24. RBC: Deterministic Neoclassical Growth Model

In this section, we are going to discuss the steady state and transitional dynamics of a deterministic Neoclassical Growth Model. In this set-up, it is deterministic how the economy would evolve starting from any point off the steady state.

24.1 The model

24.1.1 Social planner's problem

The social planner's problem is to maximize the additively time separable discounted utility of the infinitely living representative household, i.e.

$$ \max_{\{C_t,H_t,K_{t+1}\}_{t=0}^{\infty}}\sum_{t=0}^{\infty}\beta^t u(C_t,H_t) \tag{24.1} $$

$$ \text{s.t.}\quad K_{t+1}+C_t=F_t(K_t,H_t)\ \text{for }\forall t,\qquad K_0\text{ given} \tag{24.2} $$

where \(C_t\) is the consumption, \(H_t\) is the hours worked, and \(K_t\) is the capital, all in period \(t\). Also, we have the utility discounting factor \(\beta\in(0,1)\).

Note that \(F_t(K_t,H_t)\) is the output \(Y_t\) plus depreciated capital in period \(t\), i.e. \(F_t(K_t,H_t)=Y_t+(1-\delta)K_t\), and we can write the production function as

$$ Y_t=\tilde F_t(K_t,H_t)=F_t(K_t,H_t)-(1-\delta)K_t $$

and we can define the investment \(X_t=K_{t+1}-(1-\delta)K_t\).

As always, we would make the following assumptions to ensure the sufficiency of interior first order conditions:

• utility function \(u(C_t,H_t)\) is concave in \(C_t\) and convex in \(H_t\) (since \(H_t\) causes disutility, we can otherwise write the utility function as \(u(C_t,1-H_t)\), then it will be jointly concave in \((C_t,1-H_t)\)), and satisfies Inada conditions.
• production function \(\tilde F_t(K_t,H_t)\) is jointly concave in \((K_t,H_t)\), and so is \(F_t(K_t,H_t)\).

24.1.2 The Lagrangian

$$ \mathcal{L}=\sum_{t=0}^{\infty}\beta^t\big[u(C_t,H_t)+\lambda_t\big(F_t(K_t,H_t)-K_{t+1}-C_t\big)\big] \tag{24.3} $$

where \(\lambda_t\) is the Lagrangian multiplier for the budget constraint in period \(t\).

24.1.3 First order conditions

The f.o.c. of (24.3) are:

$$ [C_t]\quad u_C(C_t,H_t)=\lambda_t \tag{24.4} $$

$$ [H_t]\quad -u_H(C_t,H_t)=\lambda_t F_{H,t}(K_t,H_t) \tag{24.5} $$

$$ [K_{t+1}]\quad \lambda_t=\beta\lambda_{t+1}F_{K,t+1}(K_{t+1},H_{t+1}) \tag{24.6} $$

$$ [\lambda_t]\quad K_{t+1}+C_t=F_t(K_t,H_t) $$

Using both (24.4) and (24.5), we can get the stacked f.o.c., which is also the intra-temporal indifference condition:

$$ \frac{-u_H(C_t,H_t)}{u_C(C_t,H_t)}=F_{H,t}(K_t,H_t) \tag{24.7} $$

where the LHS of (24.7) is the marginal rate of substitution (\(MRS_t\)), i.e. the units of consumption to give up to stay indifferent when working one unit time less, and the RHS of (24.7) is the marginal product of labor (\(MPL_t\)).

24.1.4 Euler equation and transversality condition

The Euler equation (EE, or call it inter-temporal indifference condition) can be obtained by using (24.4) and (24.6):

$$ \lambda_t=\beta\lambda_{t+1}F_{K,t+1}(K_{t+1},H_{t+1})\ \Rightarrow\ u_C(C_t,H_t)=\beta u_C(C_{t+1},H_{t+1})F_{K,t+1}(K_{t+1},H_{t+1}) \tag{24.8} $$

And the transversality condition (TC):

$$ \text{TC}:\quad \lim_{T\to\infty}\beta^T\lambda_T K_T=0 $$

Note that the transversality condition is not necessary for optimality of the solution, but it ensures that the capital stock won't explode along time.

