Cournot Games with Limited Demand: From Multiple Equilibria to Stochastic Equilibrium

Abstract

We construct Cournot games with limited demand, resulting into capped sales volumes according to the respective production shares of the players. We show that such games admit three distinct equilibrium regimes, including an intermediate regime that allows for a range of possible equilibria. When information on demand is modeled by a delayed diffusion process, we also show that this intermediate regime collapses to a single equilibrium while the other regimes approximate the deterministic setting as the delay tends to zero. Moreover, as the delay approaches zero, the unique equilibrium achieved in the stochastic case provides a way to select a natural equilibrium within the range observed in the no lag setting. Numerical illustrations are presented when demand is modeled by an Ornstein–Uhlenbeck process and price is an affine function of output.

Appendix: Proofs of Propositions 2.2 and 3.3

1.1 Proof of Proposition 2.2.

We have

$$\begin{aligned}&\frac{\partial f^{(i)}}{\partial q^{(i)}} (d,q^{(1)},q^{(2)}) \nonumber \\&\quad = \left( a - 2 b q^{(i)} - b q^{(j)} \right) {\mathbf {1}}_{\{ q^{(1)} + q^{(2)}< d \}} \nonumber \\&\qquad + \left( \frac{ a q^{(j)} }{(q^{(1)} + q^{(2)})^2 } - b \right) d {\mathbf {1}}_{\{d < q^{(1)} + q^{(2)} \}} - c^{(i)}, \quad 1 \leqslant i \ne j \leqslant 2. \end{aligned}$$

(5.1)

(a) If $q^{(1)}+q^{(2)} < d$, (5.1) reads

$$\begin{aligned} \frac{\partial f^{(i)}}{\partial q^{(i)}} (d,q^{(1)},q^{(2)}) = a - 2 b q^{(i)} - b q^{(j)} - c^{(i)}, \quad 1 \leqslant i \ne j \leqslant 2, \end{aligned}$$

We distinguish two cases:

• 1. $q^{(i)*}, q^{(j)*} > 0$. In this case we must have

  $$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \frac{\partial f^{(i)}}{\partial q^{(i)}} (d,q^{(1)},q^{(2)}) = a - 2 b q^{(i)} - b q^{(j)} - c^{(i)} = 0 \\ \\ \displaystyle \frac{\partial f^{(j)}}{\partial q^{(j)}} (d,q^{(1)},q^{(2)}) = a - 2b q^{(j)} - bq^{(i)} - c^{(j)} = 0, \end{array} \right. \end{aligned}$$

  hence

  $$\begin{aligned} q^{(i)*} = \frac{a - 2c^{(i)} + c^{(j)}}{3b} \quad \text{ and } \quad q^{(j)*} = \frac{a + c^{(i)} - 2c^{(j)} }{3b}, \end{aligned}$$

  and

  $$\begin{aligned} q^{(1)*} + q^{(2)*} = \frac{2 a - c^{(1)} - c^{(2)}}{3b} < d. \end{aligned}$$

• 2. $q^{(i)*} < d$ and $q^{(j)*} = 0$. In this case we may have an inequality

  $$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \frac{\partial f^{(i)}}{\partial q^{(i)}} (d,q^{(1)},q^{(2)}) = a - 2 b q^{(i)} - b q^{(j)} - c^{(i)} = 0 \\ \\ \displaystyle \frac{\partial f^{(j)}}{\partial q^{(j)}} (d,q^{(1)},q^{(2)}) = a - 2b q^{(j)} - bq^{(i)} - c^{(j)} \leqslant 0, \end{array} \right. \end{aligned}$$

  i.e.

  $$\begin{aligned} \left\{ \begin{array}{l} \displaystyle a - 2 b q^{(i)} - c^{(i)} = 0 \\ \\ \displaystyle a - bq^{(i)} - c^{(j)} \leqslant 0, \end{array} \right. \end{aligned}$$

  or $(a-c^{(j)})/b \leqslant q^{(i)*} = (a-c^{(i)})/(2b)$, which together with the condition $a+c^{(i)}-2c^{(j)} \geqslant 0$, implies $a+c^{(i)}-2c^{(j)} = 0$ and in this case, the equilibrium is already taken care of in point 1 above.

