Why Solve the Cobb-Douglas Steady State?
In the Solow model, saving finances investment, while depreciation and population growth create the investment needed just to maintain capital per worker. A steady state is reached when these forces balance. For Cobb-Douglas production, that balance can be solved directly, giving a closed-form expression for both capital and output per worker.
This result is useful for translating model parameters into long-run levels. It also provides a direct check on calculations made from the Solow diagram: at the positive steady state, investment per worker must exactly equal break-even investment per worker.
The production function has diminishing marginal product of capital because \(0<\alpha<1\). The model’s steady-state calculation uses two per-worker quantities: investment, \(s k^\alpha\), and break-even investment, \((\delta+n)k\). The latter replaces depreciated capital and equips new workers with the same capital per worker as existing workers.
From Capital Accumulation to the Steady-State Condition
The aggregate capital stock next period equals undepreciated capital plus investment. With the stated assumptions, \(K_{t+1}=(1-\delta)K_t+sY_t\), and the workforce grows according to \(L_{t+1}=(1+n)L_t\). Dividing the first equation by the second gives the exact discrete-time law for capital per worker:
A steady state has \(k_{t+1}=k_t=k^*\). Substituting this condition into the law of motion and multiplying by \(1+n\) gives the balance equation. Equivalently, subtracting current capital per worker from next period’s capital per worker shows the same condition for zero change.
Thus the positive steady state equates saving-financed investment with break-even investment. In this discrete-time formulation, the condition is exact; the factor \(1+n\) cancels when the steady-state equation is rearranged. A common textbook expression for the change in capital per worker is \(s k^\alpha-(\delta+n)k\); it has the same zero-change condition, though the exact discrete-time change from the law above is this expression divided by \(1+n\).
Deriving the Closed-Form Solution
For a positive steady state, divide the balance equation by \((k^*)^\alpha\). The restriction \(0<\alpha<1\) makes \(1-\alpha\) positive, so raising both sides to the power \(1/(1-\alpha)\) isolates capital per worker.
\(s(k^*)^\alpha=(\delta+n)k^*\).
Divide by \((k^*)^\alpha\): \((k^*)^{1-\alpha}=s/(\delta+n)\).
Raise both sides to the appropriate power, then use \(y^*=(k^*)^\alpha\).
These formulas give the unique positive steady state under the stated parameter restrictions. Since Cobb-Douglas output is zero when capital is zero, \(k=0\) is also a steady state in the model. The closed forms above describe the positive steady state, not that zero-capital boundary case.
Worked Applications
Worked Example: Calculating Capital and Output per Worker
A hypothetical economy has \(y=k^{1/2}\), a saving rate \(s=0.16\), depreciation rate \(\delta=0.04\), and workforce growth rate \(n=0.04\). Find positive steady-state capital and output per worker, then verify the investment balance. All rates are per period.
Method. Use the closed-form expressions with \(\alpha=1/2\). First calculate the break-even rate and the ratio:
\(\delta+n=0.04+0.04=0.08\), so \(s/(\delta+n)=0.16/0.08=2\).
The capital exponent is \(1/(1-\alpha)=1/(1-0.5)=2\). Therefore,
\(k^*=2^2=4\).
The output exponent is \(\alpha/(1-\alpha)=0.5/0.5=1\), so \(y^*=2^1=2\). This also checks directly: \((k^*)^{1/2}=\sqrt{4}=2\).
Investment per worker is \(s y^*=0.16(2)=0.32\). Break-even investment per worker is \((\delta+n)k^*=0.08(4)=0.32\). The two values agree, so the steady-state balance checks. No rounding is needed.
Common mistake avoided: The ratio \(0.16/0.08\) is \(2\), not \(4\). The value \(4\) is the ratio raised to the capital exponent \(2\).
Worked Example: A Fractional Production Exponent
A hypothetical economy has \(y=k^{1/3}\), \(s=0.20\), \(\delta=0.03\), and \(n=0.02\). Calculate the positive steady-state capital and output per worker, then check the balance condition.
Method. Here \(\alpha=1/3\), so \(\delta+n=0.03+0.02=0.05\) and \(s/(\delta+n)=0.20/0.05=4\).
The exponent for capital is \(1/(1-1/3)=1/(2/3)=3/2\). Thus \(k^*=4^{3/2}=(\sqrt{4})^3=2^3=8\).
The exponent for output is \((1/3)/(2/3)=1/2\). Hence \(y^*=4^{1/2}=2\). This agrees with the production function: \((k^*)^{1/3}=8^{1/3}=2\).
Investment per worker is \(s y^*=0.20(2)=0.40\). Break-even investment per worker is \((\delta+n)k^*=0.05(8)=0.40\). The balance condition holds exactly; no rounding is needed.
