Deriving the AK Model’s Balanced Growth Rate

Why Derive the AK Growth Rate?

The AK model links long-run growth directly to capital accumulation. Its central implication is that, under the model’s assumptions, output per person can keep growing at a constant rate without the diminishing returns to capital that limit growth in the basic Solow model. Deriving that rate carefully matters: it depends on saving, productivity, depreciation, and population growth, and it is not the same as the growth rate of aggregate output.

The result is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus. The derivation below uses the standard continuous-time AK setup: aggregate output is proportional to the capital stock, a constant share of output is saved, and the population grows at a constant rate.

Model and Assumptions

Definition: In the continuous-time AK model, aggregate production is \(Y=AK\), where \(A>0\) is constant productivity and \(K\) is aggregate capital. A constant share \(s\), with \(0\leq s\leq1\), of output is invested. Capital depreciates at constant rate \(\delta\geq0\), and labor or population \(L\) grows at constant rate \(n\), so \(\dot{L}/L=n\). Define capital and output per person as \(k=K/L\) and \(y=Y/L\). Assume \(L>0\) and \(k(0)>0\).

The aggregate capital accumulation equation is \(\dot{K}=sY-\delta K\). The assumption \(k(0)>0\) ensures that the proportional growth rate \(\dot{k}/k\) is defined. All rates in this tutorial are continuous-time rates measured per year unless otherwise stated. In particular, a value of \(0.02\) means a rate of \(2\%\) per year.

Deriving Per-Capita Capital Accumulation

Because \(k=K/L\), its change reflects both the change in aggregate capital and the growth of the population sharing that capital. Applying the quotient rule gives \(\dot{k}=\dot{K}/L-k(\dot{L}/L)\). Substituting the model equations produces the per-capita accumulation equation:

\dot{k}=sAk-(\delta+n)k

The term \(n k\) is the dilution of capital per person as the population grows. This is why depreciation alone is not the full deduction from per-capita investment. The production equation per person follows directly from \(Y=AK\): \(y=Ak\).

Start from the aggregate equations.
\(\dot{K}=sY-\delta K\), \(Y=AK\), and \(\dot{L}=nL\).

Differentiate capital per person.
For \(k=K/L\), \(\dot{k}=\dot{K}/L-k(\dot{L}/L)\).

Substitute and collect terms.
Since \(Y/L=Ak\) and \(\dot{L}/L=n\), \(\dot{k}=sAk-\delta k-nk=[sA-(\delta+n)]k\).

Divide by positive capital per person.
When \(k>0\), the proportional growth rate is \(\dot{k}/k=sA-\delta-n\).

Balanced Growth Rate and Its Interpretation

Theorem: Under the stated continuous-time AK assumptions, if \(k(0)>0\), per-capita capital and per-capita output grow at the constant rate \(g=sA-\delta-n\). Their paths are \(k(t)=k(0)e^{gt}\) and \(y(t)=y(0)e^{gt}\). Aggregate capital and aggregate output grow at rate \(n+g=sA-\delta\).

To justify the paths, write the per-capita equation as \(\dot{k}=gk\). Since \(k>0\), divide both sides by \(k\) and integrate from time \(0\) to time \(t\):

\frac{\dot{k}}{k}=g \quad\Longrightarrow\quad \int_{k(0)}^{k(t)}\frac{1}{k}\,dk=\int_0^t g\,du \quad\Longrightarrow\quad \ln\!\left(\frac{k(t)}{k(0)}\right)=gt \quad\Longrightarrow\quad k(t)=k(0)e^{gt}.

Because \(y=Ak\) and \(A\) is constant, \(y(t)=Ak(0)e^{gt}=y(0)e^{gt}\), so output per person has the same growth rate as capital per person. Also \(K=kL\) and \(Y=yL\). Since \(L\) grows at rate \(n\), aggregate capital and output grow at \(g+n=sA-\delta\).

A balanced growth path here means that the relevant quantities grow at constant rates and their ratios, such as \(Y/K=A\), remain constant. It does not mean that per-capita capital is constant. If \(g=0\), per-capita capital and output are constant. If \(g>0\), they grow; if \(g<0\), they decline. In all three cases, for a positive initial \(k\), the model’s exponential path remains positive at every finite time.

If the saving rate is below one, per-capita consumption is \(c=(1-s)y\), so it grows at rate \(g\) as well. At \(s=1\), consumption is zero, and a proportional growth rate for consumption is not defined. The formulas describe the model with constant parameters; if saving, productivity, or population growth varies over time, the growth rate need not remain constant.

