The Exponential Model
The exponential model is a useful tool because in its most typical “first order” form it
traces out the path of a variable that is growing at a constant rate. By generalizing the model, the
assumption of constant growth can be dropped, allowing the model to better fit the path of a
variable that is not growing at a constant rate.
The First Order Exponential Model
The “first order” exponential model generates a path with a constant growth rate. Letting
r represent this constant rate and letting y denote the value of the variable that is growing, the
value of the variable at a point in time t is given by
(1)
y t  y 0 e rt ,
where y 0 is the initial value of the variable at point in time t  0 , and r is the constant rate at
which the variable is growing.
The continuous rate of growth of the variable y is represented by the quantity
1 dy
. The
y dt
quantity yt in equation (1) represents the value of the variable y at time t , so we can also write
yt as y (t ) .
The representation y (t ) emphasizes that we are thinking that the value of the
variable y is a function of the time t . If this function is differentiable, then it is common to
write the derivative as y ' (t )  dy / dt . Since y(t )  yt , it follows that we obtain the derivate
y ' (t ) for equation (1) by taking the derivative of the right side of the equation with respect to t.
Given data on the variable y , the least squares regression technique can be used to
obtain an estimate of the constant growth rate r. This estimate provides a characterization of the
trend followed by the variable, in particular the best fit constant growth rate. Because least
squares is a “linear regression” technique, we must “linearize” the model before we can apply
least squares. This can be accomplished by taking the natural log of both sides of equation (1).
Doing so, we obtain
(2) ln( yt )  ln( y0 )  rt .
Notice that this “logarithmic transformation” moved the variable r from a position of
being an exponent on the number e in equation (1) to being a coefficient on the variable t in
equation (2). In equation (2), the fact that r is an exponent implies that there is a non-linear
relationship between r and yt ; i.e., you would not get a straight line if you plotted the
relationship between yt and r.
However, in equation (2), the fact that r is a coefficient implies
1
that there is a linear relationship between r and ln( yt ); i.e., you would get a straight line if you
plotted the relationship between ln( yt ) and r.
Because of this linear relationship, we can obtain
a least squares estimate for r by regressing ln(yt ) on t.
The Second Order Exponential Model
The “second order” exponential model generates a path where the growth rate of the
variable may change:
(3)
yt  y0 e r1t  r2t
2
Taking the natural log of (3) we have ln( yt )  ln( y0 )  r1t  r2t 2 . Differentiating with respect to
time t , we obtain 1 / yt dyt / dt   r1  2r2t .
Letting gt denote the growth rate of the variable y
, letting a0  r1 , and letting a1  2r2 , we can rewrite the last equation as
(4) g t  a0  a1t .
Equation (4) is a first degree polynomial model. Comparing equations (3) and (4), we find that a
second degree exponential model for the level becomes a first degree polynomial model for the
growth rate.
In general, when differentiated with respect to time, an exponential model of degree k
becomes a polynomial model of degree k-1. In particular, a fourth degree exponential model
would be written
(5)
yt  y0 e r1t r2t
2
 r3t 3  r4t 4
Taking the natural log of (5) and differentiating with respect to time t , we obtain
(6) g t  a0  a1t  a2t 2  a3t 3 ,
where gt is again the growth rate of y , a0  r1 , a1  2r2 , a2  3r3 , and a1  4r3 .
Equation (6) is a relatively general model in that it allows the growth rate to vary in a
flexible manner. Note that, if a1  a2  a3  0 , then the growth rate is constant and equal to
g t  a0 . If a2  a3  0 , but a1  0 , then the growth rate either increases over time (if a1  0 ) or
decreases over time (if a1  0 ).
If a3  0 , but a1  0 and a2  0 , then the growth rate can
increase and then decrease, or vice versa.
Finally, if a1  0 , a2  0 , and a3  0 , then the
2
growth rate can change directions twice, which is about as much flexibility as you would want
for any model.
By estimating the parameters in the exponential model, it is possible to create a model
that fits the level of the variable that is growing, and this model can be used to forecast the level.
To illustrate, consider the following figure, which presents an employment measure called FullTime Equivalent Employees. Clearly, the employment level has increased over time, though it is
also evident that there are some years in which the employment level decreases.
200,000 Thousands
Employed
150,000
U.S. Full Time Equivalent
Employees
1948-2006
100,000
50,000
0
19451950195519601965197019751980198519901995200020052010
To obtain a model that fits this time series, we begin by using the general exponential
model (5). Because it is nonlinear in the parameters, we must first take the natural log of
model (5) to linerize it. Doing so, we obtain,
(7) ln( yt )  ln( y0 )  r1t  r2t 2  r3t 3  r4t 4 .
Because there is a linear relationship between the ln( yt ) and the parameters r1 , r2 , r3 , and r4 ,
estimates for these four parameters and an estimate for the constant ln( y0 ) can be obtained by
regressing the ln( yt ) on the variables t , t 2 , t 3 , and t 4 .
The data plotted in the figure starts in 1948 and runs to the year 2006. The variable t
runs from zero, which is associated with the year 1948, up to 58, which is associated with the last
year 2006. The variables t 2 , t 3 , and t 4 are constructed by squaring, cubing, and quadrupling the
variable t . The variable ln( yt ) is constructed by taking the natural log of the employment
variable.
