Why a Change in the Productivity Level Matters
In the Solow model, a higher productivity level means that a given amount of capital per worker can produce more output per worker. It also raises the return to saving when the saving rate is unchanged, so investment can support a different steady-state capital stock. The important distinction is between an instant change in the level of productivity and a continuing rate of technological progress: a one-time level shift changes output and the eventual capital stock, but does not by itself imply that output per worker keeps growing indefinitely.
This tutorial uses the per-worker production function \(y=A f(k)\), where \(y\) is output per worker, \(k\) is capital per worker, and \(A\) is a positive productivity level. The saving rate is \(s\), depreciation is \(\delta\), and the population growth rate is \(n\). As in the earlier material on the Solow model, assume saving is invested, population grows exogenously, and capital has diminishing marginal product. For the standard steady-state condition, the break-even investment per worker is \((\delta+n)k\).
A productivity shift does not instantaneously create additional physical capital. The capital stock changes through investment over time. Consequently, the economy can initially have its old \(k\) but a new \(y\); as saving and break-even investment differ, \(k\) then adjusts toward its new steady state. The precise path depends on the capital-accumulation convention being used. The examples state that convention explicitly.
Steady State Before and After a Level Shift
In the standard Solow steady-state condition, saving per worker equals investment required to maintain capital per worker:
Here \(k^*\) denotes steady-state capital per worker. At a steady state, capital per worker is constant: saving covers both depreciation and the capital needed for a growing workforce. If \(A\) rises while \(s\), \(n\), and \(\delta\) remain fixed, the saving curve \(sA f(k)\) shifts up. At the old \(k^*\), saving is now greater than break-even investment, so capital per worker tends to rise.
The direction of the capital response can be justified by differentiating the steady-state condition. Treat \(k^*\) as a function of \(A\), with the other parameters fixed:
Differentiating both sides with respect to \(A\) gives
Rearranging,
At a stable positive steady state, the marginal saving generated by a small increase in capital is less than the associated increase in break-even investment: \(sA f'(k^*)<\delta+n\). The denominator is therefore positive. Since \(s f(k^*)>0\), it follows that \(dk^*/dA>0\). Steady-state output is \(y^*=A f(k^*)\). Both the direct effect of \(A\) and the increase in \(k^*\) raise output, so \(y^*\) rises as well.
Cobb-Douglas Case: How Large Are the Level Effects?
For the Cobb-Douglas per-worker production function \(y=A k^\alpha\), where \(0<\alpha<1\), the steady-state condition becomes \(sA(k^*)^\alpha=(\delta+n)k^*\). Dividing by \((k^*)^\alpha\) and solving gives:
The exponent on \(A\) is \(1/(1-\alpha)\) in both steady-state expressions. Thus, in this particular model, the percentage change in either steady-state capital per worker or steady-state output per worker is larger than the percentage change in \(A\). This amplification arises because higher productivity raises output directly and, through higher saving at the unchanged saving rate, supports more capital per worker.
These formulas rely on the specified Cobb-Douglas production function and the standard Solow steady-state condition. Do not apply the Cobb-Douglas expressions to an arbitrary production function; for a different \(f\), solve \(sA f(k^*)=(\delta+n)k^*\) using that function.
Worked Example: Comparing the Old and New Steady States
Suppose \(y=A\sqrt{k}\), the saving rate is \(s=0.20\), depreciation is \(\delta=0.06\), and population growth is \(n=0.04\). Productivity rises once from \(A_0=1\) to \(A_1=1.5\), after which it remains constant. Find both steady states and the immediate output change at the old capital stock.
Method. Use the Solow steady-state condition, with \(\delta+n=0.06+0.04=0.10\). Since \(\alpha=1/2\), the Cobb-Douglas formula is \(k^*=[sA/(\delta+n)]^2\).
Before the shift. The steady-state capital stock is
Steady-state output is \(y_0^*=1\sqrt{4}=2\).
Immediately after the shift. Capital has not yet adjusted, so \(k=4\). Output becomes \(y=1.5\sqrt{4}=1.5(2)=3\), a change from \(2\) to \(3\), or a \(50\%\) increase.
