<p>There is no clear right and wrong about this, it's just a matter of convenience. The current-value Hamiltonian is likely to be more convenient when the objective function includes a discount factor. Following Chiang (1), suppose the problem is:</p>
<p><span class="math-container">$\qquad$</span>Maximise <span class="math-container">$V = \int_0^T G(t,y,u)e^{-\rho t}$</span></p>
<p><span class="math-container">$\qquad$</span>subject to <span class="math-container">$\dot y=f(t,y,u)$</span></p>
<p><span class="math-container">$\qquad$</span>and boundary conditions</p>
<p>The standard (present value) Hamiltonian is:</p>
<p><span class="math-container">$\qquad H=G(t,y,u)e^{-\rho t} + \lambda f(t,y,u)$</span></p>
<p>If we proceed from this Hamiltonian, the co-state equation (one of the first-order conditions) is:</p>
<p><span class="math-container">$\qquad \dot \lambda = -\dfrac{\partial H}{\partial y}= -\dfrac{\partial [G(.)e^{-\rho t}]}{\partial y}-\lambda\dfrac{\partial f}{\partial y}$</span></p>
<p>While it is possible to obtain a solution this way, the discount factor complicates the derivatives and can make interpretation more challenging.</p>
<p>Suppose instead we use the current-value Hamiltonian:</p>
<p><span class="math-container">$\qquad H_c = G(t,y,u) + mf(t,y,u)$</span></p>
<p>where <span class="math-container">$m$</span> is a current-value Lagrange multiplier defined by <span class="math-container">$m=\lambda e^{\rho t}$</span>.  The co-state equation is then:</p>
<p><span class="math-container">$\qquad \dot m -\rho m = -\dfrac{\partial H_c}{\partial y} = -\dfrac{\partial G}{\partial y} - m\dfrac{\partial f}{\partial y}$</span></p>
<p>This is simpler because it contains no discount term.</p>
<p>Reference</p>
<ol>
<li>Chiang A C (1992) <em>Elements of Dynamic Optimization</em> pp 210 ff</li>
</ol>

CC BY-SA 4.0
2020-09-08T14:15:31.967Z
2022-04-06T10:53:15.610Z
2022-04-06T10:53:15.610Z
Adam Bailey
<p>I've been looking at a number of optimal control problems and have been wondering under what conditions one should use the current value Hamiltonian over the present value Hamiltonian.</p>
<p>Does it depend on the specific result we seek to derive? (pick whichever you want) or does it depend on the nature of the problem?</p>

CC BY-SA 4.0
2020-09-08T01:03:47.507Z
Intuitvely, wouldn't it just be for instance, if we are maximizing a static problem vs a dynamic one? Ex. maximizing discounted sum of utilities would be the PV hamiltonian, whereas maximizing only current period utility would be a current value Hamiltonian?
ChinG
CC BY-SA 4.0
2020-09-08T03:09:12.950Z
@ChinG being that the current value Hamiltonian is just a transformation of the present value Hamiltonian im leaning towards that we just pick whichever is convenient, however im not an expert.
EconJohn
<mathematical-economics><optimal-control>
Current Value Hamiltonian VS Present Value Hamiltonian in Economics
EconJohn
# Current Value Hamiltonian VS Present Value Hamiltonian in Economics

(Asked by: username_0 on 2020-09-08)

I've been looking at a number of optimal control problems and have been wondering under what conditions one should use the current value Hamiltonian over the present value Hamiltonian.

Does it depend on the specific result we seek to derive? (pick whichever you want) or does it depend on the nature of the problem?

**Tags:** mathematical-economics, optimal-control
> **Comments:**
> **username_1**: Intuitvely, wouldn't it just be for instance, if we are maximizing a static problem vs a dynamic one? Ex. maximizing discounted sum of utilities would be the PV hamiltonian, whereas maximizing only current period utility would be a current value Hamiltonian?  
> **username_0**: @ChinG being that the current value Hamiltonian is just a transformation of the present value Hamiltonian im leaning towards that we just pick whichever is convenient, however im not an expert.  

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## 1 answers

### Answer by username_2 on 2020-09-08. Score: 10 | Accepted answer

There is no clear right and wrong about this, it's just a matter of convenience. The current-value Hamiltonian is likely to be more convenient when the objective function includes a discount factor. Following Chiang (1), suppose the problem is:

$\qquad$Maximise $V = \int\_0^T G(t,y,u)e^{-\rho t}$

$\qquad$subject to $\dot y=f(t,y,u)$

$\qquad$and boundary conditions

The standard (present value) Hamiltonian is:

$\qquad H=G(t,y,u)e^{-\rho t} + \lambda f(t,y,u)$

If we proceed from this Hamiltonian, the co-state equation (one of the first-order conditions) is:

$\qquad \dot \lambda = -\dfrac{\partial H}{\partial y}= -\dfrac{\partial [G(.)e^{-\rho t}]}{\partial y}-\lambda\dfrac{\partial f}{\partial y}$

While it is possible to obtain a solution this way, the discount factor complicates the derivatives and can make interpretation more challenging.

Suppose instead we use the current-value Hamiltonian:

$\qquad H\_c = G(t,y,u) + mf(t,y,u)$

where $m$ is a current-value Lagrange multiplier defined by $m=\lambda e^{\rho t}$. The co-state equation is then:

$\qquad \dot m -\rho m = -\dfrac{\partial H\_c}{\partial y} = -\dfrac{\partial G}{\partial y} - m\dfrac{\partial f}{\partial y}$

This is simpler because it contains no discount term.

Reference

1. Chiang A C (1992) *Elements of Dynamic Optimization* pp 210 ff


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