24.2 Kaldor facts: stylized facts of long run economic growth

The Kaldor facts are the statistical facts of long run economic growth summarized by Nicholas Kaldor in his famous paper published in 1957:

• The shares of national income received by labor and capital are roughly constant over long periods of time.
• The rate of growth of the capital stock per worker is roughly constant over long periods of time.
• The rate of growth of output per worker is roughly constant over long periods of time.
• The capital/output ratio is roughly constant over long periods of time.
• The rate of return on investment is roughly constant over long periods of time.
• There are appreciable variations (2 to 5 percent) in the rate of growth of labor productivity and of total output among countries.

These stylized facts are true for the long run average, but not true in the short run. In fact, growth rates and labor income share fluctuate significantly over business cycles. In this model, we assume the following version of Kaldor facts for model calibration:

1. Labor share of income \(1-\alpha\) is constant over time.
2. Capital \(K_t\), output \(Y_t\), and consumption \(C_t\) all grow at the same constant rate \(g\), i.e.

$$ K_t(1+g)^{-t}=k,\quad Y_t(1+g)^{-t}=y,\quad C_t(1+g)^{-t}=c \tag{24.9} $$

1. Hours worked \(H_t\) is constant over time (statistically \(H_t\) is slightly decreasing; the number of average hours worked is controversial because there are significant differences by gender groups, education level groups, and by countries; for simplicity, here we just assume that the \(H_t\) is constant over time), i.e. \(H_t=H\) for \(\forall t\).
2. Wage rate \(w_t\equiv MRS_t=MPL_t\) also grows at the same constant rate \(g\) (immediately from facts 1, 2 and 3; alternatively, we can drop fact 1 and CRS assumption on production function, then fact 2 and 3 imply fact 1 and 4; see the proof for proposition 24.2).

24.3 More assumptions for calibration

In addition to concavity and Inada assumptions and Kaldor facts assumptions, we are going to impose more assumptions on both \(F_t(K_t,H_t)\) and \(u(C_t,H_t)\) to more easily calibrate the model to fit the data.

24.3.1 More assumptions on \(F_t(K_t,H_t)\)

We assume that

$$ F_t(K_t,H_t)=\hat F\big(K_t,(1+g)^t H_t\big)=\hat F\big(K_t,(1+g)^t H\big) $$

• The effectiveness of constant labor supply is growing at the same constant rate \(g\).
• The time varying part of the functional form of \(F_t(\cdot,\cdot)\) comes in through the growing effectiveness of constant labor supply \(H\).
• So, function \(\hat F(\cdot,\cdot)\) is time invariant.

We also assume that function \(\hat F(\cdot,\cdot)\) is constant return to scale (CRS), i.e. h.o.d. 1. More specifically, we assume the Cobb-Douglas production function, i.e.

$$ Y_t=\tilde F_t(K_t,H_t)=AK_t^{\alpha}H_t^{1-\alpha}\quad\text{or}\quad \hat F(k,H)=Ak^{\alpha}H^{1-\alpha}+(1-\delta)k $$

24.3.2 More assumptions on \(u(C_t,H_t)\)

We assume the following specific family of utility functions

$$ u(C_t,H_t)=\frac{\big(C_t\cdot e^{-v(H_t)}\big)^{1-\sigma}-1}{1-\sigma} $$

where \(v(\cdot)\) is increasing and convex. When \(\sigma\to1\),

$$ \lim_{\sigma\to1}u(C_t,H_t)=\lim_{\sigma\to1}\frac{\big(C_t e^{-v(H_t)}\big)^{1-\sigma}-1}{1-\sigma}\overset{\text{L'Hopital}}{=}\lim_{\sigma\to1}\frac{-\big(C_t e^{-v(H_t)}\big)^{1-\sigma}\ln\big(C_t e^{-v(H_t)}\big)}{-1}=\ln\big(C_t e^{-v(H_t)}\big)=\ln C_t-v(H_t) $$

