Intertemporal Non-separability and Dynamic Oligopoly

Abstract

We construct a framework for modeling dynamic Cournot oligopoly. We consider models where utility maximizing consumers give rise to demand functions that depend on current and prior period prices. Future demand depends on the current price and consumers, and firms must take this into account when making their decisions. Focusing on problems that yield dynamic demand functions that are linear in current and prior period price, we characterize the unique Markov perfect equilibrium in linear strategies. We then demonstrate the applicability of our framework through a series of practical examples.

Appendices

Appendix 1: Proof

We only include the proofs for the case when ${\hat{d}}>0$. Suitable modifications yield proofs for ${\hat{d}}<0$.

Proof of Theorem 1

We begin with the latter part.

First, we define some additional notation. For each N let:

$$\begin{aligned} Q=\frac{{\hat{b}}}{{\hat{d}}-g}\lambda =\displaystyle \frac{N\lambda s}{1+(1-1/N)N\lambda s} \end{aligned}$$

where $s={\hat{b}}/{\hat{d}}$. Using this definition, we can rewrite (4) as:

$$\begin{aligned} \beta \lambda ^2-2Q+1=0 \end{aligned}$$

Solving this for $\lambda$ then yields:

$$\begin{aligned} \lambda =\sqrt{\frac{2Q-1}{\beta }}. \end{aligned}$$

Note that if we find a $\lambda \in (0,1)$ that satisfies (4), the above implies that . Now, define the following functions over :

$$\begin{aligned}&\displaystyle \varLambda (q)=\sqrt{\displaystyle \frac{2q-1}{\beta }}\in [0,1]\nonumber \\&\displaystyle s(q)=\left. \frac{{\hat{b}}}{{\hat{d}}}\right| _{\lambda =\varLambda (q)}\\&\displaystyle \eta (q)=\displaystyle \frac{Ns(q)}{\sqrt{\beta }}>N/\sqrt{\beta }>N\nonumber \\&\displaystyle y(q)=\eta (q)\sqrt{2q-1}=N\varLambda (q)s(q)\nonumber \\&\displaystyle \gamma (q)=\displaystyle \frac{y(q)}{1+(1-1/N)y(q)}= \frac{N\varLambda (q)s(q)}{1+(1-1/N)N\varLambda (q)s(q)}\nonumber \\&\displaystyle k(q)=q-\gamma (q).\nonumber \end{aligned}$$

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Suppose we knew an equilibrium value of $\lambda$. This would imply a value for Q so that $\lambda =\varLambda (Q)$. Then, by construction, ${\hat{b}}/{\hat{d}}=s(Q), N\lambda {\hat{b}}/{\hat{d}}=y(Q)$ and $Q=\gamma (Q)$. That is, the Q corresponding to an equilibrium will be a fixed point of $\gamma (\,\cdot \,)$. Also, if $q^*$ is a fixed point of $\gamma (\,\cdot \,)$, it is easily verified that $\varLambda (q^*)$ satisfies the equilibrium conditions. So equilibria will correspond one-to-one with fixed points of $\gamma (\,\cdot \,)$ or, equivalently, to points at which $k(q^*)=0$. Note that all the above functions are continuous over the range of q and all are nondecreasing in q except $k(\,\cdot \,)$. With this notation, we can now turn to the proof.

The strategy of the proof is to show that there is a unique such that $k(q^*)=0$ and $y(q^*)<1$. Now, we can write:

$$\begin{aligned} k(q)=\frac{q-\eta (q)\sqrt{2q-1}[1-(1-1/N)q]}{1+(1-1/N)\eta (q)\sqrt{2q-1}} \end{aligned}$$

and this will be zero if and only if the numerator is zero. Define, for and $\eta >N/\sqrt{\beta }$ (since $\eta (q)>N/\sqrt{\beta }$):

$$\begin{aligned} \mu (q,\eta )=q-\eta \sqrt{2q-1}[1-(1-1/N)q]. \end{aligned}$$

The following properties of $\mu$ are important: $\mu$ is continuous, strictly convex in q, and strictly decreasing in $\eta$. By construction, an equilibrium corresponds to a point $q^*$ such that $\mu (q^*,\eta (q^*))=0$. Note also that the composite function $\mu (q,\eta (q))$ is continuous.

