Solow Transition-Dynamics Calculation Workshop

Following Capital After a Shock

A steady state describes where capital per worker settles when model parameters remain fixed. A transition calculation asks a different question: given today's capital stock and a parameter change, how much capital will there be next period? The answer requires comparing investment with the investment needed to offset depreciation and, when present, growth in the workforce and technology.

The key discipline is to evaluate investment using the capital stock at the start of each period. After calculating the next capital stock, use that new stock to calculate the following period's output and investment. Reusing the initial investment for later periods skips the model's feedback from capital to production.

Definition: In the discrete-time Solow model, gross investment per worker or effective worker is \(s f(k_t)\), where \(s\) is the saving rate and \(f(k_t)\) is output per worker or effective worker at the start of period \(t\). Break-even investment is the investment required to keep capital per worker or effective worker unchanged. Investment above break-even raises that capital stock; investment below break-even lowers it.

The Discrete-Time Transition Equation

First consider the model with no population growth or technological progress. Let \(K_t\) be the capital stock, \(L_t\) the number of workers, and \(k_t=K_t/L_t\) capital per worker. Assume the capital stock depreciates at rate \(\delta\) per period, output per worker is \(f(k_t)\), and a constant share \(s\) of output is invested. With a constant workforce, the capital law is \(K_{t+1}=(1-\delta)K_t+sY_t\). Dividing by \(L_t\) gives the transition equation below.

k_{t+1}=(1-\delta)k_t+s f(k_t)

Thus the change in capital per worker is investment less depreciation: \(k_{t+1}-k_t=s f(k_t)-\delta k_t\). The term \(\delta k_t\) is break-even investment in this no-growth case. Rates in the discrete-time equation are per-period rates; the calculation does not treat them as continuously compounded rates.

With population growth and labor-augmenting technological progress, it is useful to measure capital per effective worker. Let \(A_t\) denote labor-augmenting technology, let \(k_t=K_t/(A_tL_t)\), and let output per effective worker be \(f(k_t)\). Assume population grows at rate \(n\), technology at rate \(g\), depreciation is \(\delta\), and the capital law is \(K_{t+1}=(1-\delta)K_t+sY_t\). The workforce measured in effective workers grows by the exact factor \((1+n)(1+g)\) per period. Dividing the capital law by the next period's effective workforce gives the generalized equation.

k_{t+1}= \frac{(1-\delta)k_t+s f(k_t)} {(1+n)(1+g)}

This form uses exact discrete-time growth factors. It is not identical to replacing the combined growth rate by \(n+g\): the product includes the \(ng\) term. Rearranging makes the required comparison explicit.

k_{t+1}-k_t = \frac{s f(k_t)-\left[(1+n)(1+g)-(1-\delta)\right]k_t} {(1+n)(1+g)}

The exact break-even investment requirement per effective worker is \(B k_t\), where \(B=(1+n)(1+g)-(1-\delta)=n+g+ng+\delta\). When \(n=g=0\), this reduces to \(\delta k_t\), as in the simpler equation. In applications that explicitly use a growth-rate approximation, the coefficient is often written \(n+g+\delta\); use the exact equation above when the problem specifies discrete-time rates and exact accounting.

Result: For a given period, calculate \(s f(k_t)\) and compare it with \(B k_t\). If investment is greater, then \(k_{t+1}>k_t\); if it is smaller, then \(k_{t+1}<k_t\); and if they are equal, \(k_{t+1}=k_t\). The comparison uses the parameters in force during that period and the capital stock at its start.

Why the Break-Even Term Has This Form

The break-even requirement follows from the capital law and the change in the number of effective workers. To keep \(k_{t+1}=k_t\), the next period's total capital must equal the target capital per effective worker multiplied by the larger effective workforce. The next-period total capital is surviving capital plus investment. Equating these quantities gives the required investment.

Set the target capital per effective worker unchanged.
If \(k_{t+1}=k_t\), the required next-period total capital is \(K_{t+1}=k_t A_{t+1}L_{t+1}\).

Write next period's effective workforce.
Since \(A_{t+1}L_{t+1}=(1+n)(1+g)A_tL_t\), the target capital stock is \((1+n)(1+g)k_tA_tL_t\).

Equate the target with surviving capital plus investment.
The capital law gives \(K_{t+1}=[(1-\delta)k_t+s f(k_t)]A_tL_t\). Equating this with the target and dividing by \(A_tL_t\) yields \(s f(k_t)=[(1+n)(1+g)-(1-\delta)]k_t\).

