9. Neoclassical Growth Model

In the neoclassical growth model, there are three basic concepts: production, consumption and capital accumulation.

For production, we assume that only two inputs are needed, which are labor \(L\) and capital \(K\). So the function can be written as \(Y=G(K,L)\) where \(Y\) is the output which can be only used as consumption good or next period capital. In this section, we will discuss the simpler case where \(Y\subseteq\mathbb R_+\). Note that we implicitly assume the consumption good and capital good are perfectly substitutable. In the neoclassical growth model, we also assume that \(G\) is homogeneous of degree 1 (h.o.d. 1) in \(K\) and \(L\), and satisfies the Inada Conditions.

For consumption, the agent has a concave utility function \(U:\mathbb R_+\to\mathbb R\). And for capital accumulation, we assume a constant depreciation rate \(\delta\) to start with, and may relax this assumption when needed.

Recall the discussion in section 4. In the neoclassical growth model, the state variable is capital \(k\), the control variable is consumption \(c\), and the period-return function is the utility function.

9.2.1 Discrete time

We only talk about the state sequence set-up. And for simplicity, in this discrete problem, let's assume the depreciation rate is \(\delta=1\). Consider the following sequence problem: $$V^\star(k_0)=\max_{\{k_{t+1}\}_{t=0}^\infty}\sum_{t=0}^\infty\beta^t U(f(k_t)-k_{t+1})\quad\text{s.t.}\quad k_{t+1}\in[0,f(k_t)],\ k_0\text{ given}$$

The relationship with the general case in §4: \(F(x,y)=U(f(x)-y)\) and \(\Gamma(x)=[0,f(x)]\).

Euler equation and transversality condition. First, calculate some derivatives: $$F_x(x,y)=U'(f(x)-y)f'(x),\qquad F_y(x,y)=-U'(f(x)-y)$$ So the EE in the discrete time neoclassical growth model with \(\delta=1\) is $$-U'(f(k_t)-k_{t+1})+\beta U'(f(k_{t+1})-k_{t+2})f'(k_{t+1})=0,\quad\text{for }\forall t\ge0$$ and the TC is $$\lim_{t\to\infty}\beta^t U'(f(k_t)-k_{t+1})f'(k_t)k_t=0$$

Steady state. By the definition of steady state, rewrite the EE: $$\begin{aligned}-U'(f(\bar k)-\bar k)+\beta U'(f(\bar k)-\bar k)f'(\bar k)&=0\\\Rightarrow U'(f(\bar k)-\bar k)(\beta f'(\bar k)-1)&=0\\\Rightarrow f'(\bar k)&=\frac1\beta\\\Rightarrow\bar k&=f'^{-1}\left(\frac1\beta\right)\end{aligned}$$ If \(k_0>\bar k\), then \(\bar k\) is feasible for \(k_1\), so the system will converge to the steady state immediately starting from \(t=1\). But if \(k_0<\bar k\), the system will never converge to the steady state since \(\bar k\) is never feasible. Note that this happens because we assumed \(\delta=1\), which means the capital cannot be accumulated.

9.2.2 Continuous time - state sequence set-up

In the continuous time setup, the law of motion of capital is $$\dot k(t)=f(k)-c(t)-\delta k(t)$$ Now consider the sequence problem: $$V^\star(k(0))=\max_{\{\dot k(t)\}_{t=0}^\infty}\int_0^\infty e^{-\rho t}U(f(k)-\delta k(t)-\dot k(t))dt\quad\text{s.t.}\quad\dot k\in\mathbb R,\ k(0)\text{ given}$$ The relationship with the general case in §4: \(F(k,\dot k)=U(f(k)-\delta k-\dot k)\) and \(\Gamma(k)=\mathbb R\).

