Why Diminishing Returns Produce Conditional Convergence

Why Convergence Depends on Diminishing Returns

An economy with little capital per effective worker can often add capital quickly relative to its existing stock. An economy with abundant capital may still invest, but each additional unit of capital produces less extra output. In the Solow model, this diminishing return helps explain why capital deepening can proceed faster in an economy that is farther below its own steady state.

The qualification its own matters. An economy’s steady state depends on its saving rate, population growth, technology growth, depreciation, and production technology. The model therefore does not imply that economies with different fundamentals must converge to the same capital stock or income level. The focus here is the logic of convergence in capital per effective worker, not a calculation of how many years convergence takes.

Definition: In the continuous-time Solow model with labor-augmenting technological progress, let $k$ be capital per effective worker, $f(k)$ output per effective worker, $s$ the saving rate, $n$ the population growth rate, $g$ the growth rate of labor efficiency, and $\delta$ the depreciation rate. The capital accumulation equation is $$ \dot{k}=s f(k)-(n+g+\delta)k. $$ The term $b=n+g+\delta$ is the break-even investment rate per unit of capital. A steady state $k^*$ satisfies $s f(k^*)=b k^*$, so $\dot{k}=0$ there.

This equation assumes a constant saving rate, labor and efficiency growing at constant rates $n$ and $g$, and capital depreciating at rate $\delta$. It describes capital per effective worker, not total capital. Investment $s f(k)$ adds to capital, while break-even investment $bk$ replaces depreciated capital and equips additional workers and effective workers with capital. The continuous-time equation is not the same as a discrete-time update equation; rates and the dot notation here are per year in a continuous-time model.

From the Capital Stock to Its Growth Rate

When $k>0$, divide the accumulation equation by $k$. The result expresses the proportional growth rate of capital per effective worker as the difference between investment per unit of capital and the break-even rate:

\frac{\dot{k}}{k} =s\frac{f(k)}{k}-(n+g+\delta) =s\frac{f(k)}{k}-b.

The ratio $f(k)/k$ is the average product of capital: output per unit of capital, with output measured per effective worker. Under diminishing returns, the marginal product $f'(k)$ declines as $k$ rises. For a differentiable, strictly concave production function with $f(0)=0$, the average product also declines with $k$. That fact links diminishing returns directly to the capital growth rate.

Key result: If $f$ is differentiable and strictly concave, $f(0)=0$, and $s>0$, then $s f(k)/k-b$ decreases as $k$ increases. Thus, for an economy with fixed fundamentals, a lower positive $k$ implies a higher proportional growth rate of capital per effective worker. Below $k^*$, capital per effective worker grows; above $k^*$, it declines.

Why the Average Product Declines

The claim can be justified by differentiating the average product. For $k>0$, the quotient rule gives

\frac{d}{dk}\left(\frac{f(k)}{k}\right) =\frac{k f'(k)-f(k)}{k^2}.

Strict concavity and $f(0)=0$ imply that the slope of the line from the origin to $(k,f(k))$, which is $f(k)/k$, exceeds the tangent slope $f'(k)$ at $k$. Equivalently, $f(k)>k f'(k)$. Therefore the numerator $k f'(k)-f(k)$ is negative, and the derivative is negative. Multiplying the average product by the positive saving rate $s$ and subtracting the constant $b$ preserves that downward relationship. This establishes why the proportional growth rate falls as the capital stock rises, under the stated assumptions.

Locate the economy relative to its steady state.
Compare current $k$ with the value satisfying $s f(k^*)=b k^*$.

Check the direction of movement.
If $k<k^*$, then $s f(k)>bk$ and $\dot{k}>0$. If $k>k^*$, then $s f(k)<bk$ and $\dot{k}<0$.

Interpret the rate, not just the sign.
For fixed fundamentals, lower $k$ means a higher $s f(k)/k-b$, so capital per effective worker grows faster proportionally.

The directional result also follows from the steady-state condition. At $k^*$, the proportional growth rate is zero. Below $k^*$, strict concavity makes $f(k)/k$ greater than $f(k^*)/k^*$, so $s f(k)/k>b$ and $\dot{k}/k>0$. Above $k^*$, the average product is lower, so the proportional growth rate is negative. This argument concerns movement toward the steady state; it does not say that the economy reaches it in a finite time.

Worked Example: Comparing Two Capital Stocks Under the Same Fundamentals

A hypothetical economy has $f(k)=\sqrt{k}$, saving rate $s=0.20$, and break-even rate $b=0.05$ per year. Find its steady-state capital per effective worker, then calculate the proportional growth rate and change in $k$ per year at $k=4$ and $k=1$. Assume the continuous-time Solow model.

Method. First solve the steady-state condition. Then use $\dot{k}/k=s f(k)/k-b$ at each current capital stock. Rates are annual.