24.4 A system of three static equations

In the steady state, we will have a balanced growth. With all the assumptions we have made and properties we have discussed so far, we can rewrite the intra-temporal condition (24.7) to obtain

$$ v'(H_t)\,c=\hat F_H(k,H) \tag{24.10} $$

We can rewrite the Euler equation (24.8) to obtain

$$ (1+g)^{\sigma}=\beta\hat F_K(k,H) \tag{24.11} $$

Finally, we can rewrite the resource constraint (24.2) to obtain

$$ (1+g)k+c=\hat F(k,H) \tag{24.12} $$

24.5 Calibrate the model

We have six items to calibrate, which are \(\alpha\), \(\beta\), \(g\), \(\sigma\), \(\delta\), \(v(H)\). Below is a calibration based on yearly data. Once we switch to quarterly data, we have to redo the calibration (the flow variables on quarterly basis (with superscript \(Q\)) will take the value a quarter of that on yearly basis (with superscript \(Y\)), e.g. \(g^Q=\frac14 g^Y\) and \(\delta^Q=\frac14\delta^Y\)).

• Preference parameter \(\sigma=1\): cannot be estimated from the data, so we have to fix it as 1. The calibration of all other parameters will depend on the value of \(\sigma\). We won't get in trouble as long as \(\sigma\) is not too big nor too small.
• Balanced growth rate \(g=0.02\): can be estimated from data. We will use the data of the U.S. after civil war: average growth rate of output (real GDP) is approximately 2%, and there it is neither trending up nor trending down over time.
• Labor share of income \(1-\alpha=0.6\) (i.e. \(\alpha=0.4\)): can be estimated from data.
• Depreciation rate \(\delta=0.06\): can be estimated from data. We obtain the estimation of \(\delta\) through the data on investment rate \(\frac{X_t}{K_t}\): empirically \(\frac{X_t}{K_t}=0.08\), and there is no significant change over time. Rewrite the law of motion of capital \(K_{t+1}=(1-\delta)K_t+X_t\Rightarrow\frac{K_{t+1}}{K_t}=(1-\delta)\frac{K_t}{K_t}+\frac{X_t}{K_t}\Rightarrow1+g=1-\delta+0.08\Rightarrow\delta=0.08-g=0.06\).
• Utility discounting rate \(\beta=0.958\): can be estimated from data. We obtain the estimation of \(\beta\) through the data on capital to output ratio \(\frac{K_t}{Y_t}\): empirically \(\frac{K_t}{Y_t}=3.2\), and there is no significant change over time. Rewrite (24.11):

$$ \begin{aligned} (1+g)^{\sigma}&=\beta\hat F_K(k,H)=\beta\frac{\partial\hat F(k,H)}{\partial k}=\beta\frac{\partial[Ak^{\alpha}H^{1-\alpha}+(1-\delta)k]}{\partial k}\\ &=\beta\big[A\alpha k^{\alpha-1}H^{1-\alpha}+(1-\delta)\big]=\beta\Big[\alpha\frac{y}{k}+(1-\delta)\Big]\ \Rightarrow\ \beta=\frac{(1+g)^{\sigma}}{\alpha\frac{y}{k}+(1-\delta)}\approx0.958 \end{aligned} $$

It is surprisingly reasonable because 0.958 is obtained indirectly and it happens to be less than 1, which makes sense economically. Since \(g\) is pinned down by the data, we can conclude that \(\beta\) increases in \(\sigma\). Recall that \(\sigma\) is arbitrarily picked, so it shows that \(\sigma\) cannot be too big, otherwise \(\beta\) would be larger than 1.

• Disutility from working \(v(H)\): cannot be estimated from the data, but \(v'(H)\) is pinned down by the data. \(y\), \(k\), \(c\) and \(H\) are directly observable from the data; \(A\) is pinned down by \(y=Ak^{\alpha}H^{1-\alpha}\); \(\alpha=0.4\) is already obtained. Rewrite (24.10): \(v'(H_t)=\dfrac{\hat F_H(k,H)}{c}=\dfrac{A(1-\alpha)k^{\alpha}H^{-\alpha}}{c}\), then plug in \(A,\alpha,k,H\) and \(c\) to obtain \(v'(H_t)\).

