Saving Shocks and Long-Run Output

Introduction

A higher saving rate means that a larger fraction of output is invested rather than consumed. In the Solow model, that extra investment raises capital per worker over time until investment once again just covers the amount needed to maintain capital per worker. The resulting long-run output per worker is higher, but that conclusion is about an output level, not a permanently higher growth rate.

This distinction matters when interpreting a lasting change in saving behavior. A temporary period of faster output growth can occur while the economy moves toward its new steady state. Once it arrives, output per worker is constant in the basic model considered here. This tutorial is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

The Steady-State Condition

Use the Solow model with output per worker \(y=f(k)\), where \(k\) is capital per worker. Let \(s\) be the constant saving rate, \(\delta\) the depreciation rate, and \(n\) the population growth rate. Technology and these parameters are held fixed. Using the standard break-even investment expression from the earlier material on population growth, capital per worker evolves according to

\dot{k}=s f(k)-(\delta+n)k.

Here \(\dot{k}\) is the change in capital per worker per unit of time. The term \(s f(k)\) is investment per worker; \((\delta+n)k\) is break-even investment per worker. The model’s steady state is a capital stock at which capital per worker no longer changes.

Definition: A steady-state capital stock per worker \(k^*\) satisfies \(s f(k^*)=(\delta+n)k^*\). Steady-state output per worker is \(y^*=f(k^*)\). With fixed technology and no change in the model’s parameters, \(k^*\) and \(y^*\) remain constant.

A saving-rate change alters the investment term while leaving the break-even term unchanged. If the saving rate rises, investment at the old steady-state capital stock exceeds break-even investment. Capital per worker then increases until diminishing returns bring investment back into balance with break-even investment. The new steady state has more capital and, when production rises with capital, more output per worker.

Deriving the Cobb-Douglas Comparison

For a precise comparison, suppose production per worker is Cobb-Douglas:

y=A k^\alpha,\qquad A>0,\quad 0<\alpha<1.

The parameter \(A\) is fixed productivity, and \(\alpha\) is capital’s output elasticity. Substitute this production function into the steady-state condition and solve for \(k^*\):

sA(k^*)^\alpha=(\delta+n)k^* \quad\Longrightarrow\quad (k^*)^{1-\alpha}=\frac{sA}{\delta+n} \quad\Longrightarrow\quad k^*=\left(\frac{sA}{\delta+n}\right)^{1/(1-\alpha)}.

To obtain steady-state output, substitute \(k^*\) into \(y=A k^\alpha\):

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

For two saving rates \(s_1\) and \(s_2\), with \(A\), \(\delta\), \(n\), and \(\alpha\) unchanged, the ratio of long-run output levels is

\frac{y_2^*}{y_1^*} =\left(\frac{s_2}{s_1}\right)^{\alpha/(1-\alpha)}.

This ratio shows why output does not generally rise in the same proportion as the saving rate. The exponent \(\alpha/(1-\alpha)\) determines how strongly the steady-state output level responds. The corresponding capital ratio is \((k_2^*/k_1^*)=(s_2/s_1)^{1/(1-\alpha)}\), so capital per worker responds more strongly than output per worker when \(0<\alpha<1\).

Worked Comparisons

Worked Example: Comparing Two Saving Rates

Suppose a hypothetical economy has production per worker \(y=\sqrt{k}\), saving rate \(s_1=0.20\), depreciation rate \(\delta=0.06\), and population growth rate \(n=0.04\). Productivity is fixed at \(A=1\). A permanent change raises the saving rate to \(s_2=0.30\). Find each steady-state capital and output level, then compare the output levels.

Method. First combine depreciation and population growth to obtain the break-even rate: \(\delta+n=0.06+0.04=0.10\). The steady-state condition is \(s\sqrt{k^*}=0.10k^*\). For positive \(k^*\), divide by \(\sqrt{k^*}\), giving \(\sqrt{k^*}=s/0.10\).

At \(s_1=0.20\), \(\sqrt{k_1^*}=0.20/0.10=2\), so \(k_1^*=2^2=4\). Output is \(y_1^*=\sqrt{4}=2\). Check: investment is \(0.20(2)=0.40\), and break-even investment is \(0.10(4)=0.40\).

