Schooling, Political Participation, and the Economy. (Online Supplementary Appendix: Not for Publication)

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1 Schooling, Political Participation, and the Economy Online Supplementary Appendix: Not for Publication) Filipe R. Campante Davin Chor July 200 Abstract In this online appendix, we present the proofs for the propositions in the model developed at the end of the paper, Schooling, Political Participation, and the Economy. Proof of Proposition Proof. he existence of a solution to the effort-allocation problem is guaranteed by the fact that the maximand P7), where P indicates a numbered equation from the model set up in the paper, is a continuous function over the compact simplex defined by P8) and the non-negativity constraints. Now, substitute x = H h A h M h S into P7). reating this as an unconstrained maximization problem, the first-order conditions with respect to h A, h M, and h S jointly imply that: and also that: αh α A α = µp M A M h µ M K µ = p S A S h S S, ) τx)) αh α A α = τ X) h α A α α µ hα A α h M α hα A α h S, his last equation can be rewritten as: τx) τ X) = α h A µ h M h S. 2) he assumption that τ 0) and the Cobb-Douglas production functions which satisfy a similar Inada condition), ensure that the non-negativity constraints do not bind in practice, since the infinite Harvard Kennedy School, Harvard University. Address: 79 JFK Street, Cambridge, MA 0238, USA. filipe Corresponding author) School of Economics, Singapore Management University. Address: 90 Stamford Road, Singapore 78903, Singapore.

2 marginal product in a neighborhood of zero guarantees that it is optimal to allocate a positive amount of effort to every activity. hus, the first-order conditions above from the unconstrained maximization problem also pin down the solution to the constrained problem. Differentiating ) yields: while differentiating 2) yields: = µ h A dh M α h M = h A dh S α h S, 3) Θ = α dh M µ dh S, 4) where Θ τ τ. Since τ < 0 and τ > 0, this implies that Θ 0. Finally, differentiating the τ τ budget constraint P8) yields: dh M dh S =. 5) Based on 3), we know that, dh M and dh S share the same sign since 0 < α, µ, < ). In addition, 4) implies that also shares this same sign because Θ 0. It immediately follows from 5) that this sign has to be positive. Proof of Proposition 2 Proof. Substituting 3) into 4) and 5), we obtain: where Λ α h A µ h M h S α α h A µ µ h M h S = ΘΛ, 6). It follows that the sign of d 2 x dθ depends on and. Our functional form assumption on τx) simplifies the problem as Θ is a positive constant equal to x ). We d can thus conclude that the sign of 2 x will be the opposite of the sign of. Now, note that: µ ) α µ h M ) α h S α α ) µ α h A ) µ h S µ α ) α h A µ ) µ h M dh M 7) dh S. 8) where denotes equality up to a positive multiplicative constant. In the proof of Proposition 3, we will show that > 0, dh M < 0, and dh S < 0. Given this and the parameter assumption 0 < α < µ < <, ) all of the terms in the expression above are negative, except for µ h S dh M µ. However, collecting the two terms involving µ) and using 6), we obtain: ) dh M h S µ ) µ) h M dh S = ) µ µ ) µ) h dh S M 2 < 0. 9)

3 It follows that < 0, and hence d2 x > 0. A similar approach signs the cross-derivative with respect to H and S. We have an expression for that mirrors 7), the only difference being that is replaced by S. he proof of Proposition 3 yields expressions for, dh M this yields:, and dh S, and we plug these into the expression for. Upon simplification, ) 2 α µ ) α µ h M ) α h S ) 2 µ α ) µ α h A ) µ h S α ) α h A µ ) µ h M h A h M α Θ Θ h A α µ Θ h M Θ µ 0) he terms in 0) involving h A h S and h M h S are all unambiguously positive. Moreover, we can collect all terms in h A h M, which are proportional up to a positive multiplicative constant) to: α α ) µ µ α ) µ µ ) α Θ Θ α ) µ Θ Θ > α α ) µ µ α ) µ µ ) α ) > 0. his last inequality follows from the restriction 0 < α < µ < <, which in turn ensures that: µ α ) < µ α ). ) hus, we can conclude that d > 0, from which it follows that 2 x < 0. We can repeat the same exercise for. his will yield an analogous equation to 7) with replaced by K, in which the term in is positive, but the term in dh S of an ambiguous sign. It turns out that this configuration implies an ambiguous sign for is negative, and the term in dh M d 2 x. is again Proof of Corollary 3 Proof. From 6), the behavior of d 2 x dθ will depend on that of and. It is easy to show that < 0. o show this, start by differentiating P8) and ) with dθ d < 0, so 2 x will be positive if respect to x, which gives us expressions analogous to 5) and 3), namely: = µ h A dh M α h M dh M and dh S = h A dh S, 2) α h S dh S = 0. 3) have the same sign, which has to be the opposite to that of. hese imply that, dh M Differentiating 2) with respect to x in turn yields the following analogous to 4)): Θ X dθ = dh M α µ 3 dh S. 4)

