilon_n, n\geq 0\}$ converging to zero such that

\begin{align*}\lim_{n\rightarrow\infty}\bigg[\sum_{k=1}^{M_t^{(\varepsilon_n)}}&[ F(\rho_{k,\varepsilon_n}^-,\rho_{k,\varepsilon_n}^-,X_{\rho_{k,\varepsilon_n}^-}-\varepsilon_n)-F(\rho_{k,\varepsilon_n}^-,\rho_{k,\varepsilon_n}^-,X_{\rho_{k,\varepsilon_n}^-})]\\&\qquad +\sum_{k=1}^{M_t^{(\varepsilon_n)}}[ F(\rho_{k,\varepsilon_n}^+,\rho_{k,\varepsilon_n}^+,X_{\rho_{k,\varepsilon_n}^+})-F(\rho_{k,\varepsilon_n}^+,\rho_{k,\varepsilon_n}^+,X_{\rho_{k,\varepsilon_n}^+}-\varepsilon_n)]\bigg] =0\end{align*}

almost surely.

Proof. For any $s>0$ and $\varepsilon>0$ , we define the stopping time

\begin{align*}\tau_{\varepsilon,s}^+\,:\!=\,\inf\{ t\geq s\,:\, X_t\geq \varepsilon\}.\end{align*}

By using that $\rho_{1,\varepsilon}^-=0$ (by definition), that N is the Poisson random measure of the jumps of X and by rearranging some of the terms, we see that

\begin{align*}\sum_{k=1}^{M_t^{(\varepsilon)}}&[ F(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,X_{\rho_{k,\varepsilon}^-}-\varepsilon)-F(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,X_{\rho_{k,\varepsilon}^-})] \nonumber\\&\quad+\sum_{k=1}^{M_t^{(\varepsilon)}}[ F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+})-F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,X_{\rho_{k,\varepsilon}^+}-\varepsilon)] \\&= F(0,0,X_0-\varepsilon)-F(0,0,X_0)+F(\rho_{1,\varepsilon}^+,\rho_{1,\varepsilon}^+,\varepsilon)-F(\rho_{1,\varepsilon}^+,\rho_{1,\varepsilon}^+,0) \\&\quad+\sum_{k=2}^{M_t^{(\varepsilon)}}[ F(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,X_{\rho_{k,\varepsilon}^-}-\varepsilon)-F(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,X_{\rho_{k,\varepsilon}^-})] \mathbb{I}_{\{ X_{\rho_{k,\varepsilon}^-}<0\}}\\&\quad +\sum_{k=2}^{M_t^{(\varepsilon)}}[ F(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,-\varepsilon)-F(\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^-,0)] \mathbb{I}_{\{ X_{\rho_{k,\varepsilon}^-}=0\}}\\&\quad +\sum_{k=2}^{M_t^{(\varepsilon)}}[ F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,\varepsilon)-F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,0)] \mathbb{I}_{\{ X_{\rho_{k,\varepsilon}^-}<0\}}\nonumber\\&\quad+[ F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,\varepsilon)-F(\rho_{k,\varepsilon}^+,\rho_{k,\varepsilon}^+,0)]\mathbb{I}_{\{ X_{\rho_{k,\varepsilon}^-}=0\}}\\&=F(0,0,X_0-\varepsilon)-F(0,0,X_0)+F(\rho_{1,\varepsilon}^+,\rho_{1,\varepsilon}^+,\varepsilon)-F(\rho_{1,\varepsilon}^+,\rho_{1,\varepsilon}^+,0)\\&\quad +\int_{[0,t]} \int_{(\!-\!\infty,0)}[ F(s,s,X_{s-}+y-\varepsilon)-F(s,s,X_{s-}+y)] \mathbb{I}_{\{ X^{(\varepsilon)}_{s-}+y<0\}} \mathbb{I}_{\{ X_{s-}^{(\varepsilon)}>0\}} N(\rm{d} s \times \rm{d} y)\\&\quad +\int_{[0,t]} \int_{(\!-\!\infty,0)}[ F(\tau_{\varepsilon,s}^+,\tau_{\varepsilon,s}^+,\varepsilon)-F(\tau_{\varepsilon,s}^+,\tau_{\varepsilon,s}^+,0)] \mathbb{I}_{\{ X^{(\varepsilon)}_{s-}+y<0\}} \mathbb{I}_{\{ X_{s-}^{(\varepsilon)}>0\}} N(\rm{d} s \times \rm{d} y)\\&\quad +\int_0^t[ F(s,s,-\varepsilon)-F(s,s,0)+ F(\tau_{\varepsilon,s}^+,\tau_{\varepsilon,s}^+,\varepsilon)-F(\tau_{\varepsilon,s}^+,\tau_{\varepsilon,s}^+,0)] \mathbb{I}_{\{ X_{s}=0\}}\rm{d} M_s^{(\varepsilon)}.\end{align*}

We then show that the integrals with respect to the Poisson random measure N converge to zero. Since $X_s^{(\varepsilon)}\leq X_{s}$ for all $s>0$ , and on the event $\{ X_{s-}^{(\varepsilon)}>0 \}$ , we have that $X_{s-}^{(\varepsilon)}=X_{s-}$ and we see that

\begin{align*}&\left|\int_{[0,t]} \int_{(\!-\!\infty,0)}[ F(s,s,X_{s-}+y-\varepsilon)-F(s,s,X_{s-}+y)] \mathbb{I}_{\{ X^{(\varepsilon)}_{s-}+y<0\}} \mathbb{I}_{\{ X_{s-}^{(\varepsilon)}>0\}} N(\rm{d} s \times \rm{d} y)\right|\\&\qquad\leq \int_{[0,t]} \int_{(\!-\!\infty,0)}\left| F(s,s,X_{s-}+y-\varepsilon)-F(s,s,X_{s-}+y)\right|\mathbb{I}_{\{ X^{(\varepsilon)}_{s-}+y<0\}} \mathbb{I}_{\{ X_{s-}^{(\varepsilon)}>0\}} N(\rm{d} s \times \rm{d} y)\\&\qquad\leq 2M\int_{[0,t]} \int_{(\!-\!\infty,0)}\mathbb{I}_{\{ X_{s-}+y<0\}} \mathbb{I}_{\{ X_{s-}>0\}} N(\rm{d} s \times \rm{d} y)\\&\qquad<\infty,\end{align*}

where we used that F is bounded, say, by a constant $M>0$ and that $\Pi(\!-\!\infty,0)<\infty$ so the last integral is finite. Hence, by the dominated convergence theorem and the continuity of F, we have that

\begin{align*}\lim_{\varepsilon\downarrow 0}\int_{[0,t]} \int_{(\!-\!\infty,0)}[ F(s,s,X_{s-}+y-\varepsilon)-F(s,s,X_{s-}+y)] \mathbb{I}_{\{ X^{(\varepsilon)}_{s-}+y<0\}} \mathbb{I}_{\{ X_{s-}^{(\varepsilon)}>0\}} N(\rm{d} s \times \rm{d} y)=0\end{align*}

almost surely. Similarly, we can see that

\begin{align*}\lim_{\varepsilon \downarrow 0}\int_{[0,t]} \int_{(\!-\!\infty,0)}[ F(\tau_{\varepsilon,s}^+,\tau_{\varepsilon,s}^+,\varepsilon)-F(\tau_{\varepsilon,s}^+,\tau_{\varepsilon,s}^+,0)] \mathbb{I}_{\{ X^{(\varepsilon)}_{s-}+y<0\}} \mathbb{I}_{\{ X_{s-}^{(\varepsilon)}>0\}} N(\rm{d} s \times \rm{d} y)=0\end{align*}

almost surely. On the other hand, since we are assuming that $x\mapsto F(t,t,x)$ is differentiable on $[0,\infty)$ and on $(\!-\!\infty,0]$ we have that

\begin{align*}&\left|\int_0^t[ F(s,s,-\varepsilon)-F(s,s,0)+ F(\tau_{\varepsilon,s}^+,\tau_{\varepsilon,s}^+,\varepsilon)-F(\tau_{\varepsilon,s}^+,\tau_{\varepsilon,s}^+,0)] \mathbb{I}_{\{ X_{s}=0\}}\rm{d} M_s^{(\varepsilon)} \right| \\&\qquad\leq \int_0^t\int_0^{\varepsilon} \left|\frac{\partial F}{\partial x} (\tau_{\varepsilon,s}^+,\tau_{\varepsilon,s}^+,y)-\frac{\partial F}{\partial x}(s,s,-y) \right|\mathbb{I}_{\{ X_{s}=0\}} \rm{d} y \rm{d} M_s^{(\varepsilon)} \\& \qquad \leq \varepsilon ( M_{t}^{(\varepsilon)}-1) \sup\nolimits{_{(s,y)\in [0,t]\times [0,\varepsilon]}}\left\{ \left|\frac{\partial F}{\partial x} (\tau_{\varepsilon,s}^+,\tau_{\varepsilon,s}^+,y)-\frac{\partial F}{\partial x}(s,s,-y) \right|\mathbb{I}_{\{ X_{s}=0\}} \right\}.\end{align*}

