An Optimal Extraction Problem with Price Impact

Abstract

A price-maker company extracts an exhaustible commodity from a reservoir, and sells it in the spot market. In absence of any actions of the company, the commodity’s spot price evolves as an Ornstein–Uhlenbeck process. While extracting, the company’s actions have an impact on the commodity’s spot price. The company aims at maximizing the total expected profits from selling the commodity, net of the total expected proportional costs of extraction. We model this problem as a two-dimensional degenerate singular stochastic control problem with finite fuel. The optimal extraction rule is triggered by a strictly decreasing smooth curve that depends on the current level of the reservoir, and for which we provide an explicit expression. Finally, our study is complemented by a theoretical and numerical analysis of the dependency of the optimal extraction strategy and value function on the model’s parameters.

Appendices

Appendix A: Proofs of Results from Sects. 4 and 5

Proof of Lemma 4.1

• (1) We refer the reader to [20], among others. Moreover, the strict convexity of $\psi$ can be checked by direct calculations on (4.10).

• (2) Define the function $f:{\mathbb {R}}_+\times {\mathbb {R}}\rightarrow {\mathbb {R}}_+$ by

  $$\begin{aligned} f(t,x)=\frac{1}{\Gamma \big (\frac{\rho }{b}\big )} t^{\big (\frac{\rho }{b}-1\big )}e^{-\frac{t^2}{2}+t\big (\frac{bx-a}{\sigma b}\big )\sqrt{2b}}, \end{aligned}$$

  that, once differentiated with respect to x, yields

  $$\begin{aligned} f_x(t,x)=\frac{\rho \sqrt{2b}}{b\sigma }\frac{1}{\Gamma \big (\frac{\rho +b}{b}\big )} t^{\big (\frac{\rho +b}{b}-1\big )}e^{-\frac{t^2}{2}+t\big (\frac{bx-a}{\sigma b}\big )\sqrt{2b}}. \end{aligned}$$

  Notice that f is the integrand appearing in (4.11) for $\beta =-\frac{\rho }{b}$. Then, differentiating (4.10) with respect to x, and invoking the dominated convergence theorem, we obtain

  $$\begin{aligned} \psi '(x)\, \propto \, e^{\frac{(bx-a)^2}{2\sigma ^2b}}D_{-\frac{\rho +b}{b}}\bigg (-\frac{bx-a}{\sigma b}\sqrt{2b}\bigg ), \end{aligned}$$

  upon noticing that $f_x(t,x)$ is the integrand of $D_{-\frac{\rho +b}{b}}\bigg (-\frac{bx-a}{\sigma b}\sqrt{2b}\bigg )$ (cf. (4.11)).

  Hence, $\psi '$ can be identified (modulo a constant) as the positive strictly increasing fundamental solution to $({{\mathcal {L}}}-(\rho +b))u=0$, and by direct calculations it can be checked that it is strictly convex. By iterating the previous argument, we see that, for any $k\in {\mathbb {N}}$, the function $\psi ^{(k)}$ is strictly convex and identifies with the positive strictly increasing fundamental solution to $({{\mathcal {L}}}-(\rho +kb))u=0$.

• (3) We define the function $f^{(k)}:{\mathbb {R}}_+\times {\mathbb {R}}\rightarrow {\mathbb {R}}_+$ by

  $$\begin{aligned} f^{(k)}(t,x)=\frac{\big (\sqrt{{2b}}/{\sigma }\big )^\frac{k}{2}}{\Gamma (\frac{\rho }{b})^{\frac{1}{2}}}t^{\frac{1}{2}\big (\frac{\rho }{b}+k-1\big )}e^{-\frac{t^2}{4}+\frac{t}{2}\big (\frac{bx-a}{\sigma b}\big )\sqrt{2b}}. \end{aligned}$$