24.6 Transitional dynamics

24.6.1 Unique steady state

In subsection 24.4 we obtained the system of three equations: (24.10), (24.11), and (24.12), which together characterized an equilibrium. If we plug in for the Cobb-Douglas production function (\(y_t=k_t^{\alpha}H_t^{1-\alpha}\), since the total factor productivity parameter \(A\) is just a constant without any dynamics, we drop it for simplicity) and for simplicity let \(\sigma=1\) in the utility function (i.e. \(\ln C_t-v(H_t)\)), then we can get that

$$ c_t v'(H_t)=(1-\alpha)k_t^{\alpha}H_t^{-\alpha} \tag{24.13} $$

$$ (1+g)c_t^{-1}=\beta c_{t+1}^{-1}\big(\alpha k_{t+1}^{\alpha-1}H_{t+1}^{1-\alpha}+1-\delta\big) \tag{24.14} $$

$$ (1+g)k_{t+1}=k_t^{\alpha}H_t^{1-\alpha}+(1-\delta)k_t-c_t \tag{24.15} $$

Note that in subsection 24.4 there is no \(t\) subscript in all expressions because we were explicitly discussing the steady state of the economy, which is a balanced growth, and thus we have \(c_t=c\), \(k_t=k\), and \(H_t=H\) for \(\forall t\). However, in this subsection, we are discussing the transitional dynamics, which means that we are interested in what happen if the economy deviates slightly from the steady state (i.e. the balanced growth), so the time subscript comes into place.

Based on this discussion, we have reached the conclusion that there is a unique steady state given that we have a specification of function \(v(H)\). Now the natural questions to ask are:

• What is the dynamics of the system starting at a point away from the steady state?
• Will the system always converge to steady state?

So, in the next two subsections, we will be discussing the transitional dynamics of the economy starting away from steady state under two special cases of \(v(H)\) function.

24.6.2 Case I: perfectly inelastic labor supply

Consider the following specification of the disutility of labor \(v(H)=\begin{cases}0 & \text{if }H\le1\\ \infty & \text{if }H>1\end{cases}\Rightarrow H_t=H=1\) for \(\forall t\).

Notice that \(v(H)\) is kinked at one and so is not differentiable. So the system of equations that characterizes the equilibrium needs to be modified. In particular, we will replace (24.13) with \(H_t=1\). Note that this is the standard version of the Neoclassical Growth Model with labor inelastically supplied. Now we have that the following equations characterize an equilibrium

$$ H_t=1 $$

$$ (1+g)c_t^{-1}=\beta c_{t+1}^{-1}\big(\alpha k_{t+1}^{\alpha-1}H_{t+1}^{1-\alpha}+1-\delta\big) \tag{24.20} $$

$$ (1+g)k_{t+1}=k_t^{\alpha}H_t^{1-\alpha}+(1-\delta)k_t-c_t \tag{24.21} $$

Now, consider the equations (24.20) and (24.21) to find the unique steady state. We will use (24.20) and (24.21) to construct a graph of two loci, of which the intersection point is the unique steady state.

• (24.20) specifies which values of \((k_{t+1},c_t)\) make \(c_{t+1}=c_t\) (i.e. a locus of all candidate steady states of \(c_t\)). When evaluated under steady state condition for \(c_t\) and plug in \(H_t=1\), (24.20) becomes

$$ 1+g=\beta(\alpha k_{t+1}^{\alpha-1}+1-\delta)\ \Rightarrow\ k_{t+1}=\left[\frac{\frac{1+g}{\beta}-(1-\delta)}{\alpha}\right]^{\frac{1}{\alpha-1}} \tag{24.22} $$

which is a vertical line, and is the steady \(c\) locus. Left and right regions dynamics: according to (24.20), \(c_t\) increases in \(k_{t+1}\). Right region to the locus (higher \(k_{t+1}\)) implies higher \(c_t\), so right region has \(c_t>c_{t+1}\), which means decreasing dynamic of \(c_t\). Left region implies lower \(c_t\), so left region has \(c_t

