Language competition: an economic theory of language learning and production

Proof of proposition 2.1

The second derivative of readership R with respect to n1 is positive so that we have extremal solutions. By n+p2≤C2, we obtain

Thus, if q1q2>Q1 holds, the representative producer employs language 1, only. If q1q2<Q1 holds, the representative producer employs language 2, only.

Proof of proposition 2.2

Assume m literary producers, among them a producer called A. Focus on producer A who is faced with the literary productions of the other authors n1others (language 1), n2others (language 2), and nothers=n1others+n2others<n. Indicating the production of producer A with A, we define

and obtain n2A=nA−n1A. The second partial derivative of RAn1A,n1others,n2others (as given in the main text) with respect to n1A is positive. We find

where the last inequality uses the assumption made in the main text that C2 is sufficiently large. Assume that the others (predominantly) choose language 1 so that n1others is large and n2others small. Then, Qcrit is small and producer A also uses language 1. Therefore, we are justified in looking for symmetric equilibria only. A separating equilibrium might exist, but it would be unstable. Thus, all producers choose language 1 if

with Q2∞:=limm→∞Q2m=1−p1+n1−α12C1+α11−p21−α22C2+α2.

In contrast, all producers choose language 2 if

with Q1∞:=limm→∞Q1m=1−p11−α12C1+α11−p2+n1−α22C2+α2.

Now, from

we obtain the results (a) though (c) of the proposition. In particular, in case of

language 2 should be adopted according to proposition 2.1, but language 1 is adopted on one of the two equilibria according to case (b) of the current proposition.

Proof of proposition 2.3

Forming the derivative of RA with respect to n1A yields

with negative second derivative. Thus,

is the first-order condition for readership maximization on the part of A. Similarly,

is the corresponding condition for B. The Nash equilibrium is the tuple of strategies n1A,n1B that fulfills both equalities.

Proof of proposition 3.2

We now solve the three-period model by backward induction. Since no decisions are made in period p, we have two stages. We begin with language learning in period f.

1. Learning language 1 is better than learning no language if

   1. c≤1−αf1+n1+α−α2f0+n0+p0=:c1f

2. Learning language 0 is better than learning no language if

   1. c≤1−α2f0+n0+p0=:c0f

3. Learning language 0 is better than learning language 1 if

   1. f0+n0+p0>f1+n1

1. Large language-0 base: f0+n0+p0>f1+n1

   1. In period f, language 0 is learned by the readership proportion r0f,la0:=c0fC=1−α2f0+n0+p0C

   2. In period f, no language is learned by the readership proportion rnof,la0:=1−c0fC

2. Small language-0 base: f0+n0+p0<f1+n1

   1. In period f, language 1 is learned by the readership proportion r1 f, sm 0:=c1 fC=(1−α)(f1+n1)+(α−α2)(f0+n0+p0)C

   2. In period f, no language is learned by the readership proportion rnof,sm0:=1−c1fC

In order to maximize period-2 readership, the producers choose the vernacular, only:

Then, we obtain:

1. Large language-0 base: n0+p0>n1 or n0>n−p02

   1. In period f, language 0 is learned by the readership proportion r0f,la0:=c0fC=1−α2p0+n0C

   2. In period f, no language is learned by the readership proportion rnof,la0:=1−c0fC

2. Small language-0 base: n0+p0<n1 or n0<n−p02

   1. In period f, language 1 is learned by the readership proportion r1f,sm0:=c1fC=1−αn1+α−α2p0+n0C

   2. In period f, no language is learned by the readership proportion rnof,sm0:=1−c1fC

We now turn to period n. Period-n learners are in the very same position as in the two-period model. Thus, a proportion of

speakers alive in period n learn language 0 while the proportion rnot 0n:=1−r0n does not learn language 0.

Since the producers of period n maximize current readership, only, they are in the same position as in the two-period model, i.e., we obtain

Proof of proposition 3.3

The proposition builds on the results obtained for current readership (see the previous appendix). In particular, we need to distinguish between large and small language-0 bases as defined above. We can safely disregard qn=qf. For large language-0 base (n0+p0>n1 or n0>n−p02), we obtain the period-n producer’s readership

The second derivative with respect to n0 is positive.

