Integrated Solow Comparative-Statics Problems

Why Integrated Comparative Statics Matter

In the Solow model, a change in one parameter can move the economy toward a different long-run position. In applied problems, several parameters may change together: saving may rise while population growth or depreciation also rises, for example. The net effect on steady-state capital is not apparent from any one change alone. A careful solution first calculates the old and new steady states, then evaluates the economy’s motion immediately after the changes.

This tutorial uses the continuous-time Solow model with labor-augmenting technological progress. All rates are annual continuous-time rates, and parameters are constant within each regime. The analysis is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus. It does not assume that capital jumps when a parameter changes: effective-worker capital is continuous at the instant of a parameter change, though its subsequent path changes.

Model and definitions: Let $K$ be physical capital, $L$ labor, and $A$ labor-augmenting technology. Production is $Y=K^\alpha(AL)^{1-\alpha}$, where $0<\alpha<1$. Define effective-worker capital $k=K/(AL)$ and output per effective worker $y=Y/(AL)=k^\alpha$. Let $s$ be the saving rate, $n$ the labor-force growth rate, $g$ the technology growth rate, and $\delta$ the depreciation rate. With $b=n+g+\delta$, capital accumulation per effective worker obeys $\dot{k}=s k^\alpha-bk$.

The term $s k^\alpha$ is investment per effective worker. The term $bk$ is the investment needed to offset depreciation and to equip the growing numbers of workers and effective workers with capital. A steady state in effective-worker units has $\dot{k}=0$. It is a balanced-growth path in levels: with constant $k$, output per worker grows at rate $g$, while aggregate output grows at rate $n+g$.

Steady-state result: For fixed $s,n,g,\delta$, the positive steady-state effective-worker capital and output are $$ k^*=\left(\frac{s}{n+g+\delta}\right)^{\frac{1}{1-\alpha}}, \qquad y^*=(k^*)^\alpha =\left(\frac{s}{n+g+\delta}\right)^{\frac{\alpha}{1-\alpha}}. $$

Comparing Old and New Steady States

When multiple parameters change, calculate the combined break-even rate $b=n+g+\delta$ in each regime. Then compare the ratio of saving to that rate, raised to the appropriate power. It is useful to write the comparison as a ratio, because that makes the combined effects visible and avoids rounding intermediate levels.

\frac{k^*_{\mathrm{new}}}{k^*_{\mathrm{old}}} = \left[ \frac{s_{\mathrm{new}}}{s_{\mathrm{old}}} \frac{n_{\mathrm{old}}+g_{\mathrm{old}}+\delta_{\mathrm{old}}} {n_{\mathrm{new}}+g_{\mathrm{new}}+\delta_{\mathrm{new}}} \right]^{\frac{1}{1-\alpha}}, \qquad \frac{y^*_{\mathrm{new}}}{y^*_{\mathrm{old}}} = \left(\frac{k^*_{\mathrm{new}}}{k^*_{\mathrm{old}}}\right)^\alpha.

These are exact finite-change ratios when $\alpha$ is unchanged. For a small change, the elasticity of $k^*$ with respect to $s$ is $1/(1-\alpha)$; its elasticities with respect to $n$, $g$, and $\delta$ are each $-b/[(1-\alpha)b]=-1/(1-\alpha)$, when one of those rates changes by itself. An elasticity describes a local percentage response, not the exact result of a finite percentage change.

Why the Comparative-Statics Formula Holds

At a steady state, investment just offsets the break-even investment requirement. Solving that condition gives the steady-state formula directly. Dividing the new formula by the old one then produces the exact ratio formula.

Set investment equal to break-even investment.
At a positive steady state, $s(k^*)^\alpha=bk^*$, where $b=n+g+\delta$.

Isolate capital.
Divide by $(k^*)^\alpha$ and by $b$: $(k^*)^{1-\alpha}=s/b$. Since $1-\alpha>0$, raise both sides to $1/(1-\alpha)$ to obtain $k^*=(s/b)^{1/(1-\alpha)}$.

Compare regimes without approximating.
Write the same expression for each regime and divide: $k^*_{\mathrm{new}}/k^*_{\mathrm{old}}=[(s_{\mathrm{new}}/s_{\mathrm{old}})(b_{\mathrm{old}}/b_{\mathrm{new}})]^{1/(1-\alpha)}$. Finally, $y^*=(k^*)^\alpha$, so the output ratio is the capital ratio raised to $\alpha$.

Taking logarithms also clarifies the local elasticities: $\ln k^*=(\ln s-\ln b)/(1-\alpha)$. Differentiating gives $d\ln k^*=[d\ln s-d\ln b]/(1-\alpha)$. This differential relationship is useful for small changes; for actual finite changes, use the exact ratio rather than treating the differential as an exact percentage.

Worked Examples: Long-Run Comparisons and Transition Effects

Worked Example: Saving Rises While Growth Rates Change

An economy has $\alpha=0.5$. Initially, $s=0.20$, $n=0.01$, $g=0.02$, and $\delta=0.02$. Then saving rises to $0.25$, population growth rises to $0.02$, technology growth falls to $0.01$, and depreciation stays at $0.02$. Find the old and new steady-state effective-worker capital, and the output-per-worker growth rate immediately after the change.

