Jacobian and the generating function

Final notes on the Ramsey-Cass-Koopmans model

This post marks the final instalment in my series on geometry in economics—for the moment.

The Ramsey-Cass-Koopmans model (Wiki) is a cornerstone of neoclassical macro-economic modelling. I introduced it last week:

Lagrangian, Hamiltonian, and Jacobian formalisms

Let me repeat the different solution approaches that originate in optics. If you are interested, I recommend this article by Bahram Houchmandzadeh. I paraphrase his wonderful exposition:

• The Lagrangian approach considers rays, modelling states q and changes \dot{q}.

• The Hamiltonian approach considers wavefronts, modelling q and moments p.

• The (Hamilton-)Jacobi approach solves for the wave-function S(q,t).

The first two were discussed last week, this week is about the Jacobi approach, and how our own is related!

The Ramsey-Cass-Koopmans model, repeated

The RCK model specification is short enough to repeat. We present a toy version of a farmer needing to decide how much grain to plant and how much to consume, to maximize their long-term welfare. Planting k produces additional f(k) in the next period (with certainty), which can be consumed or added to the seed capital¹

\(f(k) = c + \dot{k} \)

The farmer aims to optimize lifetime utility, a challenge since they live forever:

\( U(\{c\}) = \int_0^\infty e^{-\rho t} u(c(t)) dt\)

Here ρ is the intertemporal discount rate, assumed constant. We showed the optimal consumtpion/capital path satisfies the Keynes-Ramsey rule:

\(f'(k) + \frac{u''}{u'} \frac{dc}{dt} = \rho \qquad \qquad (KR)\)

The Jacobian approach

Rather than optimizing the Lagrangian over all paths, we now introduce a value function associated to states. This is another approach you find in textbooks, particularly to introduce the Bellman principle (Wiki) of dynamic programming (Wiki)², which has applications to finance. I quote Bellman:

Principle of Optimality: An optimal policy has the property that whatever the initial state and initial decision are, the remaining decisions must constitute an optimal policy with regard to the state resulting from the first decision. (See Bellman, 1957, Chap. III.3.)

Suppose there exists a function V(k₀) that gives the optimal utility U that can be derived starting at state k₀. That function depends on k₀, but not on the consumption plan {c} as it’s choosing the optimum path. So we write:

\(V(k_0) = \sup_{\{c\} } U(\{c\};k_0) = \int_0^\infty L(k^*(t),\dot{k^*}(t),t;k_0)dt = \int_0^\infty L^*(t;k_0)dt \)

Here k* is shorthand for the implied optimal capital path, from k₀. We see V is related to the Lagrangian, and so the Hamiltonian:

\(L^*= \frac{dV}{dt} = \frac{\partial V}{\partial t} + \frac{\partial V}{\partial k}\dot{k} \implies \frac{\partial V}{\partial t} + \left(\frac{\partial V}{\partial k}\dot{k}-L^*\right)=\frac{\partial V}{\partial t} + H(k, \frac{\partial V}{\partial k},t) =0\)

The latter equality is known as the Hamilton-Jacobi equation (Wiki)³ or HJE. In the RCK case we have L=e⁻ᵖᵗu(c) and so with the conditions of our model it becomes:

\(-\frac{\partial V}{\partial t} = \sup_c \left(\frac{\partial V}{\partial k}(f(k)-c) - e^{-\rho t} u(c) \right)\)

The HJE has a bad name, because it is generally hard to solve partial differential equations (PDEs). It is the difficulty of the Schrödinger equation of the wave function in quantum mechanics, after all. Fortunately for us, we know the answer we’re looking for and don’t actually need to solve for V to get it.

