# Solving a Cournot Equilibrium, the case of Q=q1+q2, Q(q1,q2)=q1+q2

I am struggling with the differentiating between when to use $Q=q_1+q_2$ and $Q(q_1,q_2)=q_1+q_2$

For a 2 player cournot game, given $$P=a-bQ, with \ MC's \ c_1 \neq c_2$$

I find the following:

FOC $$q_{i}^* = \frac{a-c_i}{2b} - \frac{q_j}{2}$$

which leads to me to a equilibrium value of $$q_i^{eq}= \frac{a-2c_i+c_j}{3b}$$

when using $$P=a-b(q_1+q_2)$$

$$\textbf{HOWEVER}$$

The supplied answer is $$q_i^{eq} = \frac{2a+c_j-2c_i}{3b}$$

with FOC

$$q_{i}^* = \frac{a-c_i}{b} - Q$$

$$P = a-bQ(q_1,q_2)$$

When am I supposed to use the different Q's? as all my lecture notes follow the method I used, and there are other questions which look identical and use $Q=q_1+q_2$ but there is obviously a time to use $Q(q_1,q_2)$.

First point: you write "I am struggling with the differentiating between when to use $Q=q_1+q_2$ and $Q(q_1,q_2)=q_1+q_2$". But these are the same thing: both define $Q$ as a function of $q_1$ and $q_2$. It's just that the second expression makes this functional dependence explicit by writing out the arguments, whereas the first does not. This is purely an issue of notation and has no bearing on the solution.

Let's start by writing out the profit function and computing the first-order condition:

$$\Pi=[a-b(q_i+q_j)]q_i-c_i q_i$$

$$\frac{\partial\Pi}{\partial q_i}=0 \iff a-b(q_i+q_j)-bq_i-c_i=0\iff q_i=\frac{a-bq_j-c_i}{2b}.$$

This is the FOC that you got in your solution (just slightly rearranged). So far, so good!

Let's repeat those same two steps but writing the profits in a slightly different way

$$\Pi=[a-bQ]q_i-c_i q_i$$

$$\frac{\partial\Pi}{\partial q_i}=0 \iff a-bQ-b\underbrace{\frac{dQ}{dq_i}}_{=1}q_i-c_i=0\iff q_i=\frac{a-c_i}{b}-Q.$$

This is the FOC from the solution you were given. But note that $Q=q_i+q_j$. Making this substitution gives

$$q_i=\frac{a-c_i}{b}-Q=\frac{a-c_i}{b}-q_i-q_j$$ $$q_i=\frac{a-bq_j-c_i}{2b}.$$

In words: your answer, and the provided solution have exactly the same first-order condition. So we can be confident that both are correct at least this far, and the discrepancy must be coming from the next step.

Now we need to solve the system of first-order conditions for equilibrium. We have

$$q_i=\frac{a-bq_j-c_i}{2b},\ q_j=\frac{a-bq_i-c_j}{2b},$$ $$q_i=\frac{a-b\frac{a-bq_i-c_j}{2b}-c_i}{2b}.$$

Solving this last equation for $q_i$ yields $$q_i^{\text{eq}}=\frac{a-2c_i+c_j}{3b}.$$

This is your answer. It is unclear where the answer in the solution came from, but the most likely situation is that there is an algebra error in the final step of computation of the solution.

• Thankyou @ubiquitous. You are right! I really appreciate the in depth feedback! – J.N Sep 14 '18 at 23:21

It seems to me that there is a mistake in the answer provided to you, or a rather unusual profit function is used.

In any case, when we use

$\pi_i = (a-bQ(q_1,q_2))q_i-c_iq_i$

as profit function, and do not further specify $Q$, the FOC is:

$\frac{d\pi_i}{dq_i}=a-c_i-b(q_i\frac{dQ(q_1,q_2)}{dq_i}+Q(q_1,q_2))$

The latter collapses to your first FOC for $Q=q_1+q_2$. The second FOC is essentially the same as ubiquitous has pointed out. Both can be obtained by evaluating the derivative of Q w.r.t. $q_i$ which is essentially 1.

• Thanks @Maarten punt, I really appreciate the fast response!! – J.N Sep 14 '18 at 23:22