We next show an equilibrium exists. First:

Also:

$$\begin{aligned} \mu \left( \displaystyle \frac{1+\beta }{2},\eta \right)&=\displaystyle \frac{1+\beta }{2}-\eta \sqrt{\beta }\left[ 1-\left( 1-\frac{1}{N}\right) \frac{1+\beta }{2}\right] \\&<\displaystyle \frac{1+\beta }{2}-N\left[ 1-\left( 1-\frac{1}{N}\right) \frac{1+\beta }{2}\right] ,~ \forall \eta >\frac{N}{\sqrt{\beta }} \end{aligned}$$

so

$$\begin{aligned} \mu \left( \frac{1+\beta }{2},\eta \left( \frac{1+\beta }{2}\right) \right)< \frac{1+\beta }{2}\left( 1+N-1\right) -N=N\left( \frac{1+\beta }{2}-1\right) <0 \end{aligned}$$

since $\beta <1$. By continuity, there exists such that $\mu (q^*,\eta (q^*))=0$.

We now show $q^*$ is unique. Take arbitrary . Since $q^*<(1+\beta )/2$, take $\alpha \in (0,1)$ such that $q^*=\alpha q+(1-\alpha )(1+\beta )/2$. By convexity:

$$\begin{aligned} \mu (q^*,\eta (q^*))=0<\alpha \mu (q,\eta (q^*))+(1-\alpha )\mu \left( \frac{1+\beta }{2},\eta (q^*)\right) . \end{aligned}$$

The second term on the right-hand side is negative (since the above argument showed this was true for arbitrary $\eta$), so the first must be positive. Therefore, $\mu (q,\eta (q^*))>0$. But $\eta (q)$ is increasing in q so $\eta (q^*)>\eta (q)$. Since $\mu$ is decreasing in $\eta$, this implies $\mu (q,\eta (q))>\mu (q,\eta (q^*))>0$, so $\mu (q,\eta (q))>0$ for all . That $\mu (q,\eta (q))$ is not zero for $q\in (q^*,(1+\beta )/2]$ is now obvious since otherwise the same argument would imply $\mu (q^*,\eta (q^*))>0$.

Turning to the former part, we first show that $N\lambda {\hat{b}}/{\hat{d}}<1$. Note that $N\lambda {\hat{b}}/{\hat{d}}=y(q^*)$ and if $q^*=\gamma (q^*)<N/(2N-1)$ it must be that $y(q^*)<1$. So it suffices to show $q^*<N/(2N-1)$. If $N/(2N-1)>(1+\beta )/2$, we are done. Otherwise:

$$\begin{aligned} \displaystyle \mu \left( \frac{N}{2N-1},\eta \left( \frac{N}{2N-1}\right) \right)&= \displaystyle \frac{N}{2N-1}-\eta \left( \frac{N}{2N-1}\right) \frac{1}{\sqrt{N-1}}\left[ 1-\frac{N-1}{N}\cdot \frac{N}{2N-1}\right] \\&=\displaystyle \frac{N}{2N-1}\frac{\displaystyle \sqrt{2N-1}-\eta \left( \frac{N}{2N-1}\right) }{\sqrt{2N-1}}\\&<\displaystyle \frac{N}{2N-1}\frac{\sqrt{2N-1}-N}{\sqrt{2N-1}}\\&\le 0 \end{aligned}$$

where the inequalities hold since $\eta >N$ and $N/(2N-1)\le 1, N\ge \sqrt{2N-1}$.⁹

Finally, solving (8) for f and using (10) yields:

$$\begin{aligned} f=\displaystyle \frac{N-1}{N}{\hat{a}}-\frac{N-1}{N}{\hat{b}}(1-\lambda )\bar{p}+ h{\bar{c}} \end{aligned}$$

Substituting this, g and h into ${\bar{p}}$ and solving yields:

$$\begin{aligned} {\bar{p}}-{\bar{c}}=\frac{{\hat{a}}-({\hat{b}}-{\hat{d}}){\bar{c}}}{\hat{b}(1+N-(N-1)\beta \lambda )-{\hat{d}}(1+\beta )} . \end{aligned}$$

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It is straightforward to see that as long as ${\hat{a}}>({\hat{b}}-{\hat{d}}){\bar{c}}$ it must be the case that ${\bar{p}}>{\bar{c}}$.

