3.2 Constant growth curves
Most economic aggregates do not grow by a constant amount each period; they grow by a roughly constant proportion. A series that grows at the rate \(g\) per period satisfies \(Y_t = Y_0(1 + g)^t\). Taking logarithms, \begin {equation} \label {eq:growth} \log Y_t = \log Y_0 + t\log (1 + g) = \alpha + \beta t , \end {equation} a linear trend in \(\log Y\). Adding a disturbance gives the constant growth curve \(\log Y_t = \alpha + \beta t + u_t\), estimated by regressing \(\log Y\) on \(t\).
Theorem 3.2.1. In the constant growth curve,
- (i)
- \(\beta = \log (1 + g)\), so the estimated growth rate per period is \(\hat g = e^b - 1\);
- (ii)
- \(\beta \) itself is the instantaneous (continuously compounded) growth rate: \(\dfrac {1}{Y}\dfrac {dY}{dt} = \beta \) for the deterministic part \(Y = e^{\alpha + \beta t}\);
- (iii)
- the growth rate over a sub-period of length \(1/m\) (a month, if periods are years and \(m = 12\)) is \((1 + g)^{1/m} - 1 = e^{\beta /m} - 1\).
Proof. (i) is (3.1). (ii) \(\frac {d}{dt}e^{\alpha + \beta t} = \beta e^{\alpha + \beta t}\). (iii) Compounding the sub-period rate \(m\) times must give one period’s growth: \((1 + g_m)^m = 1 + g = e^{\beta }\). □
For small \(g\), \(\log (1 + g) \approx g\) and the slope can be read directly as a growth rate. For the large rates of new technologies the difference is substantial, as the next example shows.
Example 3.2.2. The number of mobile telephone subscribers in Zambia, in thousands, is given for 2010–2014:
| Year | 2010 | 2011 | 2012 | 2013 | 2014 |
| \(Y\) (thousands) | 38.1 | 80.0 | 170.4 | 354.5 | 744.4 |
| \(T\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
| \(\log Y\) | 3.6402 | 4.3820 | 5.1381 | 5.8707 | 6.6126 |
Estimate the annual growth rate by fitting a constant growth curve.
Solution. By Theorem 3.1.1 with \(\sum T_i^2 = 10\), \begin {align*} b &= \frac {\sum T_i\log Y_i}{\sum T_i^2} = \frac {-2(3.6402) - 4.3820 + 5.8707 + 2(6.6126)}{10} = \frac {7.4334}{10} = 0.7433,\\ a &= \overline {\log Y} = 5.1287 . \end {align*}
The estimated annual growth rate is \(\hat g = e^{0.7433} - 1 = 1.103\): subscribers grew by about \(110\%\) a year, more than doubling each year. Reading the slope directly as \(74\%\) a year would be a serious error, since the approximation \(\log (1 + g) \approx g\) is good only for small \(g\). The instantaneous rate is \(74.3\%\) a year.
The fit is almost perfect: the residuals of \(\log Y\) are all below \(0.01\) in absolute value, and \(R^2 = 0.99998\). By contrast, a straight-line trend in \(Y\) itself has \(R^2 = 0.858\) and residuals of \(98.0, -28.8, -107.1, -91.7, 129.5\): positive at both ends and negative in the middle, the signature of fitting a straight line to a curve.
The example carries a warning. Extrapolating the fitted curve six more years gives \(744.4 \times 2.103^6 \approx 64\) million subscribers, about three times the population of Zambia. Growth at a constant rate is a description of one phase of a process, not a law. A technology that is spreading through a population must eventually saturate, and the logistic curve of Section 3.4 is the standard model for that.
Back to levels: the retransformation problem
The growth curve is estimated in logarithms, but forecasts are usually wanted in levels. The obvious forecast \(e^{a + bt}\) is not an estimate of \(E(Y_t)\).
Theorem 3.2.3. If \(\log Y = \alpha + \beta t + u\) with \(u \sim N(0, \sigma ^2)\), then \[ \operatorname {median}(Y) = e^{\alpha + \beta t}, \qquad E(Y) = e^{\alpha + \beta t + \sigma ^2/2} . \]
Proof. \(Y = e^{\alpha + \beta t}e^{u}\). The exponential is increasing, so the median of \(e^u\) is \(e^0 = 1\). The moment generating function of \(N(0, \sigma ^2)\) is \(E(e^{su}) = e^{s^2\sigma ^2/2}\), and \(s = 1\) gives \(E(e^u) = e^{\sigma ^2/2}\). □
So \(e^{a + bt}\) estimates the median, and it underestimates the mean by the factor \(e^{\sigma ^2/2}\). The usual correction multiplies the forecast by \(e^{s^2/2}\). When \(s\) is small, as in Example 3.2.2, the correction is negligible. When \(\log Y\) is noisy it is not: with \(s = 0.5\) the factor is \(e^{0.125} = 1.13\), a \(13\%\) difference.
The precision of \(\hat g\)
The growth rate \(\hat g = e^b - 1\) is a non-linear function of \(b\), so its standard error is not simply \(\operatorname {se}(b)\). The delta method (Theorem 4.3.4) gives the approximation \[ \operatorname {se}(\hat g) \approx \left |\frac {d\hat g}{db}\right |\operatorname {se}(b) = e^b\operatorname {se}(b) . \] In Example 3.2.2, \(\operatorname {se}(b) = 0.00194\), so \(\operatorname {se}(\hat g) \approx 2.103 \times 0.00194 = 0.0041\): the growth rate is \(110.3\%\) with a standard error of \(0.4\) percentage points.
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