Euler equation and transversality condition. First, calculate some derivatives: $$F_k=(f'(k)-\delta)U',\quad F_{\dot k}=-U',\quad F_{\dot k k}=-(f'(k)-\delta)U'',\quad F_{\dot k\dot k}=U''$$ So the EE in the continuous time neoclassical growth model is $$(f'(k)-\delta)U'-\rho U'=-(f'(k)-\delta)U''\dot k+U''\ddot k\Rightarrow(f'(k)-\delta-\rho)U'=-\left((f'(k)-\delta)\dot k-\ddot k\right)U''\tag{9.1}$$ and the TC is $$\lim_{t\to\infty}-e^{-\rho t}\cdot U'(f(k)-\delta k-\dot k)\cdot k(t)=0$$ Steady state. By the definition of steady state, rewrite the EE: $$(f'(k)-\delta-\rho)U'=-((f'(k)-\delta)\cdot0-0)U''\Rightarrow f'(k)=\rho+\delta$$

9.2.3 Continuous time - control-state sequence set-up and Hamiltonian

We can solve the same continuous time problem with the control-state set-up. Consider the following problem: $$V^\star(k(0))=\max_{\{c(t)\}_{t=0}^\infty}\int_0^\infty e^{-\rho t}U(c(t))dt\quad\text{s.t.}\quad\dot k(t)=f(k)-\delta k(t)-c(t),\ k(0)\text{ given}$$ The relationship with the general case in §4: \(h(k,c)=U(c)\) and \(\dot k=g(k,c)=f(k)-\delta k-c\). So we can construct the Hamiltonian function $$H(k,c,\lambda)=U(c)+\lambda(f(k)-\delta k-c)$$ Recall the maximum principle for Hamiltonian: $$\begin{aligned}H_u=0&\Rightarrow U'(c)=\lambda\\\dot\lambda=\rho\lambda-H_x&\Rightarrow\dot\lambda=\lambda(\rho-(f'(k)-\delta))\\\dot x=g&\Rightarrow\dot k=f(k)-\delta k-c\end{aligned}$$ Therefore, we can write down a set of equations for the dynamic system: $$\dot\lambda=\lambda(\rho-(f'(k)-\delta))\tag{9.2}$$ $$\dot k=f(k)-\delta k-c\tag{9.3}$$ Note that this set of dynamic equations are obtained under the optimization conditions, which means the path depicted by them is optimal. There are three variables \(\lambda\), \(k\) and \(c\) in these equations. We can get rid of either \(\lambda\) or \(c\) to form two equivalent sets of dynamic equations for the system: one is with co-state \(\lambda\) and state \(k\), the other is with control \(c\) and state \(k\).

Dynamic equations with respect to \((k,\lambda)\) space. Since \(H_u=0\Rightarrow U'(c)=\lambda\Rightarrow c=U'^{-1}(\lambda)\), we can substitute \(\lambda\) for \(c\) in (9.3): $$\dot\lambda=\lambda(\rho-(f'(k)-\delta))\tag{9.4}$$ $$\dot k=f(k)-\delta k-U'^{-1}(\lambda)\tag{9.5}$$

Phase diagram in \((k,\lambda)\) space. We want to draw the locus of \(\dot\lambda=0\) and the locus of \(\dot k=0\) (which separate the space into regions with different dynamics, and their intersection is the steady state), and based on the dynamics in each region we can draw the saddle path. - \(\dot\lambda=0\): \(f'(k)-\delta=\rho\). In \((k,\lambda)\) space, this is a vertical line \(k=\bar k\) s.t. \(f'(\bar k)=\rho+\delta\). Dynamics: by the concavity of \(f\), if \(k>\bar k\), \(f'(k)\) is lower, so that \(\dot\lambda>0\); if \(k<\bar k\), \(\dot\lambda<0\). - \(\dot k=0\): \(U'^{-1}(\lambda)=f(k)-\delta k\Rightarrow\lambda=U'(f(k)-\delta k)\). By the concavity of \(U\), \(U''<0\), so \(U'(\cdot)\) is a strictly decreasing function; \(f(k)-\delta k\) is hump-shaped (increases first and then starts to decrease), so \(\dot k=0\) is a U-shaped transformation in \((k,\lambda)\) space. Dynamics: transfer into \((k,\lambda)\) space, if \(\lambda<\lambda(k)\) then \(\dot k<0\); if \(\lambda>\lambda(k)\) then \(\dot k>0\), where \(\lambda(k)\) means the locus of \(\dot k=0\).