Note that X creeps downwards only when $\sigma>0$ , so then the first term above vanishes when $\sigma=0$ . Henceforth, we assume that $\sigma>0$ . From Lemma 5.1 we know that $\varepsilon M_t^{(\varepsilon)}$ converges to $L_t/2$ in probability when $\varepsilon\downarrow 0$ . Hence, there exists a subsequence $\{\varepsilon_n,n\geq 0 \}$ convergent to zero such that $\varepsilon_n M_t^{(\varepsilon_n)}$ converges to $L_t/2$ almost surely. Thus, we see that

\begin{align*}\lim_{n\rightarrow \infty} &\left|\int_0^t[ F(s,s,-\varepsilon_n)-F(s,s,0)+ F(\tau_{\varepsilon_n,s}^+,\tau_{\varepsilon_n,s}^+,\varepsilon_n)-F(\tau_{\varepsilon_n,s}^+,\tau_{\varepsilon_n,s}^+,0)] \mathbb{I}_{\{ X_{s}=0\}}\rm{d} M_s^{(\varepsilon)} \right|\\& \leq \lim_{n\rightarrow \infty} \varepsilon_n ( M_{t}^{(\varepsilon_n)}-1) \sup\nolimits{_{(s,y)\in [0,t]\times [0,\varepsilon_n]}}\left\{ \left|\frac{\partial F}{\partial x} (\tau_{\varepsilon_n,s}^+,\tau_{\varepsilon_n,s}^+,y)-\frac{\partial F}{\partial x}(s,s,-y) \right|\mathbb{I}_{\{ X_{s}=0\}}\right\}\\&=0\end{align*}

almost surely, where we used that for each $s>0$ , on the event $\{X_s=0 \}$ , we have $\lim_{\varepsilon \downarrow 0} \tau_{\varepsilon,s}^+=s$ almost surely, and that $(s,y)\mapsto \frac{\partial F}{\partial x}(s,s,y)$ is continuous in $[0,\infty)\times \mathbb{R}$ when $\sigma>0$ . Indeed, assumptions (i) and (ii) imply that $F(t,t,x)=F_1(t,x)$ for every $t\geq 0$ and $x> 0$ and $F(t,t,x)=F_2(t,x)$ for $t\geq 0$ and $x<0$ , where $F_1$ and $F_2$ are some $C^{1,1}$ functions on $[0,\infty)\times \mathbb{R}$ . Moreover, from the above facts and the mean value theorem we have that for any $h>0$ , there exists a value $c_h\in (0,h)$ such that

\begin{align*}\lim_{h\downarrow 0}\frac{F(t,t,h)-F(t,t,0)}{h}=\lim_{h\downarrow 0}\frac{\partial F}{\partial x}(t,t,c_h)=\frac{\partial F}{\partial x} F(t,t,0\!+\!)\end{align*}

for any $t\geq 0$ . Similarly,

\begin{align*}\lim_{h\downarrow 0}\frac{F(t,t,-h)-F(t,t,0)}{-h}=\frac{\partial F}{\partial x} F(t,t,0\!-\!).\end{align*}

Thus, from assumption (3.4), we have that $\frac{\partial F}{\partial x}(t,t,0)$ exists for all $t\geq 0$ . It follows that

\begin{align*}\frac{\partial F_1}{\partial x}(t,0)=\frac{\partial F}{\partial x} (t,t,0\!+\!) =\frac{\partial F}{\partial x} (t,t,0)=\frac{\partial F}{\partial x} (t,t,0\!-\!)=\frac{\partial F_2}{\partial x}(t,0)\end{align*}

for all $t\geq 0$ . Then we can write

\begin{align*}\frac{\partial F}{\partial x}F(t,t,x)&=\mathbb{I}_{\{x\geq 0 \}}\frac{\partial F_1}{\partial x}(t,x)+\mathbb{I}_{\{x< 0 \}}\frac{\partial F_2}{\partial x}(t,x)\\&=\mathbb{I}_{\{x> 0 \}}\frac{\partial F_1}{\partial x}(t,x)+\mathbb{I}_{\{x\leq 0 \}}\frac{\partial F_2}{\partial x}(t,x)\end{align*}

for any $t\geq 0$ and $x\in \mathbb{R}$ . Therefore, from the equation above it can be easily seen that $(s,y)\mapsto \frac{\partial F}{\partial x}(s,s,y)$ is continuous in $[0,\infty)\times \mathbb{R}$ as claimed. The conclusion holds.

Hence, when $\Pi(\!-\!\infty,0)<\infty$ , we deduce that there exists a subsequence $\{\varepsilon'_{n}, n\geq 0 \}$ , convergent to zero such that the limits in Lemmas 5.2 and 5.3 hold almost surely. Thus, following an argument similar to that in [Reference Applebaum2] (see proof of Theorem 4.4.7 on p. 226), from (5.3) along the subsequence $\{\varepsilon'_{n}, n\geq 0 \}$ and taking $n\rightarrow \infty$ , we deduce, when F and its derivatives are bounded, that

\begin{align*}F(g_{t}, t,X_t)&=F(g_0,0,X_0)+ \int_{0}^{t} \frac{\partial F_g }{\partial t } (s,X_{s-})\mathbb{I}_{\{g_{s-}=s \}}\rm{d} s+ \int_{0}^{t} \frac{\partial F }{\partial t} (g_{s-},s,X_{s-})\mathbb{I}_{\{ g_{s-} < s\}} \rm{d} s \\&\quad+\int_{0}^{t}\frac{\partial F }{\partial x} (g_{s-},s,X_{s-})\rm{d} X_s+\frac{1}{2} \sigma^2 \int_{0}^{t} \frac{\partial^2 F}{\partial x^2} (g_{s-},s,X_{s-}) \rm{d} s\\& \quad +\int_{[0,t] } \int_{(\!-\!\infty,0)} \left[ F(s,s,X_{s-}+y)\vphantom{\frac{\partial F }{\partial x}}\right. \nonumber\\&\quad\left.-F(s,s,X_{s-})-y\frac{\partial F }{\partial x} (s,s,X_{s-}) \right]\mathbb{I}_{\{g_{s-}=s \}}N(\rm{d} s\times \rm{d} y)\\&\quad+\int_{[0,t]}\int_{(\!-\!\infty,0)} \left[F(g_{s-},s,X_{s-}+y)-F(g_{s-},s,X_{s-})-y\frac{\partial F }{\partial x} (g_{s-},s,X_{s-})\right]\\&\quad\qquad\qquad \times \mathbb{I}_{\{X_{s-}+y >0 \}}\mathbb{I}_{\{ g_{s-} < s\}} N(\rm{d} s\times \rm{d} y) \\&\quad+ \int_{[0,t]}\int_{(\!-\!\infty,0)} \left[F(s,s,X_{s-}+y)-F(g_{s-},s,X_{s-})-y\frac{\partial F }{\partial x} (g_{s-},s,X_{s-}) \right]\\&\quad\qquad\qquad \times \mathbb{I}_{\{X_{s-}+y \leq 0 \}} \mathbb{I}_{\{ g_{s-} < s\}}N(\rm{d} s\times \rm{d} y)\end{align*}

with probability 1. For the case when $\Pi$ is any Lévy measure we know from Corollary 4.3.10 in [Reference Applebaum2] that there exists a sequence $\{ A_n,n\geq 1\}$ with $\Pi(A_n)<\infty$ , for each $n\geq 1$ , and $A_n \uparrow (-1,0)$ when $n\rightarrow \infty$ , such that $\lim_{n\rightarrow } X_{t}^{(n)}=X_t$ uniformly on compacts in probability, where

\begin{align*}X_t^{(n)}=\sigma B_t-\mu t+\int_{[0,t]}\int_{(\!-\!\infty,-1)} x N(\rm{d} s\times \rm{d} x)+\int_{[0,t]}\int_{(-1,0)\cap A_n} x( N(\rm{d} s\times \rm{d} x)-\rm{d} s\Pi(\rm{d} x)).\end{align*}

Hence, the result above is valid for the process $X^{(n)}=\{X_t^{(n)},t\geq 0\} $ and, by taking $n\rightarrow \infty$ along a subsequence for which $X_t^{(n)}$ converges to $X_t$ almost surely, the result follows for the process X. The general case, when F and its derivatives are not necessarily bounded, follows by a standard stopping-time argument. Moreover, from the fact that $g_t$ is continuous on the set $\{ t\geq 0\colon g_{t-}=t \text{ or } (g_{t-}<t \text{ and } X_t>0)\}$ , we obtain the first equality in (3.5). Lastly, the case when $X_t\leq 0$ , is similar and the proof is omitted.