New steady state. With \(A_1=1.5\),
The new steady-state capital stock rises from \(4\) to \(9\), and steady-state output rises from \(2\) to \(4.5\). The capital increase is \(125\%\), as is the steady-state output increase: \((9-4)/4=1.25\) and \((4.5-2)/2=1.25\). The immediate output increase is only \(50\%\) because capital is initially still \(4\).
Check and common mistake. At the new steady state, saving is \(0.20(4.5)=0.9\), and break-even investment is \(0.10(9)=0.9\), as required. A common mistake is to use \(k=4\) to calculate the new steady-state output. That gives the correct immediate output of \(3\), but not the eventual steady-state output of \(4.5\).
Capital Adjusts After the Output Shift
A useful way to keep timing straight is to separate the production effect from the accumulation effect. For a one-time increase in \(A\), evaluated at the same \(k\), output per worker changes immediately according to the production function. The saving rate then determines the amount of new investment per worker. If that investment exceeds break-even investment, capital per worker increases; if it falls short, capital per worker decreases. The capital change then affects subsequent output.
For a numerical transition, use the exact one-period law of motion with no population growth. At the start of each period, output is produced and the fraction \(s\) is saved; a fraction \(\delta\) of existing capital depreciates. The resulting capital per worker next period is
With \(n=0\), its steady-state condition is \(sA f(k^*)=\delta k^*\). This discrete-time law is stated separately because its timing is more precise than using a continuous-time or approximate change equation for a one-period calculation.
Worked Example: The First Period After a Productivity Increase
Let \(y=A\sqrt{k}\), \(s=0.20\), and \(\delta=0.10\), with no population growth. The economy begins at its steady state with \(A=1\), and productivity rises to \(A=1.5\) before the next period's production. Find output and capital per worker for that next period, then compare its capital stock with the new steady state.
Method. First find the initial steady state, then apply the stated discrete-time capital law using the new \(A\).
At \(A=1\), the steady-state condition is \(0.20\sqrt{k^*}=0.10k^*\). For a positive \(k^*\), divide by \(\sqrt{k^*}\): \(0.20=0.10\sqrt{k^*}\), so \(\sqrt{k^*}=2\) and \(k^*=4\). Initial output is \(y=1\sqrt{4}=2\).
After the productivity shift, at the unchanged \(k_0=4\), output is \(y_0=1.5\sqrt{4}=3\). Saving per worker is \(s y_0=0.20(3)=0.60\). The amount of capital remaining after depreciation is \((1-\delta)k_0=0.90(4)=3.60\). Therefore,
Output at \(k_1\), using the new productivity level, is \(y_1=1.5\sqrt{4.20}\). Since \(\sqrt{4.20}\approx2.0494\), \(y_1\approx1.5(2.0494)=3.0741\), to four decimal places.
The new steady-state capital stock is \(k_1^*=[0.20(1.5)/0.10]^2=9\). The calculated next-period capital stock \(4.20\) is above the old steady state of \(4\) but below the new one of \(9\), consistent with adjustment toward the new steady state.
Check and common mistake. The capital calculation uses \(0.60\) of new investment and \(3.60\) of retained capital, which sum to \(4.20\). Do not add the entire output \(3\) to capital; only the saved portion, \(s y=0.60\), becomes investment.
A Decrease Reverses the Adjustment
A productivity-level decrease has the reverse effects, given the same assumptions. At the old capital stock, output falls immediately. Saving falls with output, while depreciation continues to remove capital. If saving is now below break-even investment, capital per worker declines toward a lower steady state. During this transition, output need not fall by the same percentage each period: the productivity level is lower, but the capital stock also changes over time.
Worked Example: A Productivity Decline and Capital Transition
Suppose \(y=A\sqrt{k}\), \(s=0.25\), \(\delta=0.10\), and \(n=0\). The economy is initially at its steady state with \(A=1.2\). Productivity then falls to \(A=0.8\) and stays there. Find the initial and new steady states, the immediate output after the decline, and capital per worker after one period.
Method. Use \(k^*=[sA/\delta]^2\) for the steady states and the one-period law \(k_{t+1}=(1-\delta)k_t+sA f(k_t)\) for the transition.