  By direct calculations, we find

  $$\begin{aligned} \psi ^{(k+1)}(x)=\int _{0}^{\infty }f^{(k+2)}(t,x)f^{(k)}(t,x)dt,\quad x\in {\mathbb {R}}, \end{aligned}$$

  that, by the help of Hölder’s inequality (which is strict as $f^{(k)}(\cdot ,x)$ is not a multiple of $f^{(k+2)}(\cdot ,x)$), gives

  $$\begin{aligned} \Bigg (\int _{0}^{\infty }f^{(k+2)}(t,x)f^{(k)}(t,x)dt\Bigg )^2<\int _{0}^{\infty }\big (f^{(k+2)}(t,x)\big )^2 dt\int _{0}^{\infty }\big (f^{(k)}(t,x)\big )^2dt. \end{aligned}$$

  The latter is in fact equivalent to

  $$\begin{aligned} \psi ^{(k+2)}(x)\psi ^{(k)}(x)-\psi ^{(k+1)}(x)^2>0. \end{aligned}$$

$\square$

Proof of Lemma 4.2

Let $k\in {\mathbb {N}} \cup \{0\}$ be given and fixed, and define $\Lambda (x):=(x-c){\psi ^{(k+1)}(x)}-\psi ^{(k)}(x)$, $x\in {\mathbb {R}}$. We then have the following.

• (i) For $x\le c$, it is readily seen that $\Lambda (x)<0$.

• (ii) One has $\Lambda (x)>0$ for all $x>c+\frac{\psi ^{(k)}(c)}{\psi ^{(k+1)}(c)}$. To see this, rewrite $\Lambda (x)=\psi ^{(k)}(x)\bigg [(x-c)\frac{\psi ^{(k+1)}(x)}{\psi ^{(k)}(x)}-1\bigg ]$, and notice that by Lemma 4.1

  $$\begin{aligned} \left( \frac{\psi ^{(k+1)}(x)}{\psi ^{(k)}(x)}\right) '=\frac{\psi ^{(k+2)}(x) \psi ^{(k)}(x)-(\psi ^{(k+1)}(x))^2}{\big (\psi ^{(k)}(x)\big )^2}>0. \end{aligned}$$

  Hence, for all $x>c+\frac{\psi ^{(k)}(c)}{\psi ^{(k+1)}(c)}>c$ one has that $\frac{\psi ^{(k+1)}(x)}{\psi ^{(k)}(x)}>\frac{\psi ^{(k+1)}(c)}{\psi ^{(k)}(c)}$, which implies

  $$\begin{aligned} (x-c)\frac{\psi ^{(k+1)}(x)}{\psi ^{(k)}(x)}-1>(x-c)\frac{\psi ^{(k+1)}(c)}{\psi ^{(k)}(c)}-1>0, \end{aligned}$$

  for all $x>c+\frac{\psi ^{(k)}(c)}{\psi ^{(k+1)}(c)}$. The latter clearly gives $\Lambda (x)>0$ for all $x>c+\frac{\psi ^{(k)}(c)}{\psi ^{(k+1)}(c)}$.

Since $\Lambda '(x)=(x-c)\psi ^{(k+2)}(x)>0$ for all $x>c$, we conclude from (i) and (ii) that there exists a unique solution on $(c,\infty )$ to the equation $\Lambda (x)=0$ by continuity of $\Lambda$. $\square$

Proof of Lemma 4.3

We argue by contradiction, and we suppose $x_{\infty }\ge x_0$. Then by definition of $x_0$ and $x_\infty$ we have

$$\begin{aligned} x_0-x_\infty =(x_0-c)-(x_\infty -c)=\frac{\psi (x_0)}{\psi ^\prime (x_0)} -\frac{\psi ^\prime (x_\infty )}{\psi ^{\prime \prime }(x_\infty )}. \end{aligned}$$

(A-1)