5.2.2. Proof of Theorem 3.2

We recall that for any $\varepsilon>0$ and $r\geq 0$ ,

\begin{align*}U_{\varepsilon,r}=r-g_{\varepsilon,r}=r-\sup\{0\leq s\leq r\,:\, X^{(\varepsilon)}_s\leq 0 \}.\end{align*}

First, note that, since $|K(U_s,X_s)|\leq C(U_s,X_s)$ for all $s\geq 0$ and $\mathbb{E}_{u,x}\left(\int_0^{\infty} \mathrm{e}^{-qr} C(U_r,X_r+y) \rm{d} r\right)<\infty$ for all $(u,x)\in E$ and $y\in \mathbb{R}$ , we have that $K^+$ and $K^-$ are finite. Moreover, since $u\mapsto C(u,x)$ is monotone for all $x\in \mathbb{R}$ and non-negative, we have that for all $r\geq 0$ and $\varepsilon>0$ ,

\begin{align*} |K(U_{\varepsilon,r},X_r^{(\varepsilon)})|\leq C(U_{\varepsilon,r},X_r^{(\varepsilon)})\leq C(U_r,X_r)+C(U_r,X_r-\varepsilon)+C(U_r^{(\varepsilon)},X_r)+C(U_r^{(\varepsilon)},X_r-\varepsilon),\end{align*}

where $U_r^{(\varepsilon)}=r-g_t^{(\varepsilon)}=r-\sup\{0\leq s\leq r\,:\, X_s \leq \varepsilon \}$ and we used that $U_t\geq U_{\varepsilon,t} \geq U_t^{(\varepsilon)}$ , for all $t\geq 0$ . It follows from integrability of $\mathrm{e}^{-qr}C(U_r,X_r+y)$ with respect to the product measure $\mathbb{P}_{u,x} \times \rm{d} r$ , for all $(u,x)\in E$ , by dominated convergence theorem and left-continuity in each argument of K that for $x\leq 0$ ,

\begin{align*}\mathbb{E}_{x}\left(\int_0^{\infty}\mathrm{e}^{-qr} K(U_{r},X_r) \rm{d} r \right)=\lim_{\varepsilon \downarrow 0} \mathbb{E}_{x}\left(\int_0^{\infty}\mathrm{e}^{-qr} K(U_{\varepsilon,r},X_r^{(\varepsilon)}) \rm{d} r \right),\end{align*}

where we note that for $x\leq \varepsilon$ , $X^{(\varepsilon)}_0=X_0-\varepsilon=x-\varepsilon$ and $U_{\varepsilon,0}=0$ , under $\mathbb{P}_x$ . Then we calculate the right-hand side of the equation above. We define the auxiliary function

\begin{align*}J^{(\varepsilon)}(x)\,:\!=\,\mathbb{E}_x\left(\int_0^{\infty}\mathrm{e}^{-qr} K(U_{\varepsilon,r},X_r^{(\varepsilon)}) \rm{d} r \right), \quad x\leq \varepsilon.\end{align*}

Fix $\varepsilon>0$ and take any $x\leq \varepsilon$ . Then we can write

(5.5) \begin{align}J^{(\varepsilon)}(x)&=\mathbb{E}_x\left(\int_0^{\rho_{1,\varepsilon}^+} \mathrm{e}^{-qr} K(0,X_r-\varepsilon)\rm{d} r \right)+\mathbb{E}_x\left(\mathbb{I}_{\{\rho_{1,\varepsilon}^+<\infty \}}\int_{\rho_{1,\varepsilon}^+}^{\infty}\mathrm{e}^{-qr} K(U_{\varepsilon,r},X_r^{(\varepsilon)}) \rm{d} r \right)\nonumber\\&=K^-(x-\varepsilon)+\int_{0}^{\infty} \mathbb{E}_x\left(\mathbb{I}_{\{\rho_{1,\varepsilon}^+<\infty \}}\mathrm{e}^{-q(r+\rho_{1,\varepsilon}^+)} K(U_{\varepsilon,r+\rho_{1,\varepsilon}^+},X_{r+\rho_{1,\varepsilon}^+}^{(\varepsilon)})\right) \rm{d} r ,\end{align}

where we recall that $K^-$ is given in (3.8), and the last equality follows by Fubini’s theorem. By using that $g_{\varepsilon,t}=t$ , when $t\in [\rho_{k,\varepsilon}^-,\rho_{k,\varepsilon}^+]$ , and $g_{\varepsilon,t}=\rho_{k,\varepsilon}^+$ , when $t\in [\rho_{k,\varepsilon}^+,\rho_{k+1,\varepsilon}^-)$ , for some $k\geq 1$ , we obtain that for any $r> 0$ ,

\begin{align*}\mathbb{E}_x&\left(\mathbb{I}_{\{\rho_{1,\varepsilon}^+<\infty \}}\mathrm{e}^{-q(r+\rho_{1,\varepsilon}^+)} K(U_{\varepsilon,r+\rho_{1,\varepsilon}^+},X_{r+\rho_{1,\varepsilon}^+}^{(\varepsilon)})\right)\\&=\sum_{k=1}^{\infty}\mathbb{E}_x\left(\mathbb{I}_{\{\rho_{1,\varepsilon}^+<\infty \}}\mathrm{e}^{-q(r+\rho_{1,\varepsilon}^+)} K(r+\rho_{1,\varepsilon}^+-\rho_{k,\varepsilon}^+,X_{r+\rho_{1,\varepsilon}^+}) \mathbb{I}_{\{\rho_{k,\varepsilon}^+ \leq r+\rho_{1,\varepsilon}^+ < \rho_{k+1,\varepsilon}^-\}}\right)\\&\qquad+\sum_{k=1}^{\infty}\mathbb{E}_x\left(\mathbb{I}_{\{\rho_{1,\varepsilon}^+<\infty \}} \mathrm{e}^{-q(r+\rho_{1,\varepsilon}^+)} K(0,X_{r+\rho_{1,\varepsilon}^+}-\varepsilon) \mathbb{I}_{\{\rho_{k+1,\varepsilon}^- \leq r+\rho_{1,\varepsilon}^+ < \rho_{k+1,\varepsilon}^+\}}\right)\\&=\mathbb{E}_x\left(\mathbb{I}_{\{\rho_{1,\varepsilon}^+<\infty \}}\mathrm{e}^{-q \rho_{1,\varepsilon}^+} \right) \sum_{k=1}^{\infty} \mathbb{E}_{\varepsilon}\left( \mathrm{e}^{-qr} K(r-\rho_{k,\varepsilon}^+,X_{r}) \mathbb{I}_{\{\rho_{k,\varepsilon}^+ \leq r < \rho_{k+1,\varepsilon}^-\}}\right)\\&\qquad+\mathbb{E}_x\left(\mathbb{I}_{\{\rho_{1,\varepsilon}^+<\infty \}} \mathrm{e}^{-q\rho_{1,\varepsilon}^+} \right) \sum_{k=1}^{\infty} \mathbb{E}_{\varepsilon}\left( \mathrm{e}^{-qr} K(0,X_{r}-\varepsilon) \mathbb{I}_{\{\rho_{k+1,\varepsilon}^- \leq r < \rho_{k+1,\varepsilon}^+\}}\right)\\&=\mathrm{e}^{-\Phi(q) (\varepsilon-x)} \mathbb{E}_{\varepsilon}\left( \mathrm{e}^{-qr} K(U_{\varepsilon,r},X_{r}^{(\varepsilon)}) \right),\end{align*}

where the second equality follows by applying the strong Markov property at time $\rho_{1,\varepsilon}^+$ and the last from (2.2) and the definition of $U_{\varepsilon,r}$ and $X^{(\varepsilon)}_r$ . Hence, substituting the expression above into (5.5), we deduce that for any $x\leq \varepsilon$ ,

(5.6) \begin{align}J^{(\varepsilon)}(x)&=K^-(x-\varepsilon)+\mathrm{e}^{-\Phi(q) (\varepsilon-x)} \int_{0}^{\infty} \mathbb{E}_{\varepsilon}\left( \mathrm{e}^{-qr} K(U_{\varepsilon,r},X_{r}^{(\varepsilon)}) \right) \rm{d} r\nonumber\\&=K^-(x-\varepsilon)+\mathrm{e}^{-\Phi(q) (\varepsilon-x)} J^{(\varepsilon)}(\varepsilon).\end{align}

On the other hand, using a similar argument and that $\rho_{2,\varepsilon}^-=\tau_0^-$ under $\mathbb{P}_{\varepsilon}$ , we obtain that

\begin{align*}J^{(\varepsilon)}(\varepsilon)&=K^+(0,\varepsilon)+\mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}}\mathrm{e}^{-q\tau_0^-} J^{(\varepsilon)}( X_{\tau_0^-}) \right)\\&=K^+(0,\varepsilon)+\mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}}\mathrm{e}^{-q\tau_0^-} K^-(X_{\tau_0^-}-\varepsilon)\right)+\mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}}\mathrm{e}^{-q\tau_0^-}\mathrm{e}^{-\Phi(q) (\varepsilon-X_{\tau_0^-})} \right)J^{(\varepsilon)}(\varepsilon)\\&=K^+(0,\varepsilon)+\mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}}\mathrm{e}^{-q\tau_0^-} K^-(X_{\tau_0^-}-\varepsilon)\right)+\mathcal{I}^{(q,\Phi(q))}(\varepsilon)J^{(\varepsilon)}(\varepsilon),\end{align*}

where the second equality follows by (5.6) and from $X_{\tau_0^-}\leq 0$ , on the event $\{\tau_0^-<\infty \}$ , and the last equality follows from (2.5). Hence, solving for $J^{(\varepsilon)}(\varepsilon)$ in the expression above, we obtain that

\begin{align*}J^{(\varepsilon)}(\varepsilon)&=\frac{1}{1-\mathcal{I}^{(q,\Phi(q))}(\varepsilon) }\left[ K^+(0,\varepsilon)+\mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}}\mathrm{e}^{-q\tau_0^-} K^-(X_{\tau_0^-}-\varepsilon)\right)\right].\end{align*}

