Consider a representative agent model where the representative agent has standard preferences

$\int^{\infty}_te^{-\rho t}ln(c(t))dt$

where $\rho>0$ and $c(t)$ is consumption in period t as usual. And the production technology is something of the form

$y(t) = z(t)f(k(t))$

For example, to simplify the matters I take $f(k(t)) =k(t)$. By assumption take $z_t = z^*$ in even periods and $z_t= z^{**}$ for odd periods and further assume that $z^{*}-\rho >0$ and $z^{**}-\rho<0$ and $0\% (\delta= 0)$ depreciation for further simplication.

I want to solve for the planners optimal allocation and to show that the capital stock, output, and consumption increase in even periods and decrease in odd periods. Clearly the assumptions and simplifications are done with an example in mind, but I know the community guidelines such that I am not asking for the specific answer to this question. However, I am asking for a general strategy of how to proceed from the Euler's equation to the law of motion for capital. To put more clearly, let me demonstrate:

Clearly $\dot k(t) = z(t) k(t) - c(t)$. And we somehow need to link the Euler's equation with this law of motion to derive a general relationship between $\frac{\dot k(t)}{k(t)}$ and $\frac{\dot c(t)}{c(t)}$. Solving for $\frac{\dot c(t)}{c(t)}$ is easy since we can set up the current value Hamiltonian and get the result $\frac{\dot c(t)}{c(t)} = z(t) - \rho$. This gives the relationship between $z(t)-\rho$ and $\frac{\dot c(t)}{c(t)}$, so we can then plug in the assumption given by the question to derive the relationship that $\frac{\dot c(t)}{c(t)}>0$ in even periods and $\frac{\dot c(t)}{c(t)}<0$ in the odd periods. This much is clear.

However, I am consufed as to how to proceed from this relationship to the law of motion for capital. To see why we need link, we can use the production function $f(k(t))$ and differentiate with respect to time to see how it evolves. So we have

$y(t) = z(t)k(t)$ and $\frac{\dot y(t)}{y(t)} = \frac{\dot z(t)}{z(t)} + \frac{\dot k(t)}{k(t)}$

In general $\frac{\dot z(t)}{z(t)}$ will be exogeneous in simple models like this. In the context of this model it is given by the assumptions, $ \frac{\dot z(t)}{z(t)}>0$ in even periods and $ \frac{\dot z(t)}{z(t)}<0$ for the odd periods. Then we need to find out how $\frac{\dot k(t)}{k(t)}$ evolves with $\frac{\dot c(t)}{c(t)}$. Then I get stuck because we only know that

$\dot k(t) = z(t)k(t) -c(t)$

which makes

$\frac{\dot k(t)}{k(t)} = z(t) - \frac{c(t)}{k(t)}$.

If I am told to look for the steady-state then I know where to look, I can use $\frac{\dot k(t)}{k(t)}=0$ and proceed. But the question is more general. How should one proceed in questions like this where we are not concerned about the steady-state?


Well, you can actually solve $k$ analytically. This is just standard first-order linear ODE technique.

The solution of the differential equation $ \dot{c}(t) / c(t) = z(t) - \rho $ is

$$ c(t) = c_0 \exp\left( \int_0^t z(s) ds - \rho t \right) $$

Plugging this into $ \dot{k}(t) - z(t)k(t) = - c(t) $, multiplying both sides by $ \exp\left( -\int_0^t z(s) ds \right) $ and using product rule, we get

$$ \frac{d}{dt} \left[ k(t) \exp\left( -\int_0^t z(s) ds \right) \right] = -c_0 e^{-\rho t} $$

The solution is

$$ k(t) = \left[ k_0 + \frac{c_0}{\rho} (e^{-\rho t} - 1) \right] \exp\left( \int_0^t z(s) ds \right) $$

What remains is to solve for $c_0$. This is actully the trickiest part of the exercise. First, note that the steady state $(c^*, k^*)$ satisfies $ c^* = \rho k^* $. Next, the transversality condition gives us $ (c(t), k(t)) \to (c^*, k^*) $, so

$$ \begin{align} c^* &= c_0 \zeta \\ k^* &= \left( k_0 - \frac{c_0}{\rho} \right) \exp\left( \int_0^\infty z(t) dt \right) + \frac{c_0}{\rho} \zeta \\ \text{where} \quad \zeta &= \lim_{t \to \infty} \exp\left( \int_0^t z(s) ds - \rho t \right) \end{align} $$

Finally, plugging the above into $ c^* = \rho k^* $ and subtracting $c_0 \zeta$ from both sides, we get $ c_0 = \rho k_0 $. Thus, we actually have a nice closed form for $k$:

$$ k(t) = k_0 \exp\left( \int_0^t z(s) ds - \rho t \right) $$

  • $\begingroup$ Thank you for this. Much appreciated. $\endgroup$ Dec 4 '21 at 12:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.