(b) On the other hand, if $d < q^{(1)} + q^{(2)}$, (5.1) reads

$$\begin{aligned} \frac{\partial f^{(i)}}{\partial q^{(i)}} (d,q^{(1)},q^{(2)}) = \left( \frac{a q^{(j)} }{(q^{(1)} + q^{(2)})^2} - b \right) d - c^{(i)}, \quad 1 \leqslant i \ne j \leqslant 2, \end{aligned}$$

There are two kinds of candidate equilibria:

• 1. $q^{(i)*}, q^{(j)*} > 0$ and we must have:

  $$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \frac{\partial f^{(i)}}{\partial q^{(i)}} (d,q^{(1)},q^{(2)}) = \left( \frac{a q^{(j)} }{(q^{(1)} + q^{(2)})^2} - b \right) d - c^{(i)} = 0 \\ \\ \displaystyle \frac{\partial f^{(j)}}{\partial q^{(j)}} (d,q^{(1)},q^{(2)}) = \left( \frac{a q^{(i)} }{(q^{(1)} + q^{(2)})^2} - b \right) d - c^{(j)} = 0 , \end{array} \right. \end{aligned}$$

  In this case,

  $$\begin{aligned} \left\{ \begin{array}{l} \displaystyle q^{(1)} a - b (q^{(1)} + q^{(2)})^2 = \frac{c^{(2)}}{d} ( q^{(1)} + q^{(2)} )^2, \\ \\ \displaystyle q^{(2)} a - b (q^{(1)} + q^{(2)})^2 = \frac{c^{(1)}}{d} ( q^{(1)} + q^{(2)} )^2, \end{array} \right. \end{aligned}$$

  hence

  $$\begin{aligned} \displaystyle q^{(1)*}(d) = \frac{a d (c^{(2)}+bd)}{(c^{(1)}+c^{(2)}+2bd)^2}, \quad \text{ and } \quad \displaystyle q^{(2)*}(d) = \frac{a d ( c^{(1)}+bd )}{(c^{(1)}+c^{(2)}+2bd )^2}, \end{aligned}$$

  and

  $$\begin{aligned} d < q^{(1)*}(d ) + q^{(2)*}(d ) = \frac{a d }{c^{(1)}+c^{(2)}+2bd }. \end{aligned}$$

• 2. $q^{(i)*} > d$ and $q^{(j)*} = 0$ and instead we may have an inequality:

  $$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \frac{\partial f^{(i)}}{\partial q^{(i)}} (d,q^{(1)},q^{(2)}) = \left( \frac{a q^{(j)} }{(q^{(1)} + q^{(2)})^2} - b \right) d - c^{(i)} = 0 \\ \\ \displaystyle \frac{\partial f^{(j)}}{\partial q^{(j)}} (d,q^{(1)},q^{(2)}) = \left( \frac{a q^{(i)} }{(q^{(1)} + q^{(2)})^2} - b \right) d - c^{(j)} \leqslant 0 , \end{array} \right. \end{aligned}$$

And therefore,

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \frac{\partial f^{(i)}}{\partial q^{(i)}} (d,q^{(1)},q^{(2)}) = - bd - c^{(i)} = 0 \\ \\ \displaystyle \frac{\partial f^{(j)}}{\partial q^{(j)}} (d,q^{(1)},q^{(2)}) = \frac{ad}{q^{(i)}} - bd - c^{(j)} \leqslant 0 , \end{array} \right. \end{aligned}$$

which is impossible.

