Knowledge spillover entrepreneurship in an endogenous growth model

Abstract

We present a model that separates entrepreneurship from profit-motivated corporate R&D aimed at improving existing production processes. Our model embeds the core idea of the knowledge spillover theory of entrepreneurship in established knowledge-based growth models by enriching their knowledge spillover structure. Introducing knowledge spillovers drives a wedge between the optimal and market allocation of resources between new knowledge creation and commercialization. We show the first best allocation depends exclusively on the relative strength of knowledge spillovers between them and derive propositions to guide policy that can bring the market equilibrium closer to this optimum.

Appendices

Appendix 1: The full dynamic optimization problem of consumers

The Hamiltonian to this problem:

$$ H_{C} = e^{ - \rho t} \log C(t) + \mu (t)\left( {r(t)B(t) + w_{E} (t)L_{E}^{*} + w_{P} (t)L_{P}^{*} - C(t)} \right) $$

(29)

yields the first order conditions:

$$ \begin{gathered} \frac{{\partial H_{C} }}{\partial C(t)} = 0 = \frac{{e^{ - \rho t} }}{C(t)} - \mu (t) \hfill \\ \frac{{\partial H_{C} }}{\partial B(t)} = - \dot{\mu }(t) = r(t)\mu (t) \hfill \\ \mathop {\lim }\limits_{t \to \infty } \mu \left( t \right)B\left( t \right) = 0 \hfill \\ \frac{{\partial H_{C} }}{\partial \mu (t)} = \dot{B}(t) = r(t)B(t) + w_{E} (t)L_{E}^{*} + w_{P} (t)L_{P}^{*} - C(t) \hfill \\ \end{gathered} $$

(30)

Taking the first two conditions, solving the first for μ(t), taking the time derivative and substituting into the second yields:

$$ \frac{{\dot{C}(t)}}{C(t)} = r\left( t \right) - \rho $$

(31)

For any constant r(t) = r we then obtain:¹⁵

$$ C\left( t \right) = C(0)e^{(r - \rho )t} $$

(32)

Now we can use the third and fourth condition to derive C(0) and express final goods demand in variables that are given to the consumer. First rewrite condition four to:

$$ \dot{B}(t) - rB(t) = w_{E} (t)L_{E}^{*} + w_{P} (t)L_{P}^{*} - C(t) $$

(33)

Then multiply both sides with integrating factor e ^−rt and solve for C(0):

$$ \begin{aligned} \frac{{{\text{d}}B(t)}}{{{\text{d}}t}}{\text{e}}^{ - rt} - rB(t){\text{e}}^{ - rt} & = w_{E} (t)L_{E}^{*} {\text{e}}^{ - rt} + w_{P} (t)L_{P}^{*} {\text{e}}^{ - rt} - C(t){\text{e}}^{ - rt} \\ \frac{{{\text{de}}^{ - rt} B(t)}}{{{\text{d}}t}} & = w_{E} (t)L_{E}^{*} {\text{e}}^{ - rt} + w_{P} (t)L_{P}^{*} {\text{e}}^{ - rt} - C(t){\text{e}}^{ - rt} \\ {\text{de}}^{ - rt} B(t) & = w_{E} (t)L_{E}^{*} {\text{e}}^{ - rt} {\text{d}}t + w_{P} (t)L_{P}^{*} {\text{e}}^{ - rt} {\text{d}}t - C(t){\text{e}}^{ - rt} {\text{d}}t \\ \mathop \int \limits_{0}^{\infty } {\text{de}}^{ - rt} B(t) & = \mathop \int \limits_{0}^{\infty } w_{E} (t)L_{E}^{*} {\text{e}}^{ - rt} {\text{d}}t + \mathop \int \limits_{0}^{\infty } w_{P} (t)L_{P}^{*} {\text{e}}^{ - rt} {\text{d}}t - \mathop \int \limits_{0}^{\infty } C(t){\text{e}}^{ - rt} {\text{d}}t \\ \end{aligned} $$

(34)

which by using the third (transversality) condition in (30) and the expression for consumption in (32) yields:

$$ - B(0) = \mathop \int \limits_{0}^{\infty } {\text{e}}^{ - rt} w_{E} (t)L_{E}^{*} {\text{d}}t + \mathop \int \limits_{0}^{\infty } {\text{e}}^{ - rt} w_{P} (t)L_{P}^{*} {\text{d}}t - C(0)\mathop \int \limits_{0}^{\infty } {\text{e}}^{ - rt} {\text{d}}t $$

