Utility maximisation in a factor model with constant and proportional transaction costs

Abstract

We study the problem of maximising expected utility of terminal wealth under constant and proportional transaction costs in a multidimensional market with prices driven by a factor process. We show that the value function is the unique viscosity solution of the associated quasi-variational inequalities and construct optimal strategies. While the value function turns out to be truly discontinuous, we are able to establish a comparison principle for discontinuous viscosity solutions which is strong enough to argue that the value function is unique, globally upper semicontinuous, and continuous if restricted to either borrowing or non-borrowing portfolios.

Appendices

Appendix A: Auxiliary results

Lemma A.1

Let$x\in {\overline{\mathcal{S}}}$. Then the sets$\mathcal{D}(x)$and$\Xi (x)$are compact.

Proof

Clearly, it suffices to show that $\mathcal{D}(x)$ is compact. Using the definitions of $\Gamma$, $\operatorname{L}$ and ${\overline{ \mathcal{S}}}$, it is easily seen that

$$\begin{array}{rl}\mathcal{D}(x)& =\{\mathrm{\Delta }\in {\times }_{i=1}^n[-x_i,\mathrm{\infty }):x_0-\sum _{i=1}^n({\mathrm{\Delta }}_i+{\gamma }_i|{\mathrm{\Delta }}_i|)-K\\ & \phantom{=:\{\mathrm{\Delta }{\in }_{i=1}^n[-x_i,\mathrm{\infty }):}+{(\sum _{i=1}^n(1-{\gamma }_i)(x_i+{\mathrm{\Delta }}_i)-K)}^{+}\ge 0\}.\end{array}$$

Clearly, this set is closed. Moreover, whenever $\Delta _{i}\to \infty$ for some $i\in \{1,\ldots ,n\}$, the left-hand side of the above inequality tends to $-\infty$ since $\gamma _{i}>0$. Hence this set is also bounded, i.e., compact. □

Lemma A.2

Let$(x^{k})_{k\in \mathbb{N}}\subseteq {\overline{\mathcal{S}}} \setminus {\mathcal{S}_{\emptyset }}$converge to some$x\in {\overline{ \mathcal{S}}}\setminus {\mathcal{S}_{\emptyset }}$. Then:

• (1) The set$\bigcup _{k\in \mathbb{N}}\Xi (x^{k})$is bounded.

• (2) Let$\xi ^{k}\in \Xi (x^{k})$for each$k\in \mathbb{N}$. Then there exist$\xi \in \Xi (x)$and a subsequence$(\xi ^{k_{j}})_{j\in \mathbb{N}}$such that$\xi ^{k_{j}}\to \xi$as$j\to \infty$.

• (3) Let$\xi \in \Xi (x)$with$\xi \notin {\mathcal{S} _{{\mathrm{b}}}^{\mathrm{liq}}}$. Then there exist a subsequence$(k_{j})_{j\in \mathbb{N}}$and a corresponding sequence$(\xi ^{k_{j}})_{j \in \mathbb{N}}$with$\xi ^{k_{j}}\in \Xi (x^{k_{j}})$and$\xi ^{k _{j}}\to \xi$as$j\to \infty$. Moreover, if$\circ \in \{{\mathrm{b}}, {\mathrm{nb}}\}$is chosen such that$\xi \in {\widehat{\mathcal{S}} _{\circ }}$, then$(\xi ^{k_{j}})_{j\in \mathbb{N}}$can be chosen such that it is contained in${\widehat{\mathcal{S}}_{\circ }}$as well.

Proof

(1) Define $\bar{x}\in {\overline{\mathcal{S}}}$ by

$$ \bar{x}_{i} \mathrel{:=}\sup _{k\in \mathbb{N}} x_{i}^{k} \qquad \text{for all }i=0,1,\ldots ,n. $$

Since $x^{k}\leq \bar{x}$, it follows that $\Gamma (x^{k},\Delta ) \leq \Gamma (\bar{x},\Delta )$ for each $\Delta \in \mathcal{D}(x^{k})$ and $k\in \mathbb{N}$. In particular, this shows that $\mathcal{D}(x ^{k})\subseteq \mathcal{D}(x)$ for all $k\in \mathbb{N}$ and therefore

$$ \xi _{i} \leq \xi _{i}^{\max } \mathrel{:=}\sup \{\bar{\xi }_{i} : \bar{ \xi }\in \Xi (\bar{x})\} < \infty \qquad \text{for }i=0,1\ldots ,n, \xi \in \Xi (x^{k}), k\in \mathbb{N}. $$

The short-selling constraint implies moreover that $\xi _{i}^{\min } \mathrel{:=}0 \leq \xi _{i}$ for all ${i=1,\ldots ,n}$, $\xi \in \Xi (x^{k})$ and $k\in \mathbb{N}$. Hence, if we can show that

$$ -\infty < \xi _{0}^{\min } \mathrel{:=}\inf \{\bar{\xi }_{0} : \bar{ \xi }\in \Xi (\bar{x})\} \leq \xi _{0} \qquad \text{for all }\xi \in \Xi (x^{k})\text{ and }k\in \mathbb{N}, $$

(A.1)

we can conclude that $\bigcup _{k\in \mathbb{N}} \Xi (x^{k})$ is bounded since then

$$ \bigcup _{k\in \mathbb{N}} \Xi (x^{k}) \subseteq \times _{i=0}^{n} \bigl[\xi _{i}^{\min },\xi _{i}^{\max }\bigr]. $$

To verify (A.1), we fix $k\in \mathbb{N}$, $\Delta \in \mathcal{D}(x^{k})$ and set $\xi \mathrel{:=}\Gamma (x ^{k},\Delta )$. Since sell orders only increase the value of $\xi _{0}$, we may without loss of generality assume that $\Delta _{i} \geq 0$ for all $i=1,\ldots ,n$. Now define $\bar{\Delta }\in \mathbb{R}^{n+1}$ by

$$ \bar{\Delta }_{i} \mathrel{:=}\Delta _{i} + \frac{\bar{x}_{0} - x^{k} _{0}}{n(1+\gamma _{i})} \geq \Delta _{i} \qquad \text{for all }i=1,\ldots ,n. $$