The supremum condition for c imposes ∂V/∂k = -e⁻ᵖᵗuʹ. (This is exactly the co-state variable of our Hamiltonian approach last time, as can be guessed from its role in the HJE above.) Knowing that, the trick is now to take the derivative with respect to k, because we know the right-hand side and ∂/∂k (∂V/∂t) = ∂/∂t (∂V/∂k) on the left. So

\(\begin{split} \frac{\partial}{\partial k}\left(-\frac{\partial V}{\partial t}\right) =-\frac{\partial}{\partial t}\frac{\partial V}{\partial k} &= e^{-\rho t}u'' \frac{dc}{dt} -\rho e^{-\rho t}u'\\ \frac{\partial}{\partial k}\left(\frac{\partial V}{\partial k} (f(k) - c) - e^{-\rho t}u(c)\right) &= -e^{-\rho t}u' f'(k) \end{split}\)

Equating the right-hand sides and dividing throughout by e⁻ᵖᵗuʹ again gives the Keynes-Ramsey condition (KR)! Without, I repeat, having had to solve for V itself.

And finally, the generating function

These three descriptions are essentially equivalent: with a bit of math they can express the same concepts and each lead in their own way to the (KR) condition. The generation function extends this to non-equilibrium cases. What if the farmer still had to discover their utility function u(c) or production function f(k)—or had imperfect measurements of k or c, or was still learning variational calculus?—before settling on the optimal rule?

Let me recall the formulas of our geometric approach, which I never derived:

\(\dot{q} = \frac{\partial G}{\partial p}, \quad \dot{p} = -\frac{\partial G}{\partial q} -p \frac{\partial G}{\partial s}, \quad \dot{s} = p\frac{\partial G}{\partial p}-G \qquad (*)\)

Again these PDE’s are unforgiving. Rather than showing another way to deriving the (KR) conditions, let’s discuss how these equations relate to the approaches treated before.

• Like in the Hamiltonian formulation, we use state and co-state variables q and p. Additionally we have s, a state variable of unique nature. When ∂G/∂s is 0, the first two equations are the Hamiltonian equations.

• So G is like the Hamiltonian—it has the same units [€]/time. It satisfies dG/dt = -∂G/∂s G as we saw earlier, so if G=0 then dG/dt=0 also—what I’ve called equilibrium. Thus G captures the distance to the optimal path, and is constant 0 when the system finds it. G represents opportunity⁴.

• s has units of value [€]⁵. In equilibrium G=0, we will have s=V(q) like in the HJ approach, and the last equation relates the Lagrangian L* = dV/dt=\dot{s} to G(q,p,s).

• But G is not quite the Hamiltonian, out of equilibrium. A simple test case is G(q,p,s):=H(q,p) + γs, then ∂G/∂s=γ is constant and then dG/dt = -γG. γ is then the rate of improvements. In the limit when G=0, this says s settles to H(q,p)/γ which is constant even though q and p may still be varying. So exponentially G→0 , and s→H/γ.

• With the specification of H of the RCK model, ds/dt=e⁻ᵖᵗu - γs, and indeed this vanishes (s is constant) when s*=e⁻ᵖᵗu/γ. (That exponent should be a ρ, not a p—I felt like I got away with it until now.)

• Using the limit p=∂V/∂q the second equation, for \dot{p}, recovers the (KR) condition through the HJ method above.

To round off, let me repeat that the equations (*) do not assume any maximizing behavior, simply that economic processes will preserve economic equilibrium. What we can say is that equilibrium then reveals some utility function which is apparently optimized in those states. Learned improvements are not forgotten.

That’s it for now, for this series. Not because this story is complete, but because I’ve reached my self-imposed deadline and currently have no more to say. I will revisit it occasionally as I learn more in discussion with more knowledgeable people I hope to interest in these posts.

1. I am not above repeating good jokes, either.

2. Note this lemma also links to the Ramsey-Cass-Koopmans model under discussion.

3. Replacing notation S→V and q→k.

4. Or inefficiency if you take a pessimistic view, or ignorance.

5. In the RCK model, of value per unit of labor, just as k was capital intensity K/L.