3
$\begingroup$

I have been having some difficulties with recovering the same euler equation from the following optimal control problem when comparing the present valued hamiltonian to the current valued hamiltonian.

suppose we have the following Continuous time Neoclassical Growth model with a representative consumer maximizing its utility $$\int_0^\infty e^{-rt} \frac{c(t)^{1-\theta}}{1-\theta}dt$$ subject to: $$\dot{k}(t)=f(k(t))-\delta k(t)-c(t)$$ $$f(k(t))=Ak(t)^\alpha$$

it follows that the corresponding present and current valued modified lagrangians are for this problem is: $$\mathcal{J_{\textrm{pv}}}=e^{-rt}\frac{c(t)^{1-\theta}}{1-\theta}+\lambda(t)[Ak(t)^\alpha-\delta k(t)-c(t)]+\dot{\lambda}(t)k(t)$$ $$\mathcal{J_{\textrm{cv}}}=\frac{c(t)^{1-\theta}}{1-\theta}+e^{rt}\lambda(t)[Ak(t)^\alpha-\delta k(t)-c(t)]+\frac{d[e^{rt}\lambda(t)]}{dt}k(t)$$ renaming our current valued multipliers $\mu(t)=e^{rt}\lambda(t)$ our current valued Hamiltonian becomes: $$\mathcal{J_{\textrm{cv}}}=\frac{c(t)^{1-\theta}}{1-\theta}+\mu(t)[Ak(t)^\alpha-\delta k(t)-c(t)]+(\rho \mu(t)+\dot{\mu}(t)) k(t)$$

My Question: I am currently under the impression we are supposed to be able to recover the same euler equation from both types of modified lagrangian/hamiltonian set ups whether it be a present valued or current valued however in the way I currently understand the problem it is not possible to recover the same euler equation. what is going wrong here?

$\endgroup$
4
$\begingroup$

First, the Present value hamiltonian equals: $$ {\cal P} = e^{-rt} \frac{c(t)^{1-\theta}}{1 - \theta} + \lambda(t)(Ak(t)^\alpha - \delta k(t) - c(t)) $$

This gives the first order conditions:

$$ \begin{align*} &\frac{\partial{\cal I}}{\partial c} = e^{-rt} c(t)^{-\theta} - \lambda(t) = 0 \tag{1.1}\\ &\dot \lambda(t) = - \lambda(t) (\alpha A (k_t)^{\alpha-1} - \delta), \tag{1.2}\\ &\dot k(t) = Ak(t)^\alpha - \delta k(t) - c(t). \tag{1.3} \end{align*} $$ Taking the derivative of the first with respect to time and using $(1.1)$ and $(1.2)$ gives: $$ \begin{align*} &-r e^{-rt} c(t)^{-\theta} -\theta e^{-rt} c(t)^{-\theta} \frac{\dot c(t)}{c(t)} = \dot \lambda(t),\\ \to &-r - \theta \frac{\dot c(t)}{c(t)} = \frac{\dot \lambda(t)}{\lambda(t)},\\ \to & \frac{\dot c(t)}{c(t)} = \frac{\alpha A(k_t)^{\alpha-1} - (\delta + r)}{\theta} \end{align*} $$

For the current value Hamiltonian, we have: $$ {\cal C} = \frac{c(t)^{1- \theta}}{1 - \theta} + \mu(t)(Ak(t)^\alpha - \delta k(t) - c(t)) $$ The first order conditions are: $$ \begin{align*} &\frac{\partial C}{\partial c} = c(t)^{-\theta} - \mu(t) = 0, \tag{2.1}\\ &\dot \mu(t) - r\mu(t) = -\mu(t)(A\alpha k(t)^{\alpha-1}-\delta), \tag{2.2}\\ &\dot k(t) = Ak(t)^\alpha - \delta k(t) - c(t), \tag{2.3}\\ \end{align*} $$

Substituting $\mu(t) = e^{rt} \lambda(t)$ gives for $(2.1)$ the equation: $$ \begin{align*} &c(t)^{-\theta} - e^{rt} \lambda(t) = 0,\\ \to &e^{-rt} c(t)^{-\theta} - \lambda(t) = 0,\\ \end{align*} $$ Which is the same as $(1.1$).

Subsituting $\mu(t) = e^{rt} \lambda(t)$ and $\dot \mu(t) = r e^{rt} \lambda(t) + e^{rt} \dot \lambda(t)$ for $(2.2)$ gives: $$ \begin{align*} &r e^{rt}\lambda(t) + e^{rt}\dot \lambda(t) - r e^{rt} \lambda(t) = -e^{rt} \lambda(t) (A \alpha k(t)^{\alpha-1} - \delta),\\ \to &\dot \lambda(t) = \lambda(t)(A \alpha k(t)^{\alpha-1} - \delta),\\ \end{align*} $$ which is the same as $(1.2)$.

We can also solve $(2.1)$ and $(2.2)$ by taking the the derivative of $(2.1)$ with respect to time to obtain: $$ \begin{align*} &-\theta c(t)^{-\theta} \frac{\dot c(t)}{c(t)} = \dot \mu(t),\\ \to & \frac{\dot c(t)}{c(t)} = -\frac{1}{\theta} \frac{\dot \mu(t)}{\mu(t)} = \frac{A \alpha k(t)^{\alpha-1}- (\delta+r)}{\theta} \end{align*} $$

$\endgroup$

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.