Irreversible investment with fixed adjustment costs: a stochastic impulse control approach

Abstract

We consider an optimal stochastic impulse control problem over an infinite time horizon motivated by a model of irreversible investment choices with fixed adjustment costs. By employing techniques of viscosity solutions and relying on semiconvexity arguments, we prove that the value function is a classical solution to the associated quasi-variational inequality. This enables us to characterize the structure of the continuation and action regions and construct an optimal control. Finally, we focus on the linear case, discussing, by a numerical analysis, the sensitivity of the solution with respect to the relevant parameters of the problem.

A Appendix

Proposition A.1

Under Assumption 2.1 the boundaries 0 and $+\infty$ are natural in the sense of Feller’s classification for the diffusion $Z^{0,x}$.

Proof

Clearly $+\infty$ is not accessible, in the sense that $Z^{0,x}$ does not explode in finite time. It remains to show that 0 is not accessible, that is

$$\begin{aligned} x\in \mathbb {R}_{++} \ \Longrightarrow \ Z^{0,x}_{t}>0 \quad \mathbb {P}\text{-a.s. }\ \forall t\ge 0; \end{aligned}$$

(A.1)

that both 0 and $+\infty$ are not entrance, that is

$$\begin{aligned} \lim _{x\downarrow 0}\mathbb {P}\{\tau _{x,y}<t\}=0, \quad \lim _{x\uparrow \infty }\mathbb {P}\{\tau _{x,y}<t\}=0, \quad \forall t,y\in \mathbb {R}_{++}. \end{aligned}$$

(A.2)

To this end, we introduce the speed measure m of the diffusion $Z^{0,x}$ transformed to natural scale (see [19, Prop. 16.81, Th. 16.83]). Up to a multiplicative constant, we have

$$\begin{aligned} m(dy)= \frac{2}{\sigma ^2(y)} e^{\int _1^y\frac{2b(\xi )}{\sigma ^2(\xi )}d\xi } dy, \quad y\in \mathbb {R}_{++}. \end{aligned}$$

Assumption 2.1 implies that for some $C_0,C_1>0$ we have $|b(\xi )|\le C_0 \xi$ and $\sigma ^2(\xi )\le C_1\xi ^2$ for every $\xi \in \mathbb {R}_+$. According to [19, Prop. 16.43] we compute $\int _0^1 ym(dy)$. We have

$$\begin{aligned} \int _0^1 ym(dy)\ge \int _0^1 \frac{2y}{\sigma ^2(y)} e^{\int _1^y\frac{-2C_0\xi }{\sigma ^2(\xi )}d\xi } dy. \end{aligned}$$

Set $F(y):=\int _1^y\frac{-2C_0\xi }{\sigma ^2(\xi )}d\xi$. We have

$$\begin{aligned} \int _0^1 \frac{2y}{\sigma ^2(y)} e^{\int _1^y\frac{-2C_0\xi }{\sigma ^2(\xi )}d\xi } dy= & {} -\,\frac{1}{C_0} \int _0^1 F^{\prime }(y) e^{F(y)} dy= -\frac{1}{C_0}\left[ e^{F(1)}-\lim _{y\rightarrow 0^+}e^{F(y)}\right] \\= & {} -\,\frac{1}{C_0}\left[ 1-\lim _{y\rightarrow 0^+}e^{\int _1^y -\frac{2C_0\xi }{\sigma ^2(\xi )}d\xi }\right] \\= & {} -\,\frac{1}{C_0}\left[ 1-e^{\lim _{y\rightarrow 0^+}\int _y^1 \frac{2C_0}{C_1\xi }d\xi }\right] =+\infty . \end{aligned}$$

This shows, by [19, Prop. 16.43], that (A.1) holds, The fact that 0 is not-entrance, i.e. that the first limit in (A.2) holds, is then consequence of [19, Prop. 16.45(a)]. Let us show, finally, that also $+\infty$ is not-entrance, i.e. that the second limit in (A.2) holds. In this case, according to [19, Prop. 16.45(b)] we consider $\int _1^{+\infty } ym(dy)$ and see, with the same computations as above, that it is equal to $+\infty$. By the aforementioned result we conclude that $+\infty$ is not entrance. $\square$

