Deriving Growth in the Lucas Human-Capital Model

Why Human-Capital Accumulation Can Sustain Growth

In the Lucas human-capital model, workers divide their time between producing current output and acquiring skills. Time in production supports output today; time in education raises human capital, which supports output in the future. The model makes this trade-off explicit and shows how a constant allocation of time to education can sustain growth in output per worker.

The derivation uses a continuous-time model with physical capital, human capital, and a fixed share of time allocated to production. The results below describe a balanced-growth path: a path on which key quantities grow at constant rates and the ratio of physical capital to human capital remains constant. The equations are not a claim that every economy follows this path, nor do they determine an optimal education allocation.

Definition: Let $L$ denote the number of workers, $K$ aggregate physical capital, and $h$ human capital per worker. Each worker allocates a constant share $u$ of available time to production and $1-u$ to education, where $0<u<1$. Production per worker is $y=Y/L$, physical capital per worker is $k=K/L$, and aggregate production is $$ Y=A K^\alpha (u h L)^{1-\alpha}, $$ where $A>0$ is constant productivity and $0<\alpha<1$. Human capital evolves according to $\dot h=\phi(1-u)h$, where $\phi>0$. Aggregate physical capital evolves as $\dot K=sY-\delta K$, where $s$ is a constant saving rate and $\delta>0$ is the depreciation rate. The workforce grows at constant rate $n$, so $\dot L=nL$.

The time share $u$ affects production directly: only $u h L$ human-capital units are used in production. It also affects human-capital growth, because the education share is $1-u$. These two channels must be kept separate in calculations.

Production and Accumulation per Worker

Divide aggregate production by $L$. Since $k=K/L$, production per worker is

y=A k^\alpha (u h)^{1-\alpha}.

The human-capital growth rate follows directly from its accumulation equation:

g_h=\frac{\dot h}{h}=\phi(1-u).

To derive physical-capital accumulation per worker, differentiate $k=K/L$. Applying the quotient rule and using $\dot K=sY-\delta K$ and $\dot L=nL$ gives

\dot k=\frac{\dot K}{L}-\frac{K\dot L}{L^2} =s y-\delta k-nk =s y-(n+\delta)k.

Population growth therefore dilutes capital per worker: part of the aggregate capital stock must equip new workers. This is why $n$, as well as depreciation $\delta$, appears in the per-worker capital equation.

Worked Example: Finding the Balanced-Growth Ratio

Suppose $A=1$, $\alpha=0.5$, $s=0.25$, $u=0.64$, $\phi=0.05$, $n=0.010$, and $\delta=0.030$. Find the human-capital growth rate and the balanced-growth ratio $x^*=k/h$. All rates are per year.

Method. First calculate human-capital growth from the education share. Then write the growth rate of $x=k/h$, set it to zero, and solve for the ratio. This method avoids confusing the level ratio with a growth rate.

The education share is $1-u=1-0.64=0.36$, so

$$ g_h=0.05(0.36)=0.018, $$

or $1.8\%$ per year. From the per-worker production equation, $y/k=A u^{1-\alpha}(k/h)^{\alpha-1}$. The capital growth rate is $\dot k/k=s(y/k)-(n+\delta)$. Thus the ratio grows at the difference between the growth rates of $k$ and $h$. At a constant ratio, $\dot x=0$, and the equation derived below gives

x^*=\left(\frac{sA u^{1-\alpha}}{n+\delta+g_h}\right)^{1/(1-\alpha)}.

Substitute the values. Since $1-\alpha=0.5$, $u^{1-\alpha}=\sqrt{0.64}=0.8$, and $n+\delta+g_h=0.010+0.030+0.018=0.058$:

x^*=\left(\frac{0.25(1)(0.8)}{0.058}\right)^2 =\left(\frac{0.200}{0.058}\right)^2 =(3.448276)^2 \approx 11.891.

Therefore human capital grows at $1.8\%$ per year and the balanced-growth ratio is $k/h\approx11.891$. The denominator includes population growth, depreciation, and human-capital growth; omitting any of these changes the implied ratio.

Deriving the Balanced-Growth Path

Define $x=k/h$, the ratio of physical capital per worker to human capital per worker. From the production function,

\frac{y}{k} =A k^{\alpha-1}(u h)^{1-\alpha} =A u^{1-\alpha}\left(\frac{k}{h}\right)^{\alpha-1} =A u^{1-\alpha}x^{\alpha-1}.

The first equality divides output per worker by $k$; the second collects the powers of $k$ and $h$. Because $x=k/h$, its proportional growth rate is $\dot x/x=\dot k/k-\dot h/h$. Substituting the capital and human-capital growth equations yields

\frac{\dot x}{x} =sA u^{1-\alpha}x^{\alpha-1}-(n+\delta)-g_h.

