From Effective-Worker Quantities to Living Standards
In the Solow model with labor-augmenting technical progress, capital and output per effective worker can be constant even while the economy’s output per worker continues to rise. The key is that an effective worker represents one unit of labor adjusted for technology. Translating a steady-state result into a living-standard measure therefore requires restoring the technology level.
This distinction matters whenever a problem gives a steady-state capital stock or output per effective worker and asks about capital or output per person employed. The conversion is direct, but the interpretation depends on whether the economy is at its balanced-growth path and on how technology changes over time.
Assume a production function with constant returns to scale, written as \(Y_t=F(K_t,A_tL_t)\). Define \(f(k_t)=F(k_t,1)\), so output per effective worker is \(y_t=f(k_t)\). At a steady state in effective-worker terms, \(k_t=k^*\) and \(y_t=y^*=f(k^*)\) remain constant. The star denotes a steady-state value; it does not mean that the corresponding per-worker quantity is constant.
Steady-State Conversion and Growth
Suppose labor-augmenting technology grows at a constant rate \(g\) per period, so \(A_{t+1}=(1+g)A_t\). Along a balanced-growth path, capital and output per effective worker stay at their steady-state levels. Multiplying those stationary quantities by technology gives the per-worker levels:
Since \(k^*\) and \(y^*\) do not change along this path, the growth of each per-worker quantity comes from the growth of \(A_t\). Thus capital per worker and output per worker grow at rate \(g\) each period. This is a statement about growth, not about the size of the levels: the levels also depend on the current technology level and on the steady-state quantities.
For completeness, suppose the number of workers grows at rate \(n\), so \(L_{t+1}=(1+n)L_t\). Aggregate capital and output then grow at the combined factor \((1+n)(1+g)\). Their exact per-period growth rate is \(n+g+ng\), not simply \(n+g\), though \(n+g\) is a common approximation when both rates are small.
Worked Example: Converting a Solow Steady State
Consider a discrete-time Solow economy with Cobb-Douglas production \(y=k^{1/3}\), saving rate \(s=0.25\), depreciation rate \(\delta=0.0424\), worker growth \(n=0.01\), and technology growth \(g=0.01\). Find steady-state capital and output per effective worker, then convert them to per-worker quantities when \(A_t=3\).
Method. In this model, one period of capital accumulation follows \(K_{t+1}=(1-\delta)K_t+sY_t\). Effective labor grows by the factor \((1+n)(1+g)\). At the steady state, capital per effective worker is constant, so the break-even investment requirement per effective worker is \((1+n)(1+g)-(1-\delta)\) times \(k^*\).
First calculate the break-even coefficient: \((1+n)(1+g)=(1.01)(1.01)=1.0201\), and \(1-\delta=1-0.0424=0.9576\). Their difference is \(1.0201-0.9576=0.0625\). The steady-state condition becomes \(0.25(k^*)^{1/3}=0.0625k^*\).
For positive \(k^*\), divide both sides by \((k^*)^{1/3}\): \(0.25=0.0625(k^*)^{2/3}\). Therefore, \((k^*)^{2/3}=0.25/0.0625=4\), and \(k^*=4^{3/2}=8\). Output per effective worker is \(y^*=f(8)=8^{1/3}=2\).
Now convert to per-worker quantities using \(A_t=3\). Capital per worker is \(A_tk^*=3(8)=24\); output per worker is \(A_ty^*=3(2)=6\). The result is internally consistent: output per effective worker is \(2\), while output per worker is \(6\) because each worker is multiplied by technology level \(3\).
Check and common mistake. Substitution in the steady-state equation gives saving \(0.25(2)=0.5\) and break-even investment \(0.0625(8)=0.5\), as required. A common error is to report \(y^*=2\) as output per worker. That omits technology; the corrected per-worker answer is \(3(2)=6\).
Why the Conversion Works
The conversion follows from the definitions, rather than from an additional behavioral assumption. Effective labor is \(A_tL_t\), so dividing an aggregate by effective labor removes both the number of workers and the technology adjustment. To recover the aggregate per worker, multiply by \(A_t\):
\(k_t=K_t/(A_tL_t)\) and \(y_t=Y_t/(A_tL_t)\).
Multiplying the first identity by \(A_t\) gives \(A_tk_t=K_t/L_t\); multiplying the second gives \(A_ty_t=Y_t/L_t\).
On the balanced-growth path, \(k_t=k^*\) and \(y_t=y^*\). Thus the per-worker levels are \(A_tk^*\) and \(A_ty^*\), and their growth is exactly the growth of \(A_t\).
For example, the output result can also be checked by starting from \(Y_t=A_tL_ty^*\). Divide both sides by \(L_t\), which is positive, to obtain \(Y_t/L_t=A_ty^*\). If technology next period is \((1+g)A_t\), then \((Y_{t+1}/L_{t+1})/(Y_t/L_t)=((1+g)A_ty^*)/(A_ty^*)=1+g\), provided output per effective worker remains \(y^*\). The same argument applies to capital per worker.
This growth conclusion relies on being on the balanced-growth path. Away from the steady state in effective-worker terms, \(k_t\) and \(y_t=f(k_t)\) may change as well. In that case, output per worker still equals \(A_ty_t\), but its growth is not determined by technology growth alone.