(35)

such that:

$$ C(0) = r\left( {B(0) + \mathop \int \limits_{0}^{\infty } {\text{e}}^{ - rt} w_{E} (t)L_{E}^{*} {\text{d}}t + \mathop \int \limits_{0}^{\infty } {\text{e}}^{ - rt} w_{P} (t)L_{P}^{*} {\text{d}}t} \right) $$

(36)

To the consumers initial wealth, interest rate and life time wage income are given, so this determines the optimal consumption path:

$$ C(t) = r\left( {B(0) + \mathop \int \limits_{0}^{\infty } {\text{e}}^{ - rt} w_{E} (t)L_{E}^{*} {\text{d}}t + \mathop \int \limits_{0}^{\infty } {\text{e}}^{ - rt} w_{P} (t)L_{P}^{*} {\text{d}}t} \right)e^{{\left( {r - \rho } \right)t}} $$

(3)

Appendix 2: The dynamic optimization problem for the final goods producer

From the Hamiltonian in (6), we can obtain n + 5 first order conditions that characterize the profit-maximizing levels of employment in production and R&D and demand for intermediates. For production labor, we have:

$$ \frac{{\partial H_{j} }}{{\partial L_{Pj} }} = 0 = {\text{e}}^{ - rt} \left( {\beta A_{j}^{\alpha } L_{Pj}^{\beta - 1} \mathop \sum \limits_{i = 0}^{n} x_{ij}^{1 - \alpha - \beta } - w_{P} } \right) $$

(37)

which can easily be rewritten into a labor demand function:

$$ L_{Pj}^{D} = \left( {\frac{{\beta A_{j}^{\alpha } \mathop \sum \nolimits_{i = 0}^{n} x_{ij}^{1 - \alpha - \beta } }}{{w_{P} }}} \right)^{{\frac{1}{1 - \beta }}} = \frac{{\beta X_{j} }}{{w_{P} }} $$

(38)

This shows that all firms will spend exactly the same share, β, of output, X, on wages.¹⁶ Summing over all final goods producers, we obtain for total production labor demand:

$$ L_{Pj}^{D} = \frac{{\beta X_{j} }}{{w_{P} }} $$

(7)

such that the total wage sum for production workers is βX, and labor demand is stable as long as wages and production grow at the same rate in equilibrium.

For intermediates, the firm will choose their levels to satisfy:

$$ \begin{gathered} \frac{{\partial H_{j} }}{{\partial x_{0j} }} = 0 = {\text{e}}^{ - rt} \left( {(1 - \alpha - \beta )A_{j}^{\alpha } L_{Pj}^{\beta } x_{0j}^{ - \alpha - \beta } - \chi_{0} } \right) \hfill \\ \frac{{\partial H_{j} }}{{\partial x_{1j} }} = 0 = {\text{e}}^{ - rt} \left( {(1 - \alpha - \beta )A_{j}^{\alpha } L_{Pj}^{\beta } x_{1j}^{ - \alpha - \beta } - \chi_{1} } \right) \hfill \\ \ldots \hfill \\ \frac{{\partial H_{j} }}{{\partial x_{nj} }} = 0 = e^{ - rt} \left( {(1 - \alpha - \beta )A_{j}^{\alpha } L_{Pj}^{\beta } x_{nj}^{ - \alpha - \beta } - \chi_{n} } \right) \hfill \\ \end{gathered} $$

(39)

The n conditions in (39) allow one to derive the demand for intermediate good i in terms of the relative price and quantity of the nth intermediate:

$$ x_{ij}^{D} = x_{nj} \chi_{n}^{{\frac{1}{\alpha + \beta }}} \chi_{i}^{{\frac{ - 1}{\alpha + \beta }}} $$

(40)

Substituting this demand function into the production function and rewriting in terms of total output yields:

$$ \mathop \sum \limits_{i = 0}^{n} x_{ij}^{1 - \alpha - \beta } = \mathop \sum \limits_{i = 0}^{n} x_{nj}^{1 - \alpha - \beta } \chi_{n}^{{\frac{1 - \alpha - \beta }{\alpha + \beta }}} \chi_{i}^{{\frac{\alpha + \beta - 1}{\alpha + \beta }}} = x_{nj}^{1 - \alpha - \beta } \chi_{n}^{{\frac{1 - \alpha - \beta }{\alpha + \beta }}} \mathop \sum \limits_{i = 0}^{n} \chi_{i}^{{\frac{\alpha + \beta - 1}{\alpha + \beta }}} = \frac{{X_{j} }}{{A_{j}^{\alpha } L_{Pj}^{\beta } }} $$

(41)