If we further substitute the expression found for $J^{(\varepsilon)}(\varepsilon)$ into (5.6), we deduce that for any $x\leq \varepsilon$ ,

\begin{align*}J^{(\varepsilon)}(x)= K^-(x-\varepsilon)+ \frac{\mathrm{e}^{-\Phi(q) (\varepsilon-x)}}{1-\mathcal{I}^{(q,\Phi(q))}(\varepsilon) }\left[ K^+(0,\varepsilon)+\mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}}\mathrm{e}^{-q\tau_0^-} K^-(X_{\tau_0^-}-\varepsilon)\right)\right].\end{align*}

Therefore, by the dominated convergence theorem we have that for all $x\leq 0$ ,

\begin{align*}\mathbb{E}_x&\left(\int_0^{\infty}\mathrm{e}^{-qr} K(U_r,X_r) \rm{d} r \right)\\&= \lim_{\varepsilon \downarrow 0 }\left\{ K^-(x-\varepsilon)+\frac{\mathrm{e}^{-\Phi(q) (\varepsilon-x)}}{ 1-\mathcal{I}^{(q,\Phi(q))}(\varepsilon)} \left[ \mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}} \mathrm{e}^{-q \tau_0^-} K^-(X_{\tau_0^-}-\varepsilon) \right)+K^+(0,\varepsilon) \right] \right\} .\end{align*}

Using Fubini’s theorem and equation (2.10) we have that for all $x<0$ ,

(5.7) \begin{align}K^-(x)=\int_{(\!-\!\infty,0)}K(0,y) \int_0^{\infty} \mathrm{e}^{-qr} \mathbb{P}_x(X_r\in \rm{d} y, r<\tau_0^+) \rm{d} r\nonumber\\=\int_{-\infty}^0 K(0,y) [\mathrm{e}^{\Phi(q)x}W^{(q)}(-y) -W^{(q)}(x-y)] \rm{d} y.\end{align}

Recall that we are assuming that K is bounded by a non-negative integrable function C. Then we have that for any $y<0$ and $x<0$ ,

\begin{align*}|K(0,y) [\mathrm{e}^{\Phi(q)x}W^{(q)}(-y) -W^{(q)}(x-y)]|\leq C(0,y) [\mathrm{e}^{\Phi(q)x}W^{(q)}(-y) -W^{(q)}(x-y)].\end{align*}

Moreover, we have that the quantity on the right-hand side of the equation above is integrable since for any $x<0$ ,

\begin{align*}C^-(x)\,:\!=\, \mathbb{E}_x\left(\int_0^{\tau_0^+} \mathrm{e}^{-qr}C(0,X_r)\rm{d} r \right)<\mathbb{E}_x\left(\int_0^{\infty} \mathrm{e}^{-qr}C(U_r,X_r)\rm{d} r \right)<\infty\end{align*}

and

\begin{align*}C^-(x)=\int_{-\infty}^0 C(0,y) [\mathrm{e}^{\Phi(q)x}W^{(q)}(-y) -W^{(q)}(x-y)] \rm{d} y,\end{align*}

where we used Fubini’s theorem and equation (2.10). Furthermore, it can be seen (see, for example, the proof of Theorem 8.1 in [Reference Kyprianou29]) that for any $q>0$ and $y<0$ , the mapping $x\mapsto \mathrm{e}^{-\Phi(q)x}W^{(q)}(x-y)$ is non-decreasing. Thus, for any $x<0$ and any $x_1<x_2<0$ such that $x\in [x_1,x_2]$ ,

\begin{align*}C^-(x)&= \mathrm{e}^{\Phi(q) x}\int_{-\infty}^0 C(0,y) [W^{(q)}(-y) -\mathrm{e}^{-\Phi(q)x}W^{(q)}(x-y)] \rm{d} y\\&\leq \mathrm{e}^{\Phi(q) x}\int_{-\infty}^0 C(0,y) [W^{(q)}(-y) -\mathrm{e}^{-\Phi(q)x_1}W^{(q)}(x_1-y)] \rm{d} y\\&\leq \mathrm{e}^{\Phi(q) (x_2-x_1)}C^{-}(x_1).\end{align*}

Hence, since W is continuous on $(0,\infty)$ and by the dominated convergence theorem, we deduce that $K^-$ is a continuous function. Thus, we have that for any $x\leq 0$ ,

(5.8) \begin{align}\mathbb{E}_x&\left(\int_0^{\infty}\mathrm{e}^{-qr} K(U_r,X_r) \rm{d} r \right)\nonumber\\&= K^-(x)+\lim_{\varepsilon \downarrow 0 } \frac{\mathrm{e}^{-\Phi(q) (\varepsilon-x)}}{ 1-\mathcal{I}^{(q,\Phi(q))}(\varepsilon)} \left[ \mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}} \mathrm{e}^{-q \tau_0^-} K^-(X_{\tau_0^-}-\varepsilon) \right)+K^+(0,\varepsilon) \right] .\end{align}

We then proceed to find the limit on the right-hand side of the equation above. For any $x>0$ and $\varepsilon>0$ , we deduce from Fubini’s theorem and equations (2.5) and (5.7) that

\begin{align*}&\mathbb{E}_{x}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}} \mathrm{e}^{-q \tau_0^-} K^-(X_{\tau_0^-}-\varepsilon) \right)\\&= \mathrm{e}^{\Phi(q)(x-\varepsilon)}\mathcal{I}^{(q,\Phi(q))}(x) \int_{-\varepsilon}^0 K(0,y) W^{(q)}(-y)\rm{d} y \\&\quad +\int_{-\infty}^{-\varepsilon}K(0,y)\left[\mathrm{e}^{\Phi(q)(x-\varepsilon)}\mathcal{I}^{(q,\Phi(q))}(x) W^{(q)}(-y)\right.\nonumber\\ &\quad\left.- \mathbb{E}_x\left(\mathbb{I}_{\{\tau_0^-<\infty \}} \mathrm{e}^{-q\tau_0^-}W^{(q)}(X_{\tau_0^-}-\varepsilon-y)\right) \right] \rm{d} y.\end{align*}

Let $x,\varepsilon>0$ and $y<-\varepsilon$ . From the monotone convergence theorem and (2.4) we have that

(5.9) \begin{align}\mathbb{E}_x&\left(\mathbb{I}_{\{\tau_0^-<\infty \}} \mathrm{e}^{-q\tau_0^-}W^{(q)}(X_{\tau_0^-}-\varepsilon-y)\right)\nonumber\\&=\lim_{a\rightarrow \infty} \mathbb{E}_x\left(\mathbb{I}_{\{\tau_0^-<\tau_a^+ \}} \mathrm{e}^{-q\tau_0^-}W^{(q)}(X_{\tau_0^-}-\varepsilon-y)\right)\nonumber\\&=\lim_{a\rightarrow \infty} \mathbb{E}_x\left( \mathrm{e}^{-q\tau_0^-\wedge \tau_a^+ } W^{(q)}(X_{\tau_0^-\wedge \tau_a^+}-\varepsilon-y)\right)\!-\!\lim_{a\rightarrow \infty} \mathbb{E}_x\!\left( \mathbb{I}_{\{\tau_a^+ < \tau_0^- \}} \mathrm{e}^{-q \tau_a^+ } W^{(q)}(a-\varepsilon-y)\right)\nonumber\\&=\lim_{a\rightarrow \infty} \mathbb{E}_{x-\varepsilon -y}\left( \mathrm{e}^{-q\tau_{-\varepsilon -y}^-\wedge \tau_{a-\varepsilon-y}^+ } W^{(q)}(X_{\tau_{-\varepsilon-y}^-\wedge \tau_{a-\varepsilon-y}^+})\right)-\lim_{a\rightarrow \infty} W^{(q)}(a-\varepsilon-y)\frac{W^{(q)}(x)}{W^{(q)}(a)}\nonumber\\&=W^{(q)}(x-\varepsilon-y)-\mathrm{e}^{-\Phi(q)(\varepsilon+y)} W^{(q)}(x),\end{align}

where the last equality follows since, for any $a\geq 0$ , the process $\mathrm{e}^{-q(t\wedge \tau_0^-\wedge \tau_a^+)}W^{(q)}(X_{t\wedge \tau_0^-\wedge \tau_a^+})$ is a martingale, from the optional sampling theorem (note that $\tau_{-\varepsilon-y}^-\leq \tau_0^-$ ) and since $\lim_{a \rightarrow \infty} W^{(q)}(a-z)/W^{(q)}(a)=\mathrm{e}^{-\Phi(q)z}$ for $z\leq a$ and $a\geq 0$ (see Exercises 8.5 and 8.12 in [Reference Kyprianou29]).