Initially, \(sA=0.25(1.2)=0.30\), so \(k_0^*=(0.30/0.10)^2=9\). Initial output is \(1.2\sqrt{9}=3.6\). After the decline, the new steady-state capital stock is \(k_1^*=[0.25(0.8)/0.10]^2=(0.20/0.10)^2=4\). Its steady-state output is \(0.8\sqrt{4}=1.6\).
Capital is still \(9\) immediately after the productivity decline. Output at that unchanged capital stock is \(0.8\sqrt{9}=2.4\), a one-third decline from \(3.6\). Saving per worker is \(0.25(2.4)=0.60\), and retained capital after depreciation is \(0.90(9)=8.10\). Thus,
As a check, new steady-state saving at \(k=4\) is \(0.25[0.8\sqrt{4}]=0.25(1.6)=0.40\), equal to depreciation investment \(0.10(4)=0.40\). At the old capital stock of \(9\), saving after the decline is only \(0.60\), less than depreciation of \(0.10(9)=0.90\). Capital therefore falls, as the calculated \(k_1=8.70<9\) confirms.
Common mistake. Treating the new steady-state capital stock of \(4\) as the capital stock immediately after the shock skips the transition. Capital does not jump from \(9\) to \(4\); the one-period calculation gives \(8.70\).
Level Shifts Versus Ongoing Technological Progress
A constant productivity level and a continuing increase in productivity have different implications. In this tutorial, \(A\) changes once and is then held fixed. The change moves the production function and saving schedule to new levels. Once capital has adjusted to the new steady state, the model with fixed \(A\) has no continuing source of growth in output per worker: \(k\) and \(A\) are both constant there.
That statement does not mean output per worker never grows after a level shift. During the transition, capital per worker can rise, causing output per worker to rise further after its initial direct response to \(A\). It means only that this transition is not an indefinitely repeated increase in productivity. A model with ongoing technological progress requires a separate treatment of how technology evolves and how capital is measured; those are distinct from the level-shift comparison here.
This distinction matters when interpreting observed productivity changes. A one-time improvement may raise the output level and change the long-run capital stock, while the subsequent growth rate can decline as the adjustment is completed. A permanent increase in the rate of productivity growth is a different assumption and cannot be inferred merely from observing a higher level of \(A\). This treatment is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.
Check Your Understanding
For each problem, distinguish immediate output effects from steady-state or transition effects. Use the assumptions stated in the problem.
- In a Cobb-Douglas Solow model, \(y=A\sqrt{k}\), \(s=0.30\), and \(\delta+n=0.15\). Productivity rises from \(A=2\) to \(A=3\). Find the old and new steady-state capital and output per worker. (Answer: \(k_0^*=16,\ y_0^*=8;\ k_1^*=36,\ y_1^*=18\).)
- Productivity rises by \(25\%\) while capital per worker is initially unchanged. The production function is \(y=A f(k)\), with the same \(f\) before and after the shift. What is the immediate percentage change in output per worker, and does capital jump at the instant of the shift? (Answer: Output per worker rises \(25\%\); capital per worker does not jump.)
- Let \(y=A\sqrt{k}\), \(s=0.20\), and \(\delta+n=0.10\). Productivity rises from \(A=1\) to \(A=1.2\). Find the new steady-state capital and output per worker. (Answer: \(k^*=5.76,\ y^*=2.88\).)
- With no population growth, let \(y=A\sqrt{k}\), \(s=0.10\), and \(\delta=0.05\). At \(k=16\), productivity rises from \(1\) to \(1.2\) before production. Using \(k_{t+1}=(1-\delta)k_t+sA\sqrt{k_t}\), find current output at the new \(A\), next-period capital, and the new steady-state capital. (Answer: Current output \(=4.8\); \(k_{t+1}=15.68\); new \(k^*=5.76\).)
- For \(y=A k^\alpha\), with \(0<\alpha<1\), \(s\), and \(\delta+n\) fixed, productivity rises by \(10\%\). What is the ratio of new to old steady-state output when \(\alpha=0.4\), and what is the approximate percentage increase? (Answer: The ratio is \(1.1^{5/3}\approx1.1721\), an increase of approximately \(17.21\%\).)