From the nth order condition, we also know that for all i:

$$ A_{j}^{\alpha } L_{Pj}^{\beta } = x_{nj}^{\alpha + \beta } \frac{{\chi_{n} }}{1 - \alpha - \beta } $$

(42)

So combining (41) and (42) and solving for x _nj , we get:

$$ x_{nj}^{D} = \frac{{\chi_{n}^{{\frac{ - 1}{\alpha + \beta }}} }}{{\mathop \sum \nolimits_{i = 0}^{n} \chi_{i}^{{\frac{\alpha + \beta - 1}{\alpha + \beta }}} }}(1 - \alpha - \beta )X_{j} $$

(43)

And by the symmetry in the production function, this implies that all varieties i have that demand function:

$$ x_{ij}^{D} = \frac{{\chi_{i}^{{\frac{ - 1}{\alpha + \beta }}} }}{{\mathop \sum \nolimits_{i = 0}^{n} \chi_{i}^{{\frac{\alpha + \beta - 1}{\alpha + \beta }}} }}(1 - \alpha - \beta )X_{j} $$

(8)

Multiplying (8) by χ _i and summing over all varieties, i shows that total expenditure on intermediates is (1−α−β)X _j .¹⁷

Together with the result on the wage costs, this implies that the final goods producer j makes an operating profit of αX _j . We assume that final goods producers are perfectly symmetric, facing the same input and output prices, w, χ _i and 1, respectively. As they also use the same production technology, increases in the firm’s level of accumulated knowledge A _j (t) and consequently X _j (t) will cause increases in operating profit. Firms, however, have to invest labor in R&D to increase their A _j (t).

Formally the stock of knowledge is a firm specific state variable, and its optimal path is determined by choosing the optimal level of R&D labor. The final goods producer will increase R&D activity as long as the discounted future benefits of doing so exceed the current labor costs at the margin. As R&D is a deterministic process in our model, the firms can decide to spend on R&D exactly up to that point. The solution is formally characterized by two first order conditions, one transversality condition and the law of motion for A _j ¹⁸ _:

$$ \begin{gathered} \frac{{\partial H_{j} }}{{\partial L_{Ej} }} = 0 = {\text{e}}^{ - rt} w_{E} + \mu_{j} \psi A_{j}^{1 - \gamma } n^{\gamma } \hfill \\ \frac{{\partial H_{j} }}{{\partial A_{j} }} = - \dot{\mu }_{j} = {\text{e}}^{ - rt} \alpha A_{j}^{\alpha - 1} L_{Pj}^{\beta } \mathop \sum \limits_{i = 0}^{n} x_{ij}^{1 - \alpha - \beta } + (1 - \gamma )\mu_{j} \psi A_{j}^{ - \gamma } n^{\gamma } L_{Ej} \hfill \\ \mathop {\lim }\limits_{t \to \infty } \mu_{j} (t)A_{j} (t) = 0 \hfill \\ \frac{{\partial H_{j} }}{{\partial \mu_{j} }} = \dot{A}_{j} = \psi A_{j}^{1 - \gamma } n^{\gamma } L_{Ej} (t) \hfill \\ \end{gathered} $$

(44)

where the first condition sets the present value of labor costs equal to the present value of the marginal product of R&D labor times the shadow price of a marginal increase in A _j , μ _j . Solving for that shadow price yields:

$$ \mu_{j} = e^{ - rt} \frac{{w_{E} }}{{\psi A_{j}^{1 - \gamma } n^{\gamma } }} $$

(45)

Then we take the time derivative and set this expression equal to minus the right hand side in the second condition to equate the marginal return on A _j to the shadow price:

$$ \dot{\mu }_{j} = \left( {r - \frac{{\dot{w}_{E} }}{{w_{E} }} + (1 - \gamma )\frac{{\dot{A}}}{A} + \gamma \frac{{\dot{n}}}{n}} \right){\text{e}}^{ - rt} \frac{{w_{E} }}{{\psi A_{j}^{1 - \gamma } n^{\gamma } }} = {\text{e}}^{ - rt} \frac{{\alpha X_{j} }}{{A_{j} }} - (1 - \gamma ){\text{e}}^{ - rt} \frac{{w_{E} L_{Ej} }}{{A_{j} }} $$

(46)

Substituting the law of motion (5) for $\dot{A}_{j}$ into (46) and solving for w _E yields the wage level at which a positive finite amount of R&D workers will be employed by firm j:

$$ \bar{w}_{Ej} = \frac{\alpha \psi }{{r - \frac{{\dot{w}_{E} }}{{w_{E} }} + \gamma \frac{{\dot{n}}}{n}}}\left( {\frac{{A_{j} }}{n}} \right)^{ - \gamma } X_{j} $$