Remark A.2

The property (A.1) can be generalized to the case of random initial data. Let $\tau$ be a (possibly infinite) $\mathbb {F}$-stopping time and let $\xi$ be an $\mathscr {F}_\tau$-measurable random variable, clearly we have the equality in law $Z^{\tau ,\xi }_{t+\tau }= \left( Z^{0,x}_t \right) _{|_{x=\xi }}$. By (A.1), it then follows that

$$\begin{aligned} \xi \ \mathscr {F}_\tau \text{-measurable } \text{ random } \text{ variable } , \xi>0\ \mathbb {P}\text{-a.s. } \ \Longrightarrow \ Z^{\tau ,\xi }_{t+\tau }>0 \ \mathbb {P}\text{-a.s. } \text{ on } \{\tau <\infty \},\ \forall t\ge 0. \end{aligned}$$

(A.3)

Lemma A.3

Let $I\in \mathscr {I}$, $x,y\in \mathbb {R}_{++}$.

• (i) We have

  $$\begin{aligned} \mathbb {E} \left[ |X^{x,I}_s-X^{y,I}_s|^4 \right] \le |x-y|^4 e^{C_0 t}\quad \forall t \ge 0, \end{aligned}$$

  (A.4)

  where $C_0 {:}{=}4L_b+6L_\sigma ^2.$

• (ii) For each $\lambda \in [0,1]$ and $x,y\in \mathbb {R}_{++}$, define $z_\lambda {:}{=}\lambda x+(1-\lambda )y$. Then

  $$\begin{aligned} \mathbb {E} \left[ \left| X^{z_\lambda ,I}_t - \lambda X^{x,I}_t - (1- \lambda ) X^{y,I}_t\right| ^2 \right] \le A_0\lambda ^2(1-\lambda )^2 |x-y|^{4} e^{B_0t} \quad \forall \lambda \in [0,1], \ \forall t\ge 0, \end{aligned}$$

  (A.5)

  where $A_0>0$ and $B_0 {:}{=}2L_b+2L_\sigma ^2+\tilde{L}_b$.

Proof

(i) We apply Itô’s formula to $|X^{x,I}-X^{y,I}|^4$ and then—after a standar localization procedure with stopping times to let the stochastic integral term be a martingale and all the other expectations be well defined and finite; see e.g. the proof of Proposition 3.2—we take the expectation. We get, also using Assumption 2.1,

$$\begin{aligned} \mathbb {E}\left[ \left| X^{x,I}_t-X^{y,I}_t\right| ^4\right]&= |x-y|^4+4\mathbb {E}\int _0^t \left( X^{x,I}_u-X^{y,I}_u\right) ^3\left( b\left( X^{x,I}_u\right) -b\left( X^{y,I}_u\right) \right) du\\&\quad + 6\mathbb {E}\int _0^t \left( X^{x,I}_u-X^{y,I}_u\right) ^2\left( \sigma \left( X^{x,I}_u\right) -\sigma \left( X^{y,I}_u\right) \right) ^2du\\&\le |x-y|^4+\left( 4L_b+6L_\sigma ^2\right) \int _0^t \mathbb {E}\left[ |X^{x,I}_u-X^{y,I}_u|^4\right] du. \end{aligned}$$

The claim follows by Gronwall’s inequality.