Worked Example: Capital and Output Along an AK Path

A hypothetical economy has \(A=0.16\) per year, a saving rate \(s=0.25\), depreciation \(\delta=0.01\) per year, and population growth \(n=0.01\) per year. At time zero, capital per person is \(k(0)=50\). Find the per-capita growth rate and capital and output per person after five years.

Method. First derive the per-capita growth rate from the accumulation parameters, then use the exponential solution. The model has constant parameters and \(k(0)>0\).

Calculate the contribution from saving and productivity: \(sA=(0.25)(0.16)=0.04\). The total per-capita deduction is \(\delta+n=0.01+0.01=0.02\). Therefore,

g=0.04-0.02=0.02, \qquad k(5)=50e^{(0.02)(5)}=50e^{0.10}.

Using \(e^{0.10}=1.105170185\ldots\), the unrounded capital calculation is \(50(1.105170185\ldots)=55.258509\ldots\). Output per person after five years is \(y(5)=Ak(5)=0.16(50e^{0.10})=8e^{0.10}=8.841361\ldots\). Rounded to two decimal places, \(k(5)=55.26\) and \(y(5)=8.84\). As a check, initial output per person is \(y(0)=0.16(50)=8\), and \(8e^{0.10}=8.84\) to two decimals, consistent with output growing at the same rate as capital per person.

Common mistake avoided: The five-year growth factor is \(e^{gt}=e^{0.10}\), not \(1+gt\) as an exact continuous-time formula. The latter is only a first-order approximation.

How Parameter Changes Affect Growth

The balanced growth rate is \(g=sA-\delta-n\). Holding the other parameters fixed, a higher saving rate or higher productivity raises per-capita growth, while higher depreciation or faster population growth lowers it. The effect of a saving-rate change depends on productivity: raising \(s\) by \(\Delta s\) changes \(g\) by \(A\Delta s\), not simply by \(\Delta s\). Likewise, a productivity change \(\Delta A\) changes \(g\) by \(s\Delta A\) if \(s\) is fixed.

For simultaneous changes, calculate the growth rate using the complete new set of parameters, or subtract the old expression from the new one. If both \(s\) and \(A\) change, their interaction matters: the change in \(sA\) is \(A\Delta s+s\Delta A+\Delta s\Delta A\). Depreciation and population growth enter separately as deductions.

Worked Example: Comparing Saving, Depreciation, and Productivity Changes

A hypothetical economy begins with \(s=0.25\), \(A=0.20\) per year, \(\delta=0.02\) per year, and \(n=0.01\) per year. A proposed scenario changes these to \(s=0.30\), \(A=0.22\) per year, \(\delta=0.025\) per year, while \(n\) remains \(0.01\). Find the initial and new per-capita growth rates, and decompose the change using the stated order: saving first, depreciation second, and productivity last.

Method. Use \(g=sA-\delta-n\) at each stage. The order matters for attributing the productivity change because productivity is multiplied by the saving rate.

Initially, \(sA=(0.25)(0.20)=0.050\) and \(\delta+n=0.02+0.01=0.030\), so \(g_0=0.050-0.030=0.020\), or \(2.0\%\) per year.

First raise the saving rate to \(0.30\), keeping the other initial values: \(g_1=(0.30)(0.20)-0.020-0.010=0.060-0.030=0.030\). This step increases growth by \(0.030-0.020=0.010\), or 1.0 percentage point per year.

Next raise depreciation to \(0.025\): \(g_2=(0.30)(0.20)-0.025-0.010=0.060-0.035=0.025\). This reduces growth by \(0.025-0.030=-0.005\), or 0.5 percentage point per year.

Finally raise productivity to \(0.22\): \(g_3=(0.30)(0.22)-0.025-0.010=0.066-0.035=0.031\). This raises growth by \(0.031-0.025=0.006\), or 0.6 percentage point per year. The total change is \(0.010-0.005+0.006=0.011\), so growth rises from \(2.0\%\) to \(3.1\%\), a net increase of 1.1 percentage points per year. Direct substitution into the final parameters gives the same result.

Common mistake avoided: The productivity contribution is \(s\Delta A=(0.30)(0.02)=0.006\) after the saving-rate change. Using the original \(s=0.25\) for that step would not follow the specified sequence.

Worked Example: Finding Saving Needed for a Target Growth Rate

A hypothetical economy has productivity \(A=0.15\) per year, depreciation \(\delta=0.015\) per year, and population growth \(n=0.005\) per year. A policymaker seeks per-capita growth of \(g=0.01\) per year. If initial capital per person is 40, find the required saving rate and the capital and output per person after eight years.