(47)

This wage level represents a horizontal demand function or arbitrage condition. If market wages for R&D labor exceed this threshold, no R&D workers will be employed by firm j. If wages fall short, firm j will hire additional R&D workers until all are hired or wages have gone up. This so-called bang-bang equilibrium is a result of the constant returns to R&D labor assumption that we have made. Eq. (47) holds for all firms j, and this wage will be equal for all firms j as they are price takers in the market for R&D labor. We also know by the production function in (4) and Eqs. (7) and (8) that X _j is continuous and strictly proportional in A _j .¹⁹ Thus, we obtain the result that at any point in time there is a unique level of A _j that all firms hiring R&D labor must attain. The mechanism is that the firms with A _j  = A ^max also have the highest threshold wage for R&D. They will thus bid up production wages to this threshold level and employ a positive amount of R&D. Their level of A will then rise according to (5), those with A _j  < A ^max will not hire any R&D, and their A _j remains stable. The rise in A ^max pushes up the threshold and thereby the production wage. In any equilibrium with R&D, only those firms that have A _j  = A ^max can stay in the race, whereas others are forced to bring down their production employment levels to 0.²⁰ If we assume therefore that all firms start from the same initial level of A _j (0) = A ₀, the above implies that A _j (t) = A ^max(t) = A(t) for all j, and we obtain for (47) (dropping time arguments):

$$ \bar{w}_{E} = \frac{\alpha \psi }{{r - \frac{{\dot{w}_{E} }}{{w_{E} }} + \gamma \frac{{\dot{n}}}{n}}}\left( \frac{A}{n} \right)^{ - \gamma } X $$

(9)

If we could establish the total number of R&D workers, $L_{{E\dot{A}}}$, at this point, we could put $L_{{E\dot{A}}} /m$ into (5) to derive the optimal growth rate of A _j (t) = A ^max(t) = A(t). The starting condition A _j (0) = A ₀ and the law of motion in (5) then determine the optimal path for A(t), and the transversality condition helps to solve for μ _j (t). This shadow price and the exact optimal path for A(t) are not very relevant for our purpose here.

Appendix 3: The intermediate producers and entrants

To close the model, we need to model the behavior of the intermediate producers and the potential new entrants. Intermediate producers are monopolists so they will set prices to maximize profits. They solve:

$$ \mathop {\hbox{max} }\limits_{{\chi_{i} }} \pi_{i} = \chi_{i} \mathop \sum \limits_{j = 0}^{m} x_{ij}^{D} (\chi_{i} ) - rK_{i} $$

(10)

Subject to their production function and demand for intermediates:

$$ x_{i} = K_{i} $$

(48)

$$ x_{ij}^{D} = \frac{{\chi_{i}^{{\frac{ - 1}{\alpha + \beta }}} }}{{\mathop \sum \nolimits_{i = 0}^{n} \chi_{i}^{{\frac{\alpha + \beta - 1}{\alpha + \beta }}} }}(1 - \alpha - \beta )X_{j} $$

(8)

Substitution of these constraints into the profit function and setting the first derivative with respect to χ(i) to 0 yields the profit maximizing price for intermediate i:

$$ \chi_{i} = \frac{r}{1 - \alpha - \beta } $$

(11)

which does not vary over i anymore. So every intermediate producer sets his price equal to this value, and by the demand function all intermediates are demanded in the same quantity. This implies that in equilibrium the stock of raw capital is divided equally among all n varieties, and the capital share in income is given by $rK = (1 - \alpha - \beta )^{2} X$, whereas the monopoly rents in the intermediate sector are given by:

$$ \pi_{i} = \frac{(\alpha + \beta )(1 - \alpha - \beta )X}{n} $$

$$ \mathop \sum \limits_{i = 0}^{n} \pi_{i} = (\alpha + \beta )(1 - \alpha - \beta )X $$

(12)

As new firms are assumed infinitely small, the marginal entrant will also enjoy this flow of rents forever and the value of entry given by the discounted flow of rents from entry at time T to infinity. Using (12), this is given by:

$$ V_{E} (T) = \mathop \int \limits_{T}^{\infty } {\text{e}}^{ - rt} \pi_{i} (t){\text{d}}t = (\alpha + \beta )(1 - \alpha - \beta )\mathop \int \limits_{T}^{\infty } {\text{e}}^{ - rt} \frac{X(t)}{n(t)}{\text{d}}t $$

