On a Nonrenewable Resource Extraction Game Played by Asymmetric Firms

Abstract

A differential game of extraction of a nonrenewable resource is taken into account, where two firms compete over time and their two terminal times of extraction are two different random variables. The winning firm will be the only one remaining in the game after the first one retires. We explicitly compute the Hamilton–Jacobi–Bellman equations of the model and solve them in an asymmetric game with logarithmic payoff structure and linear state dynamics.

Appendix

1.1 A.1 Proof of Proposition 3.1

Since T ₁ and T ₂ are independent random variables, the p.d.f. of the random vector (T ₁,T ₂) must be the product of their p.d.f., i.e., an expression of the kind f ₁(θ)f ₂(τ). We can note that

$$\begin{aligned} \mathbb{E} \bigl[\varPsi_1^i(T_1,T_2) \bigr] =&\int_0^\omega\int_0^\omega \int_0^\theta h_1^*(t)\,dt \mathbb{I}_{[\theta<\tau]}f_2(\tau)\,d \tau f_1(\theta) \,d \theta \\ &{}+\int_0^\omega\int_0^\omega \int_0^\tau h_1^*(t)\,dt \mathbb{I}_{[\theta>\tau]}f_1(\theta)\,d \theta f_2(\tau) \,d \tau. \end{aligned}$$

(19)

From now on, call $H_{i}(\theta):=\int_{0}^{\theta}h_{i}^{*}(t)\,dt$. Hence, (19) amounts to

$$ \int_0^\omega \biggl( \int_0^\tau H_1(\theta) f_1(\theta) \,d \theta \biggr)f_2( \tau)\,d \tau+ \int_0^\omega \biggl(\int _0^\theta H_1(\tau) f_2( \tau) \,d \tau \biggr) f_1(\theta)\,d \theta. $$

(20)

Integrating by parts twice and taking into account that F ₁(ω)=F ₂(ω)=1, we obtain that the sum (20) is

$$\begin{aligned} &\int_0^\tau H_1^*(\theta) f_1(\theta) \,d \theta F_2( \omega)-\int_0^\omega H_1(\theta) f_1(\theta) F_2(\theta)\,d \theta \\ &\qquad{}+\int_0^\theta H_1( \tau)f_2(\tau)\,d \tau F_1(\omega) - \int _0^\omega H_1(\tau)f_2( \tau)F_1(\tau) \,d \tau \\ &\quad=H_1(\omega)F_1(\omega)-\int_0^\omega h_1^*(\theta)F_1(\theta)\,d \theta - H_1( \omega) F_1(\omega) F_2(\omega) \\ &\qquad{}+\int_0^\omega F_1(\theta) \bigl[h_1^*(\theta)F_2(\theta)+H_1(\theta )f_2(\theta)\bigr]\,d \theta \\ &\qquad{}+H_1(\omega)F_2(\omega)-\int _0^\omega h_1^*(\tau)F_2( \tau)\,d \tau - H_1(\omega) F_1(\omega) F_2( \omega) \\ &\qquad{}+\int_0^\omega F_2(\tau) \bigl[h_1^*(\tau)F_1(\tau)+H_1(\tau) f_1(\tau )) \bigr]\,d \tau \\ &\quad=-\int_0^\omega h_1^*(\tau) \bigl[F_1(\tau)+F_2(\tau)-2 F_1(\tau )F_2(\tau)\bigr] \,d \tau+ \int_0^\omega H_1(\theta) \bigl[F_1(\tau) F_2(\tau ) \bigr]^\prime \,d \tau \\ &\quad=-\int_0^\omega h_1^*(\tau) \bigl[F_1(\tau)+F_2(\tau)-2 F_1( \tau)F_2(\tau)\bigr] \,d \tau \\ &\qquad{}+H_1(\omega) F_1(\omega) F_2( \omega)-\int_0^\omega h_1^*(\tau ) \bigl[F_1(\tau)F_2(\tau)\bigr] \,d \tau \\ &\quad\biggl(\text{and since } H_1(\omega)=\int_0^\omega h_1^*(\tau)\,d \tau\biggr) \\ &\quad=\int_0^\omega h_1^*(\tau) \,d \tau- \int_0^\omega h_1^*(\tau) \bigl[F_1(\tau )+F_2(\tau)- F_1( \tau)F_2(\tau)\bigr] \,d \tau \\ &\quad=\int_0^\omega h_1^*(\tau) \bigl[1-F_1(\tau)-F_2(\tau)+ F_1( \tau)F_2(\tau)\bigr] \,d \tau \\ &\quad=\int_0^\omega h_1^*(\tau) \bigl[1-F_1(\tau)\bigr] \bigl[1-F_2(\tau)\bigr] \,d \tau. \end{aligned}$$