Setting $\bar{\xi }\mathrel{:=}\Gamma (\bar{x},\bar{\Delta })$ and using that $\Delta _{i},\bar{\Delta }_{i}\geq 0$, it follows that

$$ \xi _{0} = x^{k}_{0} - \sum _{i=1}^{n} (1+\gamma _{i})\Delta _{i} - K = \bar{x}_{0} - \sum _{i=1}^{n}(1+\gamma _{i})\bar{\Delta }_{i} - K = \bar{ \xi }_{0}. $$

Now, it is clear that $\bar{\xi }_{i} \geq \xi _{i}\geq 0$ for all $i=1,\ldots ,n$, and hence, in order to conclude, we are only left with verifying that $\operatorname{L}(\bar{\xi })\geq 0$. But using ${\bar{x}_{i}+\bar{\Delta }_{i} \geq x_{i} + \Delta _{i}}$ yields

$$\begin{aligned} \operatorname{L}(\bar{\xi }) & = \bar{x}_{0} - \sum _{i=1}^{n} (1+ \gamma _{i})\bar{\Delta }_{i} - K + \bigg(\sum _{i=1}^{n}(1-\gamma _{i})( \bar{x}_{i} + \bar{\Delta }_{i}) - K\bigg)^{+} \\ & \geq \bar{x}_{0} + x_{0}^{k} - x_{0}^{k} - \sum _{i=1}^{n} (1+\gamma _{i})\bigg(\Delta _{i} + \frac{\bar{x}_{0}-x_{0}^{k}}{n(1+\gamma _{i})} \bigg) - K \\ & \phantom{=:}+ \bigg(\sum _{i=1}^{n}(1-\gamma _{i})(x_{i}^{k} + \Delta _{i}) - K\bigg)^{+} \\ & = \bar{x}_{0} - x_{0}^{k} - \sum _{i=1}^{n} (1+\gamma _{i})\frac{ \bar{x}_{0}-x_{0}^{k}}{n(1+\gamma _{i})} \\ & \phantom{=:} + x_{0}^{k} - \sum _{i=1}^{n} (1+\gamma _{i})\Delta _{i} - K + \bigg( \sum _{i=1}^{n}(1-\gamma _{i})(x_{i}^{k} + \Delta _{i}) - K\bigg)^{+} \\ & = \xi _{0} + \bigg(\sum _{i=1}^{n}(1-\gamma _{i})\xi _{i} - K\bigg)^{+} = \operatorname{L}(\xi ) \geq 0, \end{aligned}$$

thus establishing (A.1).

(2) Let $\xi ^{k}\in \Xi (x^{k})$ for each $k\in \mathbb{N}$. Since $\bigcup _{k\in \mathbb{N}}\Xi (x^{k})$ is bounded and ${\overline{ \mathcal{S}}}$ is closed, it follows that $(\xi ^{k})_{k\in \mathbb{N}}$ is bounded and admits a subsequence (again indexed by $k$ for simplicity) which converges to some $\xi \in {\overline{\mathcal{S}}}$. We are left with showing that $\xi \in \Xi (x)$. For this, we first observe that for each $k\in \mathbb{N}$, there exists $\Delta ^{k} \in \mathcal{D}(x^{k})$ such that $\xi ^{k} = \Gamma (x^{k},\Delta ^{k})$, i.e.,

$$ \xi ^{k}_{0} = x_{0}^{k} - \sum _{i=1}^{n}(\Delta _{i}^{k} + \gamma _{i}| \Delta _{i}^{k}|) - K, \qquad \xi _{i}^{k} = x_{i}^{k} + \Delta _{i}^{k},\quad i=1,\ldots ,n. $$

Now since $\xi ^{k}\to \xi$ and $x^{k}\to x$, we immediately find that $\Delta _{i}^{k} = \xi _{i}^{k} - x^{k}_{i}$ converges to some $\Delta _{i}\in \mathbb{R}$ satisfying $\xi _{i}=x_{i}+\Delta _{i}$ for each $i=1,\ldots ,n$. But then we must have

$$ \xi _{0} = \lim _{k\to \infty } \xi ^{k}_{0} = \lim _{k\to \infty } x_{0} ^{k} - \sum _{i=1}^{n}(\Delta _{i}^{k} + \gamma _{i}|\Delta _{i}^{k}|) - K= x_{0} - \sum _{i=1}^{n}(\Delta _{i} + \gamma _{i}|\Delta _{i}|) - K, $$

which implies that $\xi =\Gamma (x,\Delta )$, i.e., $\xi \in \Xi (x)$.

(3) Let $\xi \in \Xi (x)$ with $\xi \notin {\mathcal{S}_{{\mathrm{b}}} ^{\mathrm{liq}}}$. Then $\xi =\Gamma (x,\Delta )$ for some $\Delta \in \mathcal{D}(x)$. We define a sequence $(\varepsilon _{k})_{k\in \mathbb{N}}$ by

$$ \varepsilon _{k} \mathrel{:=}\sum _{i=1}^{n}(1+\gamma _{i})(x_{i}-x_{i} ^{k})^{+} \qquad \text{for all }k\in \mathbb{N}. $$

