Population Growth and Capital Dilution

Why Population Growth Changes Capital per Worker

An economy can add workers without adding capital at the same rate. When that happens, the existing capital stock must be spread across a larger workforce: this is capital dilution. In the Solow model, population growth therefore affects the investment required just to keep capital per worker unchanged. A higher population growth rate raises that requirement, other things equal, and can lower the long-run levels of capital and output per worker.

The analysis below holds the saving rate and production technology fixed while population growth changes. It focuses on output and capital per worker; aggregate output may behave differently because the number of workers is changing.

Model: Let \(K\) be aggregate capital, \(L\) the number of workers, \(k=K/L\) capital per worker, and \(y=f(k)\) output per worker. Assume a constant saving rate \(s\), population growth \(\dot L/L=n\), and a fixed depreciation rate \(\delta\). There is no technological progress in this model. Aggregate output is \(Y=L f(k)\), and saving finances investment. The continuous-time capital accumulation equation per worker is \(\dot{k}=s f(k)-(n+\delta)k\). The rate \(n+\delta\) is the break-even investment rate per worker: investment of \((n+\delta)k\) is needed to keep \(k\) constant.

Here \(n\) and \(\delta\) are annual continuous-time rates, and \(\dot{k}\) is the instantaneous change in capital per worker, measured in capital units per worker per year. The model assumes that saving is invested in productive capital and that the production function is expressed in per-worker terms. When comparing population growth rates, \(s\), \(\delta\), and the production function are held fixed unless stated otherwise.

Where Capital Dilution Comes From

The break-even expression is a consequence of dividing aggregate capital accumulation by the growing workforce. Aggregate capital changes according to \(\dot K=sY-\delta K\), while the workforce changes according to \(\dot L=nL\). Since \(k=K/L\), differentiating a ratio gives

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

The \(nk\) term is the capital dilution term: even if aggregate capital does not depreciate, a growing workforce reduces capital per worker unless investment adds capital to match the increase in workers. Depreciation also removes capital, so the two requirements add. This derivation explains why the saving flow \(sf(k)\) must be compared with \((n+\delta)k\), not with \(nk\) alone.

Definition: A steady state \(k^*\) is a level of capital per worker at which \(\dot{k}=0\). Thus it satisfies \(s f(k^*)=(n+\delta)k^*\). At a positive steady state, saving per worker exactly covers capital dilution and depreciation. If \(k\) is below or above the steady state in the usual stable Solow model, the sign of \(\dot{k}\) determines whether capital per worker is rising or falling.

How a Higher Population Growth Rate Shifts the Steady State

For a given \(k\), increasing \(n\) raises break-even investment from \((n_0+\delta)k\) to \((n_1+\delta)k\), where \(n_1>n_0\). The saving curve \(sf(k)\) has not moved, but the break-even line is steeper. With the standard increasing, strictly concave per-worker production function and a stable positive steady state, the new intersection occurs at a lower value of \(k\). Since output per worker is \(f(k)\), it also falls.

The comparative-static result can be justified directly. Define the steady-state equation as \(s f(k^*)-(n+\delta)k^*=0\). Differentiating with respect to \(n\) while holding \(s\), \(\delta\), and \(f\) fixed gives

\big[s f'(k^*)-(n+\delta)\big]\frac{d k^*}{dn}-k^*=0, \qquad \frac{d k^*}{dn} =\frac{k^*}{s f'(k^*)-(n+\delta)}.

At a strictly concave production function with \(f(0)=0\), the slope \(f'(k^*)\) is below the average product \(f(k^*)/k^*\). The steady-state condition implies \(s f(k^*)/k^*=n+\delta\), so \(s f'(k^*)<n+\delta\). The denominator in the derivative is negative, and therefore \(d k^*/dn<0\). Because \(f\) is increasing, \(d y^*/dn=f'(k^*)\,d k^*/dn<0\) as well. This argument applies to the standard stable positive steady state; it is not a claim that every arbitrary production function has a unique steady state.

Cobb-Douglas Calculation

For the commonly used per-worker production function \(f(k)=A k^\alpha\), assume \(A>0\), \(0<\alpha<1\), and \(0<s<1\). At a positive steady state, if \(n+\delta>0\), solving the steady-state condition gives

sA(k^*)^\alpha=(n+\delta)k^*, \qquad k^*=\left(\frac{sA}{n+\delta}\right)^{\frac{1}{1-\alpha}}, \qquad y^*=A\left(\frac{sA}{n+\delta}\right)^{\frac{\alpha}{1-\alpha}}.

The condition \(n+\delta>0\) matters: the displayed expressions divide by this quantity. If \(n=\delta=0\), then the steady-state equation is \(sA k^\alpha=0\), whose only nonnegative solution is \(k^*=0\); the positive formula does not apply. In the usual case \(n+\delta>0\), a rise in \(n\) reduces both \(k^*\) and \(y^*\). For the special case \(f(k)=\sqrt{k}\), this becomes \(k^*=[s/(n+\delta)]^2\) and \(y^*=s/(n+\delta)\).