At the steady state, $0.20\sqrt{k^*}=0.05k^*$. For a positive steady state, divide by $\sqrt{k^*}$:

\sqrt{k^*}=\frac{0.20}{0.05}=4, \qquad k^*=16.

At $k=4$, output per effective worker is $f(4)=\sqrt{4}=2$. Investment per effective worker is $s f(4)=0.20(2)=0.40$, and break-even investment is $bk=0.05(4)=0.20$. Thus $\dot{k}=0.40-0.20=0.20$, and

\frac{\dot{k}}{k}=\frac{0.20}{4}=0.05=5\%\text{ per year}.

At $k=1$, $f(1)=1$. Investment is $0.20(1)=0.20$, and break-even investment is $0.05(1)=0.05$. Hence $\dot{k}=0.20-0.05=0.15$, giving

\frac{\dot{k}}{k}=\frac{0.15}{1}=0.15=15\%\text{ per year}.

Both capital stocks are below $k^*=16$, so both have positive capital deepening. The lower stock, $k=1$, has the higher proportional growth rate. As a check, at $k=16$, investment is $0.20(4)=0.80$ and break-even investment is $0.05(16)=0.80$, so net investment is zero. A common mistake is to compare only the absolute additions $0.15$ and $0.20$ and conclude that $k=4$ grows faster. The relevant comparison for proportional growth is $\dot{k}/k$, not $\dot{k}$ alone.

Why Convergence Is Conditional

Diminishing returns explain a tendency toward an economy’s own steady state, but they do not make all steady states equal. For the Cobb–Douglas production function $f(k)=A k^\alpha$, with $A>0$ and $0<\alpha<1$, the steady-state condition gives

sA(k^*)^\alpha=bk^*, \qquad k^*=\left(\frac{sA}{b}\right)^{\frac{1}{1-\alpha}}.

Thus, a higher saving rate or productivity level raises the steady-state capital stock, while a higher combined rate of population growth, technology growth, and depreciation lowers it, all else equal. Economies with different fundamentals can have different steady states. Even if they share the same production function, their paths and destinations depend on their own parameters.

The word conditional captures this qualification: the model predicts faster capital deepening for an economy that is relatively far below its own steady state, holding its fundamentals fixed. A comparison across economies with different saving rates, population growth, depreciation, or technology must account for those differences. Simply observing that one economy starts with less capital does not establish that it will grow faster than every other economy.

Worked Example: Different Saving Rates, Different Destinations

Two hypothetical economies have the same production function, $f(k)=\sqrt{k}$, and the same break-even rate $b=0.08$ per year. Economy C has saving rate $s_C=0.24$; Economy D has $s_D=0.16$. Calculate each steady state and each proportional growth rate when $k=2$. Assume continuous time and the given common technology.

For $f(k)=\sqrt{k}$, the positive steady state solves $s\sqrt{k^*}=bk^*$. Dividing by $\sqrt{k^*}$ gives $\sqrt{k^*}=s/b$, so $k^*=(s/b)^2$. For C,

k_C^*=\left(\frac{0.24}{0.08}\right)^2=3^2=9.

For D,

k_D^*=\left(\frac{0.16}{0.08}\right)^2=2^2=4.

At $k=2$, $\sqrt{2}\approx1.41421356$. For C, the proportional growth rate is

\frac{\dot{k}}{k} =\frac{0.24}{\sqrt{2}}-0.08 \approx0.16970563-0.08 =0.08970563.

Rounded only at the final step to four decimal places as a percentage, this is $8.9706\%$ per year. For D,

\frac{\dot{k}}{k} =\frac{0.16}{\sqrt{2}}-0.08 \approx0.11313708-0.08 =0.03313708,

or $3.3137\%$ per year to four decimal places as a percentage. Both economies are below their own steady states, but C is farther below its steady state in proportional terms: $2/9$ versus $2/4=1/2$. C also has the higher saving rate, which affects both its steady state and its rate at the given $k$. The check is that each computed growth rate is positive, as required because $2<9$ and $2<4$. Comparing starting capital alone would miss the fact that the economies have different fundamentals.

Checking the Break-Even Rate Carefully

When population, efficiency, and capital all change, the components of the break-even rate must be added with their given values. The following calculation illustrates why a small error in $b$ can change both the steady state and the growth rate.

Worked Example: Verifying the Steady State and Growth Rate

Two hypothetical economies share $f(k)=\sqrt{k}$ and $s=0.25$. Economy C has $n=0.02$, $g=0.03$, and $\delta=0.05$ per year. Economy D has $n=0.03$, $g=0.04$, and $\delta=0.05$ per year. For each economy, calculate the steady state and the proportional growth rate at $k=4$. Use the continuous-time model.