Observe that $\varepsilon _{k}\geq 0$ and $\varepsilon _{k}\to 0$ as $k\to \infty$. With this, for each $k\in \mathbb{N}$, let us now define $\Delta ^{k}\in \mathbb{R}^{n}$ by

$$ \Delta _{i}^{k} = \textstyle\begin{cases} \max \{\Delta _{i} - ((x_{0}-x_{0}^{k})^{+} + \varepsilon _{k}) / (1+ \gamma _{i}),-x_{i}^{k}\} &\quad \text{if }\Delta _{i} > 0, \\ \max \{\Delta _{i} - ((x_{0}-x_{0}^{k})^{+} + \varepsilon _{k}) / (1- \gamma _{i}),-x_{i}^{k}\} &\quad \text{if }\Delta _{i} \leq 0, \end{cases} $$

for $i=1,\ldots ,n$, and set $\xi ^{k}\mathrel{:=}\Gamma (x^{k},\Delta ^{k})$. From the definition of $\Delta ^{k}$, it is immediately clear that $\Delta _{i}^{k}\geq -x_{i}^{k}$, implying that $\xi _{i}^{k} = x _{i}^{k} + \Delta _{i}^{k} \geq 0$. Moreover, as $k\to \infty$, each $\Delta _{i}^{k}$ converges to $\max \{\Delta _{i},-x_{i}\} = \Delta _{i}$ since $\xi \in {\overline{\mathcal{S}}}$ and hence $0\leq \xi _{i} = x_{i} + \Delta _{i}$. In particular, using the continuity of $\Gamma$, this implies that

$$ \lim _{k\to \infty }\xi ^{k} = \lim _{k\to \infty } \Gamma (x^{k},\Delta ^{k}) = \Gamma (x,\Delta ) = \xi . $$

Therefore, in order to conclude, we only have to show that, after possibly passing to a subsequence, $\operatorname{L}(\xi ^{k})\geq 0$ eventually (so that $\xi ^{k}\in \Xi (x^{k})$ eventually), and $\xi ^{k}_{0}\geq 0$ eventually if $\xi _{0} = 0$ ($\xi ^{k}_{0}$ and $\xi _{0}$ have eventually the same sign if $\xi _{0}\neq 0$, so there is nothing to show in this case). Let us first consider the case $\xi _{0}\neq 0$. Then we must have $\operatorname{L}(\xi )>0$. Indeed, if $\xi \in {\widehat{\mathcal{S}}_{{\mathrm{nb}}}}$, we have $\operatorname{L}(\xi )\geq \xi _{0} > 0$ and similarly, if $\xi \in {\mathcal{S}_{{\mathrm{b}}}}$, we directly find $\operatorname{L}( \xi )>0$ since $\xi \notin {\mathcal{S}_{{\mathrm{b}}}^{ \mathrm{liq}}}$. But then by the continuity of $\operatorname{L}$, it follows that $\lim _{k\to \infty }\operatorname{L}(\xi ^{k}) = \operatorname{L}(\xi ) > 0$, and so we must eventually have $\operatorname{L}(\xi ^{k})\geq 0$. Let us therefore from now on focus on the case $\xi _{0} = 0$. In this case, it suffices to show that $\xi ^{k}_{0}\geq 0$ eventually, since this already implies $\operatorname{L}(\xi ^{k}) \geq \xi _{0}^{k}\geq 0$ eventually. Let us first suppose that

$$ \Delta _{i}^{k} = - x_{i}^{k} \qquad \text{for all }i=1,\ldots ,n\text{ and eventually all }k\in \mathbb{N}. $$

(A.2)

Then it follows that

$$ \Delta _{i} = \lim _{k\to \infty } \Delta _{i}^{k} = \lim _{k\to \infty } -x_{i}^{k} = -x_{i} \qquad \text{for all }i=1,\ldots ,n. $$

In particular, this implies $\xi = 0\in {\mathcal{S}_{{\mathrm{nb}}} ^{\mathrm{liq}}}$. But in this case, as shown in Lemma A.3 below, we have ${\Delta ^{k}=(-x_{1}^{k}, \ldots ,-x_{n}^{k})\in \mathcal{D}(x^{k})}$ for all $k\in \mathbb{N}$ and it follows that ${\xi ^{k} = \Gamma (x^{k},\Delta ^{k})\in {\overline{ \mathcal{S}}}}$. Now since $\xi ^{k}_{i} = 0$ for all $i=1,\ldots ,n$, this is only possible if $\xi ^{k}_{0}\geq 0$ and we are done. Let us now suppose that (A.2) does not hold. Then there exists a subsequence (again indexed by $k$ for simplicity) such that for each $k\in \mathbb{N}$, there exists some $i_{k}\in \{1,\ldots ,n\}$ with $\Delta _{i_{k}}^{k} > - x_{i_{k}}^{k}$. After passing to another subsequence if necessary, we may furthermore assume that ${i \mathrel{:=}i_{k}}$ does not depend on $k$. In particular, for this particular $i$, we have

Now, using that $\xi _{0} = 0$, we compute

$$ \xi _{0}^{k} = \xi _{0}^{k} - \xi _{0} = x_{0}^{k} - x_{0} - \sum _{j=1} ^{n}\big(\Delta _{j}^{k}-\Delta _{j} + \gamma _{j}(|\Delta _{j}^{k}|-| \Delta _{j}|)\big). $$

As soon as $k$ is sufficiently large, we observe that $\Delta _{j} - ((x _{0}-x_{0}^{k})^{+} +\varepsilon _{k})/ (1+\gamma _{j})$ is positive whenever $\Delta _{j}>0$ (in particular, $\Delta _{j}^{k} > -x_{j}^{k}$ and thus $\Delta _{j}^{k}\geq 0$ in this case). Moreover, we have $\Delta _{j}^{k}\leq 0$ whenever $\Delta _{j}\leq 0$. Using this, it follows that

(A.3)

for all $k\in \mathbb{N}$ sufficiently large. Now pick any $j\in \{1, \ldots ,n\}$ with $j\neq i$. If $\Delta _{j}^{k} > -x_{j}^{k}$, then it follows that $\Delta _{j}^{k} - \Delta _{j} \leq 0$ and hence

