Introduction
Two economies may begin with different capital stocks per worker. Does the initially poorer one necessarily grow faster, and should the two eventually reach the same living standard? The Solow model gives a qualified answer: the prediction depends on whether the economies share the fundamentals that determine their long-run paths.
Absolute convergence describes economies approaching a common steady state or balanced-growth path when their relevant fundamentals are the same. Conditional convergence describes economies approaching their own steady states when those fundamentals differ. A lower initial income by itself does not guarantee faster growth or eventual equality.
The Solow Model and Its Steady State
Use the continuous-time Solow model with labor-augmenting technology. Let \(K\) be capital, \(L\) labor, \(A\) labor-augmenting technology, and \(k=K/(AL)\) capital per effective worker. Let \(f(k)\) be output per effective worker, \(s\) the constant saving rate, \(n\) the constant growth rate of labor, \(g\) the constant growth rate of technology, and \(\delta\) the constant depreciation rate. Assume \(s>0\), \(n+g+\delta>0\), and a production function that is increasing and strictly concave. The model’s capital-accumulation equation is
The term \(n+g+\delta\) is break-even investment per effective worker: investment must replace depreciated capital and equip the additional effective workers created by labor and technology growth. A steady state \(k^*\) satisfies
For Cobb-Douglas production, take \(f(k)=k^\alpha\), with \(0<\alpha<1\). Solving the steady-state condition gives
Thus, the steady state in effective-worker terms depends on the production function and on \(s,n,g,\delta\). The technology level \(A\) does not appear in this expression because \(k\) and \(y=f(k)\) are already measured per effective worker. But technology levels still matter for output per worker: \(Y/L=A y\).
Why Capital Moves Toward Its Own Steady State
The direction of movement follows from the accumulation equation. Write \(m=n+g+\delta\), which is positive under the model assumptions. For Cobb-Douglas production, net investment is \(h(k)=s k^\alpha-mk\). At the positive steady state, \(h(k^*)=0\). Strict concavity implies that the production term grows less rapidly than the break-even term as \(k\) rises, so net investment is positive below \(k^*\) and negative above it.
The steady state solves \(s(k^*)^\alpha=mk^*\), giving \(k^*=(s/m)^{1/(1-\alpha)}\).
For \(0<k<k^*\), the ratio \(s k^\alpha/(mk)=(s/m)k^{\alpha-1}\) is greater than 1 because \(\alpha-1<0\). Therefore \(s k^\alpha-mk>0\), so capital rises.
For \(k>k^*\), that ratio is less than 1. Therefore \(s k^\alpha-mk<0\), so capital falls.
With positive initial capital, capital per effective worker therefore moves toward the unique positive steady state. This result explains convergence within an economy’s own Solow model. It does not, by itself, establish that two economies have the same steady state: that additionally requires their relevant fundamentals to match.
Absolute Convergence: A Common Destination Requires Common Fundamentals
If two economies have the same production function and the same \(s,n,g,\delta\), they have the same \(k^*\) and \(y^*\). If one starts below the steady state and the other starts closer to it, the former is farther from its long-run capital level and, in the Solow transition, has a stronger incentive for capital deepening. With the same technology path, their output per worker approaches the same balanced-growth path.
The precise meaning of “same path” matters. If technology is constant, \(g=0\), then \(A\) is constant and the steady-state level of output per worker is \(A y^*\). If technology grows at a positive rate, per-worker output does not settle at a constant level. At the effective-worker steady state,
When \(g>0\), both quantities grow with \(A(t)\), at rate \(g\). Economies with common fundamentals and the same technology path share a balanced-growth path, not a constant per-worker steady-state level. If they have different technology levels or technology paths, equal \(k^*\) alone does not imply equal output per worker.
Worked Example: Comparing Two Economies With the Same Fundamentals
Two hypothetical economies have Cobb-Douglas production with \(\alpha=0.5\), saving rate \(s=0.25\), labor growth \(n=0.02\), technology growth \(g=0.02\), and depreciation \(\delta=0.06\), all annual rates. They share the same technology path \(A(t)=e^{0.02t}\). Their initial capital per effective worker is \(k_0=1\) and \(k_0=4\), respectively. Find their common steady state and compare their initial capital accumulation.