Method. Calculate the break-even rate and steady state in each regime. For the transition calculation, retain the old $k$ at the instant the parameters change and substitute it into the new law of motion.

Initially, $b_{\mathrm{old}}=0.01+0.02+0.02=0.05$, so $s_{\mathrm{old}}/b_{\mathrm{old}}=0.20/0.05=4$. Because $1/(1-\alpha)=2$, $k^*_{\mathrm{old}}=4^2=16$.

After the change, $b_{\mathrm{new}}=0.02+0.01+0.02=0.05$. Thus $s_{\mathrm{new}}/b_{\mathrm{new}}=0.25/0.05=5$, and $k^*_{\mathrm{new}}=5^2=25$. The new steady state is higher even though the three growth and depreciation terms have changed: their sum is unchanged, while saving rises.

At the instant of the change, $k=16$. Under the new parameters, $\dot{k}=0.25\sqrt{16}-0.05(16)=0.25(4)-0.80=1.00-0.80=0.20$. Therefore, $\dot{k}/k=0.20/16=0.0125$, or 1.25% per year. Since output per worker is $Y/L=A k^\alpha$, its growth rate is $g+\alpha\dot{k}/k=0.01+0.5(0.0125)=0.01625$, or 1.625% per year. Check: capital is below its new steady state, so $\dot{k}>0$, as calculated. The long-run output-per-worker growth rate will be the new technology growth rate, 1% per year.

Common mistake avoided: Do not calculate transition growth using only $g$. Technology growth is part of output-per-worker growth, but so is the change in effective-worker capital during the transition.

Worked Example: Exact Ratios and Consumption per Effective Worker

Let $\alpha=0.5$. Initially, $s=0.20$, $n=0.01$, $g=0.02$, and $\delta=0.02$. In the new regime, $s=0.30$, $n=0.02$, $g=0.015$, and $\delta=0.025$. Find the exact percentage change in steady-state capital and output per effective worker, and compare steady-state consumption per effective worker.

Initially $b_{\mathrm{old}}=0.01+0.02+0.02=0.05$, so $k^*_{\mathrm{old}}=(0.20/0.05)^2=4^2=16$, and $y^*_{\mathrm{old}}=\sqrt{16}=4$.

In the new regime, $b_{\mathrm{new}}=0.02+0.015+0.025=0.060$, so $k^*_{\mathrm{new}}=(0.30/0.060)^2=5^2=25$, and $y^*_{\mathrm{new}}=\sqrt{25}=5$. The capital ratio is $25/16=1.5625$, an exact increase of $100(1.5625-1)=56.25\%$. The output ratio is $5/4=1.25$, an exact increase of 25%.

With no government or other uses of output in this model, steady-state consumption per effective worker is $c^*=(1-s)y^*$. Initially, $c^*_{\mathrm{old}}=(1-0.20)(4)=0.80(4)=3.2$. In the new regime, $c^*_{\mathrm{new}}=(1-0.30)(5)=0.70(5)=3.5$. The change is $3.5-3.2=0.3$, and the percentage increase is $100(0.3/3.2)=9.375\%$. Check: consumption rises by less than output because the fraction saved also rises.

Common mistake avoided: The 56.25% capital increase is a finite change, not an approximation from the elasticity. For example, with $\alpha=0.5$, the capital elasticity with respect to saving is 2. A 1% finite rise in saving, holding $b$ fixed, changes $k^*$ by the exact factor $1.01^2=1.0201$, or 2.01%, not exactly 2%.

Worked Example: Higher Saving Does Not Guarantee More Steady-State Capital

An economy with $\alpha=0.5$ initially has $s=0.25$, $n=0.01$, $g=0.02$, and $\delta=0.02$. A policy change raises saving to $0.36$, but the new rates are $n=0.04$, $g=0.03$, and $\delta=0.05$. Find the old and new steady-state effective-worker capital, then calculate the initial growth rate of output per worker after the change.

Initially, $b_{\mathrm{old}}=0.01+0.02+0.02=0.05$, giving $s/b=0.25/0.05=5$ and $k^*_{\mathrm{old}}=5^2=25$. In the new regime, $b_{\mathrm{new}}=0.04+0.03+0.05=0.12$, so $s/b=0.36/0.12=3$ and $k^*_{\mathrm{new}}=3^2=9$. Despite the higher saving rate, the much higher break-even rate reduces steady-state capital. Steady-state output per effective worker changes from $\sqrt{25}=5$ to $\sqrt{9}=3$, a 40% decrease.

At the instant of the change, capital remains $k=25$. The new law of motion gives $\dot{k}=0.36\sqrt{25}-0.12(25)=0.36(5)-3.00=1.80-3.00=-1.20$. Thus $\dot{k}/k=-1.20/25=-0.048$, or -4.8% per year. Output per worker initially grows at $g+\alpha\dot{k}/k=0.03+0.5(-0.048)=0.006$, or 0.6% per year. Check: effective-worker capital falls because the economy starts above its new steady state; output per worker can still grow because technology grows at 3% per year.