(c) Finally we consider what happens when ${\underline{d}} \leqslant d \leqslant {\overline{d}}$, i.e. d is between the above boundaries. The sum of derivatives

$$\begin{aligned}&\frac{\partial f^{(1)}}{\partial q^{(1)}} (d,q^{(1)},q^{(2)}) + \frac{\partial f^{(2)}}{\partial q^{(2)}} (d,q^{(1)},q^{(2)}) \nonumber \\&\quad = \left( a - 3 b (q^{(i)} + q^{(j)}) \right) {\mathbf {1}}_{\{ q^{(1)} + q^{(2)}< d \}} \nonumber \\&\qquad + \left( \frac{ a }{q^{(1)} + q^{(2)} } - b \right) d {\mathbf {1}}_{\{d < q^{(1)} + q^{(2)} \}} - c^{(i)} - c^{(j)} \end{aligned}$$

(5.2)

is a decreasing function of $q^{(1)}+q^{(2)}$ which vanishes at $q^{(1)}+q^{(2)} = {\overline{d}}$, therefore it is strictly positive when $q^{(1)}+q^{(2)} <d \leqslant {\overline{d}}$, hence at least one of the individual derivatives must be strictly positive at such points. Therefore, $q^{(1)} + q^{(2)} < d$ cannot be an equilibrium.

In the case $q^{(1)}+q^{(2)} > d$ the summation of derivatives vanishes at $q^{(1)} + q^{(2)} = {\underline{d}}$ hence it is strictly negative when $q^{(1)}+q^{(2)} >d \geqslant {\underline{d}}$ and at least one of the derivatives must be negative. Therefore $q^{(1)} + q^{(2)} > d$ cannot be an equilibrium either.

Consider a point $(q^{(1)*}, q^{(2)*}) \in [0, d] \times [0, d]$ on the line $q^{(1)} + q^{(2)} = d$. This is an equilibrium if there exists no profitable deviations. We need to show that the value function

$$\begin{aligned}&f^{(i)} (d,q^{(1)},q^{(2)}) \\&\quad = (a-b(q^{(i)} + q^{(j)*}) ) \left( q^{(i)} {\mathbf {1}}_{\{ q^{(i)} + q^{(j)*} \leqslant d \}}+\frac{q^{(i)} d }{q^{(i)} + q^{(j)*}} {\mathbf {1}}_{\{ d < q^{(i)} + q^{(j)*} \}}\right) \\&\qquad - c^{(i)}q^{(i)}, \quad 1 \leqslant i \ne j \leqslant 2, \end{aligned}$$

is increasing in $q^{(i)} \in [0,q^{(i)*})$ and decreasing in $q^{(i)} \in (q^{(i)*}, \infty )$. Given that the partial derivative

$$\begin{aligned}&\frac{\partial f^{(i)}}{\partial q^{(i)}} (d,q^{(1)},q^{(2)}) \\&\quad = \left( a - 2 b q^{(i)} - b q^{(j)*} \right) {\mathbf {1}}_{\{ q^{(i)} + q^{(j)*}< d \}} +\left( \frac{ a q^{(j)*} }{(q^{(i)} + q^{(j)*})^2 } - b \right) d {\mathbf {1}}_{\{d < q^{(i)} + q^{(j)*} \}} \\&\qquad - c^{(i)} \end{aligned}$$

is a decreasing function of $q^{(i)}$, it is necessary and sufficient to have the left derivative

$$\begin{aligned} \displaystyle \frac{\partial f^{(i)}}{\partial q^{(i)}} (d,q^{(i)*-},q^{(j)*} ) = a - b q^{(i)*} - bd - c^{(i)} = a + b q^{(j)*} - 2 bd - c^{(i)} \geqslant 0,\nonumber \\ \end{aligned}$$

(5.3)

and the right derivative satisfies

$$\begin{aligned} \displaystyle \frac{\partial f^{(i)}}{\partial q^{(i)}} (d,q^{(i)*+},q^{(j)*} ) = \frac{aq^{(j)*} }{d} - bd -c^{(i)} \leqslant 0. \end{aligned}$$

(5.4)

