Dynamic Cournot Models for Production of Exhaustible Commodities Under Stochastic Demand

Abstract

We extend the dynamic Cournot model of Ludkovski and Sircar (2011) by considering stochastic demand. We analyze a duopoly between an exhaustible producer and a “green” competitor. Both producers dynamically make decisions regarding their production rates; in addition the exhaustible producer optimizes search for new reserves. The aggregate price earned by the producers switches between high and low demand regimes with exogenously given holding rates. We study how the regime changes and the relative cost of production, which is a proxy for market competitiveness, affect game equilibria, and compare with the case of deterministic demand. A novel feature driven by stochasticity of demand is that production may shut down during low demand to conserve reserves.

Appendix

1.1 Proof of Lemma 1

Proof.

To derive the asymptotic game functions as $\lambda _{01} \rightarrow +\infty$, we set $\lambda _{01} = \frac{1} {\epsilon }$. Without loss of generality, we assume that the asymptotic expansion of v _M (x) with respect to ε is

$$\displaystyle{ v_{i}^{\epsilon } = v_{ i}^{0} + g(\epsilon )v_{ i}^{1} + o(g(\epsilon )),\qquad i = L,H }$$

(25)

where $g(\epsilon ) \rightarrow 0$, as $\epsilon \rightarrow 0$.

We substitute the v _L and v _H in the HJB ODEs with their asymptotic expansion to obtain

$$\displaystyle\begin{array}{rcl} & & \left [\frac{2} {3}\left (\frac{L + c} {2} - (v_{L}^{\epsilon })^{{\prime}}(x)\right )^{+} -\frac{1} {6}\left (2c - L - (v_{L}^{\epsilon })^{{\prime}}(x)\right )^{+}\right ]^{2} + \frac{1} {\gamma } [(\lambda \varDelta v_{L}^{\epsilon }(x)-\kappa )^{+}]^{\gamma } \\ & & \quad + \frac{1} {\epsilon } \left (v_{H}^{0}(x) - v_{ L}^{0}(x)\right ) + \frac{1} {\epsilon } \left (g(\epsilon )[v_{H}^{1}(x) - v_{ L}^{1}(x)] + o(g(\epsilon ))\right ) - rv_{ L}^{\epsilon }(x) = 0,{}\end{array}$$

(26)

$$\displaystyle\begin{array}{rcl} & & \left [\frac{2} {3}\left (\frac{H + c} {2} - (v_{H}^{\epsilon })^{{\prime}}(x)\right )^{+} -\frac{1} {6}\left (2c - H - (v_{H}^{\epsilon })^{{\prime}}(x)\right )^{+}\right ]^{2} + \frac{1} {\gamma } [(\lambda \varDelta v_{H}^{\epsilon }(x)-\kappa )^{+}]^{\gamma } \\ & & \quad +\lambda _{10}\left (v_{L}^{0}(x) - v_{ H}^{0}(x)\right ) + g(\epsilon )\lambda _{ 10}(v_{L}^{1}(x) - v_{ H}^{1}(x)) - rv_{ H}^{\epsilon }(x) + o(g(\epsilon )) = 0.{}\end{array}$$

(27)

We must have that $\lim _{\epsilon \rightarrow 0}v_{L}^{\epsilon } =\lim _{\epsilon \rightarrow 0}v_{L}^{\epsilon }$, i.e. $v_{L}^{0} = v_{H}^{0} =:\tilde{ v}$, otherwise the term $\epsilon ^{-1}(v_{H}^{0} - v_{L}^{0})$ will explode as $\epsilon \rightarrow 0$. Making that simplification, multiplying (26) by ε λ ₁₀ and adding (27) we obtain