4 his implies that, dh M, dh S < 0 and > 0. Quite intuitively, a greater human capital-intensity of political activity leads individuals to choose more political participation, and to devote less human capital to production. he rest of the proof closely follows that of Proposition 2. Again, we have an expression for that mirrors 7), with the only difference being that is replaced by x. We have three positive terms the two on, plus one on dh M ), but collecting all terms that are multiplied respectively by µ α), ) α, and µ) it is easy to show that these positive terms are dominated, so that < 0. his completes the proof. Proof of Proposition 4 Proof. Part. We proceed in a similar fashion to the proof of Proposition, to set up a system of four equations in, dh M, dh S, and. o do so, we totally differentiate P8), ), and 2) with respect to. he analogue of equation 3) is now: = h A µ h A dh M α h M, 5) µ dh M h M = dh S h S. 6) Also, 4) remains unchanged, except that all derivatives with respect to H are replaced by derivatives with respect to. Finally, the constraint P8) now implies: Substituting where we define D α Θ) this into 5) in turn yields: dh M = α h M D µ h A dh M dh S from 7) into the new version of 4) yields: = ) α D µ Θ h M µ ) µ Θ α h M µ ha Θ) α µ Θ ) α h M µ = 0. 7) Θ ) α h S Θ ) α which, with some straightforward manipulation, we can simplify as: dh M = α h M D µ α Θ < 0. Note that 6) now immediately implies that dh S dh S = h S > 0, h A to keep notation simple. Substituting h S α h M D µ < 0, since it must have the same sign as dh M. In fact: α h S D α Θ < 0. Now we can substitute into 7) the expressions for, dh M, and dh S his yields: = D α µ h M µ ) α 4 α h S ). α, that we have just obtained.

5 Since µ α < 0 and α < 0, we have > 0. Part 2. We proceed in an analogous fashion as in the proof of Part. All one has to note is that the role played by h A in Part is now played by h M, and we should replace the parameters suitably as well; what used to be α is now µ, and vice versa. Once this is done, it is easy to check that dh M > 0 just as > 0), < 0, and dh S < 0. We can also see why the sign of is ambiguous: α µ > 0, while α < 0. Part 3. A similar proof applies, with h S and replacing h M and µ respectively, in our proof of Part 2. It immediately follows that dh S α > 0, and µ > 0. Proof of Proposition 5 Proof. Manipulating ) yields: > 0, < 0, and dh M µ ) h M = α A Mp M ) h S = α A Sp S µ ha ha < 0. Now the sign of ) α µ K, ) α S. Imposing symmetry h S = S) on these two equations immediately implies: α h A = A S p S µam p M h M = A S p S ) α, ) µ K. Now we can use the functional form for τx), and equation P8), to obtain: Substituting this into 7) yields: Nx = α h A µ h M h S x N x H h S h A h M = α h A µ h M h S h S = N x N H α x = x x N H Nα x h A Nµ x h M N x µ N x α) x αn x ) h A µ) x µn x ) h M he expression for X follows immediately from the definition X Nx. is negative, since Proof of Proposition 6 Proof. Assuming an interior solution, the optimal amount of H and K from the ruler s standpoint must 5

6 satisfy the first-order conditions: where Ñ = N 2 N x Ñp S A S µ µ µam p M A S p S Ñp S A S τ X)X τx) = ) µ τ X)X τx) = F H H) 8) F K 9) K). τ X)X τx) = 0. 20) is a positive constant that depends only on model parameters. he first term in square brackets in 8) and 9), τ X)X, captures the marginal political cost of providing citizens with an extra unit of human capital: It will increase political participation, and thus reduce the share that can be captured by the ruler. On the other hand, the second term, τx), represents the marginal benefit, which stems from the additional output that is generated, part of which goes to the ruler. he marginal benefit, net of the marginal political cost, has to equal the marginal cost of factor provision, which is foregone consumption. We now characterize how the ex ante choice of H, which is implicitly defined by 20), will be affected by the key variables of the model. Define GX) τ X)X τx). It is easy to show, using the implicit function theorem applied to 8) and the second-order conditions of the optimization problem, that for any variable or parameter of interest j, we will have the sign of H j Moreover, we have: being equal to the sign of G j. G j = τ X)X 2τ X) X j = x) τ X) X j. 2) From this, since τ X) < 0, it immediately follows that any variable that increases aggregate political participation will lead to less investment in human capital by the ruler. In particular, 2) and P3) immediately yield the result. 6

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