On the other hand, if $\Delta _{j}^{k} = -x_{j}^{k}$, then we can use that $\Delta _{j} \geq -x_{j}$ to see that $\Delta _{j}^{k} - \Delta _{j} \leq -x_{j}^{k} + x_{j} \leq (x_{j}-x_{j}^{k})^{+}$ and thus

as well. Plugging the latter two estimates into (A.3) therefore gives

$$\begin{aligned} \xi _{0}^{k} & \geq x_{0}^{k} - x_{0} + (x_{0} - x_{0}^{k})^{+} + \varepsilon _{k} - \sum _{j=1,j\neq i}^{n} (1+\gamma _{j})(x_{j}-x_{j} ^{k})^{+} \\ & \geq -(x_{0}-x_{0}^{k})^{+} + (x_{0} - x_{0}^{k})^{+} + \varepsilon _{k} - \sum _{j=1}^{n} (1+\gamma _{j})(x_{j}-x_{j}^{k})^{+} = 0, \end{aligned}$$

where the last equality follows from the choice of $\varepsilon _{k}$. □

Lemma A.3

Let$x\in {\overline{\mathcal{S}}}$and set$\Delta ^{\mathrm{Liq}} \mathrel{:=}(-x_{1},\ldots ,-x_{n})$. Then$\mathcal{D}(x)\neq \emptyset$if and only if$\Delta ^{\mathrm{Liq}}\in \mathcal{D}(x)$. Moreover, if$x$satisfies

$$ x_{0} + \sum _{i=1}^{n} (1-\gamma _{i})x_{i} - K = 0, $$

then$\mathcal{D}(x) = \{\Delta ^{\mathrm{Liq}}\}$, i.e., $\Delta ^{\mathrm{Liq}}$is the only admissible transaction. In particular, we have$\Xi (x) = \{0\}$in this case.

Proof

Suppose that $\mathcal{D}(x)\neq \emptyset$, choose $\Delta \in \mathcal{D}(x)$ and set $\xi \mathrel{:=}\Gamma (x,\Delta )$. Moreover, define $\xi ^{\mathrm{Liq}}\mathrel{:=}\Gamma (x,\Delta ^{\mathrm{Liq}})$. Since $\xi ^{\mathrm{Liq}}_{i} = x_{i} + \Delta _{i}^{\mathrm{Liq}} = 0$ for all $i=1,\ldots ,n$, it suffices to show that $\operatorname{L}( \xi ^{\mathrm{Liq}})\geq 0$ to obtain $\xi ^{\mathrm{Liq}}\in \Xi (x)$, i.e., $\Delta ^{\mathrm{Liq}}\in \mathcal{D}(x)$. That is, we have to show that

$$ \operatorname{L}(\xi ^{\mathrm{Liq}}) = x_{0} + \sum _{i=1}^{n} (1- \gamma _{i})x_{i} - K \geq 0. $$

Now, $\xi \in \Xi (x)$ implies in particular that $\operatorname{L}( \xi )\geq 0$, i.e.,

$$ 0 \leq x_{0} - \sum _{i=1}^{n}(\Delta _{i} + \gamma _{i}|\Delta _{i}|) - K + \bigg(\sum _{i=1}^{n} (1-\gamma _{i})(x_{i}+\Delta _{i}) - K\bigg)^{+}. $$

(A.4)

Let us first suppose that the positive part on the right-hand side is equal to zero. Using that $0\leq \xi _{i} = x_{i}+\Delta _{i}$ implies that $-\Delta _{i}\leq x_{i}$, it follows that

(A.5)

If, on the other hand, the positive part in (A.4) is not equal to zero, we obtain

$$\begin{aligned} 0 & \leq x_{0} + \sum _{i=1}^{n}\bigl((1-\gamma _{i})(x_{i}+\Delta _{i})- \Delta _{i} - \gamma _{i}|\Delta _{i}|\bigr) - 2K \\ & = x_{0} + \sum _{i=1}^{n}(1-\gamma _{i})x_{i} - K - \sum _{i=1}^{n} \gamma _{i}(\Delta _{i}+|\Delta _{i}|) - K \\ & < x_{0} + \sum _{i=1}^{n}(1-\gamma _{i})x_{i} - K. \end{aligned}$$

(A.6)

We have therefore argued that $\Delta ^{\mathrm{Liq}}\in \mathcal{D}(x)$ whenever $\Xi (x)\neq \emptyset$. If we now suppose that $x$ is such that

$$ x_{0} + \sum _{i=1}^{n} (1-\gamma _{i})x_{i} - K = 0, $$

then it is immediately clear that $\Delta ^{\mathrm{Liq}}\in \mathcal{D}(x)$. Let us proceed to show that there exists no other $\Delta \in \mathcal{D}(x)$. By (A.5) and (A.6), we immediately find that

$$ \operatorname{L}\bigl(\Gamma (x,\Delta )\bigr) \leq \operatorname{L} \bigl(\Gamma (x,\Delta ^{\mathrm{Liq}})\bigr) = 0. $$

Since the inequality is strict if we are in the situation of (A.6), this already implies that $\Delta \notin \mathcal{D}(x)$ in this case. On the other hand, if we are in the situation of (A.5), then we can use that $\Delta \neq \Delta ^{\mathrm{Liq}}$ implies that $\Delta _{i}>-x_{i}$ for some $i\in \mathbb{N}$, and hence we find that the inequality in (A.5) must also be strict and $\Delta \notin \mathcal{D}(x)$. □

Lemma A.4

Let$(t,x,y)\in [0,T]\times {\overline{\mathcal{O}}}$and$\Lambda = ( \tau _{k},\xi _{k})_{k\in \mathbb{N}}\in \mathcal{A}(t,x,y)$. Then$\mathbb{P}[\lim _{k\to \infty }\tau _{k}>T] = 1$.