Therefore $(q^{(1)*}, q^{(2)*}) \in (0,d) \times (0,d)$ on the line $q^{(1)} + q^{(2)} = d$ is an equilibrium if and only if

$$\begin{aligned} \left( \frac{-a+c^{(i)}+2bd}{b}\right) ^+ \leqslant q^{(j)*} \leqslant \frac{(c^{(i)}+bd)d}{a}, \qquad 1 \leqslant i \ne j \leqslant 2, \end{aligned}$$

(5.5)

or equivalently

$$\begin{aligned} \left( \frac{-a+c^{(j)}+2bd}{b}\right) ^+ \leqslant d - q^{(j)*} \leqslant \frac{(c^{(j)}+bd )d }{a}, \qquad j = 1,2, \end{aligned}$$

(5.6)

which yields (2.8). Notice that from (2.8), if there exists $q^{(i)*}, q^{(j)*}$ that satisfy the inequalties, then ${\underline{d}} \leqslant d \leqslant {\overline{d}}$, or in other words, if d is outside these bounds, we do not have equilibria satisfying $q^{(i)*} + q^{(j)*} = d$. Finally, consider $(q^{(1)*}, q^{(2)*})$ on the line $q^{(1)} + q^{(2)} = d$ with $q^{(i)*} = d$ and $q^{(j)*} = 0$. In this case, the value function for i

$$\begin{aligned} f^{(i)} (d,q^{(1)},q^{(2)}) = (a-bq^{(i)}) \left( q^{(i)} {\mathbf {1}}_{\{ q^{(i)} \leqslant d \}}+ d {\mathbf {1}}_{\{ d < q^{(i)} \}}\right) - c^{(i)}q^{(i)}, \end{aligned}$$

is clearly decreasing in $q^{(i)} \in (d , \infty )$, and it suffices to show that it is increasing in $q^{(i)} \in [0,d)$, and from (5.3) for this we need

$$\begin{aligned} d \leqslant \displaystyle \frac{a-c^{(i)}}{2b}. \end{aligned}$$

On the other hand, we check that the right derivative for j, (5.4) is negative if

$$\begin{aligned} \frac{a - c^{(j)}}{b} \leqslant d, \end{aligned}$$

hence we need

$$\begin{aligned} \frac{a - c^{(j)}}{b} \leqslant d \leqslant \displaystyle \frac{a-c^{(i)}}{2b}. \end{aligned}$$

However, $(q^{(i)*}, q^{(j)*}) = (d,0)$ cannot be an equilibrium unless

$$\begin{aligned} d = \frac{a - c^{(j)}}{b} = \frac{a-c^{(i)}}{2b}, \end{aligned}$$

as we assume that $a +c^{(i)} - 2c^{(j)} \geqslant 0$. In the latter case we have $a +c^{(i)} - 2c^{(j)} = 0$ and the inequalities (2.8) read

$$\begin{aligned} d= & {} \frac{-a+c^{(j)}+2bd}{b} \leqslant q^{(i)*} \leqslant d, \text { i.e. } q^{(i)*} = d, \quad \text{ and } \quad 0 \leqslant q^{(j)*} \leqslant 0, \text{ i.e. } q^{(j)*} \\= & {} 0, \end{aligned}$$

which is compatible with the equilibrium $(q^{(i)*} , q^{(j)*}) = (d,0)$. Notice that (d, 0) is an equilibrium if and only if d is i’s monopoly quantity.

1.2 Proof of Proposition 3.3.

From (3.9) and the fact that $q^{(1)}_t + q^{(2)}_t$ is ${\mathcal {E}}_t$-measurable we have