Common mistake avoided: Do not infer the direction of the steady-state change from saving alone. Compare $s/b$ in both regimes, and keep technology growth $g$ in the output-per-worker growth calculation.

Transition Growth Versus Balanced Growth

A parameter change has two different consequences that should be reported separately. First, the economy may be away from its new steady state, so effective-worker capital changes and contributes to output-per-worker growth. Second, once the economy reaches the new steady state, $k$ is constant and output per worker grows at $g$. These statements do not say how long convergence takes; that depends on the path of capital adjustment.

Key result: At any instant with constant $\alpha$, output per worker is $Y/L=A k^\alpha$, so its growth rate is $g+\alpha\dot{k}/k$. On the balanced-growth path, $\dot{k}=0$, making output-per-worker growth $g$. If $g$ itself changes, use its new value immediately after the change, alongside the current $\dot{k}/k$.

A quick sign check is often helpful. If current $k$ is below the new $k^*$, then $s k^\alpha-bk>0$ and capital per effective worker rises. If current $k$ is above the new $k^*$, the expression is negative and $k$ falls. This check can catch an arithmetic or parameter-substitution error before it affects the output-growth calculation.

Common Errors and Exam Checks

Using a local elasticity as an exact finite-change formula. The elasticity gives an approximate response for small changes. For finite changes, use the exact ratio formula.
Mixing annual rates with rates in another time unit. The values of $n$, $g$, and $\delta$ must refer to the same period and be expressed consistently with the model’s time unit before they are added.
Holding $k$ at its old steady state forever. At the instant of a parameter change, $k$ is still at its old value; afterward it moves under the new law of motion.
Confusing effective-worker output with output per worker. Steady-state $y^*$ is output per effective worker. Output per worker is $A k^\alpha$, and its balanced-growth rate is $g$.
Ignoring a changing $g$ in transition growth. Output-per-worker growth includes the new technology growth rate as well as the contribution from changing $k$.

Key takeaway: For an integrated comparison, calculate $b=n+g+\delta$ in each regime, apply the exact steady-state formula, and then evaluate the new law of motion at the capital stock that exists immediately after the change. Report the new long-run outcome separately from the transition growth rate.

Check Your Understanding

For each problem, treat rates as annual continuous-time rates and use the Solow model and conventions defined above.

An economy has $\alpha=0.5$. Its initial parameters are $s=0.16$, $n=0.01$, $g=0.02$, and $\delta=0.01$. The new parameters are $s=0.25$, $n=0.02$, $g=0.02$, and $\delta=0.01$. Find both steady-state capital stocks and the exact percentage change in $k^*$. (Answer: $b_{\mathrm{old}}=0.04,\ b_{\mathrm{new}}=0.05;\ k^*_{\mathrm{old}}=16,\ k^*_{\mathrm{new}}=25;$ increase $=56.25\%$.)
Let $\alpha=0.5$, $s=0.24$, $n=0.02$, $g=0.01$, and $\delta=0.03$. Find $k^*$. If the economy is at that steady state, what is the output-per-worker growth rate? (Answer: $b=0.06,\ k^*=16,$ output-per-worker growth $=1\%$ per year.)
Let $\alpha=0.5$. Immediately after a parameter change, $k=16$, $s=0.30$, $n=0.02$, $g=0.025$, and $\delta=0.015$. Calculate $\dot{k}$, $\dot{k}/k$, and the current output-per-worker growth rate. (Answer: $b=0.06,\ \dot{k}=0.30(4)-0.06(16)=0.24,\ \dot{k}/k=1.5\%,$ output-per-worker growth $=2.5\%+0.5(1.5\%)=3.25\%$ per year.)
An economy has $\alpha=0.5$, current capital $k=9$, and new parameters $s=0.20$, $n=0.02$, $g=0.02$, and $\delta=0.02$. Find the new steady state and the initial output-per-worker growth rate. (Answer: $b=0.06,\ k^*=(0.20/0.06)^2=100/9,\ \dot{k}=0.20(3)-0.06(9)=0.06,\ \dot{k}/k=0.006\overline{6},$ output-per-worker growth $=2.333\ldots\%$ per year.)
With $\alpha=0.5$, an economy moves from $s=0.18,\ n=0.01,\ g=0.015,\ \delta=0.02$ to $s=0.27,\ n=0.02,\ g=0.01,\ \delta=0.03$. Find the old and new steady-state capital and output per effective worker, and their exact percentage changes. (Answer: $b_{\mathrm{old}}=0.045,\ b_{\mathrm{new}}=0.06;\ k^*_{\mathrm{old}}=16,\ k^*_{\mathrm{new}}=20.25$, a 26.5625% capital increase; $y^*_{\mathrm{old}}=4,\ y^*_{\mathrm{new}}=4.5$, a 12.5% output increase.)

What You'll Learn