Exponential Regression Newton’s Method – Advanced

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:

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Proof: The minimum is obtained when the first partial derivatives are 0. Let

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Thus we seek values for α and β such that \frac{\partial h}{\partial \alpha} = 0 and \frac{\partial h}{\partial \beta} = 0; i.e.

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

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

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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: Now

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Thus

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The proof now follows by Property 2 of Newton’s Method.

3 thoughts on “Exponential Regression Newton’s Method – Advanced”

  1. Hi Charles:
    Regarding Question 1, it is standing no more. I thought it was the Fcrit, but instead it is like the p-value, then it does have the value you presented. Sorry for my confusion.

    Question 2 still stands.

    All the best.

    Reply

  2. Hi Charles:
    I find your site very interesting and useful. Thanks for the great effort !!!

    In your “Exponential Regression using Newton’s Method” page, it does not have a “Leave a Reply” at the end, so I am posting here. Sorry for any inconvenience.

    Question 1. In Figure 3. Which is the formula for “Significance F”…is it “=F.INV(0.05, 1, 9)”? If that is the case, Excel gives the result of 0.00416, but you have 1.175E-06.

    Question 2. Which is the rationale to have the formulas in cells Q25 and Q26, for the standard error of a and b (alfa and beta)? I mean, why you take values from the JTJ-1 matrix? Please clarify the reason.

    Thanks in advance for the clarifications.

    Reply

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