Since by Lemma 4.1

$$\begin{aligned} \Big (\frac{\psi (x)}{\psi ^\prime (x)}\Big )^\prime =\frac{\psi ^\prime (x)^2 -\psi (x)\psi ^{\prime \prime }(x)}{\psi ^\prime (x)^2}<0,\quad \text {for any }x\in {\mathbb {R}}, \end{aligned}$$

we have by (A-1) that

$$\begin{aligned} x_0-x_\infty \ge \frac{\psi (x_\infty )}{\psi ^\prime (x_\infty )} -\frac{\psi ^\prime (x_\infty )}{\psi ^{\prime \prime }(x_\infty )}>0, \end{aligned}$$

again due to Lemma 4.1. But this contradicts $x_{\infty }\ge x_0$. $\square$

Proof of Lemma 4.7

First of all notice that for the existence of a solution z to (4.20) it is necessary that $y-z\ge 0$ since $F\ge 0$, and that $x-\alpha z\in (x_\infty ,x_0]$ since the domain of F is $(x_\infty ,x_0]$. Hence, if a solution to (4.20) exists, it must be such that $z(x,y) \in (\frac{x-x_0}{\alpha },\frac{x-x_\infty }{\alpha } \wedge y]$, for all $(x,y)\in {\mathbb {S}}_2$.

Let $(x,y)\in {\mathbb {S}}_2$ with $y>F(x)$ be given and fixed, and define $R(z)=y-z-F(x-\alpha z)$, for $z\in (\frac{x-x_0}{\alpha },\frac{x-x_\infty }{\alpha } \wedge y)$. Then, one has $R(0)=y-F(x)>0$ and $\lim \nolimits _{z\uparrow \left( \frac{x-x_\infty }{\alpha } \wedge y\right) }R(z)<0$. Since $z\mapsto R(z)$ is strictly decreasing (by strict monotonicity of F) it follows that there exists a unique solution to (4.20).

Finally, (4.21) follows by noticing that 0 solves (4.20) when $y=F(x)$ and by uniqueness of the solution. Analogously, (4.22) follows by noticing that $\frac{x-x_0}{\alpha }$ uniquely solves (4.20), since $F(x_0)=0$. $\square$

Proof of Lemma 5.2

The first equality in (5.4) follows from (5.2). In order to prove the last inequality in (5.4), we find by Lemma 4.1-(2) that

$$\begin{aligned} \frac{\sigma ^2}{2}\psi ^{(k+2)}(x;a,\sigma )+(a-bx)\psi ^{(k+1)}(x;a,\sigma ) -(\rho +kb)\psi ^{(k)}(x;a,\sigma )=0. \end{aligned}$$

(A-2)

From (A-2), recalling that $\psi ^{(k+1)}>0$, we obtain

$$\begin{aligned} (a-bx)=-\frac{\sigma ^2\psi ^{(k+2)}(x;a,\sigma )}{2\psi ^{(k+1)}(x;a,\sigma )} +(\rho +kb)\frac{\psi ^{(k)}(x;a,\sigma )}{\psi ^{(k+1)}(x;a,\sigma )}. \end{aligned}$$

and we thus have

$$\begin{aligned}&(a-bx)\Big [\psi ^{(k+1)}(x;a,\sigma )^2-\psi ^{(k)}(x;a,\sigma ) \psi ^{(k+2)}(x;a,\sigma )\Big ]+b\psi ^{(k+1)}(x;a,\sigma )\psi ^{(k)}(x;a,\sigma )\\ =\,&(\rho +(k+1)b)\psi ^{(k)}(x;a,\sigma )\psi ^{(k+1)}(x;a,\sigma ) -(\rho +kb)\psi ^{(k)}(x;a,\sigma )^2\frac{\psi ^{(k+2)}(x;a,\sigma )}{\psi ^{(k+1)} (x;a,\sigma )}\\&\quad + \underbrace{\frac{\sigma ^2\psi ^{(k+2)}(x;a,\sigma )}{2\psi ^{(k+1)}(x;a,\sigma )}\Big [\psi ^{(k)}(x;a,\sigma )\psi ^{(k+2)}(x,a,\sigma ) -\psi ^{(k+1)}(x;a,\sigma )^2\Big ]}_{>0\,\text { by Lemma}~4.1}\\ >&\frac{\psi ^{(k)}(x;a,\sigma )}{\psi ^{(k+1)}(x;a,\sigma )}\Big [(\rho +(k+1)b) \psi ^{(k+1)}(x;a,\sigma )^2-(\rho +kb)\psi ^{(k)}(x,a,\sigma )\psi ^{(k+2)}(x;a,\sigma ) \Big ]. \end{aligned}$$