Hence, we obtain that for any $x>0$ and $\varepsilon>0$ ,

\begin{align*}&\mathbb{E}_{x}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}} \mathrm{e}^{-q \tau_0^-} K^-(X_{\tau_0^-}-\varepsilon) \right)\\&= \mathrm{e}^{\Phi(q)(x-\varepsilon)}\mathcal{I}^{(q,\Phi(q))}(x) \int_{-\varepsilon}^0 K(0,y) W^{(q)}(-y)\rm{d} y \\&\quad +\int_{-\infty}^{-\varepsilon}K(0,y)\left[\mathrm{e}^{\Phi(q)(x-\varepsilon)}\mathcal{I}^{(q,\Phi(q))}(x) W^{(q)}(-y) -W^{(q)}(x-\varepsilon-y)\right.\nonumber\\&\quad\left.+ \,\mathrm{e}^{-\Phi(q)(\varepsilon+y)} W^{(q)}(x) \right] \rm{d} y.\end{align*}

In particular, when $x=\varepsilon$ we obtain that

\begin{align*}&\mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}} \mathrm{e}^{-q \tau_0^-} K^-(X_{\tau_0^-}-\varepsilon) \right)\\&= \mathcal{I}^{(q,\Phi(q))}(\varepsilon) \int_{-\varepsilon}^0 K(0,y) W^{(q)}(-y)\rm{d} y \\&\qquad +\int_{-\infty}^{-\varepsilon}K(0,y)\left( [\mathcal{I}^{(q,\Phi(q))}(\varepsilon)-1] W^{(q)}(-y) +\mathrm{e}^{-\Phi(q)(\varepsilon+y)} W^{(q)}(\varepsilon) \right) \rm{d} y.\end{align*}

Thus, for any $\varepsilon>0$ and $x\leq 0$ ,

\begin{align*} & \frac{\mathrm{e}^{-\Phi(q) (\varepsilon-x)}}{ 1-\mathcal{I}^{(q,\Phi(q))}(\varepsilon)} \mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}} \mathrm{e}^{-q \tau_0^-} K^-(X_{\tau_0^-}-\varepsilon) \right)\\&\quad=\frac{\mathrm{e}^{-\Phi(q) (\varepsilon-x)}}{ 1-\mathcal{I}^{(q,\Phi(q))}(\varepsilon)} \mathcal{I}^{(q,\Phi(q))}(\varepsilon) \int_{-\varepsilon}^0 K(0,y) W^{(q)}(-y)\rm{d} y \\&\qquad +\frac{\mathrm{e}^{-\Phi(q) (\varepsilon-x)}}{ 1-\mathcal{I}^{(q,\Phi(q))}(\varepsilon)} \int_{-\infty}^{-\varepsilon}K(0,y)\left( [\mathcal{I}^{(q,\Phi(q))}(\varepsilon)-1] W^{(q)}(-y) +\mathrm{e}^{-\Phi(q)(\varepsilon+y)} W^{(q)}(\varepsilon) \right) \rm{d} y\\&\quad= \Phi'(q)\frac{\mathrm{e}^{\Phi(q) x}}{ W^{(q)}(\varepsilon)} \int_{-\varepsilon}^0 K(0,y) W^{(q)}(-y)\rm{d} y-\mathrm{e}^{-\Phi(q) (\varepsilon-x)} \int_{-\varepsilon}^0 K(0,y) W^{(q)}(-y)\rm{d} y \\&\qquad- \mathrm{e}^{-\Phi(q) (\varepsilon-x)}\int_{-\infty}^{-\varepsilon}K(0,y)\left( W^{(q)}(-y) -\Phi'(q)\mathrm{e}^{-\Phi(q)y} \right) \rm{d} y,\end{align*}

where we used that $ 1-\mathcal{I}^{(q,\Phi(q))}(\varepsilon)= [\Phi'(q)]^{-1} \mathrm{e}^{-\Phi(q) \varepsilon}W^{(q)}(\varepsilon) $ . Moreover, by using that $W^{(q)}$ is increasing on $(0,\infty)$ we see that

\begin{align*}0\leq \left|\lim_{\varepsilon \downarrow 0} \Phi'(q)\frac{\mathrm{e}^{\Phi(q) x}}{ W^{(q)}(\varepsilon)} \! \int_{-\varepsilon}^0\! K(0,y) W^{(q)}(-y)\rm{d} y \right|\leq \lim_{\varepsilon \downarrow 0}\Phi'(q)\frac{\mathrm{e}^{\Phi(q) x}}{ W^{(q)}(\varepsilon)} W^{(q)}(\varepsilon)\! \int_{-\varepsilon}^0 \!|K(0,y)| \rm{d} y=0.\end{align*}

Thus, we conclude that for any $x\leq 0$ ,

\begin{align*}\lim_{\varepsilon \downarrow 0} & \frac{\mathrm{e}^{-\Phi(q) (\varepsilon-x)}}{ 1-\mathcal{I}^{(q,\Phi(q))}(\varepsilon)} \mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}} \mathrm{e}^{-q \tau_0^-} K^-(X_{\tau_0^-}-\varepsilon) \right)\\&=- \mathrm{e}^{\Phi(q) x}\int_{-\infty}^{0}K(0,y)\left( W^{(q)}(-y) -\Phi'(q)\mathrm{e}^{-\Phi(q)y} \right) \rm{d} y.\end{align*}

Substituting the expression above into (5.8) and using the expression obtained for $K^-(x)$ in (5.7), we deduce that for any $x\leq 0$ ,

(5.10) \begin{align}\mathbb{E}_x&\left(\int_0^{\infty}\mathrm{e}^{-qr} K(U_r,X_r) \rm{d} r \right)\nonumber\\&= K^-(x)+\lim_{\varepsilon \downarrow 0 } \frac{\mathrm{e}^{-\Phi(q) (\varepsilon-x)}}{ 1-\mathcal{I}^{(q,\Phi(q))}(\varepsilon)} \left[ \mathbb{E}_{\varepsilon}\left( \mathbb{I}_{\{ \tau_0^-<\infty\}} \mathrm{e}^{-q \tau_0^-} K^-(X_{\tau_0^-}-\varepsilon) \right)+K^+(0,\varepsilon) \right]\nonumber \\&=\int_{-\infty}^0 K(0,y) [\Phi'(q)\mathrm{e}^{-\Phi(q)(y-x)} -W^{(q)}(x-y)] \rm{d} y+ \Phi'(q)\mathrm{e}^{\Phi(q) x}\lim_{\varepsilon \downarrow 0 } \frac{K^+(0,\varepsilon)}{ W^{(q)}(\varepsilon)}.\end{align}

For the case when $u>0$ and $x>0$ , using the expression above and the strong Markov property, we obtain that

\begin{align*}\mathbb{E}_{u,x}&\left(\int_0^{\infty}\mathrm{e}^{-qr} K(U_r,X_r) \rm{d} r \right)\\&=\mathbb{E}_x\left(\int_0^{\tau_0^-}\mathrm{e}^{-qr} K(u+r,X_r) \rm{d} r \right)+\mathbb{E}_{x}\left(\mathbb{I}_{\{\tau_0^-<\infty \}}\int_{\tau_0^-}^{\infty}\mathrm{e}^{-qr} K(U_r,X_r) \rm{d} r \right)\\&=\mathbb{E}_x\left(\int_0^{\tau_0^-}\mathrm{e}^{-qr} K(u+r,X_r) \rm{d} r \right)+\mathbb{E}_{x}\left(\mathbb{I}_{\{\tau_0^-<\infty \}} \mathrm{e}^{q \tau_0^-} \mathbb{E}_{X_{\tau_0^-}}\left[\int_{0}^{\infty}\mathrm{e}^{-qr} K(U_r,X_r) \rm{d} r \right] \right)\\&=K^+(u,x)\\&\quad+\int_{-\infty}^0 K(0,y) \left[\Phi'(q)\mathrm{e}^{-\Phi(q)y} \mathbb{E}_{x}\left(\mathbb{I}_{\{\tau_0^-<\infty \}}\mathrm{e}^{-q\tau_0^-} \mathrm{e}^{\Phi(q) X_{\tau_0^-}} \right)\right. \nonumber\\&\quad\left.- \mathbb{E}_{x}\left(\mathbb{I}_{\{\tau_0^-<\infty \}}\mathrm{e}^{-q\tau_0^-} W^{(q)}(X_{\tau_0^-}-y) \right)\right] \rm{d} y\\&\quad+ \Phi'(q)\mathbb{E}_{x}\left(\mathbb{I}_{\{\tau_0^-<\infty \}}\mathrm{e}^{-q\tau_0^-} \mathrm{e}^{\Phi(q) X_{\tau_0^-}} \right)\lim_{\varepsilon \downarrow 0 } \frac{K^+(0,\varepsilon)}{ W^{(q)}(\varepsilon)}\\&=K^+(u,x)\\&\quad+\int_{-\infty}^0 K(0,y) \left[\Phi'(q)\mathrm{e}^{-\Phi(q)y} \mathrm{e}^{\Phi(q)x}\mathcal{I}^{(q,\Phi(q))}(x) -W^{(q)}(x-y)+\mathrm{e}^{-\Phi(q)y} W^{(q)}(x)\right] \rm{d} y\\&\quad+ \Phi'(q)\mathrm{e}^{\Phi(q)x}\mathcal{I}^{(q,\Phi(q))}(x)\lim_{\varepsilon \downarrow 0 } \frac{K^+(0,\varepsilon)}{ W^{(q)}(\varepsilon)}\\&=K^+(u,x)+\int_{-\infty}^0 K(0,y) \left[\Phi'(q)\mathrm{e}^{-\Phi(q)(y-x)} -W^{(q)}(x-y)\right] \rm{d} y\\&\quad+ [\Phi'(q)\mathrm{e}^{\Phi(q)x}-W^{(q)}(x)]\lim_{\varepsilon \downarrow 0 } \frac{K^+(0,\varepsilon)}{ W^{(q)}(\varepsilon)},\end{align*}

where the penultimate equality follows from (2.5) and (5.9) and the last by substituting the value of $\mathcal{I}^{(q,\Phi(q)}(x)$ (see (2.6)). Lastly, since $W^{(q)}(x)=0$ for $x<0$ , $W^{(q)}(0)=0$ when X has paths of infinite variation and $W^{(q)}(0)>0$ when X has paths of finite variation, we note that the expression above coincides with (5.10) when $x\leq 0$ . In other words, the expression above is valid for any $(u,x)\in E$ . The proof is now complete.