Proof

For $k\in \mathbb{N}$ fixed, we denote by $X^{k} = (X^{k,0},X^{k,1}, \ldots ,X^{k,n})$ the portfolio process obtained by following the strategy which performs only the first $k$ trades of $\Lambda$ and then lets the portfolio run uncontrolled. We write

$$ N^{k,i}_{u} \mathrel{:=}\frac{X^{k,i}_{u}}{P^{i}_{u}}, \qquad u\in [t,T], i=0,1\ldots ,n, $$

which is the number of shares of the $i$th asset $P^{i}$ held at time $u$. In the absence of transaction costs, the investor’s wealth is given by

$$ Z^{k}_{u} := \boldsymbol{1}^{\top }x + \sum _{i=0}^{n}\int _{t}^{u} N _{s}^{k,i} \,\mathrm{d}P^{i}_{s}, \qquad u\in [t,T], $$

and hence, since any transaction incurs at least a cost of $K$, it is clear that

$$ 0\leq \operatorname{L}(X^{k}_{\tau _{k}}) \leq \boldsymbol{1}^{\top }X ^{k}_{\tau _{k}} \leq Z^{k}_{\tau _{k}} - kK \qquad \text{on }A_{k}\mathrel{:=}\{\tau _{k}\leq T\}. $$

Let us now assume by way of contradiction that $A\mathrel{:=} \bigcap _{k\in \mathbb{N}} A_{k}$ has positive probability. This implies that $Z^{k}_{\tau _{k}}\to \infty$ on this event. This, however, cannot be since the price process $P$ satisfies the no-free-lunch-with-vanishing-risk property and hence the family

$$ \bigg\{ \,\sup _{u\in [t,T]}\bigg|\,\sum _{i=0}^{n}\int _{t}^{u} N_{s} ^{k,i} \,\mathrm{d}P^{i}_{s}\,\bigg| : k \in \mathbb{N}\bigg\} $$

is bounded in probability. □

Lemma A.5

For all$(t,x,y)\in [0,T]\times {\overline{\mathcal{O}}}$, there exists$M>0$such that

$$ \sup _{\Lambda \in \mathcal{A}(t,x,y)}\mathbb{E}\bigg[\sup _{u\in [t,T]}| \boldsymbol{1}^{\top }X^{t,x,y}_{u}({\Lambda })|^{2}\bigg] \leq M ( 1 + |\boldsymbol{1}^{\top }x|^{2}). $$

Proof

We write $(X,Y) \mathrel{:=}(X^{t,x,y}(\Lambda ),Y^{t,y})$ as shorthand notation. We first note that with $N \mathrel{:=}\max \{1/\gamma _{i} : i = 1,\ldots ,n\}$, it is not difficult to verify that

$$ \max _{i=0,1,\ldots ,n}x_{i} \leq (1+N)\boldsymbol{1}^{\top }x \qquad \text{for all } x\in {\overline{\mathcal{S}}}. $$

(A.7)

Now, for every fixed $u\in [t,T]$, we have the estimate

$$ |\boldsymbol{1}^{\top }X_{u}| = \sum _{i=0}^{n} X^{i}_{u} \leq \boldsymbol{1}^{\top }x + \int _{t}^{u} \sum _{i=0}^{n} \mu _{i}(Y_{s})X ^{i}_{s} \,\mathrm{d}s+ \sum _{j=1}^{d} \int _{t}^{u} \sum _{i=0}^{n} \sigma _{i,j}(Y_{s})X^{i}_{s}\,\mathrm{d}W^{j}_{s}. $$

Let $L>0$ be an upper bound of $|\mu |$. Using (A.7), we further estimate

$$ |\boldsymbol{1}^{\top }X_{u}| \leq \boldsymbol{1}^{\top }x + (n+1)(1+N)L \int _{t}^{u} |\boldsymbol{1}^{\top }X_{s}| \,\mathrm{d}s + \sum _{j=1} ^{d} \int _{t}^{u} \sum _{i=0}^{n} \sigma _{i,j}(Y_{s})X^{i}_{s}\, \mathrm{d}W^{j}_{s}. $$

Squaring both sides, estimating the square of sums by the sum of squares and using Jensen’s inequality for the Lebesgue integral then implies the existence of a constant $C>0$ such that

$$ |\boldsymbol{1}^{\top }X_{u}|^{2} \leq C\bigg(|\boldsymbol{1}^{\top }x|^{2} + \int _{t}^{u} | \boldsymbol{1}^{\top }X_{s}|^{2} \,\mathrm{d}s + \sum _{j=1}^{d}\bigg| \int _{t}^{u} \sum _{i=0}^{n} \sigma _{i,j}(Y_{s})X^{i}_{s}\,\mathrm{d}W ^{j}_{s}\bigg|^{2}\bigg). $$

Standard estimates involving Doob’s inequality then show that there exists a constant $D>0$ (which still does not depend on $\Lambda$) such that

$$ \mathbb{E}\bigg[ \sup _{r\in [t,u]} |\boldsymbol{1}^{\top }X_{r}|^{2} \bigg] \leq D\bigg(|\boldsymbol{1}^{\top }x|^{2}+ \int _{t}^{u} \mathbb{E}\bigg[\sup _{r\in [t,s]}|\boldsymbol{1}^{\top }X_{r}|^{2} \bigg] \,\mathrm{d}s\bigg), $$

and we conclude by Gronwall’s inequality. □

Lemma A.6

Let$h_{1},h_{2}:[0,T]\times {\overline{\mathcal{O}}}\to \mathbb{R}$and$\lambda _{1},\lambda _{2}\geq 0$. Then

$$ \mathcal{M}(\lambda _{1} h_{1} + \lambda _{2} h_{2}) \leq \lambda _{1} \mathcal{M}h_{1} + \lambda _{2}\mathcal{M}h_{2} \qquad \textit{on }[0,T]\times {\overline{\mathcal{O}}}. $$

Proof

This is an immediate consequence of the definition of ℳ and the sublinearity of the supremum. □

Lemma A.7

Let$h:[0,T]\times {\overline{\mathcal{O}}}\to \mathbb{R}$be nonnegative and such that the mapping$x\mapsto h(t,x,y)$is increasing on${\overline{\mathcal{S}}}$for all$(t,y)\in [0,T]\times \mathbb{R}^{m}$. Then the same is true for$x\mapsto \mathcal{M}h(t,x,y)$.