Similarly, for a small language-0 base (n0+p0<n1 or n0<n−p02), we obtain the period-n producer’s readership

again with positive second derivative,

Besides the 0-base (which can be large or small), we need the distinction between

1. increasing literary production which is given by p0<n and

2. decreasing literary production which is given by p0>n.

Assume, first, p0<n. We find:

1. for the case “incr. prod. + large 0-base”

   1. Rn,la 0n>Rn,la 0n−p02⇔

      C<1−α2+2α2α+α2+23n+p0=:Cla0,p0<n,n≻n−p02

2. for the case “incr. prod. + small 0-base”

Thus, the important comparison concerns n0=n versus n0=0. We obtain

We now turn to decreasing literary production (p0>n) which excludes n0<n−p02. Thus, we have the case “decr. prod. + large 0-base”. Here, we find

The preceding calculations imply the proposition

1. Language 0 is employed if C is sufficiently small. In particular:

   1. For p0>n (decreasing literary production), the producer employs language 0, only, in case of

      C<2α+α2+21−α1+αn+p0=:Cp0>n

      and language 1, otherwise.

   2. For p0<n (increasing literary production), the producer employs language 0, only, in case of

      C<1−α1+αnα+12+p0α+α2+2=:Cp0<n

      and language 1, otherwise.

2. Cp0>n>Cp0<n is easily confirmed. Thus, the chances for language 0 are smaller with increasing literary production than with decreasing literary production.

Proof of proposition 3.4

Assume α<α0, q1<q2 (without loss of generality), and C>p0+n. We solve a one-stage model (no language learning). Readership is then defined by

By the above assumptions, we have three cases only:

1. 0<q11−α0−q2α0−α<q21−α0−q1α0−α or, equivalently, q1q2<α0−α1−α0. Readership is maximized for n0=0,n1=0,n2=n.

2. q11−α0−q2α0−α<0<q21−α0−q1α0−α or, equivalently, 1−α0α0−α>q1q2>α0−α1−α0. Readership is maximized for n0=0,n1=0,n2=n.

3. q11−α0−q2α0−α<q21−α0−q1α0−α<0 or, equivalently, 1−α0α0−α<q1q2.

Together with q1<q2, the last inequality implies 1−α0α0−αq2<q1<q2 and α0>1+α2.

Proof of proposition 3.5

Assume α>α0, q1<q2 (without loss of generality), and C>p0+n. We solve the two-stage model by backward induction. We begin with language learning in period n. For a reader whose mother tongue is language 1, learning language 0 is better than learning language 2 if

We denote this case by “la 1” and the opposite case by “sm 1”. We now turn to the question of whether learning a language is better than learning no language. For readers with mother tongue 1, we have:

1. In case of “la 1”, learning language 0 is better than learning no language if

   1. c≤1−α0n0+p0=:c01

   holds (readership proportion r01,la 1:=c01C=1−α0n0+p0C)

2. In case of “sm 1”, learning language 2 is better than learning no language if

   1. c≤1−αn2=:c21

   holds (readership proportion r01,sm 1:=c21C=1−αn2C).

The “la 1”-“sm 2” readership (case n0+p0>1−α1−α0n2 and n0+p0<1−α1−α0n1) is

and the “sm 1”-“la 2” readership is (reversing the roles)

The readership in the “sm 1”-“sm 2” case is

In all these four cases, the Hessian with respect to n1 and n2 is positive definite. Thus, the readership functions are strictly convex and at least one of the three variables n0,n1, or n2 is zero. Letting one of these variables be zero, the restricted readership functions are convex, too. Therefore, it is sufficient to focus on the three extreme cases:

1. n0=n,n1=0,n2=0

1. n0=0,n1=n,n2=0

If, however, p0<1−α1−α0n, we have the “la 1”-“sm 2” readership

1. n0=0,n1=0,n2=n

If p0>1−α1−α0n, we have the readership

If, however, p0<1−α1−α0n, we obtain the readership

So far, we have not made use of α>α0 and q1<q2. From now on, we call p0>1−α1−α0n “p0 la” and p0<1−α1−α0n “p0 sm”. Starting with the latter, we find

and, analogously,

By q2>q1 and α>α0, we have

Thus, for “p0 sm” (p0<1−α1−α0n), we obtain:

1. for C<Cp0 sm,0≻2, all production takes place in language 0,

2. for C>Cp0 sm,0≻2, all production takes place in language 2.