(49)

Using the entry function in (13) and equating discounted future marginal rent income to marginal (opportunity) costs at the time of entry at time T, we can derive the entry arbitrage equation:

$$ \frac{{\partial \dot{n}(T)}}{{\partial L_{E} (t)}}V_{E} (T) = (\alpha + \beta )(1 - \alpha - \beta )\varphi A(T)^{\delta } n(T)^{1 - \delta } \mathop \int \limits_{T}^{\infty } {\text{e}}^{ - rt} \frac{X(t)}{n(t)}{\text{d}}t = \tilde{w}_{E} (T) $$

(50)

And as this trade-off is identical for entrants over time, we can replace T by t, and this equation can be rewritten into the arbitrage condition for entrepreneurial educated labor if we assume that at entry entrepreneurs expect that output and variety will expand at a constant rate (as they will in steady state):²¹

$$ \tilde{w}_{E} = \frac{(\alpha + \beta )(1 - \alpha - \beta )\varphi }{{r - \frac{{\dot{X}}}{X} + \frac{{\dot{n}}}{n}}}\left( \frac{A}{n} \right)^{\delta } X $$

(14)

Appendix 4: The steady state

From the production function (5) and substituting K/n for all x _i , we obtain for the growth rate of final output, X:

$$ \frac{{\dot{X}}}{X} = \alpha \frac{{\dot{A}}}{A} + (\alpha + \beta )\frac{{\dot{n}}}{n} + (1 - \alpha - \beta )\frac{{\dot{K}}}{K} $$

(51)

Using the budget constraint for the consumer, we know that wage and capital income must grow at the same rate as output and consumption in steady state. Wages will grow at the same rate by Eqs. (9), (14) and (15), so total labor income grows at the common rate. For consumption to grow at that rate, so must capital income at a constant interest rate. This implies raw capital will have to expand at the common rate. We can then rewrite (51) to:

$$ \frac{{\dot{X}}}{X} = \frac{\alpha }{\alpha + \beta }\frac{{\dot{A}}}{A} + \frac{{\dot{n}}}{n} $$

(52)

And as a stable labor allocation requires a constant ratio A/n, the steady-state growth rates will be equal to (16):

$$ \frac{{\dot{K}}}{K} = \frac{{\dot{X}}}{X} = \frac{{\dot{C}}}{C} = \frac{{\dot{B}}}{B} = \frac{{\dot{w}_{E} }}{{w_{E} }} = \frac{{\dot{w}_{P} }}{{w_{P} }} = r - \rho = \frac{{\dot{n}}}{n}\left( {\frac{2\alpha + \beta }{\alpha + \beta }} \right) $$

(17)

Appendix 5: The uniqueness and stability of the steady state

We can show the uniqueness of the steady-state equilibrium by investigating the properties of the right hand side of Eq. (22):

$$ \begin{aligned} \frac{{\dot{n}}}{n}^{\text{SS}} & = \frac{{L_{E}^{*} }}{{C_{1} \left( {\frac{{\rho + \dot{n}/n}}{{\rho + \gamma \dot{n}/n}}} \right)^{{\frac{ - \delta }{\gamma + \delta }}} + C_{2} \left( {\frac{{\rho + \dot{n}/n}}{{\rho + \gamma \dot{n}/n}}} \right)^{{\frac{\gamma }{\gamma + \delta }}} }} \\ C_{1} & \mathop = \limits^{\text{def}} \psi^{{\frac{ - \delta }{\gamma + \delta }}} \varphi^{{\frac{ - \gamma }{\gamma + \delta }}} \left( {\frac{\alpha }{(\alpha + \beta )(1 - \alpha - \beta )}} \right)^{{\frac{ - \delta }{\gamma + \delta }}} \\ C_{2} & \mathop = \limits^{\text{def}} m\left( {\frac{\alpha }{(\alpha + \beta )(1 - \alpha - \beta )}} \right)^{{\frac{\gamma }{\gamma + \delta }}} \\ \end{aligned} $$

(22)

For C ₁ and C ₂ > 0, we can calculate the right hand side of Eq. (22) in the limit for $\dot{n}/n \to \infty$ and obtain a positive constant:

$$ \mathop {\lim }\limits_{{\dot{n}/n \to \infty }} {\text{RHS}} = \frac{{L_{E}^{*} }}{{C_{1} + C_{2} }} > 0 $$

(53)