$$\displaystyle\begin{array}{rcl} 0& =& \epsilon \lambda _{10}\left [\frac{2} {3}\left (\frac{L + c} {2} - (v_{L}^{\epsilon })^{{\prime}}(x)\right )^{+} -\frac{1} {6}\left (2c - L - (v_{L}^{\epsilon })^{{\prime}}(x)\right )^{+}\right ]^{2} \\ & & +\left [\frac{2} {3}\left (\frac{H + c} {2} - (v_{H}^{\epsilon })^{{\prime}}(x)\right )^{+} -\frac{1} {6}\left (2c - H - (v_{H}^{\epsilon })^{{\prime}}(x)\right )^{+}\right ]^{2} \\ & & +\frac{\epsilon \lambda _{10}} {\gamma } [(\lambda \varDelta v_{L}^{\epsilon }(x)-\kappa )^{+}]^{\gamma } + \frac{1} {\gamma } [(\lambda \varDelta v_{H}^{\epsilon }(x)-\kappa )^{+}]^{\gamma } - r\left (\lambda _{ 10}\epsilon v_{L}^{\epsilon }(x) + v_{ H}^{\epsilon }(x)\right ).{}\end{array}$$

(28)

One can now take ε → 0 which reduces to a regular perturbation of the following ODE for $\tilde{v}(x)$ (note that the first term involving L vanishes):

$$\displaystyle\begin{array}{rcl} & & \left [\frac{2} {3}\left (\frac{H + c} {2} - (\tilde{v}_{H})^{{\prime}}(x)\right )^{+} -\frac{1} {6}\left (2c - H - (\tilde{v}_{H})^{{\prime}}(x)\right )^{+}\right ]^{2} {}\\ & & \quad + \frac{1} {\gamma } [(\lambda \varDelta \tilde{v}_{H}(x)-\kappa )^{+}]^{\gamma } - r\tilde{v}_{ H}(x) = 0, {}\\ \end{array}$$

which matches the solution of an exploration duopoly game studied in [13] with linear inverse demand p _t  = H − q _t ¹ − q _t ².

For the boundary conditions, we re-write (6) as

$$\displaystyle{ \left (v_{H}^{\epsilon }(0) - v_{ L}^{\epsilon }(0)\right )\left (\frac{1} {\epsilon } \right ) + v_{L}^{\epsilon }(\delta )\lambda a_{ L}^{\epsilon }(0) - C(a_{ L}^{\epsilon }(0)) -\left (r +\lambda a_{ L}^{\epsilon }(0)\right )v_{ L}^{\epsilon }(0) = 0, }$$

(29)

$$\displaystyle{ \left (v_{L}^{\epsilon }(0) - v_{ H}^{\epsilon }(0)\right )\lambda _{ 10} + v_{H}^{\epsilon }(\delta )\lambda a_{ H}^{\epsilon }(0) - C(a_{ H}^{\epsilon }(0)) -\left (r +\lambda a_{ H}^{\epsilon }(0)\right )v_{ H}^{\epsilon }(0) = 0. }$$

(30)

We multiply (29) by ε λ ₁₀ and add to (30) to obtain

$$\displaystyle\begin{array}{rcl} & & \epsilon \lambda _{10}\left [v_{L}^{\epsilon }(\delta )\lambda a_{ L}^{\epsilon }(0) - C(a_{ L}^{\epsilon }(0)) -\left (r +\lambda a_{ L}^{\epsilon }(0)\right )v_{ L}^{\epsilon }(0)\right ] \\ & & \quad + \left [v_{H}^{\epsilon }(\delta )\lambda a_{ H}^{\epsilon }(0) - C(a_{ H}^{\epsilon }(0)) -\left (r +\lambda a_{ H}^{\epsilon }(0)\right )v_{ H}^{\epsilon }(0)\right ] = 0.{}\end{array}$$

(31)

Letting $\epsilon \rightarrow 0$ removes the first terms and we are left with

$$\displaystyle{ \tilde{v}(\delta )\lambda \tilde{a}(0) - C(\tilde{a}(0)) -\left (r +\lambda \tilde{ a}(0)\right )\tilde{v}(0) = 0, }$$

which is equivalent to

$$\displaystyle{ \tilde{v}(0) = \frac{\tilde{v}(\delta )\lambda \tilde{a}(0) - C(\tilde{a}(0))} {\left (r +\lambda \tilde{ a}(0)\right )} =\sup _{a}\frac{\tilde{v}(\delta )\lambda a - C(a)} {\left (r +\lambda a\right )}, }$$

(32)

again matching the corresponding boundary condition in the deterministic demand setting.

