7 weeks into a quarter, I am now allergic to the word “interest rate” “saddle path” and “EIGENVALUES”.
This note presents a continuous-time neoclassical growth model, deriving household and firm optimality conditions, characterizing the steady state, and analyzing the resulting saddle-path dynamics. On top of which, per-household productivity $A$ is left out, which makes it extendable.
Consider continuous time model $t\in[0, \infty)$.
(Representative) Household’s Problem
$$ \begin{align} \max_{\lbrace c_t, x_t, k_t\rbrace_{t\ge 0}}& \int_0^\infty e^{-\rho t}U(c_t)\thinspace\text{d}t\cr \text{s.t.}& \ \dot k_t = x_t -\delta k_t, \forall t\cr & \int_0^\infty p_t(c_t + x_t)\thinspace\text{d}t = \int_0^\infty p_t(w_t + v_tk_t)\thinspace\text{d}t \cr&\text{where, } p_t :=\exp{(-\int_0^\infty r_{\tau}\thinspace\text{d}\tau)}. \end{align} $$
Now, plug $(2)$ into $(3)$ and do intergration by part on horizon $[0, T]$, we obtain
$$ \int_0^T p_t\dot k_t \thinspace\text{d}t = [p_tk_t]_0^T - \int_0^T\dot p_tk_t\thinspace\text{d}t. $$
Let $T\to \infty$ and impose transversality condition (aka no-Ponzi) $\lim_{T\to \infty}p_Tk_t =0 $. Naturally we can assume (or normalize) $p_0 = 1$. Note that $\dot p_t/p_t = -r_t$. So $(2)-(4)$ becomes $$ \begin{equation} \int_0^\infty p_tc_t\thinspace\text{d}t - k_0 + \int_0^\infty(r_t + \delta - v_t)p_tk_t\thinspace\text{d}t = \int_0^\infty p_tw_t\thinspace\text{d}t. \end{equation} $$
Now in $(5)$, because the household can arbitrarily choose $k_t$, no-arbitrage requires $$ \begin{equation} r_t + \delta - v_t = 0. \end{equation} $$ (Otw they can earn infinite profit by tilting towards the more profitable asset). With this, the households’ constraints $(5)$ further simplifies, and the household’s problem $(1) - (4)$ now simplifies to $$ \begin{equation} \max_{\lbrace c_t\rbrace_{t\ge 0}}\int_0^\infty e^{-\rho t}U(c_t)\thinspace\text{d}t \quad \text{s.t.} \quad \int_0^\infty p_tc_t\thinspace\text{d}t = \int_0^\infty p_tw_t\thinspace\text{d}t + k_0. \end{equation} $$ Note that there is only one static intergral constraint, no more differential equations. Now we take FOC of the household problem $(7)$ to characterize the optimal control $c_t$. First take the Lagrangian to $(7)$: $$ \mathcal L =\int_0^\infty \left(e^{-\rho t} U(c_t) - \lambda p_tc_t\right)\thinspace\text{d}t + \lambda\left(\int_0^\infty p_tw_t\thinspace\text{d}t + k_0\right). $$ Then we can obtain the FOC w.r.t. $c_t$ by: $$ \frac{\partial {\mathcal L}}{\partial c_t}=e^{-\rho t} U’(c_t) - \lambda p_t = 0, $$ take logarithm, then differentiate w.r.t. $t$, after a little algebra, we obtain $$ \begin{equation} \boxed{\dot c_t = -\frac{U’(c_t)}{U’’(c_t)}(r_t - \rho)\quad \forall t.} \end{equation} $$
(Competitive) Firm’s Problem
Now, assume that there is one firm, invested by and producing for $N$ households in the economy. The firm’s (aggregate) production function is: $$ Y = F(Nk, NA) $$ where $k$ is capital and $A$ is efficiency of labor (aka technology level). Assume that $F$ is homogeneous of degree 1 (constant-return-to-scale CRS production function). In other words, $F(c NK, cNA ) = cF(NK, NA), \forall c > 0$.
