Published by De Gruyter May 25, 2017

# Trade and growth in a model of allocative inefficiency

• Richard Cothren and Ravi Radhakrishnan

## Acknowledgement

We would like to thank the associate editor and two anonymous referees for their comments.

## Appendix

Proof of Proposition 1: As is clear from Equation (11), the optimal contribution of type B firms in the interval [0, θρ) is β* = 0. On the interval [θρ, ∞), the payoff πB has a derivative equal to (θρhB/β2 − 1) which is greater than, equal to, or less than 0 as θρhB is greater than, equal to, or less than β2. It follows also that if this derivative is less than zero at θρ, then it is less than zero on the entire interval [θρ, ∞). Therefore, if the derivative is negative at θρ, then the optimal value for β on [θρ, ∞) is β = θρ. In this case, the globally optimal β, as established above, is β = 0.

If the derivative at β = θρ is positive, that is, if ρ < hB/θ, then the optimal β on [θρ, ∞) is

(31)β=(θρhB)1/2.

However, this value of β is globally optimal if and only if the payoff at β = (θρhB)1/2 is greater than the payoff obtained when β = 0. This requires

(32)ρ<(hB)/4θρ^.

This establishes that the optimal value for β, β*, is

(33)β={0ifρhB/4θρ^(θρhB)1/2ifρ<ρ^

Given the values for β* in Equation (33), we can now determine the optimal allocation for the type R leader. As is clear from Equation (33), the optimal value for ρ lies in the interval [0, ρ^hB/4θ]. This is because, at ρ = ρ^, the follower’s contribution falls to zero and the type R leader receives the entire labor force so that his output is maximized. Any amount greater than ρ^ would mean a higher cost to lobbying without additional payoff in terms of output. Therefore, given Equation (33), the payoff function for the type R firms is

(34)πR={(θρ)1/2hR/hB1/2ρifρ<ρ^πR=hRρ^ifρ=ρ^

Let f(ρ) ≡ (θρ)1/2hR/hB1/2ρ and note that f′(ρ) >, <, or = 0, as ρ <, >, or = ρ* ≡ (θhR2)/4hB. From these facts it follows that πR is maximized at ρ^, if ρ^hB/4θρ* ≡ (θhR2)/4hB, that is, if θhB/hR. In this case β = 0 and all labor is allocated to the type R firms.

Suppose θ < (hB/hR) and thus ρ* is less than ρ^. Then in this case f′(ρ) = 0 at ρ = ρ* ∈ [0, ρ^) and therefore ρ* is a candidate optimal value for ρ. We say candidate because it still may pay the type R firms to generate a discrete jump in πR by increasing ρ from ρ* to ρ^.

Note that when ρ = ρ* ≡ (θhR2)/4hB, the payoff to the type R firms is πR(ρ*) = (θhR2)/2hB − (θhR2)/4hB = (θhR2)/4hB. However, this must be compared to πR(ρ^) = hRhB/4θ. One finds that when ρ* < ρ^,

(35)πR(ρ)>,<or=πR(ρ^)asg(θ)θ2hR24hBhRθ+hB2>,<,or=0.

The quadratic function g(θ) has zeros at θ = (hB/hR)(2 − 3) and θ = (hB/hR)(2 + 3) and is negative on the interval [(hB/hR)(2 − 3), (hB/hR)(2 + 3)]. Thus on this interval the optimal value for ρ is ρ^. When θ < θC ≡ (hB/hR)(2 − 3), the optimal value is ρ*. When θ > (hB/hR)(2 + 3), θ > (hB/hR), and our previous analysis has established that the optimal value is ρ^, in which case β = 0, as discussed above.

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Published Online: 2017-5-25

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