• (24.21) specifies which values of \((k_{t+1},c_t)\) make \(k_{t+1}=k_t\) (i.e. a locus of all candidate steady states of \(k_{t+1}\)). When evaluated under steady state condition for \(k_{t+1}\) and plug in \(H_t=1\), (24.21) becomes

$$ c_t=k_{t+1}^{\alpha}-(g+\delta)k_{t+1} \tag{24.23} $$

which is the steady \(k\) locus, and (24.23) implies an inverse-U shape locus. Outer (upper) region to the locus (higher \(c_t\)) implies higher \(k_t\), so outer (upper) region has \(k_t>k_{t+1}\), which means decreasing dynamic of \(k_{t+1}\). Inner (lower) region implies lower \(c_t\), so inner (lower) region has \(k_t

Figure 14 (Phase Diagram for Inelastic Labor Supply, paraphrased): horizontal axis \(k_{t+1}\), vertical axis \(c_t\). The vertical steady \(c\) locus (at \(k^*=k\)) intersects the inverse-U steady \(k\) locus at the steady state \((k^*,c^*)\). Arrows in the four regions indicate the increasing/decreasing dynamics of \(c_t\) and \(k_{t+1}\); the saddle path through the steady state slopes from southwest to northeast.

We can argue that the optimal policy should lie on the saddle path. Suppose we start at \(k_{t+1}

• If \(c_t\) is chosen to be above the saddle path, then according to the transitional dynamics, the system \((k_{t+1},c_t)\) would first go upper right and then upper left until it hits the vertical axis, which means the depletion of capital, so it cannot be optimal for the infinite life agent.
• If \(c_t\) is chosen to be below the saddle path, then according to the transitional dynamics, the system \((k_{t+1},c_t)\) would first go upper right and then lower right until it hits the horizontal axis, which means zero consumption forever thereafter, so it cannot be optimal for the infinite life agent.
• Therefore, the optimal policy choices are on the saddle path.

The next question is how to find the saddle path? Below are the two algorithms.

1. Forward Shooting Algorithm:
   • (a) Given a level of capital \(k_{t+1}\), guess different levels of consumption \(c_t\) and iterate each of those candidate \(c_t\)'s forward using (24.20) and (24.21).
   • (b) Check whether any of these candidate \(c_t\)'s allow the system \((k_{t+1},c_t)\) to converge to the steady state that is not on either axis.
     • i. Since the saddle path is uni-dimensional, you will never (probability 0) pick the true \(c_t\) to converge to the steady state.
     • ii. Instead you want to provide a small enough interval for the equilibrium level of \(c_t\), of which the upper bar \(\bar c_t\) makes the system converge to vertical axis and the lower bar \(\underline c_t\) makes the system converge to horizontal axis.
   • (c) To implement the forward shooting algorithm for this problem: i. Pick a value of \(k_{t+1}k\)). ii. Then, since we know that \(c_t\), the level of equilibrium consumption that lies on the saddle path, must be below the steady \(k\) locus, so \(c_t<\underbrace{k_{t+1}^{\alpha}-(g+\delta)k_{t+1}}_{\bar k}\) and we should search for \(c_t\) in the interval \((l_0,h_0)\) where \(l_0=0\) and \(h_0=\bar k\). iii. Bisection algorithm: find the midpoint of the interval \((0,\bar k)\), denoted as \(m_0\) where \(m_0=\frac{\bar k}{2}\). Check whether the value \(c_t=m_0\) veers the forward-iterated path off towards the \(c_t\)-axis or the \(k_{t+1}\)-axis. A. If it veers to the \(c_t\)-axis, then search for \(c_t\) in the interval \((l_0,m_0)\) and repeat this bisection algorithm for \((l_0,m_0)\). B. If it veers to the \(k_{t+1}\)-axis, then search for \(c_t\) in the interval \((m_0,h_0)\) and repeat this bisection algorithm for \((m_0,h_0)\).
   • (d) Carry out this algorithm for each \(k_{t+1}\) to obtain the saddle path.
   • (e) Finally, notice that because of issues of machine precision, even for the approximate solution to forward-iterating a sufficiently large number of steps you will still veer away from the steady state, but this is a minor thing and the solution is good enough conditional on you picked a good tolerance value for convergence test.
2. Backward Shooting Algorithm:
   • (a) This algorithm starts from the steady state \((k,c)\) by deviating from it a small amount.
   • (b) In this problem, let's find the left part of the saddle path (right part follows the same logic): for \(\varepsilon_n>0\) and \(\lim_{n\to\infty}\varepsilon_n=0\):
     • i. Start from \((k-\varepsilon_n,c)\) for each \(n\) and use (24.20) and (24.21) to trace back, i.e. trace \((k_t,c_{t-1}),(k_{t-1},c_{t-2}),\ldots\). And this sequence should approximate the left part saddle path from the above.
     • ii. Start from \((k,c-\varepsilon_n)\) for each \(n\) and use (24.20) and (24.21) to trace back, i.e. trace \((k_t,c_{t-1}),(k_{t-1},c_{t-2}),\ldots\). And this sequence should approximate the left part saddle path from the below.
     • iii. Set a tolerance level to see how large the \(n\) needs to be for the two sequences above converge to each other in any particular region, and in the region where they converge to each other, they can both be regarded as the saddle path.
   • (c) This algorithm allows us to quickly trace out the saddle path.
   • (d) But there is a drawback due to the fact that in this problem we have a discrete system. Suppose that we specified an initial level of capital \(k_0=K_0\). The path traced out will in general be discrete points and may not include \(K_0\), so what we can do is to connect each discrete points to make it a continuous path where we can find \(k_0\).

