Exponential Regression using Newton’s Method

Exponential Regression using Newton’s Method

We now show how to create a nonlinear exponential regression model using Newton’s Method.

Property 1: Given samples {x1, …, xn} and {y1, …, yn} and let ŷ = αeβx, then the value of α and β that minimize \sum{}(yi − ŷi)2 satisfy the following equations:

image9215image9216

Proof: For a proof using calculus, click here

Property 2: Under the same assumptions as Property 1, given initial guesses α0 and β0 for α  and β, let F = [f  g]T where f and g are as in Property 1 and

image9220

Now define the 2 × 1 column vectors Bn and the 2 × 2 matrices Jn  recursively as follows

image9221image9222image9223

Then provided α0 and β0 are sufficiently close to the coefficient values that minimize the sum of the deviations squared, then Bn converges to such coefficient values.

Proof: For a proof using calculus, click here

Property 3: The approximate covariance matrix for the coefficients vector is given by

image9229

whereimage9230

Example 1: We now show how to calculate the value of the α and β coefficients for the exponential regression model for the data in Example 1 of Exponential Regression using a Linear Model or Exponential Regression using Solver (repeated in range A3:B14 of Figure 0), this time using Newton’s Method (i.e. Property 2).

Exponential regression Solver output

Figure 0 – Data for Example 1

The first 5 iterations of Newton’s method are shown in Figure 1. As you can see the coefficients calculated in step 5 (range B31:B32) are the same as those in step 4 (range B28:B29) and so convergence is reached after 5 steps, with values α = 12.50475 and β = .016854.

Exponential regression Newton's Method

Figure 1 – Exponential Regression using Newton’s Method

In Figure 2 we show key formulas used in Figure 1 based on Property 1 and 2 and referencing the input X data in range A4:A14 and Y data in range B4:B14 from Figure 1 of Exponential Regression using Solver.

Item Cells Formula
B0 B19:B20 use values from Excel exponential regression
 F0 D19 =SUMPRODUCT(SUMPRODUCT($B$4:$B$14-B19*EXP(B20*$A$4:$A$14),EXP(B20*$A$4:$A$14)))
D20 =B19*SUMPRODUCT(SUMPRODUCT($B$4:$B$14-B19*EXP(B20*$A$4:$A$14),EXP(B20*$A$4:$A$14),$A$4:$A$14))
 J0 F19 =-SUMPRODUCT(EXP(2*B20*$A$4:$A$14))
G19 =SUMPRODUCT(SUMPRODUCT($B$4:$B$14-2*B19*EXP(B20*$A$4:$A$14),EXP(B20*$A$4:$A$14),$A$4:$A$14))
F20 =G19
G20 =B19*SUMPRODUCT(SUMPRODUCT($B$4:$B$14-2*B19*EXP(B20*$A$4:$A$14),EXP(B20*$A$4:$A$14),$A$4:$A$14^2))
 J0-1 I19:J20 =MINVERSE(F19:G20)
 B1 B22:B23 =B19:B20-MMULT(I19:J20,D19:D20)

Figure 2 – Formulas from Figure 1

We can now create the regression analysis as shown in Figure 3.

Exponential regression results Newton

Figure 3 – Exponential Regression results using Newton’s Method

Key formulas are shown in Figure 4, referencing the cells in Figure 1.

Item Cells Formula
α coefficient P25 =B31
β coefficient P26 =B32
α s.e. Q25 =SQRT(L31*R21)
β s.e. Q26 =SQRT(M32*R21)
SSE Q21 =SUMPRODUCT((B4:B14-B31*EXP(B32*A4:A14))^2)
MSE R21 =Q21/P21

Figure 4 – Formulas from Figure 3

Observation: The following is an alternative approach for finding the regression coefficients α and β. By Property 1

image9231

Thus, if α ≠ 0 we can solve for α to get

image9232

Thus the original two equations in two unknowns can be replaced by the following equation in one unknown:

image9233

This can be solved iteratively using Newton’s Method in one variable, as described in Newton’s Method.

Observation: There is another algorithm that is commonly used to find the regression coefficients called the Levenberg-Marquardt algorithm, which combines the advantages of Newton’s Method with those of the algorithm used by Solver. We won’t consider this algorithm further here.

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