<p>First, the Present value hamiltonian equals:
<span class="math-container">$$
{\cal P} = e^{-rt} \frac{c(t)^{1-\theta}}{1 - \theta} + \lambda(t)(Ak(t)^\alpha - \delta k(t) - c(t))
$$</span></p>
<p>This gives the first order conditions:</p>
<p><span class="math-container">$$
\begin{align*}
&amp;\frac{\partial{\cal I}}{\partial c} = e^{-rt} c(t)^{-\theta} - \lambda(t) = 0 \tag{1.1}\\
&amp;\dot \lambda(t) = - \lambda(t) (\alpha A (k_t)^{\alpha-1} - \delta), \tag{1.2}\\
&amp;\dot k(t) = Ak(t)^\alpha - \delta k(t) - c(t). \tag{1.3}
\end{align*}
$$</span>
Taking the derivative of the first with respect to time and using <span class="math-container">$(1.1)$</span> and <span class="math-container">$(1.2)$</span> gives:
<span class="math-container">$$
\begin{align*}
&amp;-r e^{-rt} c(t)^{-\theta} -\theta e^{-rt} c(t)^{-\theta} \frac{\dot c(t)}{c(t)} = \dot \lambda(t),\\
\to &amp;-r - \theta \frac{\dot c(t)}{c(t)} = \frac{\dot \lambda(t)}{\lambda(t)},\\
\to &amp; \frac{\dot c(t)}{c(t)} = \frac{\alpha A(k_t)^{\alpha-1} - (\delta + r)}{\theta} 
\end{align*}
$$</span></p>
<p>For the current value Hamiltonian, we have:
<span class="math-container">$$
{\cal C} = \frac{c(t)^{1- \theta}}{1 - \theta} + \mu(t)(Ak(t)^\alpha - \delta k(t) - c(t))
$$</span>
The first order conditions are:
<span class="math-container">$$
\begin{align*}
&amp;\frac{\partial C}{\partial c} = c(t)^{-\theta} - \mu(t) = 0, \tag{2.1}\\
&amp;\dot \mu(t) - r\mu(t) = -\mu(t)(A\alpha k(t)^{\alpha-1}-\delta), \tag{2.2}\\
&amp;\dot k(t) = Ak(t)^\alpha - \delta k(t) - c(t), \tag{2.3}\\
\end{align*}
$$</span></p>
<p>Substituting <span class="math-container">$\mu(t) = e^{rt} \lambda(t)$</span> gives for <span class="math-container">$(2.1)$</span> the equation:
<span class="math-container">$$
\begin{align*}
&amp;c(t)^{-\theta} - e^{rt} \lambda(t) = 0,\\
\to &amp;e^{-rt} c(t)^{-\theta} - \lambda(t) = 0,\\
\end{align*}
$$</span>
Which is the same as <span class="math-container">$(1.1$</span>).</p>
<p>Subsituting <span class="math-container">$\mu(t) = e^{rt} \lambda(t)$</span> and <span class="math-container">$\dot \mu(t) = r e^{rt} \lambda(t) + e^{rt} \dot \lambda(t)$</span> for <span class="math-container">$(2.2)$</span> gives:
<span class="math-container">$$
\begin{align*}
&amp;r e^{rt}\lambda(t) + e^{rt}\dot \lambda(t) - r e^{rt} \lambda(t) = -e^{rt} \lambda(t) (A \alpha k(t)^{\alpha-1} - \delta),\\
\to &amp;\dot \lambda(t) = \lambda(t)(A \alpha k(t)^{\alpha-1} - \delta),\\
\end{align*}
$$</span>
which is the same as <span class="math-container">$(1.2)$</span>.</p>
<p>We can also solve <span class="math-container">$(2.1)$</span> and <span class="math-container">$(2.2)$</span> by taking the the derivative of <span class="math-container">$(2.1)$</span> with respect to time to obtain:
<span class="math-container">$$
\begin{align*}
&amp;-\theta c(t)^{-\theta} \frac{\dot c(t)}{c(t)} = \dot \mu(t),\\
\to &amp; \frac{\dot c(t)}{c(t)} = -\frac{1}{\theta} \frac{\dot \mu(t)}{\mu(t)} = \frac{A \alpha k(t)^{\alpha-1}- (\delta+r)}{\theta}
\end{align*}
$$</span></p>

CC BY-SA 4.0
2021-05-31T05:56:52.303Z
2021-05-31T05:56:52.303Z
tdm
<p>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.</p>
<p>suppose we have the following Continuous time Neoclassical Growth model with a representative consumer maximizing its utility
<span class="math-container">$$\int_0^\infty e^{-rt} \frac{c(t)^{1-\theta}}{1-\theta}dt$$</span>
subject to:
<span class="math-container">$$\dot{k}(t)=f(k(t))-\delta k(t)-c(t)$$</span>
<span class="math-container">$$f(k(t))=Ak(t)^\alpha$$</span></p>
<p>it follows that the corresponding present and current valued modified lagrangians are for this problem is:
<span class="math-container">$$\mathcal{J_{\textrm{pv}}}=e^{-rt}\frac{c(t)^{1-\theta}}{1-\theta}+\lambda(t)[Ak(t)^\alpha-\delta k(t)-c(t)]+\dot{\lambda}(t)k(t)$$</span>
<span class="math-container">$$\mathcal{J_{\textrm{cv}}}=\frac{c(t)^{1-\theta}}{1-\theta}+e^{rt}\lambda(t)[Ak(t)^\alpha-\delta k(t)-c(t)]+\frac{d[e^{rt}\lambda(t)]}{dt}k(t)$$</span>
renaming our current valued multipliers <span class="math-container">$\mu(t)=e^{rt}\lambda(t)$</span> our current valued Hamiltonian becomes:
<span class="math-container">$$\mathcal{J_{\textrm{cv}}}=\frac{c(t)^{1-\theta}}{1-\theta}+\mu(t)[Ak(t)^\alpha-\delta k(t)-c(t)]+(\rho \mu(t)+\dot{\mu}(t)) k(t)$$</span></p>
<p><strong>My Question:</strong> 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?</p>

<mathematical-economics><optimal-control>
Current valued VS Present Valued Hamiltonian Differing Euler equations
EconJohn
# Current valued VS Present Valued Hamiltonian Differing Euler equations

(Asked by: username_0 on 2021-05-31)

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?

**Tags:** mathematical-economics, optimal-control

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## 1 answers

### Answer by username_1 on 2021-05-31. Score: 6 | Accepted answer

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:
$$
\begin{align\*}
&-\theta c(t)^{-\theta} \frac{\dot c(t)}{c(t)} = \dot \mu(t),\\
\to & \frac{\dot c(t)}{c(t)} = -\frac{1}{\theta} \frac{\dot \mu(t)}{\mu(t)} = \frac{A \alpha k(t)^{\alpha-1}- (\delta+r)}{\theta}
\end{align\*}
$$


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