$$\begin{aligned}&\left( P(q^{(1)}_t + q^{(2)}_t ) + q^{(i)}_t P'(q^{(1)}_t + q^{(2)}_t ) \right) \mathbb {E}[ {\mathbf {1}}_{\{q \leqslant D_t\}} \mid {\mathcal {E}}_t ]_{q = q^{(1)}_t + q^{(2)}_t} \nonumber \\&\qquad + \left( q^{(j)}_t \frac{ P(q^{(1)}_t + q^{(2)}_t ) }{(q^{(1)}_t + q^{(2)}_t)^2 } + q^{(i)}_t \frac{ P'(q^{(1)}_t + q^{(2)}_t )}{ q^{(1)}_t + q^{(2)}_t } \right) \mathbb {E}[ D_t {\mathbf {1}}_{\{q > D_t\}} \mid {\mathcal {E}}_t ]_{q = q^{(1)}_t + q^{(2)}_t} \nonumber \\&\qquad \quad = c^{(i)},\quad 1 \leqslant i \ne j \leqslant 2. \end{aligned}$$

(5.7)

From (3.18) we write $D_t =d_{t-\delta } e^{-\alpha \delta } + \beta (1-e^{-\alpha \delta })+X$ where $X \simeq {\mathcal {N}}\left( 0, \varepsilon ^2 (t) \right)$ and

$$\begin{aligned} \varepsilon ^2 (t) = \mathbb {E}\left[ \left( \int _{t-\delta }^t e^{\alpha (s-t)} \sigma (s) {\mathrm {\mathrm{d}}}B_s \right) ^2 \right] = \int _{t-\delta }^t e^{2\alpha (s-t)} \sigma ^2 (s) ds = \sigma ^2\frac{1-e^{-2\alpha \delta }}{2\alpha }, \end{aligned}$$

hence by the independence between ${\mathcal {E}}_t := {\mathcal {F}}_{t-\delta }$ and $\int _{t-\delta }^t e^{\alpha (s-t)} \sigma (s) {\mathrm {\mathrm{d}}}B_s$ in (3.18) we have

$$\begin{aligned} \mathbb {E}[ {\mathbf {1}}_{\{q \leqslant D_t\}} \mid {\mathcal {E}}_t ]_{q = q^{(1)}_t + q^{(2)}_t}= & {} {\mathord {{\mathbb {P}}}}( q \leqslant d e^{-\alpha \delta } + \beta (1-e^{-\alpha \delta })+X )_{d = d_{t-\delta }, \ q = q^{(1)}_t + q^{(2)}_t} \\= & {} \int _{q^{(1)}_t+q^{(2)}_t - \varphi ( D_{t-\delta }) }^\infty e^{ -x^2 / ( 2\varepsilon ^2 (t) ) } \frac{dx}{\sqrt{2 \pi \varepsilon ^2 (t)}} \\= & {} \Phi \left( \frac{ \varphi ( D_{t-\delta }) -q^{(1)}_t - q^{(2)}_t}{\varepsilon (t)} \right) . \end{aligned}$$

For the second expectation we have

$$\begin{aligned}&\mathbb {E}[ D_t {\mathbf {1}}_{\{q> D_t\}} ]_{q = q^{(1)}_t + q^{(2)}_t} \\&\quad = \mathbb {E}[(d e^{-\alpha \delta } + \beta (1-e^{-\alpha \delta })+X ){\mathbf {1}}_{\{q> d e^{-\alpha \delta } + \beta (1-e^{-\alpha \delta })+X\}}]_{d = d_{t-\delta }, \ q = q^{(1)}_t + q^{(2)}_t} \\&\quad = \varphi ( D_{t-\delta }) \mathbb {E}[ {\mathbf {1}}_{q> \varphi ( D_{t-\delta }) + X} ]_{q = q^{(1)}_t + q^{(2)}_t} + \mathbb {E}[ X {\mathbf {1}}_{q > \varphi ( D_{t-\delta }) + X}]_{q = q^{(1)}_t + q^{(2)}_t} \\&\quad = \varphi ( D_{t-\delta }) {\mathord {{\mathbb {P}}}}(X < q - \varphi ( D_{t-\delta }) )_{q = q^{(1)}_t + q^{(2)}_t}\\&\qquad + \int _{-\infty }^{q^{(1)}_t + q^{(2)}_t- \varphi ( D_{t-\delta }) } x e^{ -x^2 / ( 2\varepsilon ^2 (t) ) }\frac{dx}{\sqrt{2 \pi \varepsilon ^2 (t)}} \\&\quad = \varphi ( D_{t-\delta }) \Phi \left( \frac{q^{(1)}_t + q^{(2)}_t- \varphi ( D_{t-\delta }) }{\varepsilon (t)}\right) - \varepsilon (t) \phi \left( \frac{q^{(1)}_t + q^{(2)}_t- \varphi ( D_{t-\delta }) }{\varepsilon (t)}\right) . \end{aligned}$$