Proof

Let $(t,y)\in [0,T]\times \mathbb{R}^{m}$ and fix $x_{1},x_{2}\in {\overline{\mathcal{S}}}$ with $x_{1} \leq x_{2}$. If $\Xi (x_{1}) = \emptyset$, there is nothing to show, so let us suppose that $\Xi (x_{1})\neq \emptyset$. Now fix an arbitrary $\xi _{1}=\Gamma (x _{1},\Delta )\in \Xi (x_{1})$ for some $\Delta \in \mathcal{D}(x_{1})$, define $\xi _{2}\mathrel{:=}\Gamma (x_{2},\Delta )$ and observe that $x_{1}\leq x_{2}$ implies that $\xi _{1}\leq \xi _{2}$. In particular, $\xi _{2}\in \Xi (x_{2})$. Using the monotonicity of $h$, this shows that $h(t,\xi _{1},y) \leq h(t,\xi _{2},y) \leq \mathcal{M}h(t,x_{2},y)$, and since $\xi _{1}$ was chosen arbitrarily, the result follows. □

Lemma A.8

Let$h:[0,T]\times {\overline{\mathcal{O}}}\to \mathbb{R}$be nonnegative and upper semicontinuous. Then$\mathcal{M}h$is upper semicontinuous as well.

Proof

Since ${\overline{\mathcal{S}}}\setminus {\mathcal{S}_{\emptyset }}$ is closed, $\mathcal{M}h(t,x,y) = -1$ whenever $x\in {\mathcal{S}_{ \emptyset }}$, and $h\geq 0$, it suffices to show that $\mathcal{M}h$ is upper semicontinuous on ${[0,T]\times ({\overline{\mathcal{S}}} \setminus {\mathcal{S}_{\emptyset }})\times \mathbb{R}^{m}}$. Let therefore $(t,x,y)\in [0,T]\times ({\overline{\mathcal{S}}}\setminus {\mathcal{S}_{\emptyset }})\times \mathbb{R}^{m}$ and choose an arbitrary sequence ${(t_{k},x_{k},y_{k})_{k\in \mathbb{N}}\subseteq [0,T] \times ({\overline{\mathcal{S}}}\setminus {\mathcal{S}_{\emptyset }}) \times \mathbb{R}^{m}}$ converging to $(t,x,y)$. Since every $\Xi (x_{k})$ is compact and $h$ is upper semicontinuous, there exists $\xi _{k}\in \Xi (x_{k})$ with

$$ \mathcal{M}h(t_{k},x_{k},y_{k}) = h(t_{k},\xi _{k},y_{k}) \qquad \text{for all }k\in \mathbb{N}. $$

Moreover, by Lemma A.2(2) and after possibly passing to a subsequence, there exists $\xi \in \Xi (x)$ such that $\xi _{k}\to \xi$. But then, by upper semicontinuity of $h$,

$$\begin{aligned} \limsup _{k\to \infty } \mathcal{M}h(t_{k},x_{k},y_{k}) &= \lim _{k\to \infty } \mathcal{M}h(t_{k},x_{k},y_{k}) \\ &\leq \limsup _{k\to \infty } h(t_{k},\xi _{k},y_{k}) \leq h(t,\xi ,y) \leq \mathcal{M}h(t,x,y), \end{aligned}$$

showing that $\mathcal{M}h$ is upper semicontinuous. □

Appendix B: The local viscosity property of the optimal stopping problem

In this section, we show that the value function of the abstract optimal stopping problem studied in Sect. 6.1 is a local viscosity solution of the VIs (6.3). It is readily checked that the abstract optimal stopping problem can be embedded into the setting of [10, Theorem 4.1], hence yielding the following weak dynamic programming principle.

Theorem B.1

(Bouchard and Touzi [10])

Let$(t,x,y)\in [0,T) \times {\mathcal{O}}$and$\circ \in \{{\mathrm{b}},{\mathrm{nb}}\}$be such that$x\in {\mathcal{S}_{\circ }}$. Let furthermore$F\subseteq [0,T]\times {\mathcal{S}_{\circ }}\times \mathbb{R}^{m}$be compact such that$(t,x,y)$is in the interior of$F$relative to$[0,T)\times {\mathcal{O}}$. Next, let$\theta \in \mathcal{T}^{t}_{t,T}$be such that the process$(\cdot ,\overline{X}^{t,x,y},Y^{t,y})$never leaves$F$on$[t,\theta ]$. Then

(B.1)

Moreover, it holds that

(B.2)

for any upper semicontinuous function$\varphi :[0,T)\times {\mathcal{O} _{\circ }^{\star }}\to \mathbb{R}$with$\varphi \leq \mathrm{V}$.

Some brief remarks are in order here. First, we observe that the assumptions ensure that $\theta <\overline{\tau }_{{\mathcal{S}}}^{t,x,y}$, i.e., the state process never reaches the boundary of the state space. One difference in the conclusion of the above theorem and the result in [10] is that we write $\mathrm{V}^{\circ }$ on the right-hand side of (B.1) instead of the global upper semicontinuous envelope of $\mathrm{V}$. This is justified since the process $\overline{X}^{t,x,y}$ never switches from ${\widehat{\mathcal{S}} _{{\mathrm{nb}}}}$ to ${\widehat{\mathcal{S}}_{{\mathrm{b}}}}$ and vice versa. The second difference is that the test functions $\varphi$ in the second dynamic programming inequality (B.2) are only defined on $[0,T)\times {\mathcal{O}_{\circ }^{\star }}$ instead of the entire state space. This is justified by the assumption that $x\in {\mathcal{S}_{\circ }}$ and since $(\cdot \wedge \theta , \overline{X}^{t,x,y}_{\cdot \wedge \theta },Y^{t,y}_{\cdot \wedge \theta })$ never leaves $F\subseteq [0,T)\times {\mathcal{O}_{\circ } ^{\star }}$.

