Hi vitaminc,
Thank you for your questions! Looks like this whole economics grad
school is paying off after all. I assume these questions are from some
sort of problem set, so I’ll try to walk you through the solution path
as opposed to just state the final answers. I’d also encourage you to
work through this problem based on my solution. That way you can also
recheck my algebra… :-)
(1) In Stackelberg leader examples, the basic notion is that the firm
who gets to move first, has the so-called “first-mover” advantage.
This will also be the case here.
We are given that p = a – b*q. Firm 1’s MC is c, while firm 2’s MC is
cH with probability p and cL with prob. 1-p. We can split up q into
q1, the quantity produced by firm 1 and q2, where q = q1 + q2.
The problem now boils down to maximizing the two firms’ profit
functions, where firm 1 maximizes given the result from firm 2’s
maximization. This is also what gives it the aforementioned
first-mover advantage.
Okay, so here we go with the algebra:
max pi2 (=profit of the 2nd firm) with respect to q2, where pi2 = q2
(a - b*q1 - b*q2), which will yield Q2(q1), the quantity firm 2
produces, given firm 1's production of q1. Since we are dealing with
probabilities here, it's actually the expected value of pi2 we are
maximizing:
max E(pi2) = max (p)(a-b*q1-b*q2-cH)(q2) + (1-p)(a-b*q1-b*q2-cL)(q2)
We maximize that by taking the derivative with respect to q2, setting
it equal to zero, and solving for q2. That yields
Q2(q1) = q2 = (1/(2b))(a-b*q1-p*cH-(1-p)*cL)
Now we can maximize firm 1's profit function, given this value of
Q2(q1):
max pi1 = (q1)(a-a-b*q1-b*Q2(q1)-c) =
(q1)(a-a-b*q1-b*(1/(2b))(a-b*q1-p*cH-(1-p)*cL)-c)
Once again, we take derivatives, this time with respect to q1, set it
equal to 0 and - voilą - we get q1, our final result, after some
algebraic manipulation:
q1 = (1/(2b))(a+p*cH+(1-p)cL-2c)
That's the optimum quantity for firm 1 to produce, given it is the
Stackelberg leader.
(2) Shapley value might at first appear to be a pretty arcane concept.
It is indeed, although from a purely mechanical perspective, it's
actually quite simple. I won't try to explain here what Shapley was
thinking when he came up with this stuff in his Ph.D. dissertation,
but I'd strongly urge you to open your favorite game theory textbook
for some theoretical background. Mas-Colell, Whinston, and Green's
Microeconomic Theory would do as well. They spend about 5 pages trying
to explain the concept (p. 679-684).
Okay, here's the mechanics for your specific example. There are six
possible orderings in this 3 person cooperative game:
{1,2,3}, {1,3,2}, {2,1,3}, {2,3,1}, {3,1,2}, {3,2,1}
We'll now compute three values, Sh1, Sh2 and Sh3, where Sh1 is the
marginal contribution of player 1, averaged over these 6
possibilities. That is, in case 1, player 1 contributes v(1)=10 at the
margin. The same in case 2. In case 3, player 1 contributes v(2,1) -
v(2) = 15 - 10 = 5 [note that v(2,1) = v(1,2)]. In case 4, player 1
contributes v(2,3,1) - v(2,3) = 50 - 25 = 25 [again, note that
v(2,3,1) = v(1,2,3)]. In case 5, player 1 then contributes v(1,3) -
v(3) = 20 - 10 = 10. And finally in case 6, it contributes again
v(3,2,1) - v(2,3) = 50 - 25 = 25. Averaging over these 6 values, we
get (10 + 10 + 5 + 25 + 10 + 25)/6 = 85/6 = 14.17.
Following the same logic for Sh2 and Sh3, we get Sh2 = (5 + 30 + 10 +
10 + 30 + 15)/6 = 100/6 = 16.67 and Sh3 = (35 + 10 + 35 + 15 + 10 +
10)/6 = 115/6 = 19.17.
The overall Shapley value for this game is then denoted as Sh =
(Sh1,Sh2,Sh3) = (14.17, 16.67, 19.17).
Again, please check out your favorite Game Theory book for what these
Shapley values really mean. In essence, it's a method to come up with
a reasonable way to divide gains from cooperation, but there's of
course much more to this concept then this simple example might
indicate.
I hope my answers are useful to you. Good luck studying!
gwagner-ga |