Existence and Uniqueness Of

Appendix 2

Existence and uniqueness of

From conditions (i) and (ii) and surrounding discussion in the main text, we assume that there is a continuous, 2-variable benefit function s(z,q), defined for , and satisfying:

s(z,q) > 0 / for all z and p, / (A1a)
s(z,q) / is monotonically increasing in z for fixed p, / (A1b)
s(z,q) / is monotonically decreasing in p for fixed z, / (A1c)
/ as , for each z. / (A1d)

It follows from (A1c) that, for each z, s(z,q) < s(z,0) for all q > 0. Further, from (A1b), for all .

Let a be a constant, with 0 < a < s(a,0). Then, by (A1c) and (A1d), for each there is a unique popularity value such that

/ (A2)

Figure A1 below shows the intuition behind this. It follows from the continuity of s(z,q) that is continuous in z and a. Further, from (A1b), is monotonically increasing in z for fixed a, and from (A1c) is decreasing in a for fixed z. It also follows from (A1d) that as for each z.

Now suppose that . Then it follows from (A1b) that there exists a unique such that

/ (A2)

Further, it follows from (A1b) that is monotonically increasing in a. Also, s(z,q) < a for all and . On the other hand, if , then s(z,0) > a, and it follows from (A1c) and (A1d) that there is a unique value such that (A2) holds (see Figure A1). We also have s(z,q) > a for and s(z,q) < a for .

Clearly, by construction, as . We therefore define for . With this definition, it follows that for all , and from the continuity of s(z,q), that is continuous in both z and a. Further, from (A1b), (A1c) and (A3), it follows that is monotonically increasing in for fixed a. Also, for fixed , decreases as a increases.

Now suppose that, for 0 < a < s(b,0), we set

/ (A4)

Then it follows from the construction of that k(a) is a continuous, monotonically decreasing function of a. Clearly, if a = s(b,0), then , and for all z. Hence, .

At the other extreme, since as , then given any constant we may find an (depending on ), such that for all and . Thus, for such an a, we have

Since is arbitrary, this shows that as .

Since increases monotonically from 0 to as a decreases from s(b,0) to 0, it now follows that there is a unique , with , such that . Then, setting if , and c = a if , we define . We therefore have that

/ (A5)

This shows that is a probability density function on [a,b] satisfying: (i) is continuous; (ii) for z < c (if c > a); (iii) is monotonically increasing for . It completes the construction of the unique candidate ESS equilibrium .

Figure A1. The construction of the function . The red curve is the function as a function of q for fixed z. If z is such that , then exists, as shown. For any with , then the curve (blue curve) lies below for all . In this case we set . Thus, is defined for all z in the range , and all .

Proof that is an ESS

We have constructed the probability density function so that

/ is continuous on [a,b], / (A6a)
/ for (applicable if c > a), / (A6b)
/ is monotonically increasing for , / (A6c)
, / a constant, for , / (A6d)
/ for . / (A6e)

We require to prove that

/ (A7)

for all and -probability densities .

To prove this result, we need to make a stronger assumption than (A1c), namely that

is differentiable in q for each z, and for all q 0. / (A8)

For any -probability density function p(z) defined for , we let and with . Let . Then

, and / (A9a)
. / (A9b)

Thus, from (A6b), (A6d), (A6e), (A9a), and (A9b), we have

That is

/ (A10)

with equality if and only if m = 0; i.e. if and only if v(z) = - p(z) = 0 almost everywhere on [a,c).

Now observe that . Using the assumption (A8), we apply the Mean Value Theorem in the form

/ (A11)

Then, , and by (A11) and (A10),

for any , since by assumption (A8), and implies that almost everywhere on [a,b]. This shows that (A7) holds, and hence that is an ESS.

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