For C, the break-even rate is $b_C=0.02+0.03+0.05=0.10$. For D, it is $b_D=0.03+0.04+0.05=0.12$. Since the production function is $f(k)=\sqrt{k}$, the positive steady state is $k^*=(s/b)^2$. Therefore,

k_C^*=\left(\frac{0.25}{0.10}\right)^2=2.5^2=6.25, \qquad k_D^*=\left(\frac{0.25}{0.12}\right)^2 =\left(\frac{25}{12}\right)^2 =\frac{625}{144}\approx4.3403.

At $k=4$, $\sqrt{k}=2$. For C, the proportional growth rate is $0.25/2-0.10=0.125-0.10=0.025$, or $2.500\%$ per year. For D, it is $0.25/2-0.12=0.125-0.12=0.005$, or $0.500\%$ per year. Both rates are positive because $4<6.25$ and $4<4.3403$.

As a steady-state check for D, $s\sqrt{k_D^*}=0.25(25/12)=25/48$, while $b_D k_D^*=0.12(625/144)=25/48$; investment equals break-even investment. The common mistake is to add D’s rates incorrectly and use $b_D=0.125$. The correct sum is $0.03+0.04+0.05=0.12$, so neither the steady state nor the growth rate should be calculated using $0.125$.

What the Result Does and Does Not Say

The model’s convergence mechanism is about capital per effective worker. When that measure is below its steady state, investment exceeds break-even investment and $k$ rises. Because diminishing returns make output per unit of capital higher at lower $k$, the proportional rate of capital deepening is higher there, all else equal. As $k$ rises, the average product of capital falls and the net proportional growth rate declines toward zero.

This statement does not mean that every low-capital economy must grow faster than every high-capital economy. Their saving rates, technology, population growth, and depreciation may differ, and those differences affect both the steady state and the growth rate at a given $k$. Nor does a decline in capital per effective worker above its steady state necessarily imply a fall in output per worker: labor-augmenting technology can continue to improve. Keep the unit of analysis explicit when interpreting a growth rate.

Common Exam Traps

Confusing a level with a growth rate. A larger $k$ is a larger capital stock per effective worker, not necessarily a higher proportional growth rate.
Using $\dot{k}$ when asked for proportional growth. Divide net investment by $k$ to calculate $\dot{k}/k$.
Forgetting one break-even component. In this model $b=n+g+\delta$, so omit none of population growth, technology growth, or depreciation.
Assuming common destinations. Different saving rates or other fundamentals can give economies different steady states.
Overstating the prediction. Diminishing returns imply a higher growth rate at lower $k$ when fundamentals are held fixed; a comparison across economies with different fundamentals needs more information.

Key takeaway: With diminishing returns and fixed fundamentals, the average product of capital falls as capital per effective worker rises. Consequently, an economy below its own Solow steady state has positive capital deepening, and the farther below that steady state it is, the higher its proportional growth rate under the same model parameters. Its steady state is determined by its own saving, population growth, technology growth, depreciation, and production technology.

Check Your Understanding

For each question, use the continuous-time Solow model. Rates are annual, and any requested percentage is rounded to four decimal places at the final step.

A hypothetical economy has $f(k)=1.5\sqrt{k}$, saving rate $s=0.20$, and $n=0.01$, $g=0.02$, $\delta=0.07$. Find $k^*$ and the proportional growth rate at $k=4$. (Answer: $b=0.10$, $k^*=9$, growth $=5.0000\%$ per year.)
For $f(k)=\sqrt{k}$, let $s=0.24$, $n=0.02$, $g=0.03$, and $\delta=0.05$. Find the steady state and the proportional growth rate at $k=1$. (Answer: $b=0.10$, $k^*=5.76$, growth $=14.0000\%$ per year.)
A hypothetical economy has $f(k)=3\sqrt{k}$, $s=0.12$, and $b=0.06$. Find its positive steady state and the proportional growth rate at $k=9$, then state whether $k$ is rising or falling there. (Answer: $k^*=36$, growth $=6.0000\%$ per year, so $k$ is rising.)
Economies C and D both have $f(k)=\sqrt{k}$ and $b=0.10$. Their saving rates are $s_C=0.30$ and $s_D=0.20$. Find both steady states and compare their proportional growth rates at $k=1$. (Answer: $k_C^*=9$, $k_D^*=4$; growth rates are $20.0000\%$ and $10.0000\%$ per year, respectively.)
A hypothetical economy has $f(k)=2\sqrt{k}$, $s=0.25$, $n=0.01$, $g=0.04$, and $\delta=0.05$. Calculate $b$, $k^*$, and the proportional growth rate at $k=36$; interpret the sign. (Answer: $b=0.10$, $k^*=25$, growth $=-1.6667\%$ per year, so capital per effective worker is falling.)

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