Method. First calculate the break-even rate: \(m=n+g+\delta=0.02+0.02+0.06=0.10\). The steady-state condition is \(0.25\sqrt{k}=0.10k\). For positive \(k\), divide by \(\sqrt{k}\) to obtain \(\sqrt{k}=2.5\), so \(k^*=6.25\). Steady-state output per effective worker is \(y^*=\sqrt{6.25}=2.5\).
At \(k=1\), net investment is \(0.25\sqrt{1}-0.10(1)=0.25-0.10=0.15\). At \(k=4\), it is \(0.25\sqrt{4}-0.10(4)=0.50-0.40=0.10\). Both values are positive, so both economies accumulate capital per effective worker toward \(6.25\). Their initial proportional capital growth rates are \(0.15/1=0.15\), or 15% per year, and \(0.10/4=0.025\), or 2.5% per year.
At the effective-worker steady state, output per worker is \(2.5e^{0.02t}\) in both economies. It grows at 2% per year rather than remaining constant. Check: substituting \(k^*=6.25\) gives saving per effective worker \(0.25(2.5)=0.625\), exactly equal to break-even investment \(0.10(6.25)=0.625\). The common-fundamentals conclusion is a shared balanced-growth path, not a fixed per-worker level. A common mistake is to call \(2.5\) the steady-state output per worker; it is output per effective worker.
Conditional Convergence: Different Fundamentals, Different Destinations
When economies differ in saving, population growth, or technology growth, their steady states may differ. A higher saving rate raises steady-state capital per effective worker: it increases investment at any given \(k\). Higher population growth or technology growth raises break-even investment and lowers steady-state capital per effective worker, all else equal. A higher depreciation rate has the same direction of effect as higher break-even requirements.
Conditional convergence means that initial capital matters for the direction and pace of transition relative to the economy’s own steady state. It does not say that every initially poorer economy must grow faster than every richer one. Differences in saving, population growth, or technology can make a low-income economy’s steady state lower, or make its transition different. A persistent difference in the level of technology can also leave output per worker different even when capital per effective worker is the same.
Worked Example: Saving Rates Create Different Steady States
Two hypothetical economies share Cobb-Douglas production with \(\alpha=0.5\), labor growth \(n=0.02\), technology growth \(g=0.02\), and depreciation \(\delta=0.06\). The first saves \(s_1=0.16\); the second saves \(s_2=0.36\). Both start at \(k=1\). Find each steady state and compare initial capital accumulation.
Method. The common break-even rate is \(m=0.02+0.02+0.06=0.10\). For \(\alpha=0.5\), \(k^*=(s/m)^2\). The first economy has \(k_1^*=(0.16/0.10)^2=1.6^2=2.56\), and \(y_1^*=\sqrt{2.56}=1.6\). The second has \(k_2^*=(0.36/0.10)^2=3.6^2=12.96\), and \(y_2^*=\sqrt{12.96}=3.6\).
At their shared initial \(k=1\), the first economy’s net investment is \(0.16\sqrt{1}-0.10(1)=0.06\), while the second’s is \(0.36\sqrt{1}-0.10(1)=0.26\). Both begin below their own steady states, but the second accumulates capital faster at that point. Its larger saving rate gives it a different destination and a different transition.
Check: at \(k_1^*=2.56\), saving is \(0.16(1.6)=0.256\) and break-even investment is \(0.10(2.56)=0.256\). At \(k_2^*=12.96\), the corresponding amounts are \(0.36(3.6)=1.296\) and \(0.10(12.96)=1.296\). The common mistake is to infer that the economy with lower initial capital must grow faster. Here initial capital is equal, and the different saving rates produce different growth rates and steady states.
Worked Example: Population and Technology Growth Alter the Comparison
Two hypothetical economies have \(f(k)=\sqrt{k}\), saving rate \(s=0.25\), and depreciation \(\delta=0.06\). Economy A has \(n_A=0.02\) and \(g_A=0.02\); economy B has \(n_B=0.04\) and \(g_B=0.04\). Both have \(A(0)=1\). Calculate their steady states in effective-worker terms and their output per worker at the effective-worker steady state after 10 years.