With the weak dynamic programming principle at hand, one can show that $\mathrm{V}$ is a local viscosity solution of the VIs (6.3).

Theorem B.2

The value function$\mathrm{V}$of the abstract optimal stopping problem (6.2) is a local viscosity solution of the VIs (6.3).

Proof

Step 1. We show that $\mathrm{V}$ is a local viscosity subsolution. For this, let $\circ \in \{{\mathrm{b}},{\mathrm{nb}}\}$, fix $(\bar{t}, \bar{x},\bar{y})\in [0,T)\times {\mathcal{O}_{\circ }^{\star }}$ and let $\varphi \in C^{1,2}([0,T)\times {\mathcal{O}_{\circ }^{\star }})$ be such that $\mathrm{V}^{\circ }-\varphi$ has a strict global maximum at $(\bar{t},\bar{x},\bar{y})$ with $\mathrm{V}^{\circ }(\bar{t},\bar{x}, \bar{y}) = \varphi (\bar{t},\bar{x},\bar{y})$. Assume by way of contradiction that there exists $\kappa >0$ such that

$$ \min \{\mathcal{L}\varphi (\bar{t},\bar{x},\bar{y}), \varphi (\bar{t}, \bar{x},\bar{y}) - w^{\circ }(\bar{t},\bar{x},\bar{y})\} \geq 2\kappa > 0. $$

Since $\varphi$ and $\mathcal{L}\varphi$ are continuous and $w^{\circ }$ is upper semicontinuous, there exists some $\varepsilon >0$ such that ${\overline{\mathcal{D}}^{\varepsilon }_{{\overline{ \mathcal{O}}}_{\circ }}(\bar{t},\bar{x},\bar{y})}\subseteq [0,T) \times {\mathcal{O}_{\circ }^{\star }}$ and

$$ \min \{\mathcal{L}\varphi , \varphi - w^{\circ }\} \geq \kappa > 0 \qquad \text{on }{\overline{\mathcal{D}}^{\varepsilon }_{{\overline{ \mathcal{O}}}_{\circ }}(\bar{t},\bar{x},\bar{y})}. $$

(B.3)

Since ${{\overline{\mathcal{D}}^{\varepsilon }_{{\overline{ \mathcal{O}}}_{\circ }}(\bar{t},\bar{x},\bar{y})}\setminus {\mathcal{D} ^{\varepsilon }_{{\overline{\mathcal{O}}}_{\circ }}(\bar{t},\bar{x}, \bar{y})}}$ is compact and the maximum of $\mathrm{V}^{\circ }-\varphi$ is strict, we see that

$$ - \gamma \mathrel{:=} \max _{(t,x,y)\in {{\overline{\mathcal{D}}^{\varepsilon }_{{\overline{ \mathcal{O}}}_{\circ }}(\bar{t},\bar{x},\bar{y})}\setminus {\mathcal{D} ^{\varepsilon }_{{\overline{\mathcal{O}}}_{\circ }}(\bar{t},\bar{x}, \bar{y})}}}\bigl(\mathrm{V}^{\circ }(t,x,y)-\varphi (t,x,y)\bigr) < 0. $$

(B.4)

Now choose $(t_{k},x_{k},y_{k})_{k\in \mathbb{N}}\subseteq [0,T) \times {\mathcal{O}_{\circ }}$ converging to $(\bar{t},\bar{x}, \bar{y})$ such that

$$ \mathrm{V}^{\circ }(\bar{t},\bar{x},\bar{y}) = \lim _{k\to \infty } \mathrm{V}(t_{k},x_{k},y_{k}). $$

Clearly, we can assume that $(t_{k},x_{k},y_{k})_{k\in \mathbb{N}} \subseteq {\overline{\mathcal{D}}^{\varepsilon /2}_{ \widehat{{\mathcal{O}}}_{\circ }}(\bar{t},\bar{x},\bar{y})}$. Now define

$$ \theta _{k} \mathrel{:=}\inf \bigl\{ t>t_{k} : (t,X^{k}_{t},Y^{k}_{t}) \notin {\overline{\mathcal{D}}^{\varepsilon }_{{\overline{ \mathcal{O}}}_{\circ }}(\bar{t},\bar{x},\bar{y})}\bigr\} , $$

where we have set $X^{k} \mathrel{:=}\overline{X}^{t_{k},x_{k},y_{k}}$ and $Y^{k}\mathrel{:=}Y^{t_{k},y_{k}}$ as shorthand notation. Finally, we write $\eta _{k}\mathrel{:=}\mathrm{V}(t_{k},x_{k},y_{k}) - \varphi (t_{k},x_{k},y_{k})\leq 0$ for each $k\in \mathbb{N}$. An application of Itô’s formula, as well as (B.3) and (B.4), now shows that

Now $\eta _{k}\to 0$ and hence $\eta _{k} + \min \{\kappa ,\gamma \}>0$ for eventually all $k\in \mathbb{N}$. Using that $\tau$ was chosen arbitrarily, this shows that

for $k$ sufficiently large. But this is a contradiction to the first weak dynamic programming inequality (B.1).