Common mistake avoided: The capital exponent is \(3/2\), not \(2/3\). The exponent \(2/3\) is \(1-\alpha\), and the solution requires its reciprocal.
Worked Example: Finding the Saving Rate for a Target Output
A hypothetical economy has \(y=k^{1/2}\), depreciation \(\delta=0.03\), and workforce growth \(n=0.02\). What saving rate is required for positive steady-state output per worker to equal \(4\)? Verify that the required saving rate is feasible.
Method. Work backward from the production function. Since \(y^*=(k^*)^{1/2}=4\), squaring both sides gives \(k^*=16\).
The steady-state condition is \(s y^*=(\delta+n)k^*\). Substitute the target and the given rates:
\(s(4)=(0.03+0.02)(16)=0.05(16)=0.80\).
Therefore \(s=0.80/4=0.20\). This saving rate is between zero and one, so it satisfies the model’s saving-rate restriction.
As a second check, the closed-form capital formula gives \([0.20/0.05]^{1/(1-1/2)}=4^2=16\), and the production function then gives \(\sqrt{16}=4\). The target is consistent with the required rate; no rounding is needed.
Common mistake avoided: Do not set the target output equal to the capital stock. Output \(4\) implies capital \(16\) under \(y=k^{1/2}\); the balance equation must use both quantities in their correct places.
Interpreting the Formula and Avoiding Traps
The formula makes clear that the positive steady state depends on the saving rate relative to the combined depreciation and population-growth rate. The exponent magnifies this ratio for capital when \(\alpha\) is close to one. Output has a smaller exponent than capital because output is \(k^\alpha\), not \(k\).
For example, if \(s/(\delta+n)=4\) and \(\alpha=1/3\), capital is \(4^{3/2}=8\), while output is \(4^{1/2}=2\). Reporting \(8\) as output would confuse the capital result with the output result. A reliable check is always to substitute calculated capital into \(y=(k)^\alpha\).
Several exam traps follow directly from the derivation:
- Using only depreciation. Break-even investment includes population growth as well as depreciation, so use \(\delta+n\), not just \(\delta\).
- Reversing the exponent. The capital exponent is \(1/(1-\alpha)\), the reciprocal of \(1-\alpha\).
- Using the capital formula for output. Calculate \(y^*=(k^*)^\alpha\), or use its own exponent \(\alpha/(1-\alpha)\).
- Skipping the balance check. Verify that \(s y^*=(\delta+n)k^*\); a mismatch often reveals an arithmetic or exponent error.
The closed form is specific to the stated production function and model assumptions. If production is not \(y=k^\alpha\), this particular exponent formula does not follow. Likewise, a model with technological progress requires a different choice of per-worker variable and is outside the assumptions used here.
Check Your Understanding
For each problem, use the positive steady-state formulas and check the investment balance where requested. Rates are per period.
- An economy has \(y=k^{1/2}\), \(s=0.09\), \(\delta=0.03\), and \(n=0.03\). Find positive steady-state capital and output per worker and the common investment amount. (Answer: \(k^*=2.25,\ y^*=1.5\), and investment and break-even investment are both \(0.135\).)
- An economy has \(y=k^{1/3}\), \(s=0.15\), \(\delta=0.04\), and \(n=0.01\). Find positive steady-state capital and output per worker. (Answer: \(k^*=3\sqrt{3}\approx5.1962,\ y^*=\sqrt{3}\approx1.7321\).)
- An economy has \(y=k^{0.6}\), \(s=0.18\), \(\delta=0.04\), and \(n=0.02\). Calculate positive steady-state capital and output per worker, rounding each to four decimal places. (Answer: \(k^*=3^{2.5}\approx15.5885,\ y^*=3^{1.5}\approx5.1962\).)
- An economy has \(y=k^{1/2}\), \(\delta=0.04\), and \(n=0.02\). Find the saving rate that supports target positive steady-state output \(y^*=3\), and check whether it is between zero and one. (Answer: \(k^*=9,\ s=0.18\), which is feasible.)
- Two economies have \(\alpha=0.4\). In economy A, \(s=0.12\) and \(\delta+n=0.06\); in economy B, \(s=0.27\) and \(\delta+n=0.09\). Find the ratio of their positive steady-state capital levels and the ratio of their output levels, \(k_B^*/k_A^*\) and \(y_B^*/y_A^*\), rounding to four decimals. (Answer: \(k_B^*/k_A^*=(3/2)^{5/3}\approx1.9656,\ y_B^*/y_A^*=(3/2)^{2/3}\approx1.3104\).)