Figure (\((k,\lambda)\) phase diagram, paraphrased): the vertical axis is \(\lambda\), horizontal axis \(k\). The "steady \(\lambda\) locus" (\(\dot\lambda=0\)) is a vertical line through \(\bar k\); the "steady \(k\) locus" (\(\dot k=0\)) is U-shaped; they intersect at the steady state \((\bar k,\bar\lambda)\), with a saddle path converging to the steady state. (The optimal path is called the saddle path because any deviation leads to an unwanted destination instead of the steady state; the agent with perfect knowledge of the future chooses the exact path on the saddle path.)

Dynamic equations with respect to \((k,c)\) space. Since \(H_u=0\Rightarrow U'(c)=\lambda\), we have \(\dot\lambda=U''(c)\dot c\) (9.6). Substitute \(U''(c)\dot c\) for \(\dot\lambda\) and \(U'(c)\) for \(\lambda\) in (9.2): $$\dot c=\frac{U'(c)}{-U''(c)}((f'(k)-\delta)-\rho)\tag{9.7}$$ $$\dot k=f(k)-\delta k-c\tag{9.8}$$

Phase diagram in \((k,c)\) space. - \(\dot c=0\): \(f'(k)-\delta=\rho\). In \((k,c)\) space, this is also a vertical line \(k=\bar k\) s.t. \(f'(\bar k)=\rho+\delta\). Dynamics: by the concavity of \(f\), if \(k>\bar k\), \(f'(k)\) is lower, so that \(\dot c<0\); if \(k<\bar k\), \(\dot c>0\). - \(\dot k=0\): \(c=f(k)-\delta k\), which is hump-shaped (increases first and then starts to decrease). Dynamics: if \(c>c(k)\) then \(\dot k<0\); if \(c0\), where \(c(k)\) means the locus of \(\dot k=0\).

Figure (\((k,c)\) phase diagram, paraphrased): the vertical axis is \(c\), horizontal axis \(k\). The "steady \(c\) locus" (\(\dot c=0\)) is a vertical line through \(\bar k\); the "steady \(k\) locus" (\(\dot k=0\)) is hump-shaped; they intersect at the steady state \((\bar k,\bar c)\), with a saddle path converging to the steady state.

9.3.1 Discrete time

First, calculate some derivatives of \(F(x,y)=U(f(x)-y)\): $$\begin{aligned}F_x&=f'(x)U'(f(x)-y)=f'U'\\F_y&=-U'(f(x)-y)=-U'\\F_{xx}&=f''(x)U'+[f'(x)]^2 U''=f''U'+f'^2 U''\\F_{yy}&=U''(f(x)-y)=U''\\F_{xy}&=-f'(x)U''=-f'U''\end{aligned}$$ Recall (5.2): \(Q(\lambda)\equiv\beta F_{xy}\lambda^2+(F_{yy}+\beta F_{xx})\lambda+F_{yx}=0\), where \(\lambda=g'(x^\star)\) and \(y=g(x)\). Plug in the derivatives and use \(\beta f'(k^\star)=1\) to simplify \(Q(\lambda)\) to \(\tilde Q(\lambda)\) for the neoclassical growth model: $$\tilde Q(\lambda)\equiv\lambda^2-\left(1+\frac1\beta+\frac{f''/f'}{U''/U'}\right)\lambda+\frac1\beta=0$$ Solve for the two roots: $$\lambda_1=\frac12\left[\left(1+\frac1\beta+\frac{f''/f'}{U''/U'}\right)-\sqrt{\left(1+\frac1\beta+\frac{f''/f'}{U''/U'}\right)^2-\frac4\beta}\right]$$ $$\lambda_2=\frac12\left[\left(1+\frac1\beta+\frac{f''/f'}{U''/U'}\right)+\sqrt{\left(1+\frac1\beta+\frac{f''/f'}{U''/U'}\right)^2-\frac4\beta}\right]$$ Both roots are larger than 0, and \(\lambda_1\) is the only possible root to be less than 1 (we can only have one such root). To check whether the smaller root \(\lambda_1\) makes the system converge, simply plug \(\lambda=1\) into \(\tilde Q(\lambda)=0\) and find \(\tilde Q(1)<0\). Therefore \(\lambda_1<1\).