(21)

Because T ₁ and T ₂ are independent, if we define T=min{T ₁, T ₂}, we can see that its p.d.f. F(t) is given by

$$\begin{aligned} F(t)&=1-\mathrm{Prob} \{T \geq t \}=1-\mathrm{Prob} \{T_1 \geq t \} \mathrm{Prob} \{T_2 \geq t \} \\ &=1-\bigl(1-F_1(t)\bigr) \bigl(1-F_2(t)\bigr) \quad \Longleftrightarrow\quad \bigl(1-F_1(t)\bigr) \bigl(1-F_2(t) \bigr)=1-F(t), \end{aligned}$$

hence (21) becomes

$$ \int_0^\omega h_1^*(\tau)\bigl[1-F( \tau)\bigr]\, d \tau=\mathbb{E} \biggl[ \int_0^{\min\{T_1, T_2\}} h_i^*(t) \,dt \biggr], $$

completing the proof.

1.2 A.2 Proof of Proposition 3.2

Integrating by parts and taking into account that F ₁(ω)=1, we have that

$$\begin{aligned} &\mathbb{E} \bigl[\varPhi_i\bigl(x^*(T)\bigr)\mathbb{I}_{[T_i>T_j]} \bigr] \\&\quad= \int_0^\omega \biggl( \int_0^\omega\varPhi_i\bigl(x^*(\tau)\bigr) \mathbb{I}_{[\theta >\tau]} f_2(\tau) \,d \tau \biggr) f_1(\theta) \,d \theta \\ &\quad=F_1(\omega) \int_0^\omega\varPhi_1\bigl(x^*(\tau)\bigr) f_2(\tau) \,d \tau- \int_0^\omega F_1(\theta) \varPhi_i\bigl(x^*(\theta)\bigr)f_2(\theta)\,d \theta \\ &\quad=\int_0^\omega\varPhi_1\bigl(x^*(\tau)\bigr) f_2(\tau) \,d \tau- \int_0^\omega F_1(\theta) \varPhi_i\bigl(x^*(\theta)\bigr)f_2(\theta)\,d \theta, \end{aligned}$$

(22)

then, by considering a unique variable for integration, we conclude that

$$ \mathbb{E} \bigl[\varPhi_i\bigl(x^*(T)\bigr)\mathbb{I}_{[T_i>T_j]} \bigr]=\int_0^\omega\varPhi_i \bigl(x^*(\tau)\bigr)f_2(\tau) \bigl(1-F_1(\tau)\bigr)\,d \tau. $$

1.3 A.3 Proof of Proposition 4.1

We just consider the Cauchy problem for A _i (t) because the explicit calculation of B _i (t) can be avoided in that B _i (t) does not appear in the expression of $u_{i}^{*}$:

$$ \begin{cases} \dot{A}_i(t)= A_i(t) [\lambda_i(t)+\lambda_j(t) ]- 1 - c_i \lambda_j(t), \\ \lim_{t \to\omega} A_i(t) = 0, \end{cases} $$

whose general solution is given by

$$\begin{aligned} A_i(t) = e^{\int_0^t [\lambda_i(\tau) + \lambda_j(\tau)] \,d \tau} \biggl(C - \int_0^t \bigl(1 + c_i \lambda_j(\tau)\bigr) e^{-\int_0^\tau [\lambda_i(s) + \lambda_j(s)] \,ds} \,d \tau \biggr), \end{aligned}$$

(23)

where the constant C is determined by employing the transversality condition on A _i (t)

$$C=\int_0^\omega\bigl(1 + c_i \lambda_j(\tau) \bigr) e^{-\int_0^\tau [\lambda_i(s) + \lambda_j(s)] \,ds} \,d \tau, $$

leading to the solution

$$\begin{aligned} A_i^*(t) = e^{\int_0^t (\lambda_i(\tau) + \lambda_j(\tau)) \,d \tau} \biggl[\int_t^\omega \bigl(1 + c_i \lambda_j(\tau) \bigr) e^{- \int_0^\tau (\lambda_i(s) + \lambda_j(s)) \,ds}\, d \tau \biggr]. \end{aligned}$$

(24)

Finally, the expression of the optimal feedback strategy for the ith firm can be achieved from the FOCs of the model

$$ u_i^*(t,x) = \frac{x}{A_i^*(t)}=\frac{x e^{-t-\int_0^t (\lambda _i(\tau) + \lambda_j(\tau)) \,d\tau} }{ \int_t^\omega (1 + c_i \lambda _j(\tau)) e^{-\tau-\int_0^\tau (\lambda_i(\theta) + \lambda_j(\theta)) \,d\theta }\,d\tau}. $$

(25)