5.3. Proof of Theorem 4.1

We first state a verification lemma that provides sufficient conditions for the optimality of a given candidate solution $\tau^*$ .

Proof. From the definition of V, we deduce that $V\geq V^*$ . On the other hand, due to the optimal sampling theorem we have that, for any $t\geq 0$ , $N>0$ and any stopping time $\tau \in \mathcal{T}$ , the stopped process $Z_{t\wedge \tau \wedge \tau_N^+ }$ is a supermartingale. This implies that for any $t\geq 0$ , $N>0$ and $\tau \in \mathcal{T}$ ,

\begin{align*}V^*(u,x) &\geq \mathbb{E}_{u,x}\left(\mathrm{e}^{-r(T\wedge \tau)} V^*(U_{T\wedge \tau },X_{T\wedge \tau})+\int_0^{T\wedge \tau } \mathrm{e}^{-rs}G(U_s,X_s)\rm{d} s\right)\\ &\geq \mathbb{E}_{u,x}\left( \int_0^{T\wedge \tau}\mathrm{e}^{-rs} G(U_s,X_s)\rm{d} s\right),\end{align*}

where $T=t\wedge \tau_N^+$ and the last inequality follows since $V^*\geq 0$ , by assumption. From the dominated convergence theorem we conclude (see (4.1)), by taking $t,N\rightarrow \infty$ in the equation above, that

\begin{align*}V^*(u,x) \geq \mathbb{E}_{u,x}\left( \int_0^{\tau}\mathrm{e}^{-rs} G(U_s,X_s)\rm{d} s\right)\end{align*}

for all $(u,x)\in E$ and $\tau \in \mathcal{T}$ . Hence, we have that $V\leq V^*$ , implying that $V=V^*$ . Therefore, the supremum in (4.2) is attained by $\tau^*$ as claimed.

For $z\leq 0$ fixed, we define the function

\begin{align*}V_z(u,x)\,:\!=\,\mathbb{E}_{x,u}\left(\int_0^{\tau_z^-}\mathrm{e}^{-rs}G(U_s,X_s)\rm{d} s \right),\end{align*}

for $(u,x)\in E$ . The following lemma gives a semi-explicit expression for $V_z$ in terms of the scale functions.

Lemma 5.5. For any $z< 0$ and $(u,x)\in E$ such that $x\geq z$ we have that

(5.11) \begin{align}V_z(u,x)&=K^+(u,x) + \left[\mathrm{e}^{\Phi(r)z}W^{(r)}(x-z)-W^{(r)}(x) \right]\int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v\nonumber\\&\qquad+\mathrm{e}^{\Phi(r) z}W^{(r)}(x-z)\int_{z}^{0} G(0,y) \mathrm{e}^{-\Phi(r) y}\rm{d} y -\int_{z}^{0} G(0,y)W^{(r)}(x-y)\rm{d} y.\end{align}

Proof. Note that for any $(u,x)\in E$ ,

\begin{align*}V_z(u,x)=\mathbb{E}_{x,u}\left(\int_0^{\tau_z^-}\mathrm{e}^{-rs}G(U_s,X_s)\mathbb{I}_{\{X_s>0 \}}\rm{d} s \right)+\mathbb{E}_{x}\left(\int_0^{\tau_z^-}\mathrm{e}^{-rs}G(0,X_s)\mathbb{I}_{\{X_s\leq 0 \}}\rm{d} s \right),\end{align*}

where the two terms on the right-hand side are finite due to equation (4.1). Using equation (2.11) and Fubini’s theorem, we deduce that for any $x\geq z$ ,

\begin{align*}\mathbb{E}_{x}\left(\int_0^{\tau_z^-}\mathrm{e}^{-rs}G(0,X_s)\mathbb{I}_{\{X_s\leq 0 \}}\rm{d} s \right)&=\mathbb{E}_{x-z}\left(\int_0^{\tau_0^-}\mathrm{e}^{-rs}G(0,X_s+z)\mathbb{I}_{\{X_s+z\leq 0 \}}\rm{d} s \right)\\&=\int_{(0,-z]} \!\! G(0,y+z) \int_0^{\infty} \mathrm{e}^{-rs}\mathbb{P}_{x-z}\left(X_s\in \rm{d} y, s<\tau_0^- \right)\rm{d} s\\&=\int_{0}^{-z} \!\! G(0,y+z) \left[ \mathrm{e}^{-\Phi(r) y}W^{(r)}(x-z) \!-\!W^{(r)}(x-z-y) \right]\rm{d} y\\&=\mathrm{e}^{\Phi(r) z}W^{(r)}(x-z)\int_{z}^{0} G(0,y) \mathrm{e}^{-\Phi(r) y}\rm{d} y \nonumber\\&\quad-\int_{z}^{0} G(0,y)W^{(r)}(x-y)\rm{d} y.\end{align*}

On the other hand, from the strong Markov property, we have that for $(u,x)\in E$ such that $x\geq z$ ,

\begin{align*}\mathbb{E}_{x,u}\left(\int_0^{\tau_z^-}\mathrm{e}^{-rs}G(U_s,X_s)\mathbb{I}_{\{X_s>0 \}}\rm{d} s \right)&=H(u,x)-\mathbb{E}_{x}(\mathrm{e}^{-r \tau_z^-}\mathbb{I}_{\{\tau_z^- <\infty \}} H(0,X_{\tau_z^-})),\end{align*}

where

(5.12) \begin{align}H(u,x)&\,:\!=\,\mathbb{E}_{u,x}\left(\int_0^{\infty} \mathrm{e}^{-rs}G(U_s,X_s)\mathbb{I}_{\{X_s>0 \}}\rm{d} s \right).\end{align}

It follows from (4.1) that $|H(u,x)|<\infty$ for all $(u,x)\in E$ . Hence, by using the potential measure of (U, X) given in Corollary 3.3, we see that for any $(u,x)\in E$ ,

(5.13) \begin{align}H(u,x)&=\int_{(0,\infty)}\int_0^{\infty} G(v, y) \int_0^{\infty}\mathrm{e}^{-rs}\mathbb{P}_{u,x}(X_s\in \rm{d} y, U_s\in \rm{d} v) \rm{d} s \nonumber\\&=\int_{(0,\infty)}\int_u^{\infty} G(v, y)\mathrm{e}^{-r(v-u)}\mathbb{P}_x(X_{v-u}\in \rm{d} y,v-u<\tau_0^-)\rm{d} v\nonumber\\&\qquad + \left[\mathrm{e}^{\Phi(r) x}\Phi'(r)-W^{(r)}(x) \right]\int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v \nonumber\\&=K^+(u,x) + \left[\mathrm{e}^{\Phi(r) x}\Phi'(r)-W^{(r)}(x) \right]\int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v . \end{align}

Thus, by using (2.5) and since $K^+(0,x)=0=W^{(q)}(x)$ when $x<0$ , we deduce that for any $x\in \mathbb{R}$ ,

(5.14) \begin{align*}&\mathbb{E}_{x}(\mathrm{e}^{-r \tau_z^-}\mathbb{I}_{\{\tau_z^- <\infty \}} H(0,X_{\tau_z^-}))\nonumber\\&\quad=\Phi'(r)\mathbb{E}_x\left(\mathrm{e}^{-r\tau_z^-+\Phi(r) X_{\tau_z^-}} \mathbb{I}_{\{\tau_z^-<\infty \}}\right) \int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v \end{align*}\begin{align}&\quad=\Phi'(r)\mathbb{E}_{x-z}\left(\mathrm{e}^{-r \tau_0^-+ \Phi(r) (X_{\tau_0^-}+z)}\mathbb{I}_{\{\tau_0^-<\infty \}} \right)\nonumber\\&\qquad\times\left[ \int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v \right]\nonumber\\&\quad=\left(\Phi'(r) \mathrm{e}^{\Phi(r)x} -\mathrm{e}^{\Phi(r)z}W^{(r)}(x-z) \right) \int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v.\end{align}

Therefore, we get that

\begin{align*}\mathbb{E}_{x,u}&\left(\int_0^{\tau_z^-}\mathrm{e}^{-rs}G(U_s,X_s)\mathbb{I}_{\{X_s>0 \}}\rm{d} s \right)\\&=H(u,x)-\mathbb{E}_{x}(\mathrm{e}^{-r \tau_z^-}\mathbb{I}_{\{\tau_z^- <\infty \}} H(0,X_{\tau_z^-}))\\&=K^+(u,x) + \left[\mathrm{e}^{\Phi(r)z}W^{(r)}(x-z)-W^{(r)}(x) \right]\int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v\end{align*}

for any $(u,x)\in E$ . The result follows.