Hence (5.7) reads

$$\begin{aligned}&\left( q^{(i)}_t \frac{ P(q^{(1)}_t + q^{(2)}_t ) }{(q^{(1)}_t + q^{(2)}_t)^2 } + q^{(j)}_t \frac{ P'(q^{(1)}_t + q^{(2)}_t )}{ q^{(1)}_t + q^{(2)}_t } \right) \nonumber \\&\qquad \times \left( \varepsilon (t) \phi \left( \frac{q^{(1)}_t + q^{(2)}_t- \varphi ( D_{t-\delta }) }{\varepsilon (t)}\right) -\varphi ( D_{t-\delta }) \Phi \left( \frac{q^{(1)}_t + q^{(2)}_t- \varphi ( D_{t-\delta }) }{\varepsilon (t)}\right) \right) \nonumber \\&\qquad - \left( P(q^{(1)}_t + q^{(2)}_t ) + q^{(j)}_t P'(q^{(1)}_t + q^{(2)}_t ) \right) \Phi \left( \frac{\varphi ( D_{t-\delta }) - q^{(1)}_t - q^{(2)}_t }{\varepsilon (t)}\right) \nonumber \\&\quad = -c^{(j)}, \qquad 1 \leqslant i \ne j \leqslant 2. \end{aligned}$$

(5.8)

By summing the above two equations we find that the sum $S^*_t : = q^{(1)}_t + q^{(2)}_t$ satisfies the equation

$$\begin{aligned}&\left( \frac{ P(q^{(1)}_t + q^{(2)}_t ) }{q^{(1)}_t + q^{(2)}_t } + P'(q^{(1)}_t + q^{(2)}_t ) \right) \nonumber \\&\qquad \times \left( \varepsilon (t) \phi \left( \frac{q^{(1)}_t + q^{(2)}_t- \varphi ( D_{t-\delta }) }{\varepsilon (t)}\right) -\varphi ( D_{t-\delta }) \Phi \left( \frac{q^{(1)}_t + q^{(2)}_t- \varphi ( D_{t-\delta }) }{\varepsilon (t)}\right) \right) \nonumber \\&\qquad - \left( 2P(q^{(1)}_t + q^{(2)}_t ) + ( q^{(1)}_t + q^{(2)}_t )P'(q^{(1)}_t + q^{(2)}_t ) \right) \Phi \left( \frac{ \varphi ( D_{t-\delta }) - q^{(1)}_t - q^{(2)}_t}{\varepsilon (t)}\right) \nonumber \\&\qquad \quad = -c^{(1)}-c^{(2)}, \qquad 1 \leqslant i \ne j \leqslant 2. \end{aligned}$$

(5.9)