A balanced-growth path has a constant positive $x$, so $\dot x/x=0$. Rearranging the equation gives

sA u^{1-\alpha}x^{\alpha-1}=n+\delta+g_h, \qquad x^{1-\alpha}=\frac{sA u^{1-\alpha}}{n+\delta+g_h}.

Since $0<\alpha<1$, the exponent $1-\alpha$ is positive. Raising both sides to the power $1/(1-\alpha)$ gives the balanced-growth ratio:

x^*=\left(\frac{sA u^{1-\alpha}}{n+\delta+g_h}\right)^{1/(1-\alpha)} =\left(\frac{sA u^{1-\alpha}}{n+\delta+\phi(1-u)}\right)^{1/(1-\alpha)}.

This expression has a useful interpretation. Higher saving or productivity raises the ratio of physical to human capital needed on the balanced path. Faster population growth or depreciation lowers it. A higher human-capital growth rate also lowers $k/h$: human capital is expanding faster, so the ratio can remain constant only if physical capital per worker grows at the same rate from a lower relative position.

Theorem: For constant $u$, $s$, and model parameters, with $A,s,\phi>0$, $n+\delta+g_h>0$, and $0<\alpha<1$, the positive constant ratio $x^*$ is given by the formula above. On that path, $k$ and $h$ grow at the same rate $g_h=\phi(1-u)$. Because $y=A u^{1-\alpha}(k/h)^\alpha h$, output per worker grows at $g_h$ as well. Aggregate output grows at $n+g_h$.

The output-growth result follows without assuming that output grows at the same rate as aggregate capital. On the balanced path, $x=k/h=x^*$ and $u$ are constant, so

y=A u^{1-\alpha}(x^*)^\alpha h.

The coefficient multiplying $h$ is constant. Therefore $y$ grows at the same rate as $h$, namely $g_h$. Since $Y=yL$ and $L$ grows at rate $n$, aggregate output grows at $n+g_h$. This distinction matters: output per worker growth is not the same as aggregate output growth when the workforce is growing.

Compute education-driven human-capital growth.
Use $g_h=\phi(1-u)$; do not use $u$, the production-time share, in place of the education-time share.

Find the constant capital ratio.
Substitute $g_h$ into the balanced-growth formula and include population growth and depreciation in its denominator.

Report the relevant growth rate.
Output per worker grows at $g_h$; aggregate output grows at $n+g_h$.

How Education Time Changes Growth and Levels

With the other parameters fixed, increasing education time means reducing $u$. This raises $g_h=\phi(1-u)$, so it raises balanced-growth output per worker growth. But it also reduces the share of human capital used directly in production through $u^{1-\alpha}$. Consequently, the balanced ratio $x^*$ generally changes too. A comparison of education allocations should therefore distinguish growth rates from the levels implied by the balanced path.

In this model, a constant change in $u$ changes the long-run growth rate, not merely the level of output per worker. That statement depends on the human-capital accumulation equation and the assumption that $u$ remains constant. It is not a general claim about every education policy or every model of growth.

Worked Example: Comparing Two Time Allocations

Take $A=1$, $\alpha=0.5$, $s=0.2$, $\phi=0.04$, $n=0.01$, and $\delta=0.03$. Compare $u=0.75$ with $u=0.50$. For each allocation, calculate human-capital growth and $x^*$, then compare output-per-worker growth. Rates are per year.

Method. Apply the same balanced-growth formula to each allocation, changing both the direct production term and the education-driven growth term. A common mistake is to account for the higher education share in $g_h$ but leave the $u^{1-\alpha}$ production term unchanged.

For $u=0.75$, the education share is $0.25$, so $g_h=0.04(0.25)=0.01$, or $1\%$. The denominator is $n+\delta+g_h=0.01+0.03+0.01=0.05$. Since $\alpha=0.5$, the exponent in the ratio formula is $2$, and

x^*=\left(\frac{0.2\sqrt{0.75}}{0.05}\right)^2 =\left(\frac{0.2(0.866025)}{0.05}\right)^2 =(3.46410)^2 =12.

For $u=0.50$, the education share is also $0.50$, giving $g_h=0.04(0.50)=0.02$, or $2\%$. The denominator becomes $0.01+0.03+0.02=0.06$, so

x^*=\left(\frac{0.2\sqrt{0.50}}{0.06}\right)^2 =\left(\frac{0.2(0.707107)}{0.06}\right)^2 =(2.35702)^2 \approx 5.5556.