Levels Across Time and Economies
Worked Example: Tracking Per-Worker Levels Over Time
An economy is on its balanced-growth path with steady-state capital per effective worker \(k^*=8\), steady-state output per effective worker \(y^*=2\), and initial technology \(A_0=2\). Technology grows by 5% each year. Find capital and output per worker initially and after three years.
Method. First find technology in each period using \(A_t=A_0(1+g)^t\). Then multiply by the steady-state quantities. Initially, \(K_0/L_0=2(8)=16\) and \(Y_0/L_0=2(2)=4\).
After three years, the technology level is \(A_3=2(1.05)^3\). Since \(1.05^2=1.1025\), \(1.05^3=1.1025(1.05)=1.157625\), so \(A_3=2.31525\).
Capital per worker after three years is \(A_3k^*=2.31525(8)=18.522\). Output per worker is \(A_3y^*=2.31525(2)=4.6305\). Reporting to three decimals gives \(18.522\) units of capital per worker and \(4.631\) units of output per worker.
Check and common mistake. Both per-worker quantities have risen by the same factor, \(A_3/A_0=1.157625\). For instance, \(16(1.157625)=18.522\). Treating a 5% annual increase as a one-time 15% increase would instead use the wrong three-year technology factor; the correct factor is compounded, \((1.05)^3\).
A steady-state value per effective worker is not by itself a living-standard comparison across economies. The technology level must also be known. Consider two economies with the same production function \(y=k^{1/3}\), but different steady-state capital per effective worker and technology levels.
Worked Example: Comparing Two Economies’ Per-Worker Levels
Economy A has \(k_A^*=8\), \(A_A=1.5\), and production per effective worker \(y=k^{1/3}\). Economy B has \(k_B^*=27\), \(A_B=1\), and the same production function. Calculate and compare their steady-state output and capital per worker.
Method. First calculate output per effective worker from the production function. For A, \(y_A^*=8^{1/3}=2\); for B, \(y_B^*=27^{1/3}=3\).
Convert the levels for A: \(K_A/L_A=A_Ak_A^*=1.5(8)=12\), and \(Y_A/L_A=A_Ay_A^*=1.5(2)=3\). For B, \(K_B/L_B=1(27)=27\), and \(Y_B/L_B=1(3)=3\).
The economies have equal output per worker, \(3\), despite different steady-state capital per effective worker and technology levels. Economy B has more capital per worker, \(27\) versus \(12\). The equality in output per worker follows from the particular values: B’s higher output per effective worker is offset by A’s higher technology level.
Check and common mistake. The production relation also gives \(Y_A/L_A=(K_A/L_A)^{1/3}A_A^{2/3}\) and \(Y_B/L_B=(K_B/L_B)^{1/3}A_B^{2/3}\), consistent with the calculated outputs. Comparing \(y_A^*=2\) with \(y_B^*=3\) alone would incorrectly suggest that B has higher output per worker; those figures are per effective worker, not per worker.
Common Interpretation Traps
- Leaving technology out of the level conversion. Output per effective worker is \(y^*\); output per worker is \(A_ty^*\). They are equal only if \(A_t=1\).
- Calling the per-worker level constant on the balanced-growth path. The steady quantities are \(k^*\) and \(y^*\). If \(g>0\), capital and output per worker rise at rate \(g\).
- Mixing units of time or growth rates. If \(g\) is an annual rate, the number of periods in \(A_t=A_0(1+g)^t\) must also be in years. Apply the growth factor once per period, not once over the entire horizon.
- Using a balanced-growth conclusion during transition. The identity \(Y_t/L_t=A_ty_t\) always holds, but the claim that output per worker grows at rate \(g\) requires \(y_t\) to be constant.
- Confusing aggregate growth with per-worker growth. With growing population, aggregate output grows by the factor \((1+n)(1+g)\), while output per worker grows by \(1+g\) on the balanced-growth path.
Check Your Understanding
For each problem, distinguish effective-worker quantities from per-worker quantities and show the relevant conversion.
- An economy has \(k^*=25\), production \(y=\sqrt{k}\), and technology \(A=1.2\). Find steady-state capital and output per worker. (Answer: capital per worker \(=30\); output per worker \(=6\).)
- An economy on its balanced-growth path has \(k^*=8\), \(y^*=2\), \(A_0=1.25\), and technology growth of 4% per year. Find output per worker after two years. (Answer: \(2.704\).)
- A discrete-time economy has Cobb-Douglas production \(y=k^{1/2}\), saving rate \(s=0.20\), depreciation \(\delta=0.05\), worker growth \(n=0.01\), and technology growth \(g=0.02\). Find \(k^*\), \(y^*\), and output per worker when \(A=2\), rounding final values to four decimals. (Answer: \(k^*=6.2189\), \(y^*=2.4938\), output per worker \(=4.9875\).)
- On a balanced-growth path, workers grow by 1% per period and technology by 3% per period. Find the exact growth rate of aggregate output and the growth rate of output per worker. (Answer: aggregate output growth \(=4.03\%\); output per worker growth \(=3\%\).)
- Two economies have steady-state output per effective worker of 2 and 3, respectively. Their technology levels are 1.5 and 1. Find each economy’s output per worker and explain whether the ranking matches the ranking in output per effective worker. (Answer: both have output per worker \(=3\); the effective-worker ranking does not imply a per-worker ranking.)