Method. Rearrange the balanced growth equation to solve for saving, check that the implied rate is feasible, and then apply the exponential path.

Starting from \(g=sA-\delta-n\), add depreciation and population growth to both sides and divide by \(A\):

s=\frac{g+\delta+n}{A} =\frac{0.010+0.015+0.005}{0.15} =\frac{0.030}{0.15} =0.20.

The required saving rate is \(20\%\), which is feasible because it lies between zero and one. With \(g=0.01\) and \(t=8\), \(k(8)=40e^{(0.01)(8)}=40e^{0.08}\). Using \(e^{0.08}=1.083287068\ldots\), this gives \(k(8)=43.331483\ldots\). Output per person is \(y(8)=Ak(8)=0.15(40e^{0.08})=6e^{0.08}=6.499722\ldots\). Rounded to two decimal places, capital per person is \(43.33\) and output per person is \(6.50\). Initially, \(y(0)=0.15(40)=6\), and \(6e^{0.08}\) confirms the same 1% annual growth over eight years.

Common mistake avoided: The target is per-capita growth, so population growth must be included when solving for saving. Omitting \(n\) would give \(s=(g+\delta)/A=0.025/0.15\), which is not the saving rate that delivers the stated per-capita target.

Common Growth-Rate and Unit Traps

A frequent error is to report \(sA-\delta\) as the per-capita growth rate. That expression is the growth rate of aggregate capital and output in this model. For example, if \(sA=0.04\), \(\delta=0.01\), and \(n=0.01\), then aggregate output grows at \(0.04-0.01=0.03\), or \(3\%\) per year, while output per person grows at \(0.03-0.01=0.02\), or \(2\%\) per year.

Another trap is mixing a continuous-time rate with a discrete-time approximation. Over \(t\) years, the exact factor is \(e^{gt}\). The expression \(1+gt\) is generally only an approximation, and it can be materially inaccurate for a large rate or a long horizon. Also, parameter rates must use consistent time units: if \(A\), \(\delta\), or \(n\) are given per month while the horizon is in years, convert them before combining rates or calculating \(e^{gt}\).

Finally, positive initial capital is not a cosmetic condition for the growth-rate statement. If \(k(0)=0\), the equation \(\dot{k}=gk\) gives \(k(t)=0\), and \(\dot{k}/k\) is undefined. The formula for \(g\) describes the proportional growth rate along a positive-capital path.

Key takeaway: In the continuous-time AK model, per-capita growth is \(sA-\delta-n\), while aggregate growth is \(sA-\delta\). Derive the per-capita equation before applying the formula, check that rates share the same time unit, and use \(e^{gt}\) for exact growth over time.

Check Your Understanding

For each hypothetical economy, assume the continuous-time AK model, constant parameters, and positive initial capital per person. Use the stated units consistently.

An economy has \(s=0.30\), \(A=0.12\) per year, \(\delta=0.015\) per year, and \(n=0.005\) per year. Find the per-capita growth rate and the aggregate output growth rate. (Answer: Per-capita growth is \(0.016\), or \(1.6\%\) per year; aggregate output growth is \(0.021\), or \(2.1\%\) per year.)
Productivity is \(A=0.20\) per year, depreciation is \(0.03\) per year, and population growth is \(0.01\) per year. What saving rate gives zero per-capita growth? (Answer: \(s=(0+0.03+0.01)/0.20=0.20\), or \(20\%\).)
Capital per person is initially 25, and parameters imply \(g=-0.01\) per year. Find capital per person after ten years, rounded to two decimals. (Answer: \(25e^{-0.10}=22.6209\ldots\), or \(22.62\).)
An economy has \(s=0.20\), \(A=0.25\) per year, \(\delta=0.02\) per year, and \(n=0.01\) per year. If saving rises to \(0.24\) with all other parameters unchanged, find the growth rate before and after the change and the increase in percentage points. (Answer: Initially \(0.20(0.25)-0.02-0.01=0.02\), or \(2.0\%\); afterward \(0.24(0.25)-0.02-0.01=0.03\), or \(3.0\%\); the increase is 1.0 percentage point per year.)
An economy has \(s=0.40\), \(A=0.10\) per year, \(\delta=0.02\) per year, and \(n=0.015\) per year. Calculate per-capita and aggregate output growth. Then check that the aggregate rate also equals \(sA-\delta\). (Answer: Per-capita growth is \(0.040-0.020-0.015=0.005\), or \(0.5\%\); aggregate growth is \(0.005+0.015=0.020\), or \(2.0\%\), which equals \(0.040-0.020\).)

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