The firm then optimize over capital investments $\lbrace k_t\rbrace_{t\ge 0}$ at each time to maximize profit:
$$ \max_{\lbrace k_t\rbrace_{t\ge 0}}\int_0^\infty p_t[F(Nk_t, NA) -\underbrace{w_tNA}_\text{constant} - v_tNk_t]\thinspace\text{d}t. $$
Two important observations (i) because there is no adjustment cost, the firm can optimize pointwise in time. (ii) we can without loss normalize $N=1$. Now, the firm’s problem reduces to $$ \begin{align} & \max_{k_t, n_t}\ p_tf(k_t, A) - v_t k_t, \quad \forall t\cr \Rightarrow &\ f_k(k_t, A) = \underbrace{v_t = r_t + \delta}_\text{Equation $(6)$}, \quad \forall t && \text{(FOC)} \end{align} $$
Aggregate System Dynamic and Steady State Property
From goods market clearing per effective worker, require that output = consumption + investment + depreciation, i.e. $$ \begin{equation} f(k_t, A) = c_t + x_t + \delta k_t \end{equation} $$ Plug $(2)$ into $(11)$, plug $(10)$ into $(8)$, light rearrangement, we obtain the following system dynamic: $$ \begin{equation} \begin{cases} \dot k_t = f(k_t, A) -c_t -\delta k_t && [(2) + (11)]\cr \dot c_t=-\frac{U’(c_t)}{U’’(c_t)}(f_k(k_t, A) -\delta - \rho) && [(10) + (8)] \end{cases} \end{equation} $$ With $(12)$ we can study the property of steady state $(\bar k, \bar c)$. Now, with a little abuse of notation, allow steady state be contingent with $A$. At steady state, $\dot k_t=\dot c_t = 0$. We then use $(12)$ to show that the ratio $\bar k(A)/\bar c(A)$ is constant:
Because $f(\cdot, \cdot)$ is CRS, there exists a function $\phi(\cdot)$ such that $$ f(k, A) = A\phi(\frac kA). $$ Then, $f_k(k_t, A) = \phi’(\frac{k}A)$. So from $(12.2)$, letting $\dot c = 0$ determines the ratio betwee steady state $\bar k$ and $A$: $$ \frac{\bar k(A)}{A} = (\phi’)^{-1}(\delta + \rho) := \kappa. $$ From $(12.1)$, letting $\dot k_t = 0$ and sub in $\bar k(A) = A\kappa$: $$ f(\bar k(A), A) - \bar c(A) - \delta \bar k(A) = 0\Rightarrow \bar c(A) = (\phi(\kappa) - \delta \kappa)A. $$ Which implies $$ \frac{\bar c(A)}{\bar k(A)} = \frac{\phi(\kappa)}{\kappa} - \delta, \quad \forall A. $$
Saddle Path
Fix $A$ as a constant parameter, an dformallly, consider the dynamic system: $$ \begin{align*} & \dot k = g(k,c) := f(k,A) - c - \delta k, \qquad\cr & \dot c = h(k,c) := -\frac{U’(c)}{U’’(c)}\big(f_k(k,A)-\delta-\rho\big). \end{align*} $$ This is a 2D system $\dot z = F(z)$, where $z = \begin{bmatrix}k\cr c\end{bmatrix}, F=\begin{pmatrix}g\cr h\end{pmatrix}$. A steady state $(\bar k, \bar c)$ solves $g(\bar k, \bar c) = 0, f(\bar k, \bar c) = 0$. Let the Jacobian at the steady state be
$$ \begin{equation} \mathcal J := \nabla F= \begin{pmatrix} g_k(\bar k,\bar c) & g_c(\bar k,\bar c)\cr h_k(\bar k,\bar c) & h_c(\bar k,\bar c) \end{pmatrix}=\begin{pmatrix} \rho & -1\cr -\frac{U’(\bar c)}{U’’(\bar c)}f_{kk}(\bar k, A) & 0 \end{pmatrix}. \end{equation} $$
(Note: the 3nd equality above plugged in the steady state condition $g(\bar k, \bar c) = 0, f(\bar k, \bar c) = 0$ and did a little bit of algebra…)
As $\mathcal J$ is 2x2, and because of its special structure, it has exactly one stable eigenvalue $\mu_s <0$ (and another unstable $\mu_{us} >0$). Let $\begin{bmatrix}v_k^s\cr v_c^s\end{bmatrix}$ be the eigenvector associated with the stable eigenvalue $\mu_s$.
$$ \mathcal J =\begin{pmatrix} \rho & -1\cr -\frac{U’(\bar c)}{U’’(\bar c)}f_{kk}(\bar k, A) & 0 \end{pmatrix}\Rightarrow \begin{cases} \lambda_s = \frac{\rho - \sqrt{\rho^2 -4\frac{U’(\bar c)}{U’’(\bar c)}f_{kk}(\bar k, A)}}{2}\cr \frac{v_c^s}{v_k^s} = \rho - \lambda_s = \frac{\rho + \sqrt{\rho^2 + 4\big(-\frac{U’(\bar c)}{U’’(\bar c)} f*{kk}(\bar k,A)\big)}}{2}. \end{cases} $$
Now, consider the manifold $$ W^s(\bar z) = \lbrace z_0 : \text{solution } z(t;z_0) \to \bar z \text{ as } t\to \infty\rbrace. $$ Because $\mathcal J$ has a pair of opposite sign eigenvectors, $(\bar k,\bar c)$ is a saddle point, and $W^s(\bar z) $ a stable manifold (saddle path). Now,
The Stable Manifold Theorem from dynamical systems) says: If $\mathcal J$ has eigenvalues with strictly negative real parts (stable) and strictly positive real parts (unstable), then there exists a unique smooth stable manifold ($W^s(\bar z) $), tangent at $\bar z$ to the stable eigenspace of $\mathcal J$.
Therefore, on the Saddle path at the steady state:
$$ \frac{\text{d}c}{\text{d} k}{(\bar k, \bar c)} = \frac{\rho + \sqrt{\rho^2 + 4\big(-\dfrac{U’(\bar c)}{U’’(\bar c)} f_{kk}(\bar k,A)\big)}}{2}. $$
Therefore, locally around the steady state, say a shock moves capital away from $\bar k$ to $k_0$, the consumption would jumps to
$$ c_0 = \bar c + \frac{\rho + \sqrt{\rho^2 + 4\big(-\dfrac{U’(\bar c)}{U’’(\bar c)} f_{kk}(\bar k,A)\big)}}{2}(k_0 - \bar k) $$
then gradually recovers back to steady state.