Now we show that there exist $\varepsilon _L,\varepsilon _H$ where $p_t,X_t>0$. Define

$$\begin{aligned} p_L= & {} {\bar{p}}-N\frac{\lambda }{1-\lambda }\frac{{\hat{b}}}{\hat{d}}\varepsilon _L \end{aligned}$$

(48)

$$\begin{aligned} p_H= & {} {\bar{p}}+N\frac{\lambda }{1-\lambda }\frac{{\hat{b}}}{\hat{d}}\varepsilon _H. \end{aligned}$$

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It is easy to show that if $p_t\in [p_L,p_H]$, then $p_{t+1}\in (p_L,p_H)$. By induction, if $p_t\in [p_L,p_H]$, then $p_{t+k}\in (p_L,p_H)$ for any $k>0$. Thus, by bounding the errors we can ensure that prices set by the Euler equation will never be negative and will be lower than some price at which the households would always choose positive consumption in both periods. $\square$

Proof of Corollary 1

Since must tend to as N tends to infinity. Since $\lambda =\varLambda (Q)$ and this function is continuous, $\lambda$ tends to .

Note Q is a one-to-one, continuous function of $y=N\lambda s$, so the fact that Q converges implies y converges to some limit point $y'$. But then it must be that . Solving this yields $y'=1$.

Rearranging (8) and substituting (13), we get

$$\begin{aligned} f=\frac{N-1}{N}{\hat{a}}-\frac{N-1}{N}{\hat{b}}(1-\lambda )\bar{p}+\frac{N-1}{N}\frac{N\lambda {\hat{b}}}{{\hat{d}}}{\hat{b}}{\bar{c}}. \end{aligned}$$

Since we have already shown that $\lambda ^*\rightarrow 0$ and $N\lambda ^*\hat{b}/{\hat{d}}\rightarrow 1$, in the limit as $N\rightarrow \infty , f\rightarrow \hat{a}-{\hat{b}}({\bar{p}}-{\bar{c}})$ or in other words, ${\hat{a}}-f\rightarrow {\hat{b}}(\bar{p}-{\bar{c}})$. It is straightforward to see that $h\rightarrow {\hat{b}}$ and $\hat{d}-g\rightarrow 0$ and therefore ${\bar{p}}\rightarrow {\bar{c}}$. $\square$

Proof of Proposition 1

Consider the case where $d>0$.¹⁰ The determinant of the matrix on the left-hand side of (23) is $D=1-\theta _1(\theta _3\theta _6+\theta _5)$. Now:

$$\begin{aligned} \theta _1=\frac{m}{\sqrt{2q-1}}>\sqrt{2N-1} \end{aligned}$$

since $m>1$ and $q<N/(2N-1)$.

$$\begin{aligned} \theta _5=\frac{Ns}{[1+(1-1/N)N\lambda s]^2}>\frac{N}{[(2N-1)/N]^2}=\frac{N^3}{(2N-1)^2} \end{aligned}$$

since $s>1$ and $N\lambda s<1$. Therefore, $\theta _1\theta _5>N^3/(2N-1)^{3/2}$. It is possible to show that the last expression equals one when $N=1$ and is greater than one for $N>1$. This guarantees that D is negative.

We can now sign the partial derivatives of $\lambda$ using Cramer’s rule:

$$\begin{aligned}&\displaystyle \frac{d\lambda }{dN}=\frac{1}{D}\left| \begin{array}{ccc}0&{}\quad 0&{}\quad -\theta _1\\ 0&{}\quad 1&{}\quad 0\\ \theta _7&{}\quad -\theta _6&{}\quad 1\end{array}\right| =\frac{\theta _1\theta _7}{D}<0\\&\displaystyle \frac{d\lambda }{dm}=\frac{1}{D}\left| \begin{array}{ccc}\theta _2&{}\quad 0&{}\quad -\theta _1\\ 0&{}\quad 1&{}\quad 0\\ 0&{}\quad -\theta _6&{}\quad 1\end{array}\right| =\frac{\theta _2}{D}<0\\&\displaystyle \frac{d\lambda }{dr}=\frac{1}{D}\left| \begin{array}{ccc}0&{}\quad 0&{}\quad -\theta _1\\ \theta _4&{}\quad 1&{}\quad 0\\ 0&{}\quad -\theta _6&{}\quad 1\end{array}\right| =\frac{\theta _1\theta _4\theta _6}{D}<0 \end{aligned}$$