Method. For A, \(m_A=0.02+0.02+0.06=0.10\). Thus \(k_A^*=(0.25/0.10)^2=2.5^2=6.25\), and \(y_A^*=2.5\). For B, \(m_B=0.04+0.04+0.06=0.14\). Thus \(k_B^*=(0.25/0.14)^2=(25/14)^2=625/196\approx3.1888\), and \(y_B^*=25/14\approx1.7857\).
At year 10, A’s technology level is \(A_A(10)=e^{0.02(10)}=e^{0.2}\approx1.2214\). Its steady-path output per worker is \(A_A(10)y_A^*\approx1.2214(2.5)=3.0535\). B’s technology level is \(A_B(10)=e^{0.04(10)}=e^{0.4}\approx1.4918\). Its steady-path output per worker is \(A_B(10)y_B^*\approx1.4918(1.7857)=2.6640\). Values are rounded to four decimals at the final step.
Check: A’s steady-state saving and break-even investment are both \(0.25(2.5)=0.625\). B’s are both \(0.25(25/14)=25/56\approx0.4464\), also equal to \(0.14(625/196)=25/56\). The economies have different steady-state capital per effective worker and different per-worker growth rates. A common mistake is to omit technology growth from break-even investment or to treat \(y^*\) as output per worker. Here \(y^*\) is per effective worker; output per worker also depends on \(A(t)\).
What the Distinction Does—and Does Not—Establish
The distinction is useful whenever comparing long-run outcomes across economies. The model’s prediction is not simply “poor economies catch up.” It is that capital per effective worker moves toward the steady state implied by the economy’s parameters. If those parameters match, the destination matches; if they differ, convergence can still occur, but toward different destinations.
Keep three comparisons separate:
- Same steady state: requires the same production function and the same saving, labor-growth, technology-growth, and depreciation parameters.
- Same per-worker path: additionally requires the same technology level and path. With positive technology growth, the common path grows rather than settling at a fixed level.
- Different steady states: imply that a low initial level alone cannot determine which economy grows faster or whether their per-worker incomes eventually become equal.
This framework is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus. For an exam-style comparison, state the model assumptions, calculate each economy’s own break-even term and steady state, and only then compare initial positions with those destinations. In particular, distinguish capital and output per effective worker from capital and output per worker.
Check Your Understanding
For each problem, use the continuous-time Solow model and show how the fundamentals determine the relevant comparison.
- Two economies have \(f(k)=\sqrt{k}\), \(s=0.20\), and \(n+g+\delta=0.10\). Find their common \(k^*\) and \(y^*\). If their technology is constant and \(A=2\), what is steady-state output per worker? (Answer: \(k^*=4\), \(y^*=2\), and output per worker \(=4\).)
- Economy C has \(s=0.30\) and \(n+g+\delta=0.10\); economy D has \(s=0.20\) and the same break-even rate. Both have \(f(k)=\sqrt{k}\). Find each steady-state capital per effective worker and identify whether they satisfy absolute or conditional convergence. (Answer: \(k_C^*=9\), \(k_D^*=4\); they have different steady states, so the comparison is conditional convergence.)
- An economy has \(s=0.25\), \(\alpha=0.5\), \(n=0.03\), \(g=0.01\), and \(\delta=0.06\). Calculate \(k^*\) and \(y^*\). (Answer: \(n+g+\delta=0.10\), \(k^*=6.25\), and \(y^*=2.5\).)
- Two economies share \(k^*=5\) in effective-worker terms and have \(y^*=2\). Their technology paths are \(A_1(t)=e^{0.02t}\) and \(A_2(t)=1.5e^{0.02t}\). Find their output per worker on their steady paths at \(t=0\), and state whether equal \(k^*\) implies equal output per worker. (Answer: \(2\) and \(3\), respectively; no, the different technology levels imply different output per worker.)
- With \(f(k)=\sqrt{k}\), \(s=0.25\), and \(\delta=0.05\), economy E has \(n=0.02,g=0.03\), while economy F has \(n=0.04,g=0.01\). Find each \(k^*\), then compare their steady-path per-worker growth rates. (Answer: both have \(n+g+\delta=0.10\), so both have \(k^*=6.25\); their per-worker output growth rates are 3% and 1%, respectively.)