We now aim at establishing that the last term on the right-hand side of the latter equation is positive. With regard to (5.4), this would clearly imply that $\frac{\partial (\psi ^{(k)}(x;a,\sigma )/\psi ^{(k+1)}(x;a,\sigma ))}{\partial \sigma }>0$. From (A-2) we have

$$\begin{aligned} (\rho +(k+1)b)\psi ^{(k+1)}(x;a,\sigma )=\frac{\sigma ^2}{2}\psi ^{(k+3)}(x;a,\sigma ) +(a-bx)\psi ^{(k+2)}(x;a,\sigma ), \end{aligned}$$

which then yields

$$\begin{aligned}&\frac{\psi ^{(k)}(x;a,\sigma )}{\psi ^{(k+1)}(x;a,\sigma )} \Big [(\rho +(k+1)b)\psi ^{(k+1)}(x;a,\sigma )^2-(\rho +kb) \psi ^{(k)}(x;a,\sigma )\psi ^{(k+2)}(x;a,\sigma )\Big ]\\ =&\frac{\psi ^{(k)}(x;\sigma )}{\psi ^{(k+1)}(x;a,\sigma )} \Big [\frac{\sigma ^2}{2}\psi ^{(k+3)}(x;a,\sigma )\psi ^{(k+1)}(x;a,\sigma )\\&+\psi ^{(k+2)}(x;a,\sigma )\big ((a-bx)\psi ^{(k+1)}(x;a,\sigma ) -(\rho +kb)\psi ^{(k)}(x;a,\sigma )\big )\Big ]\\ =&\frac{\sigma ^2}{2}\frac{\psi ^{(k)}(x;\sigma )}{\psi ^{(k+1)} (x;a,\sigma )}\Big [\psi ^{(k+3)}(x;a,\sigma )\psi ^{(k+1)}(x;a,\sigma ) -\psi ^{(k+2)}(x;a,\sigma )^2\Big ] > 0, \end{aligned}$$

where the last equality follows again by an application of (A-2), and the last inequality by Lemma 4.1. Hence $\frac{\partial (\psi ^{(k)}(x;a,\sigma )/\psi ^{(k+1)}(x;a,\sigma ))}{\partial \sigma }>0$ and the proof is completed. $\square$

Appendix B: An Auxiliary Result

Lemma B.1

Let $x_0$ be the solution to (4.13) and

$$\begin{aligned} {\bar{x}}:=\frac{a + \rho c}{\rho + b}. \end{aligned}$$

(B-1)

We have

$$\begin{aligned} {\bar{x}} < x_0. \end{aligned}$$

Proof

Define $H(x):=(x-c)\psi ^\prime (x)-\psi (x),$ $x\in {\mathbb {R}}$. Since $\psi$ satisfies

$$\begin{aligned} \frac{\sigma ^2}{2}\psi ''(x)+(a-bx)\psi '(x)-\rho \psi (x)=0,\quad \hbox { for all}\ x\in {\mathbb {R}}, \end{aligned}$$

and $\frac{\sigma ^2}{2}\psi ''(x)>0$, we find $-\psi (x)<-\frac{(a-bx)}{\rho }\psi ^\prime (x),\,\forall x\in {\mathbb {R}}.$ Thus, we have

$$\begin{aligned} H({\bar{x}})<({\bar{x}}-c)\psi '({\bar{x}})-\frac{(a-b{\bar{x}})}{\rho } \psi '({\bar{x}})=\Big [({\bar{x}}-c)\rho -(a-b{\bar{x}})\Big ] \frac{\psi ^\prime ({\bar{x}})}{\rho }=0, \end{aligned}$$

by the definition of ${\bar{x}}$. Since $H(x_0)=0$, $H(x)<0$ for all $x<x_0$ and $H(x)>0$ for all $x>x_0$, it must necessarily be ${\bar{x}}< x_0$. $\square$