In order to get some idea as to the behavior of the term $N\lambda {\hat{b}}/\hat{d}$, we will need to get similar comparative static results on s. These are:

$$\begin{aligned}&\displaystyle \frac{ds}{dm}=\frac{1}{D}\left| \begin{array}{ccc}1&{}\quad \theta _2&{}\quad -\theta _1\\ -\theta _3&{}\quad 0&{}\quad 0\\ -\theta _5&{}\quad 0&{}\quad 1\end{array}\right| =\frac{\theta _2\theta _3}{D}<0\\&\displaystyle \frac{ds}{dr}=\frac{1}{D}\left| \begin{array}{ccc}1&{}\quad 0&{}\quad -\theta _1\\ -\theta _3&{}\quad \theta _4&{}\quad 0\\ -\theta _5&{}\quad 0&{}\quad 1\end{array}\right| =\frac{\theta _4(1-\theta _1\theta _5)}{D}>0 \end{aligned}$$

follows from the fact that $\theta _1\theta _5>1$.

$$\begin{aligned} \frac{ds}{dN}=\frac{1}{D}\left| \begin{array}{ccc}1&{}\quad 0&{}\quad -\theta _1\\ -\theta _3&{}\quad 0&{}\quad 0\\ -\theta _5&{}\quad \theta _7&{}\quad 1\end{array}\right| =\frac{\theta _1\theta _3\theta _7}{D}<0 \end{aligned}$$

Now, differentiating $y=N\lambda s$ with respect to m, e and N yields:

$$\begin{aligned}&\displaystyle \displaystyle \frac{dy}{dm}=Ns\frac{d\lambda }{dm}+N\lambda \frac{ds}{dm}<0\\&\displaystyle \displaystyle \frac{dy}{dr}=Ns\frac{d\lambda }{dr}+N\lambda \frac{ds}{dr}<0 \end{aligned}$$

Finally, to sign dy / dN, first note that:

$$\begin{aligned} \begin{aligned} \frac{dy}{dN}&=\lambda s+N\lambda \frac{ds}{dN}+Ns\frac{d\lambda }{dN} \\&=\lambda s-N\lambda \frac{\theta _1\theta _3\theta _7}{|D|}-Ns\frac{\theta _1\theta _7}{|D|} \end{aligned} \end{aligned}$$

where $|D|=\theta _1(\theta _3\theta _6+\theta _5)-1>0$. Factoring out |D|, we get:

$$\begin{aligned} \frac{dy}{dN} =\frac{1}{|D|}\left( \theta _1\left[ \theta _3\left( \theta _6\lambda s-\theta _7N\lambda \right) +\theta _5\lambda s-\theta _7Ns\right] -\lambda s\right) \end{aligned}$$

Using the definition of the $\theta$s:

$$\begin{aligned} \frac{dy}{dN}=\frac{1}{|D|}\left( \theta _1\left[ \theta _3\left( \frac{N\lambda ^2 s}{H^2}- \frac{N\lambda ^2s(1-\lambda s)}{H^2}\right) +\frac{N\lambda s^2}{H^2}-\frac{N\lambda s^2(1-\lambda s)}{H^2}\right] -\lambda s\right) \end{aligned}$$

where $H=1+(N-1)\lambda s$. So,

$$\begin{aligned} \begin{aligned} \frac{dy}{dN}&=\frac{1}{|D|}\left( \theta _1\left[ \theta _3\frac{N\lambda ^3s^2}{ H^2}+ \frac{N\lambda ^2s^3}{H^2}\right] -\lambda s\right) \\&=\frac{1}{|D|}\left( \theta _3\frac{N\lambda ^2s^2}{\beta H^2}+\frac{N\lambda s^3}{\beta H^2}-\lambda s\right) \end{aligned} \end{aligned}$$

since $\theta _1=1/\beta \lambda$. Factoring out $\lambda s/H^2$ yields:

$$\begin{aligned} \frac{dy}{dN}=\frac{\lambda s}{|D|H^2}\left( \frac{\theta _3N\lambda s}{\beta }+\frac{Ns^2}{\beta }-H^2\right) \end{aligned}$$

This is positive whenever:

$$\begin{aligned} \frac{Ns^2}{\beta }>H^2 \end{aligned}$$

or

$$\begin{aligned} \frac{s}{\sqrt{\beta }}\sqrt{N}>H. \end{aligned}$$

Now:

$$\begin{aligned} \frac{s}{\sqrt{\beta }}\sqrt{N}-H&>\sqrt{N}-H~(\hbox {since } \displaystyle \frac{s}{\sqrt{\beta }}>1)\\&=\sqrt{N}-1-\frac{N-1}{N}N\lambda s \\&=\sqrt{N}-1-\frac{N-1}{N}y \\&>\sqrt{N}-1-\frac{N-1}{N}~(\hbox {since }y<1) \\&=\frac{N\sqrt{N}-2N+1}{N} \end{aligned}$$

The term in the numerator is negative for N between one and two. However, it is positive for $N=3$ and is strictly increasing in N for $N\ge 3$, so dy / dN is positive for $N\ge 3$. $\square$

Appendix 2: Correlated Shocks

We have considered shocks which are i.i.d. over time. We now consider the case in which there is correlation in the cost shocks. As a simple case, take the overlapping generations example of Sect. 3.1 and suppose there is first-order autoregression in the cost-shock series.