Worked Example: Comparing Two Population-Growth Steady States

A hypothetical economy has \(f(k)=\sqrt{k}\), saving rate \(s=0.20\), and depreciation rate \(\delta=0.02\). Population growth rises from \(n_0=0.03\) to \(n_1=0.06\). Calculate steady-state capital and output per worker before and after the change, and find the percentage reductions.

Method. Use the steady-state formula for square-root production, keeping the depreciation rate fixed in both calculations.

Initially, the break-even rate is \(n_0+\delta=0.03+0.02=0.05\). Therefore \(k_0^*=(0.20/0.05)^2=4^2=16\), and \(y_0^*=\sqrt{16}=4\).

After the increase, the break-even rate is \(n_1+\delta=0.06+0.02=0.08\). The new steady state is \(k_1^*=(0.20/0.08)^2=2.5^2=6.25\), and \(y_1^*=\sqrt{6.25}=2.5\).

The percentage reduction in capital per worker is \((16-6.25)/16\times100=9.75/16\times100=60.9375\%\). The percentage reduction in output per worker is \((4-2.5)/4\times100=1.5/4\times100=37.5\%\). Rounded only at the final step, the reductions are 60.94% in capital per worker and 37.50% in output per worker. As a check, the initial steady-state investment is \(0.20\sqrt{16}=0.80\), equal to \(0.05(16)=0.80\); the new investment is \(0.20\sqrt{6.25}=0.50\), equal to \(0.08(6.25)=0.50\). The common mistake this method avoids is adding the old population growth rate to the new one instead of replacing it.

The Transition After a Population-Growth Shock

A change in \(n\) does not instantly change the economy’s existing capital per worker. It changes the rate at which that capital must be replenished relative to the workforce. Immediately after a rise from \(n_0\) to \(n_1\), if \(k\) has not yet moved, the change in capital accumulation is

\dot{k}_{\text{after}}-\dot{k}_{\text{before}} =-(n_1-n_0)k.

Thus, at any positive \(k\), a rise in population growth lowers \(\dot{k}\) by \((n_1-n_0)k\). If the economy was exactly at its old steady state, then \(\dot{k}_{\text{before}}=0\), so \(\dot{k}_{\text{after}}<0\). Capital per worker begins to fall. Under the standard stable model, it continues toward the lower new steady state; output per worker falls as \(k\) falls. A decrease in population growth reverses the immediate effect, but the direction of subsequent adjustment still depends on the economy’s current position relative to its new steady state.

Worked Example: The Immediate Effect of a Population-Growth Increase

Suppose a hypothetical economy has \(f(k)=\sqrt{k}\), \(s=0.20\), and \(\delta=0.02\). It is initially at its steady state with \(n_0=0.02\). Population growth then rises to \(n_1=0.03\). Find the old steady state, the new steady state, and the immediate value of \(\dot{k}\) after the shock.

Method. First find the old steady state using the old combined break-even rate. Then hold \(k\) at that value for the instant of the shock and substitute the new rate into the accumulation equation.

The old break-even rate is \(0.02+0.02=0.04\), so \(k_0^*=(0.20/0.04)^2=5^2=25\), and \(y_0^*=\sqrt{25}=5\). The new break-even rate is \(0.03+0.02=0.05\), giving \(k_1^*=(0.20/0.05)^2=4^2=16\), and \(y_1^*=\sqrt{16}=4\).

At the instant after the change, \(k\) is still 25, so current output per worker is still \(\sqrt{25}=5\), and saving per worker is \(0.20(5)=1\). New break-even investment is \(0.05(25)=1.25\). Hence \(\dot{k}=1-1.25=-0.25\) capital units per worker per year. Equivalently, because the old steady-state accumulation rate was zero, \(\dot{k}=-(0.03-0.02)(25)=-0.25\). The check is that new saving falls short of new break-even investment by exactly \(0.25\). Do not substitute the new steady-state capital 16 when calculating the immediate change: the stock cannot jump to its new long-run value at the instant the growth rate changes.

The Current Capital Stock Determines the Adjustment Direction

The claim that a population-growth decrease raises capital per worker is immediate only for an economy starting from its old steady state. More generally, the shock shifts the steady state, and the sign of \(\dot{k}\) at the economy’s current \(k\) determines the initial movement. For a stable steady state, capital tends to rise below the new steady state and fall above it. This is why a lower \(n\) does not guarantee an increase in \(k\) at every instant: an economy that begins sufficiently far above its new steady state can still be moving downward.

Worked Example: A Population-Growth Decrease at a Non-Steady-State Capital Stock

A hypothetical economy has \(f(k)=\sqrt{k}\), \(s=0.30\), \(\delta=0.02\), and current capital per worker \(k=36\). Population growth falls from \(n_0=0.04\) to \(n_1=0.01\). Determine the old and new steady states and the direction of capital accumulation immediately before and after the change.