(ii) Define $\Sigma ^{\lambda ,x,y,I}{:}{=}\lambda X^{x,I} + (1- \lambda ) X^{y,I}$. We apply Itô’s formula to the process $(X^{z_\lambda ,I}-\Sigma ^{\lambda ,x,y,I})^2$ and then—after a standar localization procedure with stopping times to let the stochastic integral term be a martingale and all the other expectations are well defined and finite; see e.g. the proof of Proposition 3.2—take the expectation, obtaining, also using Assumption 2.1,

$$\begin{aligned}&\mathbb {E} \left[ \left( X^{z_\lambda ,I}_t-\Sigma ^{\lambda ,x,y,I}_t\right) ^2 \right] \nonumber \\&\quad = 2 \int _0^t \mathbb {E} \left[ \left( X^{z_\lambda ,I}_u-\Sigma ^{\lambda ,x,y,I}_u\right) \left( b\left( X^{z_\lambda ,I}_u\right) - \lambda b\left( X^{x,I}_u\right) - (1-\lambda ) b\left( X^{y,I}_u\right) \right) \right] du\nonumber \\&\qquad + \int _0^t \mathbb {E} \left[ \left( \sigma \left( X^{z_\lambda ,I}_u\right) - \lambda \sigma \left( X^{x,I}_u\right) - (1-\lambda ) \sigma \left( X^{y,I}_u\right) \right) ^2 \right] du\nonumber \\&\quad \le 2 \int _0^t \mathbb {E} \left[ \left| X^{z_\lambda ,I}_u-\Sigma ^{\lambda ,x,y,I}_u\right| \cdot \left| b\left( X^{z_\lambda ,I}_u\right) - b\left( \Sigma ^{\lambda ,x,y,I}_u\right) \right| \right] du\nonumber \\&\qquad + 2 \int _0^t \mathbb {E} \left[ \left| X^{z_\lambda ,I}_u-\Sigma ^{\lambda ,x,y,I}_u \right| \cdot \left| b\left( \Sigma ^{\lambda ,x,y,I}_u\right) - \lambda b\left( X^{t,\xi ,I}_u\right) - (1-\lambda ) b\left( X^{t,\xi ^{\prime },I}_u\right) \right| \right] du\nonumber \\&\qquad +2\int _0^t \mathbb {E} \left[ \left| \sigma \left( X^{z_\lambda ,I}_u\right) - \sigma \left( \Sigma ^{\lambda ,x,y,I}_u\right) \right| ^2 \right] du\nonumber \\&\qquad + 2\int _0^t \mathbb {E} \left[ \left| \sigma \left( \Sigma ^{\lambda ,x,y,I}_u\right) - \lambda \sigma \left( X^{x,I}_u\right) - (1-\lambda ) \sigma \left( X^{y,I}_u\right) \right| ^2 \right] du\nonumber \\&\quad \le 2\left( L_b+L_\sigma ^2 \right) \int _0^t \mathbb {E} \left[ \left| X^{z_\lambda ,I}_u-\Sigma ^{\lambda ,x,y,I}_u\right| ^2 \right] du\nonumber \\&\qquad + 2 \int _0^t \mathbb {E} \left[ \left| X^{z_\lambda ,I}_u-\Sigma ^{\lambda ,x,y,I}_u \right| \cdot \left| b\left( \Sigma ^{\lambda ,x,y,I}_u\right) - \lambda b\left( X^{t,\xi ,I}_u\right) - (1-\lambda ) b\left( X^{t,\xi ^{\prime },I}_u\right) \right| \right] du\nonumber \\&\qquad +2 \int _0^t \mathbb {E} \left[ \left| \sigma \left( \Sigma ^{\lambda ,x,y,I}_u\right) - \lambda \sigma \left( X^{x,I}_u\right) - (1-\lambda ) \sigma \left( X^{y,I}_u\right) \right| ^2 \right] du. \end{aligned}$$

(A.6)

By doing the same computations as in [72, p. 188] in order to obtain [72, p. 188, formulae (4.22) and (4.23)], we have

$$\begin{aligned} \left| b\left( \lambda x^{\prime }+(1-\lambda )x^{\prime \prime }\right) -\lambda b(x^{\prime }) -(1-\lambda ) b(x^{\prime \prime })\right|\le & {} \tilde{ L}_b\lambda (1-\lambda )|x^{\prime }-x^{\prime \prime }|^2 \quad \forall x^{\prime },x^{\prime \prime }\in \mathbb {R}_{++}, \nonumber \\\end{aligned}$$