So break-even investment covers more than worn-out capital when effective workers are growing: it must also provide enough capital to equip the additional effective workers. This derivation also explains why simply subtracting depreciation is insufficient when \(n\) or \(g\) is positive.

Worked Transition Calculations

Worked Example: Two Periods After a Saving-Rate Increase

Suppose output per worker is \(f(k)=\sqrt{k}\), depreciation is \(\delta=0.10\) per year, and population and technology are constant. The saving rate rises from \(s=0.20\) to \(s=0.40\). Immediately before the change, capital per worker is \(k_0=4\). Calculate capital per worker after one and two years under the new saving rate, and compare the result with the new steady state.

Method. With no population growth or technological progress, use \(k_{t+1}=k_t+s\sqrt{k_t}-0.10k_t\). Apply the new saving rate in each period. The pre-shock steady state is consistent with the given starting stock because \(0.20\sqrt{4}=0.40=0.10(4)\).

First year. At \(k_0=4\), output per worker is \(\sqrt{4}=2\). Investment is \(0.40(2)=0.80\), and break-even investment is depreciation, \(0.10(4)=0.40\). The net addition is \(0.80-0.40=0.40\), so \(k_1=4+0.40=4.40\).

Second year. Use \(k_1=4.40\), not \(k_0=4\). Output is \(\sqrt{4.40}\approx2.0976\), investment is \(0.40(2.0976)\approx0.8390\), and depreciation is \(0.10(4.40)=0.4400\). The net addition is approximately \(0.8390-0.4400=0.3990\), giving \(k_2\approx4.40+0.3990=4.7990\).

Steady-state check. A steady state satisfies \(s\sqrt{k^*}=\delta k^*\). For positive \(k^*\), this gives \(\sqrt{k^*}=s/\delta=4\), so \(k^*=16\). Both calculated stocks are below 16, and both steps raise capital, consistent with the investment comparison. The results are reported to four decimal places.

Common mistake avoided: Holding first-year investment fixed in the second year would ignore that output changes with capital. Recalculate output and investment at \(k_1\) for every transition.

Worked Example: Two Periods After a Saving-Rate Decrease

Use the same production function, \(f(k)=\sqrt{k}\), and annual depreciation rate \(\delta=0.10\), with no population growth or technological progress. The saving rate falls from \(0.40\) to \(0.20\) when capital per worker is \(k_0=16\). Calculate the next two capital stocks and determine whether the economy is above or below its post-change steady state.

The new steady state solves \(0.20\sqrt{k^*}=0.10k^*\). Dividing by \(0.10\sqrt{k^*}\) for positive capital gives \(\sqrt{k^*}=2\), hence \(k^*=4\). The starting stock of 16 is above this target.

First year. Output at \(k_0=16\) is \(\sqrt{16}=4\). New investment is \(0.20(4)=0.80\), while depreciation is \(0.10(16)=1.60\). Net investment is \(0.80-1.60=-0.80\), so \(k_1=16-0.80=15.20\).

Second year. Output at \(k_1=15.20\) is \(\sqrt{15.20}\approx3.8987\). Investment is \(0.20(3.8987)\approx0.7797\); depreciation is \(0.10(15.20)=1.5200\). Net investment is approximately \(0.7797-1.5200=-0.7403\), so \(k_2\approx15.20-0.7403=14.4597\).

Both stocks remain above \(k^*=4\), and investment is below depreciation at each calculated stock, so capital per worker falls. The calculation is reported to four decimal places. Finding the new steady state first helps check the direction, but it does not replace calculating the period-by-period changes.

Common mistake avoided: The old steady state no longer determines the direction of change after the saving-rate shock. Compare investment with break-even investment under the new saving rate.

Worked Example: Including Population Growth in Successive Updates

Let output per worker be \(f(k)=\sqrt{k}\), with \(s=0.30\), annual depreciation \(\delta=0.10\), population growth \(n=0.05\), and no technological progress (\(g=0\)). All rates are exact discrete-time annual rates. Starting from \(k_0=1\), calculate \(k_1\) and \(k_2\), then check the direction against the steady state.

Here the denominator in the transition equation is \((1+n)(1+g)=1.05\). The exact break-even coefficient is \(B=1.05-(1-0.10)=0.15\). Thus break-even investment at any \(k_t\) is \(0.15k_t\), and the change is \([0.30\sqrt{k_t}-0.15k_t]/1.05\).

First period. At \(k_0=1\), investment is \(0.30\sqrt{1}=0.30\) and break-even investment is \(0.15(1)=0.15\). Net investment above break-even is \(0.30-0.15=0.15\). Therefore \(k_1=1+0.15/1.05\approx1.1429\).