For optimal stopping problems it is common to choose candidate solutions to satisfy the principle of smooth fit. Recall that we are assuming that $\sigma>0$ so that, in this case, $W^{(r)}$ is $C^2$ on $(0,\infty)$ with $W^{(r)\prime}(0\!+\!)=2/\sigma^2$ (see, for example, Theorem 3.10 and Lemma 3.2 in [Reference Kyprianou, Kuznetsov and Rivero26]). Then, by differentiating $V_z(u,x)$ with respect to x, we obtain for $z<x<0$ that

\begin{align*}\frac{\partial}{\partial x} V_z(0,x)&=\mathrm{e}^{\Phi(r)z}W^{(r)\prime}(x-z)\left[\int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v\right. \nonumber\\ &\qquad\left.+\int_{z}^0 G(0,y) \mathrm{e}^{-\Phi(r) y}\rm{d} y\right]\\&\qquad -\int_{z}^x G(0,y)W^{(r)\prime}(x-y) \rm{d} y.\end{align*}

Then, by letting $x\downarrow z$ , we see that the equation

\begin{align*}\frac{\partial}{\partial x} V_z(0,z\!+\!)=0\end{align*}

is satisfied if and only if z solves the equation

\begin{align*}\int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v +\int_{z}^0 G(0,y) \mathrm{e}^{-\Phi(r) y}=0.\end{align*}

That is, if $z=z^*$ . In the following lemma, we verify that the characterization of $z^*$ given in the statement of Theorem 4.1 indeed holds and that condition (i) given in Lemma 5.4 holds when $z\geq z^*$ .

Lemma 5.6. For $z\leq 0$ , we define the function

\begin{align*}f(z)=\int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v+\int_{z}^0 G(0,y) \mathrm{e}^{-\Phi(r ) y}\rm{d} y.\end{align*}

Then the equation $f(z)=0$ has a unique solution $z^*$ on $(\!-\!\infty,0)$ such that $z^*\leq y_0$ . Moreover, we have that $V_z(u,x)\geq 0$ for all $z\geq z^*$ and $(u,x)\in E$ .

Proof. From Corollary 3.3 and by assumption (4.1) we know that

\begin{align*}0\leq\, f(0)&=\int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v\\&=\frac{1}{\Phi'(r)}\mathbb{E}\left(\int_0^{\infty} \mathrm{e}^{-rs}G(U_s,X_s)\mathbb{I}_{\{X_s>0 \}}\rm{d} s \right)<\infty.\end{align*}

On the other hand, since G(0,y) is non-positive on $(\!-\!\infty,y_0)$ , we have that f(z) is increasing on $(\!-\!\infty,y_0)$ with $f(z)>0$ for all $ y_0\leq z\leq 0$ and $\lim_{y\rightarrow -\infty}f(z)=-\infty$ , where the latter follows due to the assumption $\lim_{y\rightarrow -\infty }G(0,y)<0$ . Then, due to the continuity of f, we see that the equation $f(z)=0$ has a unique solution $z^*$ on $(\!-\!\infty,y_0)$ .

Next, we proceed to show the statement on $V_z$ . Since G(u, x) is non-negative for all $(u,x)\in E$ such that $x\geq y_0$ , we see that $V_z(x,u)\geq 0$ for all $(u,x)\in E$ and $z\geq y_0$ . Taking $z<0$ and $(u,x)\in E$ such that $x>z$ , we see from (5.11) that

\begin{align*}\frac{\partial }{\partial z} V_z(u,x)=f(z)\frac{\partial }{\partial z} (\mathrm{e}^{\Phi(r)z}W^{(r)}(x-z)).\end{align*}

Note that we can write $\mathrm{e}^{\Phi(r)z}W^{(r)}(x-z)=\mathrm{e}^{\Phi(r)x}W_{\Phi(r)}(x-z)$ , where $W_{\Phi(r)}$ is the r-scale function under the measure $\mathbb{P}^{\Phi(r)}$ (see, for example, the proof of Theorem 8.1 in [Reference Kyprianou29]). Then we see that the mapping $z \mapsto \mathrm{e}^{\Phi(r)z}W^{(r)}(x-z)$ is non-increasing on $\mathbb{R}$ , and then $\frac{\partial }{\partial z} V_z(u,x)\leq 0$ for all $(u,x) \in E$ and $ z^*\leq z < 0$ such that $x>z$ . We conclude that, for $(u,x)\in E$ fixed such that $x>z^*$ , the mapping $z\mapsto V_z(x,u)$ is non-increasing on $[z^*,x\wedge 0)$ . Hence,

\begin{align*} V_z(x,u)\geq \lim_{z\uparrow x\wedge 0}V_z(u,x)\geq 0\end{align*}

for any $z^*\leq z\leq 0$ and $(u,x)\in E$ such that $x>z$ . The proof is now complete.

For ease of notation, we denote $V^*=V_{z^*}$ . Note that for any $(u,x)\in E$ ,

(5.15) \begin{align}V^*(u,x)= K^+(u,x) \!-\!W^{(r)}(x)\!\int_{(0,\infty)}\int_0^{\infty} \!G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v-\int_{z^*}^0 \!G(0,y)W^{(r)}(x-y) \rm{d} y.\end{align}

Next, we show that the supermartingale property holds for $V^*$ .

Lemma 5.7. For any $N>0$ we have that the process $\{ Z_{t\wedge \tau_N^+}^*, t\geq 0\}$ is a supermartingale under $\mathbb{P}_{u,x}$ , for each $(u,x)\in E$ , where

\begin{align*}Z_t^{*}=\mathrm{e}^{-rt}V^*(U_t,X_t)+\int_0^t \mathrm{e}^{-rs} G(U_s,X_s)\rm{d} s.\end{align*}

Proof. Due to the fact that X is of infinite variation, we have that $\mathbb{P}(\tau_0^-=0)=1$ and $W^{(r)}$ is continuous on $\mathbb{R}$ . Thus, $V_z$ is continuous on E and $\lim_{h\downarrow 0} V^*(u,h)=V^*(0, 0)$ for any $u\geq 0$ . Moreover, since we are assuming that $\sigma>0$ , we have that $W^{(r)}\in C^2(0,\infty)$ with $W^{(r)\prime}(0\!+\!)=2/\sigma^2$ (see Lemma 3.2 and Theorem 3.10 in [Reference Kyprianou, Kuznetsov and Rivero26]). Hence, we have that $V_z(u,x)$ is a $C^{1,1}$ function on $[0,\infty)\times [0,\infty)$ and the second derivative $\frac{\partial^2}{\partial x^2} V^*(u,x)$ exists and is continuous on $(0,\infty)$ for all $u\geq 0$ (recall that we are assuming that $K^+$ is a $C^{1,2}$ function on $[0,\infty)\times [0,\infty)$ ). On the other hand, for $z^*<x<0$ we have that

(5.16) \begin{align}\frac{\partial}{\partial x} V^*(0,x)&=\int_{z^*}^x G(0,y)W^{(r)\prime}(x-y) \rm{d} y,\end{align}

(5.17) \begin{align}\frac{\partial^2}{\partial x^2} V^*(0,x)&=\int_{z^*}^x G(0,y)W^{(r)\prime \prime}(x-y) \rm{d} y+G(0,x)W^{(r)\prime}(0\!+\!).\end{align}

Hence, we see that $V_z$ is a $C^1$ function on $(\!-\!\infty,0]$ and its second derivative exists and is continuous on $(\!-\!\infty,0]\setminus \{z^*\}$ . Furthermore, by applying formula (3.7) to H(0, 0) (see equation (5.12)) and from (5.13) we see that

(5.18) \begin{align}\Phi'(r)\int_{(0,\infty)}\int_0^{\infty} G(v, y) \mathrm{e}^{-rv} \frac{y}{v}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v &=H(0, 0)=\lim_{\varepsilon\downarrow 0} \frac{K^+(0,\varepsilon)}{\psi'(\Phi(r)) W^{(r)}(\varepsilon)}\nonumber\\ &=\Phi'(r)\frac{\sigma^2}{2}\frac{\partial }{\partial x} K^+(0,0\!+\!).\end{align}

Hence, from the equality above and (5.15) we deduce that

\begin{align*}\frac{\partial}{\partial x} V^*(0,0\!+\!)=\frac{\partial}{\partial x} V^*(0,0\!-\!).\end{align*}

It can easily be seen that the process $\{ Z_{t\wedge \tau_{z^*}^-}^*,t\geq 0 \}$ is a martingale. Hence, by using standard arguments (cf. Section III.7.2 in [Reference Peskir and Shiryaev38] or Proposition 2.4 in [Reference Lamberton and Mikou31]), we deduce that