For growth rates of 0, the right hand side of (22) simplifies to:

$$ \mathop {\lim }\limits_{{\dot{n}/n \to \infty }} {\text{RHS}} = \frac{{L_{E}^{*} }}{{C_{1} \gamma^{{\frac{\delta }{\gamma + \delta }}} + C_{2} \gamma^{{\frac{ - \gamma }{\gamma + \delta }}} }} > 0 $$

(54)

The right hand side of Eq. (22) will increase or decrease monotonically from one positive constant $C_{3} \mathop = \limits^{\text{def}} \frac{{L_{E}^{*} }}{{C_{1} + C_{2} }}$ to another $C_{4} \mathop = \limits^{\text{def}} \frac{{L_{E}^{*} }}{{C_{1} \gamma^{{\frac{\delta }{\gamma + \delta }}} + C_{2} \gamma^{{\frac{ - \gamma }{\gamma + \delta }}} }}$ as the growth rate, also on the left hand side, goes from 0 to infinity. This implies there is a single steady state in the model. As it is not a priory clear that C ₃ <> C ₄, we can draw the right and left hand side of Eq. (22) in the two panels in Fig. 3.

Fig. 3

Stability of the steady state

The arrows indicate that in both cases the steady state is stable if the growth rate of n increases when the right hand side of Eq. (22) exceeds the left hand side. We can establish this by considering the alternative. If the steady state is not stable, this implies that variety expansion would go to 0 or infinity out of steady state. Both cannot be an equilibrium outcome. Variety expansion can only be zero when all educated labor is allocated to R&D, in which case A/n would inevitably rise and educated labor would switch to entrepreneurship as soon as this drives the wage ratio in Eq. (16) above 1, turning the growth rate of n positive. Likewise, the rate of variety expansion cannot go to infinity for any finite level of employment in entrepreneurship unless the ratio A/n goes to infinity. Consequently, A would have to outgrow n, and its growth rate would have to rise faster than that of n, in which case Eq. (16) would inevitably drop below 1 and (all) educated labor would reallocate to R&D. This implies there is a unique and stable steady-state growth rate of n for which (22) holds. We cannot derive the transitional dynamics toward the steady state because we have assumed constant returns to educated labor in both entrepreneurship and R&D. This implies that the labor market equilibrium is a bang-bang equilibrium, and this discontinuity prevents us from analytically solving for out of equilibrium dynamics. We have also assumed perfect foresight and abstracted from uncertainty, such that the effective discount rates are assumed to immediately adjust to their long-run equilibrium values (which depend on the rates of innovation). If instead we had assumed that, more realistically, these effective discount rates are set and updated if proven wrong, then convergence toward the single steady state would also be driven by that process. That assumption, however, would needlessly complicate our model in the steady state.

Appendix 6: The central planner

The problem is a dynamic optimization problem with two control and three state variables that can be solved by constructing a Hamiltonian and deriving the first order conditions for that problem. The Hamiltonian is given by:

$$ \begin{aligned} H_{C} & = e^{ - \rho t} \log C(t) + \lambda (t)\left( {A(t)^{\alpha } L_{P}^{*} (t)^{\beta } K(t)^{1 - \alpha - \beta } n(t)^{\alpha + \beta } - C(t)} \right) \\ \quad + \mu (t)\left( {\psi A(t)^{1 - \gamma } n(t)^{\gamma } L_{{E\dot{A}}} (t)} \right) + \nu (t)\left( {\varphi A(t)^{\delta } n(t)^{1 - \delta } \left( {L_{E}^{*} - L_{{E\dot{A}}} } \right)} \right) \\ \end{aligned} $$

(55)

In this setup, the optimal rate of capital accumulation is computed in a standard fashion. To save on notation, we will drop time arguments where this causes no confusion. Taking the first derivative of the Hamiltonian with respect to consumption and setting equal to zero yields an expression for the shadow price of consumption in terms of utility. This is given by:

$$ \lambda = \frac{{e^{ - \rho t} }}{C} $$

(56)

Taking the derivative of the Hamiltonian with respect to capital, K, and setting it equal to minus the change in the shadow price of consumption yields the optimal consumption path:

$$ - \dot{\lambda } = \frac{(1 - \alpha - \beta )X}{K}\lambda $$

(57)

Solving this first order differential equation, using (56), we obtain:

$$ \frac{{\dot{C}}}{C} = gC = \frac{(1 - \alpha - \beta )X}{K} - \rho $$

(58)

which shows that the central planner will save, invest and consume to keep the growth rate of consumption equal to the marginal product of capital minus the discount rate. This is a standard result in the literature [see, e.g., Barro and Sala-I-Martin (2004)] that our model replicates. The central planner invests up to the point where the marginal loss of utility due to lower immediate consumption is exactly offset by the additional discounted marginal utility increase due to higher future consumption (cf. Eq. 31 above).