Output per worker grows at $1\%$ in the first case and $2\%$ in the second. As a level comparison on each balanced path, $y/h=A u^{1-\alpha}(x^*)^\alpha=\sqrt{u x^*}$. Thus the ratios are $\sqrt{0.75(12)}=\sqrt{9}=3$ and $\sqrt{0.50(5.5556)}\approx\sqrt{2.7778}\approx1.6667$. The higher-education allocation has faster balanced growth here, but its output per unit of human capital on the path is lower. These calculations compare balanced paths; they do not establish which allocation maximizes welfare.

Check. In both cases, the computed growth rate of $h$ is the growth rate of $y$, because $u$ and $x^*$ are constant on each path.

Worked Example: Meeting a Target Growth Rate

Suppose $A=1$, $\alpha=0.5$, $s=0.25$, $\phi=0.05$, $n=0.01$, and $\delta=0.03$. Find the constant production-time share $u$ that gives output per worker a balanced-growth rate of $1.5\%$ per year. Then calculate $x^*$ and aggregate output growth.

Method. First use the target growth rate to solve the human-capital accumulation equation for $u$. Then substitute that value in the balanced-growth ratio. This avoids trying to infer $u$ from the ratio equation alone.

On a balanced path, output per worker grows at $g_h$, so the target requires $g_h=0.015$. From $g_h=\phi(1-u)$,

0.015=0.05(1-u), \qquad 1-u=\frac{0.015}{0.05}=0.30, \qquad u=0.70.

The denominator of the ratio formula is $n+\delta+g_h=0.01+0.03+0.015=0.055$. Also, $u^{1-\alpha}=\sqrt{0.70}\approx0.836660$, and the exponent $1/(1-\alpha)$ is $2$. Therefore,

x^*=\left(\frac{0.25(0.836660)}{0.055}\right)^2 =(3.80300)^2 \approx14.4628.

Aggregate output growth is population growth plus output-per-worker growth:

$$ n+g_h=0.01+0.015=0.025, $$

or $2.5\%$ per year. The result is $u=0.70$, $x^*\approx14.4628$, and aggregate output growth of $2.5\%$ per year. This is a calculation for a specified target, not a derivation of an optimal time allocation.

Common Exam Traps and Interpretation

Using the wrong time share. The human-capital growth equation uses $1-u$, the education share. Production uses $u$, the production share.
Dropping population growth. The denominator for $x^*$ contains $n+\delta+g_h$, because population growth dilutes physical capital per worker.
Confusing per-worker and aggregate growth. On the balanced path, $y$ grows at $g_h$, while $Y$ grows at $n+g_h$.
Treating the balanced ratio as a growth rate. $x^*=k/h$ is a level ratio. It is constant on the path; its constancy implies that $k$ and $h$ grow at the same rate.
Changing $u$ in only one equation. A change in $u$ affects both human-capital growth and the amount of human capital used in current production.

The Lucas model is useful because it makes the source of sustained per-worker growth explicit: with a constant positive share of time devoted to education, human capital can keep growing, and output per worker grows with it on the balanced path. For calculations, the reliable sequence is to find $g_h$, derive or use the constant-ratio condition, and then state whether the requested result concerns output per worker or aggregate output.

Check Your Understanding

Work through each problem using the production-time share and education-time share distinctly. Rates are per year unless stated otherwise.

Let $A=2$, $\alpha=0.5$, $s=0.10$, $u=0.75$, $\phi=0.04$, $n=0.02$, and $\delta=0.02$. Find $g_h$, $x^*$, and aggregate output growth. (Answer: $g_h=1.0\%$, $x^*=12$, and aggregate output growth $=3.0\%$.)
Let $A=1$, $\alpha=0.5$, $s=0.25$, $u=0.64$, $\phi=0.05$, $n=0.01$, and $\delta=0.03$. Calculate $g_h$ and $x^*$. (Answer: $g_h=1.8\%$ and $x^*\approx11.891$.)
In a Lucas model with $\phi=0.04$, what constant $u$ produces balanced output-per-worker growth of $1.2\%$? (Answer: $u=0.70$.)
Suppose balanced output per worker grows at $1.4\%$ and the workforce grows at $0.6\%$. What is aggregate output growth? Explain which two rates are being combined. (Answer: $2.0\%$ per year, the sum of workforce growth and output-per-worker growth.)
Set $A=1$, $\alpha=0.5$, $s=0.20$, $\phi=0.06$, $n=0.01$, and $\delta=0.03$. Compare $u=0.50$ and $u=0.75$. Which allocation gives faster balanced output-per-worker growth, and what is that rate in each case? (Answer: $u=0.50$ gives $3.0\%$ growth; $u=0.75$ gives $1.5\%$ growth.)

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