(5.19) \begin{align}\mathcal{A}_{U,X}(V^*)(u,x)+G=rV^*\end{align}

for all $(u,x)\in E$ such that $x\geq z^*$ , where from Corollary 3.1 we obtain that

\begin{align*}\mathcal{A}_{(U,X)}(V^*)(u,x)&=\frac{\partial}{\partial u}V^*(u,x)\mathbb{I}_{\{x>0 \}}-\mu \frac{\partial}{\partial x} V^*(u,x)+\frac{1}{2}\sigma^2 \frac{\partial^2}{\partial x^2} V^*(u,x) \\&\quad +\int_{(\!-\!\infty,0)} \left( V^*(u,x+y) -V^*(u,x)-y\mathbb{I}_{\{y>-1\}}\frac{\partial }{\partial x} V^*(u,x) \right)\mathbb{I}_{\{x+y >0 \}}\Pi(\rm{d} y)\\& \quad + \int_{(\!-\!\infty,0)} \left( V^*(0,x+y) -V^*(0,x)-y\mathbb{I}_{\{y>-1\}}\frac{\partial }{\partial x} V^*(0,x) \right)\mathbb{I}_{\{x\leq 0 \}}\Pi(\rm{d} y)\\& \quad + \! \int_{(\!-\!\infty,0)}\! \left( V^*(0,x+y)-V^*(0,x)-y\mathbb{I}_{\{y>-1\}} \frac{\partial }{\partial x}V^*(u,x) \right)\mathbb{I}_{\{x>0 \}}\mathbb{I}_{\{x+y <0 \}}\Pi(\rm{d} y).\end{align*}

Hence, for any $(u,x)\in E$ , $t\geq 0$ and $N>0$ , by applying the version of Itô’s formula derived in Theorem 3.1 and letting $T=t \wedge \tau_N^+$ , we deduce that, under $\mathbb{P}_{u,x}$ ,

\begin{align*}\mathrm{e}^{-rT}&V^*(U_T,X_T)\\&=V^*(u,x)-\int_0^T \mathrm{e}^{-rs}rV^*(U_s,X_s)\rm{d} s + \int_{0}^{T } \mathrm{e}^{-rs}\frac{\partial }{\partial u} V^*(U_s,X_{s})\mathbb{I}_{\{ X_s>0\}} \rm{d} s\\&\qquad+\int_{0}^{T}\mathrm{e}^{-rs} \frac{\partial }{\partial x} V^*(U_{s-},X_{s-})\rm{d} X_s+\frac{1}{2} \sigma^2 \int_{0}^{T } \mathrm{e}^{-rs}\frac{\partial^2 }{\partial x^2} V^*(U_{s},X_{s})\rm{d} s\\\end{align*}

\begin{align*}& \qquad +\int_{[0,T] }\mathrm{e}^{-rs} \int_{(\!-\!\infty,0)} \left( V^*(U_{s},X_{s^-}+y) -V^*(U_{s-},X_{s-})\right.\nonumber\\&\qquad \left.-y\frac{\partial }{\partial x} V^*(U_{s-},X_{s-}) \right)N(\rm{d} s\times \rm{d} y)\\&=M_{T}-\int_0^T \mathrm{e}^{-rs}[\mathcal{A}_{(U,X)}(V^*)(U_s,X_s)-rV^*(U_s,X_s)]\rm{d} s \\&=M_{T}-\int_0^T \mathrm{e}^{-rs}G(U_s,X_s)\mathbb{I}_{\{X_s>z^* \}}\rm{d} s,\end{align*}

where $\{ M_{t\wedge \tau_N^+}, t\geq 0\}$ is a martingale and the last equality follows since $V^*(0,x)=0$ for all $x\leq z^*$ and then $\mathcal{A}_{U,X}(V^*)(0,x)=0$ for all $x\leq z^*$ . Hence, we deduce that, for each $t\geq 0$ and $N>0$ ,

\begin{align*}Z^*_{t\wedge \tau_N^+}&=\mathrm{e}^{-r(t\wedge \tau_N^+)}V^*(U_{t\wedge \tau_N^+},X_{t\wedge \tau_N^+})+\int_0^{t\wedge \tau_N^+}\mathrm{e}^{-rs} G(U_s,X_s) \rm{d} s\\ &=M_{t\wedge \tau_N^+}+\int_0^{t\wedge \tau_N^+} \mathrm{e}^{-rs}G(0,X_s)\mathbb{I}_{\{X_s\leq z^* \}}\rm{d} s.\end{align*}

Hence, since $G(0,x)\leq 0$ for all $x\leq z^*\leq y_0$ , we conclude that $\{Z_{t\wedge \tau_N^+}^*,t\geq 0\}$ is a supermartingale as claimed.

Then the statements in Theorem 4.1 follow from Lemmas 5.4, 5.6 and 5.7. Finally, note from (5.16) that the smooth fit property holds in this case.

5.4. Proof of Proposition 4.1

Note that the optimal stopping problem is of the form (4.2), with $G(u,x)=\mathrm{e}^{x+\beta u}-K$ . From the definition of $\psi$ and since $U_s\leq u+s$ under $\mathbb{P}_{u,x}$ , for any $(u,x)\in E$ , we have that the assumption $r>\psi(1)+\beta$ implies that

\begin{align*}\mathbb{E}_{u,x}\left(\int_0^{\infty} \mathrm{e}^{-rs}|G(U_s,X_s)|\rm{d} s \right)&\leq \mathrm{e}^x\mathbb{E}\left(\int_0^{\infty} \mathrm{e}^{-rs}\mathrm{e}^{X_s+\beta(u+s)}\rm{d} s \right)+\frac{K}{r}\\&=\frac{\mathrm{e}^{x+\beta u}}{r-\psi(1)-\beta} +\frac{K}{r} < \infty.\end{align*}

Due to (2.11) we see that for any $x>0$ and $u>0$ ,

\begin{align*}K^+(u,x)&=\mathbb{E}_{x}\left(\int_0^{\tau_0^-} \mathrm{e}^{-rs}[\mathrm{e}^{X_s+\beta(u+s) }-K]\rm{d} s\right)\\&=\int_{(0,\infty)} \mathrm{e}^{\beta u}\mathrm{e}^{y} \int_{0}^{\infty}\mathrm{e}^{-(r-\beta) s} \mathbb{P}_x(X_s \in \rm{d} y,s<\tau_0^-)\rm{d} s-\frac{K}{r}[1-\mathbb{E}_x(\mathrm{e}^{-r\tau_0^-}\mathbb{I}_{\{\tau_0^-<\infty\}})]\\&=\mathrm{e}^{\beta u} \left[ \frac{W^{(r-\beta)}(x)}{\Phi(r-\beta)-1} -\int_0^x \mathrm{e}^y W^{(r-\beta)}(x-y)\rm{d} y \right] +K\int_0^xW^{(r)}(y)\rm{d} y-\frac{K}{\Phi(r)}W^{(r)}(x),\end{align*}

where we used that $\Phi(r-\beta)>1$ due to the assumption $r>\psi(1)+\beta$ and since $\Phi$ is the right-inverse of $\psi$ . On the other hand,

\begin{align*}\int_z^{0}G(0,y) \mathrm{e}^{-\Phi(r)y}\rm{d} y= \int_z^{0}[\mathrm{e}^y-K] \mathrm{e}^{-\Phi(r)y}\rm{d} y=\frac{\mathrm{e}^{-(\Phi(r)-1)z}-1}{\Phi(r)-1}-K\frac{\mathrm{e}^{-\Phi(r)z}-1}{\Phi(r)}.\end{align*}

By differentiating $K^+$ (see (5.18)) or by using Kendall’s identity (see (3.11)), we can easily see that

\begin{align*}\int_{(0,\infty)}\int_0^{\infty} G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v &=\frac{1}{\Phi(r-\beta)-1}-\frac{K}{\Phi(r)}.\end{align*}

Then from Theorem 4.1 we know that $\tau_{z^*}^-$ is optimal, where in this case $z^*$ is the unique solution on $(\!-\!\infty,0)$ to the equation

\begin{align*}\frac{\mathrm{e}^{-(\Phi(r)-1)z}}{\Phi(r)-1}-\frac{K\mathrm{e}^{-\Phi(r)z}}{\Phi(r)}+\frac{1}{\Phi(r-\beta)-1}-\frac{1}{\Phi(r)-1}=0.\end{align*}

Lastly, from Theorem 4.1, we see that the value function is given by

\begin{align*}V(u,x)&= K^+(u,x) -W^{(r)}(x)\int_{(0,\infty)}\int_0^{\infty} \! G(v, y) \frac{y}{v}\mathrm{e}^{-rv}\mathbb{P}(X_{v} \in \rm{d} y) \rm{d} v-\int_{z^*}^0 \!G(0,y)W^{(r)}(x-y) \rm{d} y\\&=\mathrm{e}^{\beta u} \left[ \frac{W^{(r-\beta)}(x)}{\Phi(r-\beta)-1} -\int_0^x \mathrm{e}^y W^{(r-\beta)}(x-y)\rm{d} y \right] +K\int_0^xW^{(r)}(y)\rm{d} y-\frac{K}{\Phi(r)}W^{(r)}(x)\\&\qquad-W^{(r)}(x)\left[\frac{1}{\Phi(r-\beta)-1}-\frac{K}{\Phi(r)} \right]-\int_{z^*}^0 [\mathrm{e}^{y}-K]W^{(r)}(x-y) \rm{d} y,\end{align*}

for any $(u,x)\in E$ . The proof is now complete.