The name is absent

If I holds, since all active players at n are active at n+1 and x = ∑ x
we have a contradiction. Therefore part a) is proved. Part b) is proved
noticing that (1) implies that if x(n) = x(n+l), then x (n) = x (n+1) V i ∈ N
1                   i

∩ N+1. But since all active players at n will be active at n+1 and y > O we

reach a contradiction. Therefore x(n) < x(n+l), A.2. plus (1) show that x (n)
i

> X (n+1) V i ∈ N ∩ N+1. Finally if i & N+1 but i ∈ N x (n) > x (n+1) = O .—
i                                                                                       i            i                 ■

If A. 2 does not hold, Proposition 1 fails as the following examples

-which refer to the Cournot model- show.

Example 1.- (Seade (1980)). Let p = x °⁸, C = x . Using the first order
----------------------------------- i i

conditions of profit maximization, it is easily seen that x (1) < x (2).
i                i

2

Example 2.- p = a - bx, C = ex + d/2 x with a > c, d < O, d + 2b > O and
--------------------------------- i i                     i

d + b < O. (Total costs will be negative for x large enough, but this can be
i

(a - c)n

fixed). Then x = ------------ so x is decreasing with n if b + d < O. On the

b+d+nb

other hand second order conditions are fulfilled if d+2b > O. A graphical

argument similar to Example 2 can be found in Mc Manus (1964).

We now turn to study how payoffs change with entry.

Proposition 2: Under A.l, A.2 and A.3

a) U (n) ≥ U (n+1).
i                    i

b) If y > O the above inequalities are strict.

17

<<   13   14   15   16   17   18   19   >>

More intriguing information