The other first order conditions are more interesting. With respect to the control variable $L_{{E\dot{A}}}$, we obtain that the central planner equates their marginal contribution to discounted utility in the two alternative activities for skilled labor, invention and commercialization:

$$ A^{1 - \gamma } n^{\gamma } \mu = \varphi A^{\delta } n^{1 - \delta } \nu $$

(59)

where μ and ν represent the shadow prices of the marginal process; an increase in A, and product; an increase in n, innovation, respectively. It will be convenient to solve (59) for µ and compute the growth rate of this shadow price in the optimum as:

$$ \dot{\mu }/\mu = \dot{\nu }/\nu + (1 - \gamma - \delta )(\dot{n}/n - \dot{A}/A) $$

(60)

The first order conditions with respect to A are given by:

$$ \frac{{\partial H_{C} }}{\partial A} = - \dot{\mu } = \alpha \lambda \frac{X}{A} + \mu (1 - \gamma )\frac{{\dot{A}}}{A} + \nu \frac{n}{A}\delta \frac{{\dot{n}}}{n} $$

(61)

Rearranging this can be rewritten as:

$$ \frac{{\dot{\mu }}}{\mu } = - \alpha \frac{X}{A}\frac{\lambda }{\mu } - \delta \frac{n}{A}\frac{\nu }{\mu }\frac{{\dot{n}}}{n} - (1 - \gamma )\frac{{\dot{A}}}{A} $$

(62)

The first order condition with respect to n is given by:

$$ \frac{{\partial H_{C} }}{\partial n} = - \dot{\nu } = (\alpha + \beta )\lambda \frac{X}{n} + \mu \gamma \frac{A}{n}\frac{{\dot{A}}}{A} + \nu (1 - \delta )\frac{{\dot{n}}}{n} $$

(63)

This expression can be rewritten into:

$$ \frac{{\dot{\nu }}}{\nu } = - (\alpha + \beta )\frac{X}{n}\frac{\lambda }{\nu } - (1 - \delta )\frac{{\dot{n}}}{n} - \gamma \frac{A}{n}\frac{\mu }{\nu }\frac{{\dot{A}}}{A} $$

(64)

When we equate Eq. (62) to (60) and substituting for the implied ratio between μ and ν in Eq. (59), we obtain:

$$ \frac{{\dot{\nu }}}{\nu } = - \delta \frac{{\dot{A}}}{A} - \alpha \frac{X}{A}\frac{\lambda }{\mu } - \left( {1 - \gamma - \delta - \delta \frac{\psi }{\varphi }\left( \frac{A}{n} \right)^{ - \gamma - \delta } } \right)\frac{{\dot{n}}}{n} $$

(65)

Equating this to Eq. (64), taking all terms including X to the left hand side, rearranging and using the laws of motion to eliminate $\dot{A}$ and $\dot{n}$ from the right hand side allows us to write:

$$ \frac{\lambda X((\alpha + \beta )\mu A - \alpha \nu n)}{\mu A\nu n} = \delta \psi \left( \frac{A}{n} \right)^{ - \gamma } - \gamma \varphi \left( \frac{A}{n} \right)^{\delta } $$

(66)

We can show that a steady state exists for $\frac{{\dot{A}}}{A} = \frac{{\dot{n}}}{n}$ and $\frac{{\dot{X}}}{X} = \frac{{\dot{K}}}{K} = \frac{{\dot{C}}}{C} = \left( {1 - \alpha - \beta } \right)\frac{X}{K} - \rho$ by guessing it exists, and then show it is indeed a steady-state equilibrium that satisfies the first order and transversality conditions:

$$ \begin{gathered} \mathop {\lim }\limits_{t \to \infty } \lambda \left( t \right)K\left( t \right) = 0 \hfill \\ \mathop {\lim }\limits_{t \to \infty } \mu \left( t \right)A\left( t \right) = 0 \hfill \\ \mathop {\lim }\limits_{t \to \infty } \nu \left( t \right)n\left( t \right) = 0 \hfill \\ \end{gathered} $$

(67)