1.2 Proof of Lemma 2

Proof.

We set $\lambda _{01} = \frac{b_{L}} {\epsilon },\lambda _{10} = \frac{b_{H}} {\epsilon }$, where b _L , b _H are some constants, and ε > 0 can be arbitrarily small. The stationary distribution $\boldsymbol{\pi }$ given by $\pi _{L} = \frac{\lambda _{10}} {\lambda _{01}+\lambda _{10}} = \frac{b_{H}} {b_{L}+b_{H}}$, $\pi _{H} = \frac{b_{L}} {b_{L}+b_{H}}$ is unchanged as $\epsilon \rightarrow 0$.

We consider the asymptotic expansions of v _M in terms of ε:

$$\displaystyle{ v_{i}^{\epsilon } = v_{ i}^{0} + f(\epsilon )v_{ i}^{1} + o(\,f(\epsilon )),\qquad i = L,H, }$$

(33)

where $f(\epsilon ) \rightarrow 0$, as $\epsilon \rightarrow 0$. Substituting (33) into (15) yields

$$\displaystyle\begin{array}{rcl} & & \left [\frac{2} {3}\left (\frac{L + c} {2} - (v_{L}^{\epsilon })^{{\prime}}(x)\right )^{+} -\frac{1} {6}\left (2c - L - (v_{L}^{\epsilon })^{{\prime}}(x)\right )^{+}\right ]^{2} + \frac{1} {\gamma } [(\lambda \varDelta v_{L}^{\epsilon }(x)-\kappa )^{+}]^{\gamma } \\ & & \quad + \frac{b_{L}} {\epsilon } \left (v_{H}^{0}(x) + f(\epsilon )v_{ H}^{1}(x) - v_{ L}^{0}(x) - f(\epsilon )v_{ L}^{1}(x)\right ) - rv_{ L}^{\epsilon }(x) + o(\,f(\epsilon )) = 0;{}\end{array}$$

(34)

$$\displaystyle\begin{array}{rcl} & & \left [\frac{2} {3}\left (\frac{H + c} {2} - (v_{H}^{\epsilon })^{{\prime}}(x)\right )^{+} -\frac{1} {6}\left (2c - H - (v_{H}^{\epsilon })^{{\prime}}(x)\right )^{+}\right ]^{2} + \frac{1} {\gamma } [(\lambda \varDelta v_{H}^{\epsilon }(x)-\kappa )^{+}]^{\gamma } \\ & & \quad + \frac{b_{H}} {\epsilon } \left (v_{L}^{0}(x) + f(\epsilon )v_{ L}^{1}(x) - v_{ H}^{0}(x) - f(\epsilon )v_{ H}^{1}(x)\right ) - rv_{ H}^{\epsilon }(x) + o(\,f(\epsilon )) = 0.{}\end{array}$$

(35)

We must have that $\lim _{\epsilon \rightarrow 0}v_{L}^{\epsilon } =\lim _{\epsilon \rightarrow 0}v_{L}^{\epsilon }$, i.e. $v_{L}^{0} = v_{H}^{0} =\bar{ v}$, otherwise the terms $\frac{b_{L}} {\epsilon } \left (v_{H}^{\epsilon } - v_{L}^{\epsilon }\right )$ and $\frac{b_{H}} {\epsilon } \left (v_{L}^{\epsilon } - v_{H}^{\epsilon }\right )$ above would explode as $\epsilon \rightarrow 0$. Indeed, it is clear that | v _L (x) − v _H (x) | → 0 as ε → 0 due to the fast switching of the regimes, making the initial macroeconomic conditions irrelevant.