Note that \(\dfrac{f''/f'}{U''/U'}\) determines the speed of convergence. Higher \(\left|\dfrac{f''}{f'}\right|\) will yield a smaller \(\lambda_1\), which makes the system converge faster around its stable steady state. So a larger curvature of the production function (i.e. larger \(-f''/f'\)) makes the system converge faster, while a larger curvature of the utility function (i.e. larger \(-U''/U'\)) makes the system converge slower.

Economic intuition: a larger curvature in the production function means the economy hopes to smooth its production, so in the case of \(k<\bar k\), it will not consume so much to keep the capital level come up back again to make production smooth; and in the case of \(k>\bar k\), it will consume more to keep the capital level come down again to make production smooth. On the other hand, a larger curvature in the utility function means the economy hopes to smooth its consumption, which is against smoothing production, so the exactly opposite argument holds.

9.3.2 Continuous time

First, calculate some derivatives of \(F(k,\dot k)=U(f(k)-\delta k-\dot k)\): $$\begin{aligned}F_k&=(f'(k)-\delta)U',\quad F_{\dot k}=-U'\\F_{\dot k k}&=-(f'(k)-\delta)U'',\quad F_{\dot k\dot k}=U''\\F_{kk}&=f''(k)U'+(f'(k)-\delta)^2 U''\end{aligned}$$ Denote \(k^\star\) as the steady state and \(\dot k=g(k)\) as the optimal decision rule. Evaluate the above derivatives at \(k=k^\star\) (using \(f'(k^\star)=\rho+\delta\)): $$F_k(k^\star,0)=\rho U',\quad F_{\dot k}(k^\star,0)=-U',\quad F_{\dot k k}(k^\star,0)=-\rho U'',\quad F_{\dot k\dot k}(k^\star,0)=U'',\quad F_{kk}(k^\star,0)=f''(\bar k)U'+\rho^2 U''$$ Recall (5.6): \(Q(\lambda)\equiv-F_{\dot k\dot k}\lambda^2+\rho F_{\dot k\dot k}\lambda+\rho F_{\dot k k}+F_{kk}=0\). Plug in the derivatives: $$\begin{aligned}0=Q(\lambda)&=(-U'')\lambda^2+(\rho U'')\lambda+(-\rho^2 U''+f''U'+\rho^2 U'')\\&=(-U'')\lambda^2+(\rho U'')\lambda+f''U'\\&=(-U'')\left(\lambda^2-\rho\lambda-\frac{f''U'}{U''}\right)\\&=(-U'')\left(\lambda^2-\rho\lambda-\frac{f''/f'}{U''/U'}(\rho+\delta)\right)\end{aligned}$$ Solve for the two roots: $$\lambda_1=\frac12\left[\rho-\sqrt{\rho^2+4\frac{f''/f'}{U''/U'}(\rho+\delta)}\right],\qquad\lambda_2=\frac12\left[\rho+\sqrt{\rho^2+4\frac{f''/f'}{U''/U'}(\rho+\delta)}\right]$$ Clearly, \(\lambda_1<0\), i.e. there is one stable steady state \(k^\star\) with \(g'(k^\star)<0\) for the neoclassical growth model.

