I am currently reading the book "Microeconomics: Principles and Analysis" by Cowell on my own. I'm reading the externalities chapter, and i found an interesting example:

There are just two firms: firm 1 is a polluter and firm 2 the victim. Firm 2 (the victim) makes an offer of a side-payment or bribe to firm 1. The bribe is an amount that is made conditional upon the amount of output that firm 1 generates: the greater the pollution, the smaller is the bribe; so we model the bribe as a decreasing function β(⋅).

The optimization problem is

enter image description here

My question is how did they arrive at those FOC's?

UPDATE:The second part of this optimization is to look at the problem from firm 1 perspective, it follows like this: Now look at the problem from the point of view of firm 1. Once the victim firm makes its offer of a conditional bribe, firm 1 should take account of it. So its profits must look like this

enter image description here

This is from F.A.Cowell - Microeconomics - Principles and Analysis p.444-445

  • 1
    $\begingroup$ I deleted my answer because it seems like I misinterpreted what some of the variables in the problem are. Could you include details of what the $i$ and $j$ in $q_i^j$ are? $\endgroup$
    – Ubiquitous
    Dec 1 '16 at 11:56
  • $\begingroup$ j=1,2 depending on if its firm 1 producing the output q or firm 2 $\endgroup$
    – user10699
    Dec 1 '16 at 12:01
  • $\begingroup$ i=1 or 2, first order condition wrt to firm 1 and first order wrt to firm 2 , i believe $\endgroup$
    – user10699
    Dec 1 '16 at 12:08
  • $\begingroup$ OP are you familiar with partial derivatives at all? $\endgroup$
    – Lee Sin
    Dec 1 '16 at 14:37

To make sure it's completely explicit: superscripts below are indices referring to either firm $1$ or firm $2$.

The choice variables in this problem are $\mathbf{q}^2$ and $\beta$. Notice that $\mathbf{q}^2$ is a vector of $n$ quantities. That is to say, $\mathbf{q}^2=\left(q_1^2,q_2^2,\ldots,q_n^2\right)$.

$(13.9)$ is just the derivative of the objective function $(13.8)$ with respect to $q_i^2$. Notice that you can rewrite the objective function $(13.8)$ as $$ p_1q_1^2+p_2q_2^2+\cdots+p_iq_i^2+\cdots+p_nq_n^2-\beta\left(q_1^1\right)-\mu_2\Phi^2\left(q_1^2,\ldots,q_i^2,\ldots,q_n^2 ,q_1^1\right) $$

Differentiating this with respect to $q_i^2$ and setting that equal to $0$ gives us $$ p_i - \mu_2 \frac{\partial\Phi^2 \left( \mathbf{q}^2,q_1^1 \right)}{\partial q_i^2} = 0 $$

In Cowell's notation, $\Phi^2_i$ is just the derivative of $\Phi^2$ with respect to $q_i^2$.

The second first-order condition is the derivative of the objective function with respect to $\beta$. Since $\beta(\cdot)$ is a decreasing function of $q_1^1$, we can also think of $q_1^1$ as a decreasing function of $\beta$. (Formally, $q_1^1$ is the inverse of $\beta$, which is well-defined since $\beta$ is decreasing. Intuitively, if firm $2$ conditions their bribe on firm $1$'s level of output, then firm $1$'s output choice also depends on the amount of the bribe.)

Thus, applying the chain rule, the derivative of the objective with respect to $\beta$ is $$ -1 -\mu_2\frac{\partial\Phi^2 \left( \mathbf{q}^2,q_1^1 \right)}{\partial q_1^1} \frac{dq_1^1}{d\beta}=0 $$ which, unfortunately, isn't quite the same as in Cowell. Notice however that $\frac{dq_1^1}{d\beta}<0$, so perhaps he is using the absolute value of that derivative to get rid of the minus sign in front of $\mu_2$.

  • $\begingroup$ can you do the same but now from firm 1's perspective? $\endgroup$
    – user10699
    Dec 1 '16 at 15:34
  • $\begingroup$ I could, but it's essentially exactly the same. If you're actually serious about learning this, you should do it on your own -- you'll learn more that way. $\endgroup$ Dec 1 '16 at 15:41
  • $\begingroup$ but i want to know if im doing it right , so if you cud that would be great $\endgroup$
    – user10699
    Dec 1 '16 at 15:45
  • $\begingroup$ Why don't you post what you've done (possibly in a separate question) so that other people can comment on whether you've done it correctly? $\endgroup$ Dec 1 '16 at 15:47
  • $\begingroup$ just by looking at the foc's for firm 1, in ( 13:14 ) ∂q11 and in (13:15) ∂q12 , although im not sure how that would give p1q1i in (13:14) $\endgroup$
    – user10699
    Dec 1 '16 at 16:22

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