Step 2. We show that $\mathrm{V}$ is a local viscosity supersolution. For this, let $\circ \in \{{\mathrm{b}},{\mathrm{nb}}\}$, fix $(\bar{t},\bar{x},\bar{y})\in [0,T)\times {\mathcal{O}_{\circ }^{ \star }}$ and let $\varphi \in C^{1,2}([0,T)\times {\mathcal{O}_{ \circ }^{\star }})$ be such that $\mathrm{V}_{\circ }-\varphi$ has a strict global minimum at $(\bar{t},\bar{x},\bar{y})$ with $\mathrm{V} _{\circ }(\bar{t},\bar{x},\bar{y}) = \varphi (\bar{t},\bar{x},\bar{y})$. Let us first show that

$$ \mathrm{V}_{\circ }(\bar{t},\bar{x},\bar{y}) - w_{\circ }(\bar{t}, \bar{x},\bar{y}) \geq 0. $$

(B.5)

For this, we pick $(t_{k},x_{k},y_{k})_{k\in \mathbb{N}}\in [0,T) \times {\mathcal{O}_{\circ }}$ converging to $(\bar{t},\bar{x}, \bar{y})$ and with

$$ \mathrm{V}_{\circ }(\bar{t},\bar{x},\bar{y}) = \lim _{k\to \infty } \mathrm{V}(t_{k},x_{k},y_{k}). $$

By the very definition of the optimal stopping problem, we have $\mathrm{V}\geq w$ on $[0,T)\times {\mathcal{O}_{\circ }}$. Thus, the lower semicontinuity of $w_{\circ }$ immediately yields

$$ \mathrm{V}_{\circ }(\bar{t},\bar{x},\bar{y}) = \lim _{k\to \infty } \mathrm{V}(t_{k},x_{k},y_{k}) \geq \liminf _{k\to \infty } w(t_{k},x _{k},y_{k}) \geq w_{\circ }(\bar{t},\bar{x},\bar{y}). $$

Now fix $\varepsilon >0$ such that ${\overline{\mathcal{D}}^{\varepsilon }_{{\overline{\mathcal{O}}}_{\circ }}(\bar{t},\bar{x},\bar{y})}\subseteq [0,T)\times {\mathcal{O}_{\circ }^{\star }}$ and, after possibly passing to a subsequence, $(t_{k},x_{k},y_{k})_{k\in \mathbb{N}}\subseteq {\overline{\mathcal{D}}^{\varepsilon }_{\widehat{{\mathcal{O}}}_{ \circ }}(\bar{t},\bar{x},\bar{y})}$. Next, define

$$ \eta _{k} \mathrel{:=}\mathrm{V}(t_{k},x_{k},y_{k}) - \varphi (t_{k},x _{k},y_{k}) \geq 0 \qquad \text{for all }k\in \mathbb{N}, $$

as well as

We observe that $\varepsilon _{k}>0$ for all $k\in \mathbb{N}$ and

$$ \lim _{k\to \infty } \eta _{k} = \lim _{k\to \infty } \varepsilon _{k} = 0 \qquad \text{and} \qquad \lim _{k\to \infty } { \frac{\eta _{k}}{\varepsilon _{k}}} = 0. $$

Now introduce the stopping times

$$ \theta _{k} \mathrel{:=}\inf \bigl\{ t>t_{k} : (t,X^{k}_{t},Y^{k}_{t}) \notin {\overline{\mathcal{D}}^{\varepsilon }_{{\overline{ \mathcal{O}}}_{\circ }}(\bar{t},\bar{x},\bar{y})}\text{ or }t\notin [\bar{t}-\varepsilon _{k},\bar{t}+\varepsilon _{k}]\bigr\} . $$

It is apparent that $\theta _{k}\in \mathcal{T}^{t_{k}}_{t_{k},T}$ for $k$ sufficiently large, and hence the second weak dynamic programming inequality (B.2) with $\tau = T$ implies

$$ \mathrm{V}(t_{k},x_{k},y_{k}) \geq \mathbb{E}[\varphi (\theta _{k},X ^{k}_{\theta _{k}},Y^{k}_{\theta _{k}})] $$

for eventually all $k\in \mathbb{N}$. Using the definition of $\eta _{k}$ and applying Itô’s formula, it follows that

$$\begin{aligned} \frac{\eta _{k}}{\varepsilon _{k}} & = \frac{1}{\varepsilon _{k}}\bigl( \mathrm{V}(t_{k},x_{k},y_{k}) - \varphi (t_{k},x_{k},y_{k})\bigr) \\ & \geq \frac{1}{\varepsilon _{k}}\mathbb{E}[\varphi (\theta _{k},X^{k} _{\theta _{k}},Y^{k}_{\theta _{k}}) - \varphi (t_{k},x_{k},y_{k})] \\ & = \mathbb{E}\bigg[-\frac{1}{\varepsilon _{k}}\int _{t_{k}}^{\theta _{k}}\mathcal{L}\varphi (u,X^{k}_{u},Y^{k}_{u}) \,\mathrm{d}u\bigg]. \end{aligned}$$

Now for each $\omega \in \Omega$, there exists some $K(\omega ) \in \mathbb{N}$ such that $\theta _{k}(\omega ) = \varepsilon _{k}$ for all $k\geq K(\omega )$. Thus the mean value theorem and dominated convergence imply

$$ 0 = \lim _{k\to \infty } { \frac{\eta _{k}}{\varepsilon _{k}}} \geq \lim _{k\to \infty }\mathbb{E}\bigg[{ -\frac{1}{\varepsilon _{k}} \int _{t_{k}}^{\theta _{k}}\mathcal{L}\varphi (u,X^{k}_{u},Y^{k}_{u}) \,\mathrm{d}u}\bigg] = -\mathcal{L}\varphi (\bar{t},\bar{x},\bar{y}). $$

Combining this with (B.5) yields

$$ \min \{\mathcal{L}\varphi (\bar{t},\bar{x},\bar{y}), \mathrm{V}_{ \circ }(\bar{t},\bar{x},\bar{y}) - w_{\circ }(\bar{t},\bar{x},\bar{y}) \} \geq 0, $$

and the proof is complete. □