Combining (5.8) with (5.9), we get

$$\begin{aligned}&\left( \frac{ q^{(i)}_t P(q^{(1)}_t + q^{(2)}_t ) + q^{(j)}_t (q^{(1)}_t + q^{(2)}_t)P'(q^{(1)}_t + q^{(2)}_t ) }{ (q^{(1)}_t + q^{(2)}_t) P(q^{(1)}_t + q^{(2)}_t ) + (q^{(1)}_t + q^{(2)}_t)^2 P'(q^{(1)}_t + q^{(2)}_t ) } \right) \nonumber \\&\qquad \times \left( c^{(1)}+c^{(2)} -\left( 2P(q^{(1)}_t + q^{(2)}_t ) \right. \right. \nonumber \\&\qquad \left. \left. + ( q^{(1)}_t + q^{(2)}_t )P'(q^{(1)}_t + q^{(2)}_t ) \right) \Phi \left( \frac{ \varphi ( D_{t-\delta }) - q^{(1)}_t - q^{(2)}_t}{\varepsilon (t)}\right) \right) \nonumber \\&\qquad + \left( P(q^{(1)}_t + q^{(2)}_t ) + q^{(j)}_t P'(q^{(1)}_t + q^{(2)}_t ) \right) \Phi \left( \frac{\varphi ( D_{t-\delta }) - q^{(1)}_t - q^{(2)}_t }{\varepsilon (t)}\right) \nonumber \\&\qquad \quad = c^{(j)}, \qquad 1 \leqslant i \ne j \leqslant 2, \end{aligned}$$

(5.10)

or

$$\begin{aligned}&- q^{(j)}_t ( P(S^*_t ) - S_t P'(S^*_t ) ) \left( c^{(1)}+c^{(2)} \right. \\&\qquad \left. -\left( 2P(S^*_t ) + S^*_t P'(S^*_t ) \right) \Phi \left( \frac{ \varphi ( D_{t-\delta }) - S^*_t}{\varepsilon (t)}\right) \right) \\&\qquad + q^{(j)}_t P'(S^*_t ) ( S^*_t P(S^*_t )+ {S^*_t}^2 P'(S^*_t ) ) \Phi \left( \frac{\varphi ( D_{t-\delta }) - S^*_t }{\varepsilon (t)}\right) \\&\quad = c^{(j)}( S^*_t P(S^*_t )+ {S^*_t}^2 P'(S^*_t ) ) - P(S^*_t ) ( S^*_t P(S^*_t )\\&\qquad + {S^*_t}^2 P'(S^*_t ) ) \Phi \left( \frac{\varphi ( D_{t-\delta }) - S^*_t }{\varepsilon (t)}\right) \\&\qquad - S^*_t P(S^*_t ) \left( c^{(1)}+c^{(2)} -\left( 2P(S^*_t ) + S^*_t P'(S^*_t ) \right) \Phi \left( \frac{ \varphi ( D_{t-\delta }) - S^*_t}{\varepsilon (t)}\right) \right) , \\&\quad \qquad 1 \leqslant i \ne j \leqslant 2, \end{aligned}$$

which shows that

$$\begin{aligned} q^{(j)*}_t = S^*_t \frac{ P^2(S^*_t ) \Phi \left( \frac{\varphi ( D_{t-\delta }) - S^*_t }{\varepsilon (t)}\right) + ( c^{(j)} - c^{(i)} ) P(S^*_t ) - c^{(j)} ( P(S^*_t) - S^*_t P'(S^*_t ) ) }{ 2 P^2(S^*_t ) \Phi \left( \frac{\varphi ( D_{t-\delta }) - S^*_t }{\varepsilon (t)}\right) - ( P(S^*_t ) - S_t P'(S^*_t ) ) ( c^{(1)}+c^{(2)} ) } . \end{aligned}$$

In the limit when $d_t$ is large, (5.10) yields the system of equations

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle 2 b q^{(1)}_t + b q^{(2)}_t + c^{(1)} = a \\ \\ \displaystyle b q^{(1)}_t + 2 b q^{(2)}_t + c^{(2)} = a, \end{array} \right. \end{aligned}$$

with solution

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle q^{(1)}_t = \frac{a -2 c^{(1)} + c^{(2)} }{3b} \\ \\ \displaystyle q^{(2)}_t = \frac{a + c^{(1)} -2 c^{(2)}}{3b}, \end{array} \right. \end{aligned}$$

and sum

$$\begin{aligned} q^{(1)}_t + q^{(2)}_t = \frac{2a - c^{(1)} - c^{(2)} }{3b}, \qquad t\in [\delta , T], \end{aligned}$$

which recover the no lag setting of Proposition 2.2.