# Example

We can now completely solve an example.

## Utility and production functions

We assume that utility is logarithmic and we take a Cobb-Douglas production function.

$$\begin{eqnarray} u(c_{t}) = \log (c_{t}), \\\ F(K_{t},L_{t}) = K_{t}^{\alpha}L_{t}^{1-\alpha}, \\\ \alpha \in (0,1). \end{eqnarray}$$ We can check that both functions satisfy the Inada conditions:

$$\begin{eqnarray}\lim_{c_{t} \rightarrow 0}u^{\prime}(c_{t}) = \lim_{c_{t} \rightarrow 0} \frac{1}{c_{t}} = +\infty, \\\ \lim_{c_{t} \rightarrow +\infty}u^{\prime}(c_{t}) = \lim_{c_{t} \rightarrow +\infty} \frac{1}{c_{t}} = 0, \\\ \lim_{K_{t} \rightarrow 0} F^{\prime}_{K_{t}}(K_{t}, L_{t}) = \lim_{K_{t} \rightarrow 0} \alpha K_{t}^{\alpha - 1}L_{t}^{1-\alpha} = +\infty, \\\ \lim_{K_{t} \rightarrow +\infty} F^{\prime}_{K_{t}}(K_{t}, L_{t}) = \lim_{K_{t} \rightarrow +\infty} \alpha K_{t}^{\alpha - 1}L_{t}^{1-\alpha} = 0, \\\ \lim_{L_{t} \rightarrow 0} F^{\prime}_{L_{t}}(K_{t}, L_{t}) = \lim_{L_{t} \rightarrow 0} (1-\alpha) K_{t}^{\alpha}L_{t}^{\alpha} = +\infty, \\\ \lim_{L_{t} \rightarrow +\infty} F^{\prime}_{L_{t}}(K_{t}, L_{t}) = \lim_{L_{t} \rightarrow +\infty} (1- \alpha) K_{t}^{\alpha}L_{t}^{\alpha} = 0. \end{eqnarray}$$

The production function, expressed in intensive terms, becomes:

$$F(K_{t}, L_{t}) = L_{T} F\left(\frac{K_{t}}{L_{t}},1\right) = L_{t} f(k_{t}) = L_{t} k_{t}^{\alpha}, k_{t} \equiv \frac{K_{t}}{L_{t}}.$$ with total production $Y_{t} = F(K_{t},L_{t})$ and production per capita is equal to $y_{t} \equiv \frac{Y_{t}}{L_{t}} = f(k_{t}).$

## Household’s optimisation

Instead of using the results from previous sections, we develop again the utility maximisation. First, we write down the household’s budget constraint:

### Household’s budget constraint

At each period, the total received income is composed of wages and interest. It can be spent on consumption or saving. Remember that capital depreciates at a rate $\delta \in [0,1].$ Hence, we receive back from firms $1-\delta$ times the capital we lend. Therefore:

$$w_{t} + r_{t}k_{t} + (1-\delta)k_{t} = c_{t} + k _{t+1}.$$

### Intertemporal utility maximisation: Lagrangian

The intertemporal utility maximisation problem is:

$$\begin{eqnarray} \max_{c_{t}, k_{t+1}} \sum_{t=0}^{\infty} \beta^{t} u(c_{t}) & \\\ \mathrm{s.t.} \quad w_{t} + r_{t} k_{t} + (1-\delta)k_{t} = c_{t} + k_{t+1}, \\\ c_{t}, k_{t+1} >0, k_{t=0} = k_{0} > 0. \end{eqnarray}$$

The Lagrangian becomes:

$$\mathcal{L} = \sum_{t=0}^{\infty} \beta^{t} u(c_{t}) + \sum_{t=0}^{\infty} \lambda_{t}(w_{t} + (r_{t} + 1 - \delta)k_{t} - c_{t} - k_{t+1}).$$

We use the first-order conditions with respect to $c_{t}$ and $k_{t+1}$ to find the optimal consumption path: the Euler equation, which we combine later with the transversality condition.

$$\frac{\partial \mathcal{L}}{\partial c_{t}} = \beta^{t} u^{\prime}(c_{t}) - \lambda_{t} = \beta^{t}\frac{1}{c_{t}} - \lambda_{t} = 0$$ $$\frac{\partial \mathcal{L}}{\partial k_{t+1}} = -\lambda_{t} + \lambda_{t+1} (r_{t+1} + 1 - \delta) = 0.$$

Combining both, we obtain the Euler equation:

$$\beta^{t}\frac{1}{c_{t}} = r_{t+1} \beta^{t+1} \frac{1}{c_{t+1}} \implies c_{t+1} = \beta (r_{t} + 1 - \delta)c_{t}.$$

Finally, we should remember to impose the transversality condition:

$$\lim_{t \rightarrow +\infty} \beta^{t}u^{\prime}(c_{t})k_{t+1} = \lim_{t \rightarrow +\infty}\beta^{t}\frac{1}{c_{t}}k_{t+1} = 0.$$

## Firm’s optimisation

In this model, firms operate under perfect competition, making zero profits. Moreover, factors are paid their marginal productivity.

$$r_{t} = F^{\prime}_{K_{t}}(K_{t}, L_{t}) = \alpha K_{t}^{\alpha - 1}L_{t}^{1-\alpha} = \alpha \left(\frac{K_{t}}{L_{t}}\right)^{\alpha - 1} = \alpha k_{t}^{\alpha -1},$$ $$w_{t} = F^{\prime}_{L_{t}}(K_{t}, L_{t}) = (1-\alpha) K_{t}^{\alpha} L_{t}^{-\alpha} = (1-\alpha)\left(\frac{K_{t}}{L_{t}}\right)^{\alpha} = (1-\alpha) k_{t}^{\alpha}.$$