24.6.3 Case II: elastic labor supply

Assume that \(v(H)=\gamma H\). This is a case where we have infinite Frisch elasticity of labor supply (the Frisch elasticity of labor supply is the elasticity of hours worked to the wage rate given a constant marginal utility of wealth; so, Frisch elasticity measures the (compensated) substitution effect of a change in the wage rate on labor supply). To see why the Frisch elasticity is infinite, consider

$$ cv'(H)=(1-\alpha)k^{\alpha}H^{-\alpha}=w\ \Rightarrow\ c\gamma=w\ \Rightarrow\ \frac{dw}{dH}=0\ \Rightarrow\ \frac{dH}{dw}=\infty\ \Rightarrow\ \varepsilon^{\text{Frisch}}=\frac{dH/H}{dw/w}=\frac{dH}{dw}\cdot\frac{w}{H}=\infty $$

where the third line is true because the consumption level \(V\equiv wH\) is held constant so we have \(c=V\) held constant. Now, the system of equations (24.13), (24.14) and (24.15) becomes

$$ c_t v'(H_t)=(1-\alpha)k_t^{\alpha}H_t^{-\alpha}\ \Rightarrow\ c_t\gamma=(1-\alpha)k_t^{\alpha}H_t^{-\alpha}\ \Rightarrow\ H_t=\left[\frac{1-\alpha}{c_t\gamma}\right]^{\frac{1}{\alpha}}k_t \tag{24.24} $$

$$ (1+g)c_t^{-1}=\beta\left(\alpha\left[\frac{1-\alpha}{\gamma}\right]^{\frac{1-\alpha}{\alpha}}c_{t+1}^{-\frac{1}{\alpha}}+c_{t+1}^{-1}(1-\delta)\right) \tag{24.25} $$

$$ (1+g)k_{t+1}=k_t\left[\frac{1-\alpha}{c_t\gamma}\right]^{\frac{1-\alpha}{\alpha}}+(1-\delta)k_t-c_t \tag{24.26} $$

• (24.25) specifies which values of \((k_{t+1},c_t)\) make \(c_{t+1}=c_t\). When evaluated under steady state condition for \(c_t\), (24.25) becomes (the steady \(c\) locus, a horizontal line)

$$ c=\left[\frac{\frac{1+g}{\beta}-(1-\delta)}{\alpha}\right]^{\frac{\alpha}{\alpha-1}}\cdot\frac{1-\alpha}{\gamma} \tag{24.27} $$