(A.7)

$$\begin{aligned} \left| \sigma \left( \lambda x^{\prime }+(1-\lambda )x^{\prime \prime }\right) -\lambda \sigma (x^{\prime }) -(1-\lambda ) \sigma (x^{\prime \prime })\right|\le & {} \tilde{ L}_\sigma \lambda (1-\lambda )|x^{\prime }-x^{\prime \prime }|^2 \quad \forall x^{\prime },x^{\prime \prime }\in \mathbb {R}_{++}, \nonumber \\ \end{aligned}$$

(A.8)

where $\tilde{L}_b, \tilde{L}_\sigma$ are as in Assumption 2.1. Then, by using (A.7) and (A.8) in (A.6), we get

$$\begin{aligned} \mathbb {E} \left[ \left| X^{z_\lambda ,I}_s-\Sigma ^{\lambda ,x,y,I}_s\right| ^2 \right] \le&2\left( L_b+L_\sigma ^2 \right) \int _0^t \mathbb {E} \left[ \left| X^{z_\lambda ,I}_u-\Sigma ^{\lambda ,x,y,I}_u\right| ^2 \right] du\nonumber \\&+ 2\lambda (1-\lambda ) \tilde{ L}_b \int _0^t \mathbb {E} \left[ \left| X^{z_\lambda ,I}_u-\Sigma ^{\lambda ,x,y,I}_u\right| \cdot \left| X^{x,I}_u - X^{y,I}_u \right| ^2 \right] du\nonumber \\&+ 2 \lambda ^2(1-\lambda )^2 \tilde{ L}_\sigma ^2 \int _0^t \mathbb {E} \left[ \left| X^{x,I}_u -X^{y,I}_u \right| ^4 \right] du. \end{aligned}$$

(A.9)

Using the inequality

$$\begin{aligned} 2\lambda (1-\lambda ) ab \le a^2 + {\lambda ^2(1-\lambda )^2}b^{2} \quad \forall a,b\in \mathbb {R}, \end{aligned}$$

and (A.4) into (A.9), we obtain

$$\begin{aligned} \mathbb {E} \left[ \left| X^{z_\lambda ,I}_t-\Sigma ^{\lambda ,x,y,I}_t\right| ^2 \right] \le&\left( 2L_b+2L_\sigma ^2 +\tilde{L}_b\right) \int _0^t \mathbb {E} \left[ \left| X^{z_\lambda ,I}_u-\Sigma ^{\lambda ,x,y,I}_u\right| ^2 \right] du\\&+ \lambda ^2(1-\lambda )^2 \left( \tilde{L}_b+2\tilde{ L}_\sigma ^2\right) \int _0^t \mathbb {E} \left[ \left| X^{x,I}_u -X^{y,I}_u \right| ^4 \right] du\\ \le&\left( 2L_b+2L_\sigma ^2 +\tilde{L}_b\right) \int _0^t \mathbb {E} \left[ \left| X^{z_\lambda ,I}_u-\Sigma ^{\lambda ,x,y,I}_u\right| ^2 \right] du\\&+ (\tilde{L}_b+2\tilde{ L}_\sigma ^2) \lambda ^2(1-\lambda )^2 \int _0^t e^{C_0u} |x-y|^4 du\\ \le&\left( 2L_b+2L_\sigma ^2+\tilde{L}_b \right) \int _0^t \mathbb {E} \left[ \left| X^{z_\lambda ,I}_u-\Sigma ^{\lambda ,x,y,I}_u\right| ^2 \right] du\\&+ \frac{\tilde{L}_b+2\tilde{ L}_\sigma ^2}{C_0}\left( e^{C_0t}-1\right) \lambda ^2(1-\lambda )^2|x-y|^4, \end{aligned}$$

where $C_0$ is the constant of (A.4). We conclude by Gronwall’s inequality. $\square$