In this case, the same method of solving for a Markov perfect equilibrium with linear pricing strategies does not work. If young consumers form expectations of the next period’s price as a linear function of the current price, the firms’ problem would be the same as in Sect. 2 and Eq. (14) would still give the equilibrium response of firms to such a strategy by consumers. However, the last term in Eq. (14) involves the term $\varepsilon _{t+1}$ which will not have expectation of zero unless the current cost shock is zero. Thus, if consumers were to form expectations of the future price assuming the price sequence is first-order autoregressed, the price sequence firms would choose would be second-order autoregressed and those expectations would be inconsistent.

To solve this problem, we instead assume the young consumers observe only the current price and not the cost shock or the history of prices that occurred before they were born. Even with this simple information set, the expectation of the next period’s price will not generally be linear (this depends on the distribution that generates the shocks), so we assume consumers use a least-squares projection to form forecasts of the future price.

Given linear consumer forecasts, the firms’ problem remains the same and has a solution of as in (14). Let $z_t$ denote the deviation in the price at time t from the long-run expected price. That is, $z_{t+1}\equiv p_{t+1}-{\bar{p}}=\lambda (p_t-{\bar{p}})+\lambda (N{\hat{b}}/{\hat{d}})\varepsilon _{t+1}$. This can be rewritten in the form:

$$\begin{aligned} z_{t+1}=\lambda z_t+ \lambda e_{t+1} \end{aligned}$$

where $e_t$ is proportional to $\varepsilon _t$. Since the cost shock follows an AR(1) process:

$$\begin{aligned} e_{t+1}=\rho e_t+u_{t+1} \end{aligned}$$

where $\rho$ is less than one in absolute value and $u_t$ is white noise. Let $P(z_{t+1}|z_t)$ be the projection of $z_{t+1}$ given $z_t$. Since $z_t$ is known by consumers born on date t, to compute $P(z_{t+1}|z_t)$, we need to find the projection of $e_{t+1}$ given $z_t, P(e_{t+1}|z_t)$. This takes the form:

$$\begin{aligned} P(e_{t+1}|z_t)=\phi z_t \end{aligned}$$

where

$$\begin{aligned} \phi = \frac{\mathrm{COV}(z_t,e_{t+1})}{\mathrm{VAR}(z_t)} \end{aligned}$$

It can be shown that

$$\begin{aligned} \phi =\frac{(1-\lambda ^2)\rho }{\lambda (1+\rho \lambda )} \end{aligned}$$

so that the projection of $z_{t+1}$ on $z_t$ is:

$$\begin{aligned} P(z_{t+1}|z_t)=\lambda (1+\phi ) z_t=\frac{\lambda +\rho }{1+\lambda \rho } z_t \equiv \zeta z_t \end{aligned}$$

It is easy to show that $\zeta$ as defined above is in the interval $[-1,1]$ whenever $\rho \in [$-1, 1] and $\lambda \in [$-1, 1]. To show existence of an equilibrium when $d>0$, define the function $G:[-1,1]^2\rightarrow [-1,1]^2$ as follows. For any given $\rho$, let:

$$\begin{aligned} G_1(\zeta ,\lambda )=\frac{\lambda +\rho }{1+\lambda \rho } \end{aligned}$$

and let $G_2$ be the right-hand side of (6) where for the solution to the consumers utility maximization problem, we have replaced the $\lambda$’s appearing in (22) with $\zeta$’s. Here, the subscripts index the two arguments of G. It is straightforward to show that G is continuous and maps the compact, convex set $[-1,1]^2$ back into itself. Therefore, G has a fixed point. By construction, fixed points of G correspond to Markov perfect equilibria, so an equilibrium exists. The primary difference is that prices now follow a second-order autoregression process.

Using similar informational and behavioral assumptions, other stochastic processes can lead to similar results. For example, if shocks are instead assumed to follow a first-order moving average process, it can be shown that prices will then follow an ARMA(1,1) process.