Canceling the terms $v_{L}^{0} - v_{H}^{0} \equiv 0$ in (34)–(35), multiplying (34) by b _H ∕(b _L + b _H ), (35) by $b_{L}/(b_{L} + b_{H})$, and adding them up we obtain

$$\displaystyle\begin{array}{rcl} 0& =& \pi _{L}\left [\frac{2} {3}\left (\frac{L + c} {2} - (v_{L}^{\epsilon })^{{\prime}}(x)\right )^{+} -\frac{1} {6}\left (2c - L - (v_{L}^{\epsilon })^{{\prime}}(x)\right )^{+}\right ]^{2} {}\\ & & \quad +\pi _{H}\left [\frac{2} {3}\left (\frac{H + c} {2} - (v_{H}^{\epsilon })^{{\prime}}(x)\right )^{+} -\frac{1} {6}\left (2c - H - (v_{H}^{\epsilon })^{{\prime}}(x)\right )^{+}\right ]^{2} {}\\ & & \quad + \frac{\pi _{L}} {\gamma } [(\lambda \varDelta v_{L}^{\epsilon }(x)-\kappa )^{+}]^{\gamma } + \frac{\pi _{H}} {\gamma } [(\lambda \varDelta v_{H}^{\epsilon }(x)-\kappa )^{+}]^{\gamma } {}\\ & & \quad - r\left (\pi _{L}v_{L}^{\epsilon }(x) +\pi _{ H}v_{H}^{\epsilon }(x)\right ) + o(\,f(\epsilon )). {}\\ \end{array}$$

Note that all the terms involving ε ⁻¹ have cancelled out. Once again plugging in (33) we can now take ε → 0 since this just amounts to a regular perturbation; the result is precisely (21).

For the boundary conditions, we re-write the original

$$\displaystyle\begin{array}{rcl} v_{i}^{\epsilon }(0)& =& \frac{v_{j}^{\epsilon }(0)(\frac{b_{i}} {\epsilon } ) + v_{i}^{\epsilon }(\delta )\lambda a_{i}^{\epsilon }(0) - C(a_{i}^{\epsilon }(0))} {r + \frac{b_{i}} {\epsilon } +\lambda a_{i}^{\epsilon }(0)},\quad i,j = L,H {}\\ \end{array}$$

as

$$\displaystyle{ \left (v_{H}^{\epsilon }(0) - v_{ L}^{\epsilon }(0)\right )\left (\frac{b_{L}} {\epsilon } \right ) + v_{L}^{\epsilon }(\delta )\lambda a_{ L}^{\epsilon }(0) - C(a_{ L}^{\epsilon }(0)) -\left (r +\lambda a_{ L}^{\epsilon }(0)\right )v_{ L}^{\epsilon }(0) = 0, }$$

(36)

$$\displaystyle{ \left (v_{L}^{\epsilon }(0) - v_{ H}^{\epsilon }(0)\right )\left (\frac{b_{H}} {\epsilon } \right ) + v_{H}^{\epsilon }(\delta )\lambda a_{ H}^{\epsilon }(0) - C(a_{ H}^{\epsilon }(0)) -\left (r +\lambda a_{ H}^{\epsilon }(0)\right )v_{ H}^{\epsilon }(0) = 0. }$$

(37)

Once again multiplying (36) by $b_{H}/(b_{L} + b_{H})$ and (37) by $b_{L}/(b_{L} + b_{H})$, and summing produces

$$\displaystyle\begin{array}{rcl} & & \pi _{L}v_{L}^{\epsilon }(\delta )\lambda a_{ L}^{\epsilon }(0) +\pi _{ H}v_{H}^{\epsilon }(\delta )\lambda a_{ H}^{\epsilon }(0) -\pi _{ L}C(a_{L}^{\epsilon }(0)) -\pi _{ H}C(a_{H}^{\epsilon }(0)) \\ & & \quad -\pi _{L}\left (r +\lambda a_{L}^{\epsilon }(0)\right )v_{ L}^{\epsilon }(0) -\pi _{ H}\left (r +\lambda a_{H}^{\epsilon }(0)\right )v_{ H}^{\epsilon }(0) = 0. {}\end{array}$$

(38)

As ε → 0, $a_{M}^{\epsilon }(0) = [(\lambda \varDelta v_{M}^{\epsilon }(0)-\kappa )^{+}]^{\gamma -1} \rightarrow [(\lambda \varDelta \bar{v}(0)-\kappa )^{+}]^{\gamma -1} =\bar{ a}(0),$ and we find $\bar{v}(\delta )\lambda \bar{a}(0) - C(\bar{a}(0)) -\left (r +\lambda \bar{ a}(0)\right )\bar{v}(0) = 0$, which is equivalent to (22).