TITLE: Basic exponential regression
QUESTION [5 upvotes]: Background:
I'm attempting to learn about basic statistics for the infrastructure asset management industry:

Basic math explanation (related to estimating linear regression with no intercept)


Problem:
I would like to learn how to generate an exponential regression equation for road condition data, just like it was done for me here:

Development of a Flexible Framework for Deterioration Modelling in Infrastructure Asset Management
Hint: Skip to page 53 (the printed page number, not the PDF page number)

In other words, I want to learn how to generate an exponential regression equation, so I can eventually update the coefficient in the existing model (on the full/real dataset).

My soloution (mock data):
I've mocked up a sample dataset here:
+--------------+---------------+
|    X (AGE)   | Y (CONDITION) |
+--------------+---------------+
|       0      |       20      |
|       1      |       20      |
|       2      |       20      |
|       3      |       20      |
|       4      |       20      |
|       5      |       20      |
|       6      |       18      |
|       7      |       18      |
|       8      |       18      |
|       9      |       18      |
|       10     |       16      |
|       11     |       16      |
|       12     |       14      |
|       13     |       14      |
|       14     |       12      |
|       15     |       12      |
|       16     |       10      |
|       17     |        8      |
|       18     |        6      |
|       19     |        4      |
|       20     |        2      |
+--------------+---------------+


Steps in Excel:
Column C: Convert Y to be more linear using the natural logarithm function
Column D: Calculate a straight line that best fits the data, and then return an array that describes the line (using the LINEST function).
Column E: Generate a trend-line on D, and use the coefficient from that trend-line to create an exponential regression equation: =21-exp(0.14723*x)



Question:
How successful was I? Was my approach mathematically correct?

Related:

Normal equation in Excel (statistics)
Tune an exponential regression estimate using calculus
Options for tuning exponential regression?
Mimic Excel Solver nonlinear regression? (to reduce ESS of exponential regression)
Excel: data table (hit and trial)

REPLY [8 votes]: You want to fit the model
$$y=21-e^{a x}\tag 1$$
For sure, you can have an extimate writing
$$21-y=e^{a x}\implies \log(21-y)=ax\implies z=a x\tag 2$$ and a preliminary linear regression gives $a=0.147233$ (just as you did).
In fact, you do not need to use regression since you can get $a$ directly from the normal equation
$$a=\frac{\sum_{i=1}^n x_iz_i } { \sum_{i=1}^n x_i^2 }=\frac{\sum_{i=1}^n x_i \log(21-y_i)} { \sum_{i=1}^n x_i^2 }$$
But this is only the preliminary step since what is measured is $y$ and not $\log(21-y)$. So, you need to continue with a nonlinear regression using this estimate. This would lead to $a=0.149140$.
Let us compare the results for $y$ using both models
$$\left(
\begin{array}{cccc}
 x & y & (2) & (1) \\
 0 & 20 & 20.0000 & 20.0000 \\
 1 & 20 & 19.8414 & 19.8392 \\
 2 & 20 & 19.6576 & 19.6525 \\
 3 & 20 & 19.4447 & 19.4357 \\
 4 & 20 & 19.1979 & 19.1841 \\
 5 & 20 & 18.9121 & 18.8921 \\
 6 & 18 & 18.5809 & 18.5531 \\
 7 & 18 & 18.1972 & 18.1595 \\
 8 & 18 & 17.7526 & 17.7027 \\
 9 & 18 & 17.2374 & 17.1723 \\
 10 & 16 & 16.6406 & 16.5567 \\
 11 & 16 & 15.9491 & 15.8421 \\
 12 & 14 & 15.1479 & 15.0125 \\
 13 & 14 & 14.2196 & 14.0495 \\
 14 & 12 & 13.1441 & 12.9316 \\
 15 & 12 & 11.8980 & 11.6339 \\
 16 & 10 & 10.4542 & 10.1275 \\
 17 & 8 & 8.78136 & 8.37881 \\
 18 & 6 & 6.84319 & 6.34887 \\
 19 & 4 & 4.59758 & 3.99245 \\
 20 & 2 & 1.99576 & 1.25704
\end{array}
\right)$$
Using model $(2)$ and back to the $y$'s, the sum of squares is $8.28$ while using model $(1)$ lead to a sum of squares equal to $6.66$ which is quite better.
Moreover, it is interesting to look at the statistics.
For model $(2)$, we have
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 0.147233 & 0.005034 & \{0.136698,0.157769\} \\
\end{array}$$ while for model $(1)$
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 0.149140 & 0.000873 & \{0.147312,0.150967\} \\
\end{array}$$ showing that, using the "true" model, the standard error is basically divided by a factor of almost $6$.
If you do not want to use nonlinear regression, you could use Excel to solve for $a$ the equation
$$f(a)=\sum_{i=1}^n e^{ax_i}\left(21-e^{ax_i}-y_i \right)=0$$ strating from the preliminary guess. Even graphing the function could be sufficient.
For solving the equation, you could also use Newton method
$$f'(a)=a\sum_{i=1}^n e^{ax_i}\left(21-2e^{ax_i}-y_i \right)$$ and use
$$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}$$ using for $a_0$ the value obtained from the preliminary step.
For your problem, Newton iterates would be
$$\left(
\begin{array}{cc}
 n & a_n \\
 0 & 0.1472330000 \\
 1 & 0.1492437955 \\
 2 & 0.1491401458 \\
 3 & 0.1491398530
\end{array}
\right)$$
Edit
If we consider the data set outside its specific context, we could have obtained a better fit using
$$y=a-b\, e^{cx}\tag 3$$ which leads to a sum of squares equal to $4.97$ with the following parameters
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 22.1098 & 0.5276 & \{20.9968,23.2229\} \\
 b & 1.57255 & 0.3101 & \{0.91830,2.22680\} \\
 c & 0.12823 & 0.0092 & \{0.10875,0.14771\} \\
\end{array}$$ leading to the following results
$$\left(
\begin{array}{ccc}
 x & y & (3) \\
 0 & 20 & 20.5373 \\
 1 & 20 & 20.3221 \\
 2 & 20 & 20.0775 \\
 3 & 20 & 19.7995 \\
 4 & 20 & 19.4834 \\
 5 & 20 & 19.1241 \\
 6 & 18 & 18.7156 \\
 7 & 18 & 18.2513 \\
 8 & 18 & 17.7234 \\
 9 & 18 & 17.1233 \\
 10 & 16 & 16.4410 \\
 11 & 16 & 15.6655 \\
 12 & 14 & 14.7838 \\
 13 & 14 & 13.7816 \\
 14 & 12 & 12.6422 \\
 15 & 12 & 11.3469 \\
 16 & 10 & 9.87440 \\
 17 & 8 & 8.20046 \\
 18 & 6 & 6.29750 \\
 19 & 4 & 4.13420 \\
 20 & 2 & 1.67494
\end{array}
\right)$$