Alternatively, using the information in the Appendix and working directly with the intensive-form production function:

$$r_{t} = f^{\prime}(k_{t}) = \frac{\partial f(k_{t})}{\partial k_{t}} = \alpha k_{t}^{\alpha -1},$$ $$w_{t} = f(k_{t}) - f^{\prime}(k_{t})k_{t} = k_{t}^{\alpha} - \alpha k_{t}^{\alpha - 1} k_{t} = (1-\alpha)k_{t}^{\alpha}.$$

## The dynamic system

We are now in a position to solve the model. We have the following equations:

$$c_{t+1} = \beta (r_{t+1} + 1 - \delta) c_{t}, \\\ w_{t} + r_{t}k_{t} + (1-\delta) k_{t} = c_{t} + k_{t+1}, \\\ w_{t} = (1-\alpha) k^{\alpha}_{t}, \\\ r_{t} = \alpha k^{\alpha-1}, \\\ \lim_{t \rightarrow +\infty} \beta^{t}\frac{1}{c_{t}} k_{t+1} = 0.$$

Since markets clear, we substitute the wage and interest rate into the budget constraint.

$$c_{t+1} = \beta (\alpha k^{\alpha-1} + 1 - \delta) c_{t},$$ $$k_{t}^{\alpha} + (1-\delta) k_{t} = c_{t} + k_{t+1},$$ $$\lim_{t \rightarrow +\infty} \beta^{t}\frac{1}{c_{t}} k_{t+1} = 0.$$

Before trying to solve for the optimal trajectory, we analyse the steady states of this economy. A steady-state $(\bar{k}, \bar{c})$ is such that capital and consumption remain constant over time: $c_{t+1} = c_{t}$ and $k_{t+1} = k_{t}.$ Applying this idea to the equations above we get —here we can temporarily forget about the transversality condition.

$$\bar{c} = \beta(\alpha \bar{k}^{\alpha-1} + 1 - \delta) \bar{c},$$ $$\bar{k}^{\alpha} + (1-\delta)\bar{k} = \bar{c} + \bar{k}.$$

Using the first equation we can easily get the steady level of capital:

$$\bar{k} = \left(\frac{\alpha \beta}{1-\beta(1-\delta)}\right)^{\frac{1}{1-\alpha}}.$$

Applying $\bar{k}$ in the second equation yields the steady-state level of consumption:

$$\bar{c} = \left(\frac{\alpha \beta}{1-\beta (1-\delta)}\right)^{\frac{\alpha}{1-\alpha}} - \delta \left(\frac{\alpha \beta}{1-\beta (1- \delta)}\right)^{\frac{1}{1-\alpha}}.$$

## Stability

First, we analyse the stability of the steady state, as we did in the general case Since we are using log-utility, the coefficient of relative risk aversion $\epsilon_{c} = 1.$

Instead of plugging the correct values in the Jacobian matrix $\bar{A}$, for the sake of clarity, we derive everything again:

$$\bar{A} = \begin{pmatrix} \frac{\partial c_{t+1}}{\partial c_{t}} & \frac{\partial c_{t+1}}{\partial k_{t}} \\\ \frac{\partial k_{t+1}}{\partial c_{t}} & \frac{\partial k_{t+1}}{\partial k_{t}} \end{pmatrix}$$ $$=$$ $$\begin{pmatrix} \beta \left( \alpha k_{t+1}^{\alpha -1} + 1 - \delta \right) - \frac{\beta c_{t} \alpha (\alpha-1) }{k_{t+1}^{2-\alpha}} & \frac{\beta c_{t} \alpha (\alpha -1)}{k_{t+1}^{2-\alpha}} \left( \alpha k_{t}^{\alpha-1} + 1 - \delta \right) \\\ -1 & \alpha k_{t}^{\alpha -1} + 1 - \delta \end{pmatrix}$$

Evaluating it at the steady state: $$\bar{A}\bigr\rvert_{\substack{ k_{t} = k_{t+1} = \bar{k} \\\ c_{t} = c_{t+1} = \bar{c} } } = \begin{pmatrix} 1 - \beta \bar{c} \alpha (\alpha -1) \bar{k}^{\alpha -2} & \alpha (\alpha -1 ) \bar{c} \bar{k}^{\alpha -2} \\\ -1 & \frac{1}{\beta} \end{pmatrix},$$

where we have used $1 = \beta \left( \alpha \bar{k}^{\alpha -1} + 1 - \delta \right)$ at the steady state.

### Eigenvalues

We can use the fact $\mathrm{tr}(\bar{A}) = \lambda_{1} + \lambda_{2}, \mathrm{det}(\bar{A}) = \lambda_{1} \lambda_{2}.$ Therefore:

$$\lambda_{1} \lambda_{2} = \frac{1}{\beta}$$ $$\lambda_{1} + \lambda_{2} = \frac{1}{\beta} + 1 - \beta \bar{c} \alpha (\alpha-1) \bar{k}^{\alpha -2}.$$ The first equation implies that either a) $\lambda_{1} > 0, \lambda_{2} > 0$ or b) $\lambda_{1} < 0, \lambda_{2} < 0.$ However, since $\lambda_{1} + \lambda_{2} > 1$ b) is impossible. Then, $\lambda_{1} > 0, \lambda_{2} > 0.$ Substitute $\lambda_{1} \lambda_{2} = \frac{1}{\beta}$ in the second equation:

$$\lambda_{1} + \lambda_{2} = 1+ \lambda_{1} \lambda_{2} - \beta ( \alpha (\alpha - 1) \bar{k}^{\alpha -2}\frac{\bar{c}}{\epsilon_{\bar{c}}}.$$

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