Upper and lower regions dynamics: according to (24.26), \(k_{t+1}\) decreases in \(c_t\) and \(c_{t+1}\) increases in \(k_{t+1}\). Upper region to the locus (higher \(c_t\)) implies lower \(k_{t+1}\) and thus lower \(c_{t+1}\), so upper region has \(c_t>c_{t+1}\), which means decreasing dynamic of \(c_t\). Lower region implies higher \(k_{t+1}\) and higher \(c_{t+1}\), so lower region has \(c_t

• (24.26) specifies which values of \((k_{t+1},c_t)\) make \(k_{t+1}=k_t\). When evaluated under steady state condition for \(k_{t+1}\), (24.26) becomes (the steady \(k\) locus, an increasing with a diminishing rate shape)

$$ k_{t+1}=\frac{1}{\left[\frac{1-\alpha}{\gamma}\right]^{\frac{1-\alpha}{\alpha}}c_t^{-\frac{1}{\alpha}}-c_t^{-1}(g+\delta)} \tag{24.28} $$

Outer (upper) region to the locus (higher \(c_t\)) implies higher \(k_t\), so \(k_t>k_{t+1}\), which means decreasing dynamic of \(k_{t+1}\). Inner (lower) region implies lower \(c_t\), so \(k_t

Figure 15 (Phase Diagram for Elastic Labor Supply, paraphrased): horizontal axis \(k_{t+1}\), vertical axis \(c_t\). The horizontal steady \(c\) locus intersects the increasing steady \(k\) locus at the steady state; the saddle path through the steady state slopes from southwest to northeast. The argument for the optimality of saddle path and the two shooting algorithms are the same as in Case I.

24.6.4 Log linearization

In this subsection, we hope to linearize the saddle path in the small neighborhood of the steady state. There are two reasons that we want to do this kind of linearization:

1. Reason for linearization: it simplifies the dynamics of the system and it is accurate enough in the sufficiently small neighborhood of the steady state.
2. Reason for linearization in the small neighborhood of steady state: it is because the system will spend most of the time in that neighborhood (early stage convergence is fast and the speed of convergence decreases as the system approximates the steady state).

Empirically the log linearization turns out to be more accurate than direct linearization, so we will focus on log linearization. Assume

$$ v(H)=\frac{\gamma\varepsilon}{1+\varepsilon}H^{\frac{1+\varepsilon}{\varepsilon}},\quad\varepsilon>0,\qquad v'(H)=\gamma H^{\frac{1}{\varepsilon}} $$

where \(\varepsilon\) is the Frisch elasticity of labor supply because \(cv'(H)=(1-\alpha)k^{\alpha}H^{-\alpha}=w\Rightarrow c\gamma H^{1/\varepsilon}=w\), so \(\varepsilon^{\text{Frisch}}=\frac{dH/H}{dw/w}=\varepsilon\). We want to log-linearize the following system of equations

$$ c_t\gamma H_t^{\frac{1}{\varepsilon}}=(1-\alpha)k_t^{\alpha}H_t^{-\alpha} \tag{24.29} $$

$$ (1+g)c_t^{-1}=\beta c_{t+1}^{-1}\big(\alpha k_{t+1}^{\alpha-1}H_{t+1}^{1-\alpha}+1-\delta\big) \tag{24.30} $$

$$ (1+g)k_{t+1}=k_t^{\alpha}H_t^{1-\alpha}+(1-\delta)k_t-c_t \tag{24.31} $$

After first-order Taylor approximation, we can characterize transitional dynamics as follows.

• First, define \(\phi_t=\big(\hat c_t,\hat H_t,\hat k_t\big)'\) where \(\hat c_t\equiv\ln c_t-\ln c\), \(\hat H_t\equiv\ln H_t-\ln H\), and \(\hat k_t\equiv\ln k_t-\ln k\) and \(c,H,k\) are steady state values. Then, (24.29), (24.30), and (24.31) gives us

$$ \phi_{t+1}=\mathbf{A}\phi_t \tag{24.32} $$

where \(\mathbf{A}\) is a \(3\times3\) matrix. Note that the solution to the steady state satisfies \(\phi_t=\mathbf{0}\).