At \(s_2=0.30\), \(\sqrt{k_2^*}=0.30/0.10=3\), so \(k_2^*=3^2=9\). Output is \(y_2^*=\sqrt{9}=3\). Check: investment is \(0.30(3)=0.90\), and break-even investment is \(0.10(9)=0.90\).

The output ratio is \(3/2=1.5\), so steady-state output per worker is 50% higher. Capital per worker is \(9/4=2.25\), or 125% higher. The saving rate rises by 10 percentage points, from 20% to 30%; that is a 50% increase relative to its initial value.

Common mistake avoided: Treating the 50% relative increase in the saving rate as a 50% increase in steady-state capital. Here capital rises by 125%, while output rises by 50%; the steady-state equations determine each response.

Worked Example: Using the Cobb-Douglas Exponent

A hypothetical economy has production per worker \(y=4k^{1/3}\), with fixed productivity \(A=4\) and capital elasticity \(\alpha=1/3\). The break-even rate is \(\delta+n=0.08\). Compare steady states when the saving rate rises from \(0.18\) to \(0.32\).

Method. Apply \(k^*=(sA/(\delta+n))^{1/(1-\alpha)}\). Since \(1/(1-\alpha)=1/(2/3)=3/2\), first calculate the expression inside the power.

At \(s_1=0.18\), \(s_1A/(\delta+n)=(0.18)(4)/0.08=0.72/0.08=9\). Thus \(k_1^*=9^{3/2}=(\sqrt{9})^3=27\). Then \(y_1^*=4(27^{1/3})=4(3)=12\).

At \(s_2=0.32\), \(s_2A/(\delta+n)=(0.32)(4)/0.08=1.28/0.08=16\). Thus \(k_2^*=16^{3/2}=(\sqrt{16})^3=64\). Then \(y_2^*=4(64^{1/3})=4(4)=16\).

The output ratio is \(16/12=4/3\), so long-run output per worker rises by one-third, or about 33.33%. The capital ratio is \(64/27\), about 2.3704. Check the steady-state condition directly: initially, \(s_1y_1^*=(0.18)(12)=2.16\) and \(0.08k_1^*=(0.08)(27)=2.16\). After the change, \(s_2y_2^*=(0.32)(16)=5.12\) and \(0.08k_2^*=(0.08)(64)=5.12\).

Common mistake avoided: Using \(1/(1-\alpha)\) as the exponent for output. That exponent applies to capital; output uses \(\alpha/(1-\alpha)=(1/3)/(2/3)=1/2\). The output ratio is therefore \((0.32/0.18)^{1/2}=4/3\), consistent with the computed levels.

Worked Example: Faster Growth During Adjustment, Not Forever

Suppose a hypothetical economy has \(y=3\sqrt{k}\), with \(\delta+n=0.10\). Its saving rate rises permanently from \(s_1=0.20\) to \(s_2=0.30\). Find the old and new steady states, then calculate the instantaneous growth rates of capital and output per worker just after the change, when capital is still at its old steady-state level.

Method. For \(y=A\sqrt{k}\), the steady-state condition \(sA\sqrt{k^*}=0.10k^*\) implies \(\sqrt{k^*}=sA/0.10\). At \(s_1=0.20\), this gives \(\sqrt{k_1^*}=(0.20)(3)/0.10=6\), so \(k_1^*=36\) and \(y_1^*=3(6)=18\).

At \(s_2=0.30\), \(\sqrt{k_2^*}=(0.30)(3)/0.10=9\), so \(k_2^*=81\) and \(y_2^*=3(9)=27\). The new steady-state output level is 50% higher: \(27/18=1.5\).

Immediately after the saving-rate change, \(k=36\) and \(y=18\). Investment per worker is \(s_2y=(0.30)(18)=5.4\), while break-even investment is \((0.10)(36)=3.6\). Thus \(\dot{k}=5.4-3.6=1.8\). The instantaneous capital growth rate is \(\dot{k}/k=1.8/36=0.05\), or 5% per unit of time.