Second period. Using the unrounded first-period result \(k_1=8/7\), output is \(\sqrt{8/7}\approx1.0690\). Investment is \(0.30(1.0690)\approx0.3207\), and break-even investment is \(0.15(8/7)\approx0.1714\). The difference is approximately \(0.1493\); dividing by \(1.05\) gives a capital increase of approximately \(0.1422\). Thus \(k_2\approx1.1429+0.1422=1.2850\). The result is reported to four decimal places.

Steady-state check. The steady-state equation is \(0.30\sqrt{k^*}=0.15k^*\), giving \(\sqrt{k^*}=2\) and \(k^*=4\). Both calculated stocks are below 4, and the updates are positive, as expected. The exact discrete-time population growth factor is included in both the break-even requirement and the denominator of the update.

Common mistake avoided: Counting depreciation alone would give break-even investment of \(0.10k_t\). At \(k_0=1\), that would incorrectly produce a change of \((0.30-0.10)/1.05\approx0.1905\), instead of the correct \(0.15/1.05\approx0.1429\). Population growth also requires capital per worker to be maintained.

Reading the Transition and Avoiding Traps

For Cobb–Douglas production per effective worker, \(f(k)=A k^\alpha\), with \(A>0\) and \(0<\alpha<1\), a positive steady state solves \(sA(k^*)^\alpha=Bk^*\). Rearranging gives \(k^*=(sA/B)^{1/(1-\alpha)}\). Below this stock, investment exceeds break-even investment; above it, investment is lower. This comparison helps check a transition calculation, but the transition equation is still needed to find the size of each change.

A shock changes the relevant parameters for the transition that follows it. For example, after a saving-rate change, use the new \(s\) in the next period's investment calculation. Capital is a stock, however, so it does not instantly jump to its new steady-state value. Calculate the next stock from the previous stock and the new period's investment, then repeat.

Do not confuse a flow and a stock. Investment is a period flow; \(k_t\) is the capital stock at the start of the period.
Do not reuse old output. Because \(f(k)\) changes as capital changes, recompute output and investment at every step.
Do not omit dilution. With growing effective labor, break-even investment includes both depreciation and the resources needed to maintain capital per effective worker.
Do not mix exact and approximate rates. If the question specifies exact discrete-time rates, use \((1+n)(1+g)\), not an unannounced approximation.
Do not infer the size of a step from the steady state alone. The steady state indicates the no-change point; the accumulation equation determines the actual change.

Practising these steps builds the ability to move between a model's long-run benchmark and its short-run arithmetic. The calculation skills are aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

Check Your Understanding

For each problem, use the stated period rates and round calculated capital stocks to four decimal places where needed.

With \(f(k)=\sqrt{k}\), \(s=0.25\), \(\delta=0.10\), and no population or technology growth, start from \(k_0=4\). Calculate \(k_1\), and state whether capital rises or falls. (Answer: Investment \(=0.5000\), depreciation \(=0.4000\), and \(k_1=4.1000\); capital rises.)
With \(f(k)=\sqrt{k}\), \(s=0.30\), \(\delta=0.10\), population growth \(n=0.05\), and \(g=0\), start from \(k_0=1\). Using exact discrete-time rates, find \(k_1\). (Answer: Break-even investment \(=0.1500\), denominator \(=1.0500\), and \(k_1=1.1429\).)
With \(f(k)=\sqrt{k}\), \(\delta=0.10\), and no population or technology growth, the saving rate falls from \(0.20\) to \(0.10\) at \(k_0=9\). Calculate \(k_1\) and the new steady state. (Answer: New investment \(=0.3000\), depreciation \(=0.9000\), \(k_1=8.4000\), and the new steady state is \(k^*=1.0000\).)
With \(f(k)=A\sqrt{k}\), \(s=0.20\), \(\delta=0.10\), and no population or technology growth, technology rises from \(A=1\) to \(A=1.5\) when \(k_0=4\). Find the next capital stock and the new steady state. (Answer: New investment \(=0.6000\), depreciation \(=0.4000\), \(k_1=4.2000\), and \(k^*=9.0000\).)
With \(f(k)=\sqrt{k}\), \(s=0.40\), \(\delta=0.10\), and \(g=0\), population growth rises from \(n=0\) to \(n=0.10\) when \(k_0=16\). Using the new population-growth rate, calculate \(k_1\) and the new steady state. (Answer: Investment \(=1.6000\), break-even investment \(=0.2000(16)=3.2000\), denominator \(=1.1000\), \(k_1=14.5455\), and the new steady state is \(k^*=4.0000\).)

What You'll Learn