Method. Compare the current \(k\) with each steady state, then verify the direction by calculating saving minus break-even investment.

Before the change, the combined break-even rate is \(0.04+0.02=0.06\), so \(k_0^*=(0.30/0.06)^2=5^2=25\). Since current capital 36 is above 25, capital per worker should be falling. Directly, output per worker is \(\sqrt{36}=6\), saving is \(0.30(6)=1.80\), and break-even investment is \(0.06(36)=2.16\). Thus \(\dot{k}=1.80-2.16=-0.36\).

After the change, the combined break-even rate is \(0.01+0.02=0.03\), giving \(k_1^*=(0.30/0.03)^2=10^2=100\). The current \(k=36\) is now below the new steady state. Saving remains \(1.80\), while new break-even investment is \(0.03(36)=1.08\). Therefore \(\dot{k}=1.80-1.08=0.72\), so capital per worker begins to rise toward 100. Both calculations check the phase-diagram direction: above the old steady state meant falling capital, and below the new one means rising capital. This example avoids assuming that the economy was initially at a steady state.

What the Long-Run Comparison Does—and Does Not—Say

A higher \(n\) raises break-even investment at each positive level of \(k\), shifting the steady state to lower capital and output per worker under the standard Solow assumptions. It does not imply that aggregate output \(Y=L f(k)\) must fall: the workforce \(L\) is growing faster, while output per worker is lower in the new steady state. The net effect on aggregate output depends on the time horizon and the paths of both employment and output per worker.

Nor does the steady-state comparison describe an instantaneous change in living standards. At the moment of a population-growth shock, \(k\) and \(y=f(k)\) are inherited from the previous period. The accumulation rate changes first; capital and output per worker adjust over time. The model determines the direction of this transition, but the exact speed requires solving the transition dynamics and depends on the production function and parameter values.

Common Mistakes

Using population growth alone as break-even investment. The required rate is \(n+\delta\), because both workforce growth and depreciation must be offset.
Using the new steady-state stock for the immediate shock calculation. At the instant of a shock, the existing \(k\) is unchanged; evaluate \(\dot{k}\) at that inherited stock.
Adding growth rates incorrectly. When \(n\) changes from \(n_0\) to \(n_1\), replace \(n_0\) with \(n_1\); do not add the two rates together.
Confusing a rate with a level. A calculation of \(\dot{k}=-0.25\) gives an instantaneous rate of decline, not a claim that \(k\) falls by exactly \(0.25\) over a full year.
Interpreting per-worker results as aggregate results. The steady-state result concerns capital and output per worker. Aggregate output also depends on the number of workers.

Key takeaway: Population growth creates capital dilution. Holding saving, depreciation, and technology fixed, a higher population growth rate raises break-even investment, lowers the stable steady-state levels of capital and output per worker, and—if the economy starts at its old steady state—makes capital per worker begin falling immediately.

Check Your Understanding

For each problem, use the continuous-time Solow model and keep the stated depreciation rate fixed when population growth changes.

An economy has \(f(k)=\sqrt{k}\), \(s=0.15\), and \(\delta=0.01\). Find \(k^*\) and \(y^*\) when \(n=0.02\), then when \(n=0.04\). (Answer: At \(n=0.02\), \(k^*=25\) and \(y^*=5\); at \(n=0.04\), \(k^*=9\) and \(y^*=3\).)
A hypothetical economy has \(f(k)=2\sqrt{k}\), \(s=0.25\), and \(\delta=0.03\). Population growth rises from \(0.02\) to \(0.07\). Find steady-state capital and output per worker before and after. (Answer: Initially \(k^*=100\), \(y^*=20\); afterward \(k^*=25\), \(y^*=10\).)
An economy with \(f(k)=\sqrt{k}\), \(s=0.24\), and \(\delta=0.01\) is at its steady state when \(n=0.03\). If \(n\) rises to \(0.05\), calculate the old steady-state capital and the immediate \(\dot{k}\) after the shock. (Answer: \(k_0^*=36\), and \(\dot{k}=-0.72\) capital units per worker per year.)
An economy has \(f(k)=\sqrt{k}\), \(s=0.20\), \(\delta=0.02\), and current \(k=49\). Population growth falls from \(0.03\) to \(0.01\). Calculate the new steady-state capital and the immediate \(\dot{k}\). Is capital per worker initially rising or falling? (Answer: The new steady state is \(k_1^*=(0.20/0.03)^2\approx44.4444\); immediate \(\dot{k}=0.20(7)-0.03(49)=1.40-1.47=-0.07\), so capital per worker is initially falling.)
For \(f(k)=3\sqrt{k}\), \(s=0.20\), \(\delta=0.02\), and population growth rising from \(0.03\) to \(0.06\), calculate steady-state capital and output per worker before and after, then give the percentage reduction in each. (Answer: Initially \(k^*=144\), \(y^*=36\); afterward \(k^*=56.25\), \(y^*=22.5\); capital per worker falls by \(60.9375\%\) and output per worker by \(37.5\%\).)

What You'll Learn