9.4.1 Slope of the saddle path in \((k,\lambda)\) space

Recall that in \((k,\lambda)\) space, we have the dynamic functions \(\dot\lambda=\lambda(\rho-(f'(k)-\delta))\) and \(\dot k=f(k)-\delta k-c\). Since \(U'(c)=\lambda\), replace \(c\) with \(U'^{-1}(\lambda)\): \(\dot k=f(k)-\delta k-U'^{-1}(\lambda)\). Assume the saddle path has the function \(\lambda=\phi(k)\), then the slope of the saddle path is $$\phi'(k)=\frac{d\lambda}{dk}=\frac{\dot\lambda}{\dot k}\overset{\text{L'Hopital}}{=}\frac{d\dot\lambda/dk}{d\dot k/dk}=\frac{\frac{d\dot\lambda}{d\lambda}\phi'+\frac{d\dot\lambda}{dk}}{\frac{d\dot k}{d\lambda}\phi'+\frac{d\dot k}{dk}}$$ where $$\frac{d\dot\lambda}{d\lambda}=\rho-(f'(k)-\delta),\quad\frac{d\dot\lambda}{dk}=\phi'(k)(\rho-(f'(k)-\delta))-\lambda f''(k),\quad\frac{d\dot k}{d\lambda}=-\frac{d(U'^{-1}(\lambda))}{d\lambda},\quad\frac{d\dot k}{dk}=f'(k)-\delta-\frac{d(U'^{-1}(\lambda))}{d\lambda}\phi'(k)$$ Evaluate at \(k=k^\star\) and simplify using \(f'(\bar k)=\rho+\delta\) to obtain a quadratic equation of \(\phi'(k^\star)\): $$0=2\frac{d(U'^{-1}(\lambda))}{d\lambda}\phi'(k^\star)^2-\rho\phi'(k^\star)-\lambda f''(k^\star)$$ Solve it: $$\phi'_1(k^\star)=\frac{\rho-\sqrt{\rho^2+8\frac{d(U'^{-1}(\lambda))}{d\lambda}\lambda f''(k^\star)}}{4\frac{d(U'^{-1}(\lambda))}{d\lambda}},\qquad\phi'_2(k^\star)=\frac{\rho+\sqrt{\rho^2+8\frac{d(U'^{-1}(\lambda))}{d\lambda}\lambda f''(k^\star)}}{4\frac{d(U'^{-1}(\lambda))}{d\lambda}}$$ In §9.2.3 we showed by graph that the slope of the saddle path in \((k,\lambda)\) space is negative. We know that \(U''<0\) and \(d(U'^{-1}(\lambda))/d\lambda<0\), so \(\phi'_1(k^\star)<0\) is the slope of the saddle path.

9.4.2 Slope of the saddle path in \((k,c)\) space

Recall \(\dot c=\dfrac{U'(c)}{-U''(c)}(f'(k)-\delta-\rho)\) and \(\dot k=f(k)-\delta k-c\). Define \(c=\psi(k)\) as the saddle path and \(\psi'(k^\star)\) the slope evaluated at \(k=k^\star\). Similar as in \((k,\lambda)\) space, evaluate at \(k^\star\) to obtain a quadratic equation of \(\psi'(k^\star)\): $$(\psi')^2-\rho\psi'-\frac{f''/f'}{U''/U'}(\rho+\delta)=0$$ Solve it: $$\psi'_1(k^\star)=\frac{\rho-\sqrt{\rho^2+4\frac{f''/f'}{U''/U'}(\rho+\delta)}}{2},\qquad\psi'_2(k^\star)=\frac{\rho+\sqrt{\rho^2+4\frac{f''/f'}{U''/U'}(\rho+\delta)}}{2}$$ Clearly, \(\psi'_2(k^\star)>\rho>0\) is the slope of the saddle path.

9.4.3 The relationship between the slope of the saddle path and the slope of the optimal decision rule

To show the close relationship between the slope of the saddle path and the slope of the optimal decision rule, we will focus on the saddle path in \((k,c)\) space.

Recall that in §9.3.2 we obtained a stable steady state \(k^\star\) with $$g'(k^\star)=\frac{\rho-\sqrt{\rho^2+4\frac{f''/f'}{U''/U'}(\rho+\delta)}}{2}<0$$ where \(\dot k=g(k(t))\) is the optimal decision rule. In §9.4.2 we obtained the slope of the saddle path around the stable steady state $$\psi'(k^\star)=\frac{\rho+\sqrt{\rho^2+4\frac{f''/f'}{U''/U'}(\rho+\delta)}}{2}$$ where \(c=\psi(k)\) is the function of the saddle path. Notice that $$\psi'(k^\star)+g'(k^\star)=\rho$$ which is not a coincidence. To see where this relationship comes from, consider the law of motion of capital $$\begin{aligned}g(k)=\dot k=f(k)-\delta k-c(k)&\Rightarrow g'(k)+c'(k)=f'(k)-\delta\\&\Rightarrow g'(k^\star)+c'(k^\star)=f'(k^\star)-\delta=\rho\end{aligned}$$ where in the first equality we write \(c\) as \(c(k)\) because now we are considering the optimal decision rule and points on the saddle path satisfy the optimal decision rule. So around a stable steady state, the slope of the optimal decision rule and the slope of the saddle path are closely related by \(\psi'(k^\star)+g'(k^\star)=\rho\).