• Since we are interested in the linearization of the saddle path, we can define \(\psi_t=\big(\hat k_t,\hat c_t\big)'\). Then, (24.29), (24.30), and (24.31) gives us \(\psi_{t+1}=\mathbf{B}\psi_t\) where \(\mathbf{B}\) is also a \(2\times2\) matrix. Denote the two eigenvalues of \(\mathbf{B}\) by \(\lambda_1\) and \(\lambda_2\), and the corresponding eigenvectors by \(\mathbf{v}_1\) and \(\mathbf{v}_2\), then

$$ \mathbf{B}=\begin{bmatrix}\mathbf{v}_1 & \mathbf{v}_2\end{bmatrix}\begin{bmatrix}\lambda_1 & 0\\ 0 & \lambda_2\end{bmatrix}\begin{bmatrix}\mathbf{v}_1 & \mathbf{v}_2\end{bmatrix}^{-1} $$

So, given \(k_0\) (and thus \(\psi_0\)), we have \(\psi_t=\mathbf{B}^t\psi_0=\begin{bmatrix}\mathbf{v}_1 & \mathbf{v}_2\end{bmatrix}\begin{bmatrix}\lambda_1^t & 0\\ 0 & \lambda_2^t\end{bmatrix}\begin{bmatrix}\mathbf{v}_1 & \mathbf{v}_2\end{bmatrix}^{-1}\psi_0\). Denote \(\begin{bmatrix}\mu_1 & \mu_2\end{bmatrix}\equiv\begin{bmatrix}\mathbf{v}_1 & \mathbf{v}_2\end{bmatrix}^{-1}\psi_0\), then

$$ \psi_t=\begin{pmatrix}\hat k_t\\ \hat c_t\end{pmatrix}=\mu_1\lambda_1^t\mathbf{v}_1+\mu_2\lambda_2^t\mathbf{v}_2 $$

• Plug in the data. Assume that we have the following observed data \(g=0.005\), \(\beta=0.989\), \(\alpha=0.4\), \(\delta=0.014\), \(\varepsilon=1\), \(\gamma=0.00152\Rightarrow H=23\). And then we get that

$$ \begin{pmatrix}\hat k_{t+1}\\ \hat c_{t+1}\end{pmatrix}=\begin{bmatrix}1.024 & -0.088\\ -0.013 & 0.989\end{bmatrix}\begin{pmatrix}\hat k_t\\ \hat c_t\end{pmatrix} $$

The two eigenvalues are \(\lambda_1=0.968\) and \(\lambda_2=1.044\), and we can also solve for the associated eigenvectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\). Recall the general solution \(\psi_t=\mu_1\lambda_1^t\mathbf{v}_1+\mu_2\lambda_2^t\mathbf{v}_2\). It has to be the case that \(\mu_2=0\) if we want to get to the steady state, which is achieved by having appropriate \(\hat c_0\) for given \(\hat k_0\) (this also pins down \(\mu_1\)). Otherwise the solution will explode as the eigenvalue \(\lambda_2\) associated with \(\mu_2\) is greater than one. We also must have that \(\mu_1\mathbf{v}_1\ne0\). Since we also know \(\lambda_1\) and \(\mu_1\), we have that \(\hat k_{t+1}=0.968\hat k_t\). Since we already know \(\mathbf{v}_1\), we have that

$$ \hat c_t=0.632\hat k_t \tag{24.33} $$

Plug (24.33) into (24.32), we get \(\hat H_t=0.268\hat k_t-0.714\hat c_t=-0.160\hat k_t\). Clearly, the labor supply is higher to offset the lower-than-steady-state capital. And plug into the production function, we get \(\hat y_t=0.301\hat k_t\). We can see that the output is low when capital is low, but less so because the higher labor supply partially offset low capital. Finally, we can also get \(\hat x_t=-0.653\hat k_t\), which comes from the law of motion of capital \(x_t=k_{t+1}-(1-\delta)k_t\).