By Eq. (56) and the “guess” that the growth rate of K is the same as that of C, we can derive that the product $\lambda (t)K(t)$ will grow at $- \rho$ in the steady state. As this is negative, the first transversality condition is satisfied. By Eq. (60) we can immediately derive that if $\frac{{\dot{A}}}{A} = \frac{{\dot{n}}}{n}$, the shadow prices for A and n also grow at the same rate, and we obtain $\frac{{\dot{\mu }}}{\mu } = \frac{{\dot{\nu }}}{\nu }$. Therefore, the growth rate of the products $\mu (t)A(t)$ and $\nu (t)n(t)$ are equal and have to be negative to satisfy the transversality conditions, such that:

$$ - \frac{{\dot{\mu }}}{\mu } = - \frac{{\dot{\nu }}}{\nu } > \frac{{\dot{A}}}{A} = \frac{{\dot{n}}}{n} $$

(68)

Taking the growth rate of the left hand side in Eq. (66) and using $\frac{{\dot{\lambda }}}{\lambda } + \frac{{\dot{K}}}{K} = \frac{{\dot{\lambda }}}{\lambda } + \frac{{\dot{X}}}{X} = - \rho$, $\frac{{\dot{A}}}{A} = \frac{{\dot{n}}}{n}$ and $\frac{{\dot{\mu }}}{\mu } = \frac{{\dot{\nu }}}{\nu }$ the left hand side growth rate is given by:

$$ \frac{{{\text{L}}\mathop {\text{H}}\limits^{.} {\text{S}}}}{\text{LHS}} = - \frac{{\dot{n}}}{n} - \frac{{\dot{\nu }}}{\nu } - \rho $$

(69)

If this growth rate is positive, the left hand side will go to infinity in the steady state. As the right hand side of Eq. (66) is a function of $\frac{A}{n}$ and parameters only, we know that in that case $\frac{A}{n}$ should go to infinity (or zero depending on γ, δ, φ and ψ) as well. As that cannot be a steady state, we assume the growth rate in (69) is negative. With the constraint in inequality (68), this implies $= \frac{{\dot{n}}}{n} < - \frac{{\dot{\nu }}}{\nu } < \frac{{\dot{n}}}{n} + \rho$ and we know the right hand side of Eq. (66) will be zero in the steady state. This allows us to obtain for the steady state:

$$ \frac{A}{n}^{\text{SS}} = \left( {\frac{\delta }{\gamma }\frac{\psi }{\varphi }} \right)^{{\frac{1}{\gamma + \delta }}} $$

(70)

By Eqs. (5′) and (13′) and $\frac{{\dot{A}}}{A} = \frac{{\dot{n}}}{n}$, we can also derive that in steady state the share of total educated labor allocated to R&D is given by:

$$ \frac{{L_{{E\dot{A}}} }}{{L_{E}^{*} }} = \frac{\varphi }{{\varphi + \psi \left( \frac{A}{n} \right)^{ - \gamma - \delta } }} $$

(71)

And using Eqs. (71) and (70) in (5′) and (13′), we can derive the steady-state rate of knowledge accumulation and variety expansion:

$$ \frac{{\dot{A}}}{A}^{\text{SS}} = \frac{{\dot{n}}}{n}^{\text{SS}} = \frac{{\left( {\gamma^{\gamma } \delta^{\delta } \varphi^{\gamma } \psi^{\delta } } \right)^{{\frac{1}{\gamma + \delta }}} }}{\gamma + \delta }L_{E}^{*} $$

(72)

Taking growth rates of Eq. (23) and solving for growth in output using $\frac{{\dot{X}}}{X} = \frac{{\dot{K}}}{K}$ yields:

$$ \frac{{\dot{X}}}{X}^{\text{SS}} = \frac{{\dot{K}}}{K}^{\text{SS}} = \frac{{\dot{C}}}{C}^{\text{SS}} = \frac{2\alpha + \beta }{\alpha + \beta }\frac{{\dot{A}}}{A}^{\text{SS}} = \frac{2\alpha + \beta }{\alpha + \beta }\frac{{\left( {\gamma^{\gamma } \delta^{\delta } \varphi^{\gamma } \psi^{\delta } } \right)^{{\frac{1}{\gamma + \delta }}} }}{\gamma + \delta }L_{E}^{*} $$

(73)

and plugging this into the utility function yields lifetime utility at:

$$ U^{\text{CP}} = \frac{{\left( {2\alpha + \beta } \right)\left( {\gamma^{\gamma } \delta^{\delta } \varphi^{\gamma } \psi^{\delta } } \right)^{{\frac{1}{\gamma + \delta }}} L_{E}^{*} + (\alpha + \beta )(\gamma + \delta )\rho \log C_{0} }}{{(\alpha + \beta )(\gamma + \delta )\rho^{2} }} $$

(74)

where C ₀ is the initial level of consumption that is chosen to exhaust total discounted lifetime production (cf. Eq. 36).