The following graphic displays the output obtained from Excel from the regression. The
R Square, which ranges between 0 and 1, measures the percentage of of the variance in the
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dependent variable ln( yt ) that is explained by the regression. R 2  .9955 indicates that 99.55
percent of the variance is explained. This is a very high R square, but is typical for this type of
regression where the model is fitting a trend. It is important to understand that, while it does
provide information about how well the model fits the data, the R square should generally not be
used to determine the quality of the model.
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.997743599
R Square
0.995492289
Adjusted R Square
0.995158385
Standard Error
2016.212786
Observations
59
ANOVA
df
Regression
Residual
Total
Intercept
t
t2
t3
t4
4
54
58
SS
MS
F
48478406922 12119601731 2981.368
219516156
4065114
48697923078
Coefficients Standard Error
t Stat
51863.38621
1189.562957 43.59868966
804.5338063
289.3534157 2.780453807
15.01072125
20.54046926
0.73078765
0.278632597
0.534288411 0.521502227
-0.005357015
0.004568738 -1.172537265
P-value
9E-44
0.007457
0.468068
0.604149
0.246128
The coefficient column presents the estimates of the parameters. The intercept, which is
also referred to as the constant, is the estimate of ln( y0 ) in equation (7). The other four
numbers in the column are the estimates of r1 , r2 , r3 , and r4 . The summary output presented in
this graphic is not typically presented. More typically, the estimated equations is presented in
something like the following manner:
(8) ln( yt )  51,863  805t  15.0t 2
(1,190)*** (289)*** (20.5)
 0.28t 3
(0.53)
 0.005t 4 ,
(0.005)
R 2  0.9955
Underneath each estimate, the standard error associated with the estimate is presented in
parenthesis. Some researchers would present the t-stat rather than the standard error. The two
are interchangable in a sense because there is a relationship between them: The t-stat is equal to
the coefficeint divided by the standard error. The t-stat is a test statistic associated with the null
hypothesis that the coefficient is zero. The estimate for the coefficient will (almost surely) never
actually be precisely zero. The question is how close the coefficient is to zero. The standard
error is a measure of the lack of fit associated with the coefficient. The t-stat will be near zero
when the coefficient itself is near zero, or when the standard error is large. When the t-stat is
far from zero, we would tend to want to reject the hypothesis that the coefficient is zero.
We say the coefficient is statistically insignificant if it is close enough to zero that we fail
to reject the null hypothesis that the coefficient is zero. The p-value is the level of significance
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at which we can reject the null hypothesis. For example, p-value of 0.246128 for the coefficient
estimate -0.005 for r4 , the coefficient on t 4 , indicates we can reject the hypothesis that r4  0 at a
25% level of significance. This is the same as saying we can reject the hypothesis with 75
percent confidence. Typically, at least 90 percent confidence is desirable to reject. Thus, in this
case the p-value of 0.246128 indicates we should fail to reject the hypothesis, meaning we
conclude the r4 estimate is insignificant, which is the same as saying it is not significantly
different than zero (even though it actually is -.005).
The commonly used levels of significance are 10 percent, 5 percent, and 1 percent. Note
that the p-value for the intercept estimate is 9E-44. This is a very small number, a decimal point
followed by 43 zeros until you reach a 9. Thus, we are 99.99999…. percent confident that the
intercept is not zero, so we confidently reject the null hypothesis that it is zero. One common
way of indicating the level of confidence at which a null hypothesis can be rejected is by
attaching asterisks to the standard errors as shown in (8). Three asterisks (***) indicates 99
percent confidence, or rejection at a p-value less than or equal to 0.01. Two asterisks (**)
indicates 95 percent confidence, or rejection at a p-value less than or equal to 0.05 but greater
than 0.01. One asterisk (*) indicates 90 percent confidence, or rejection at a p-value less than or
equal to 0.10 but greater than 0.05. Applying this method to (8), the lack of asterisks on the last
three standard errors indicates the estimates for r2 , r3 , and r4 are insignificant, while the three
asterisks for the intercept and for r1 indicate that we are very confident these estimates are not
zero.
Because we find that our estimate for r4 is insignificant, the common practice is to drop
the variable t 4 from the regression and re-estimate. Doing so, we obtain,
(9) ln( yt )  52,643  513t  38.1t 2  0.34t 3 ,
(990)*** (149)*** (6.00)*** (0.68)***
R 2  0.9954
In (9), the asterisks indicate that all of the estimates are significant at the 1 percent level of
signficance, meaning we have at least 99 percent confidence that all of the estimates are
signficantly different than zero. Notice the R square dropped, which will happen when you drop
an explanatory variable. However, the R square typically will not drop much when you drop
and insignificant variable because the variable had close to zero explanatory power. Here,
indeed we see the R square did not drop much. The following figure plots the model (9) along
with the actual data.
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U.S. Full Time Equivalent
Employees
1948-2006
200,000Thousands
Employed
150,000
100,000
50,000
0
19451950195519601965
197019751980
198519901995200020052010
Actual
Model
Because the model (9) is based only upon the time variables, it can be used to forecast.
This is accomplished by extending the time variables and then extending the model. The
following figure does this for 10 additional years, so we obtain a forecast to the year 2016.
From our model, we find that the estimated employment level for 2016 is 155,721,500. One can
see cubic of the polynomial (9) at work. The third order of the polynomial is what allows the
model to first increase at an increasing rate, but than increase at a decreasing rate.
U.S. Full Time Equivalent
Employees
1948-2006
200,000Thousands
Employed
150,000
100,000
50,000
0
1940
1945
1950
1955
1960
1965
1970
1975
1980
1985
1990
1995
2000
2005
2010
2015
2020
Actual
Model
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