Since \(y=3\sqrt{k}\), differentiating gives \(\dot{y}/y=(1/2)(\dot{k}/k)\). Therefore the instantaneous output growth rate is \(0.5(5\%)=2.5\%\) per unit of time. As capital approaches 81, investment and break-even investment converge; at the new steady state \(\dot{k}=0\), so output per worker’s growth rate is also zero.

Common mistake avoided: Reporting the initial 2.5% as a permanent growth rate. It applies just after the change at the old capital stock. The new steady state has a higher output level, but no continuing output-per-worker growth under these fixed-technology assumptions.

What Changes—and What Does Not

The derivation and examples rely on the model’s assumptions: the saving rate changes permanently, production has diminishing returns to capital, and productivity, depreciation, and population growth remain fixed. Under these conditions, a higher saving rate shifts investment upward at each given capital stock. The new steady state has higher capital and output per worker, while the economy’s long-run growth rate of output per worker remains zero.

This is a level effect, not a permanent growth-rate effect. During the adjustment, capital per worker rises and output per worker can grow. At the new steady state, however, capital per worker stops changing. With productivity fixed, output per worker then stops changing as well. Aggregate output may still grow as the number of workers grows; that is distinct from growth in output per worker.

The model does not imply that any increase in saving is automatically a welfare improvement. More saving means less of current output is consumed, even as it supports a higher steady-state capital and output level. A full welfare comparison requires additional analysis. It is also important not to apply these conclusions unchanged when the assumptions differ: for example, if productivity changes over time, the steady-state growth implications require a different model specification.

Common Exam Traps

Confusing percentage points with percent change. A saving rate rising from 20% to 30% increases by 10 percentage points, but by 50% relative to its initial value.
Comparing saving rates without the production exponent. Under Cobb-Douglas production, the output ratio is \((s_2/s_1)^{\alpha/(1-\alpha)}\), not simply \(s_2/s_1\).
Using the wrong exponent. The exponent for capital is \(1/(1-\alpha)\); the exponent for output is \(\alpha/(1-\alpha)\).
Calling a level increase a lasting growth-rate increase. Transitional growth fades as the economy approaches its new steady state.
Skipping the equilibrium check. At every computed steady state, verify that \(s y^*=(\delta+n)k^*\).

Key takeaway: In the fixed-technology Solow model, a permanent rise in the saving rate raises steady-state capital and output per worker. It can produce faster growth during the transition, but it does not by itself create permanently higher growth in output per worker.

Check Your Understanding

For each problem, use the steady-state condition and distinguish the long-run output level from the growth rate.

A hypothetical economy has \(y=2\sqrt{k}\), break-even rate \(\delta+n=0.20\), and saving rates \(0.25\) and \(0.40\). Find both steady-state capital and output levels, and the percentage change in output. (Answer: \(k_1^*=6.25,\ y_1^*=5;\ k_2^*=16,\ y_2^*=8;\) output rises by 60%, while its steady-state growth rate is zero.)
Production is \(y=5k^{1/3}\), and \(\delta+n=0.15\). Compare saving rates \(0.12\) and \(0.27\): calculate both steady-state capital and output levels, and verify the new steady-state investment equality. (Answer: \(k_1^*=8,\ y_1^*=10;\ k_2^*=27,\ y_2^*=15;\) new investment and break-even investment are both \(0.15(27)=4.05\).)
With Cobb-Douglas production, \(\alpha=0.40\), and all other parameters fixed, the saving rate doubles. Find the ratio of new to old steady-state output and state the percentage increase. (Answer: \(2^{2/3}\approx1.5874\), so output per worker rises by about 58.74%.)
A hypothetical economy has \(y=2\sqrt{k}\), \(\delta+n=0.08\), and an initial saving rate of \(0.18\). What saving rate would produce a steady-state output level 25% above the initial level? (Answer: \(s=0.225\), or 22.5%.)
A hypothetical economy has \(y=3\sqrt{k}\), \(\delta+n=0.10\), and saving rate \(0.20\). The saving rate then rises to \(0.30\). Find the old steady-state capital, the initial change in capital per worker at that old capital stock, and the initial output-per-worker growth rate. (Answer: \(k_1^*=36;\ \dot{k}=1.8;\) initial output-per-worker growth is 2.5% per unit of time, and it eventually falls to zero at the new steady state.)

What You'll Learn