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I have to do reproduce a power regression but I don't have any experience in procedures like that.

I read a little bit about power fit/power regression and that a formula like $y = ax^b$ is used for this. But in the formula which was used in this calculation, there is a constant $C$ included.

A formula $y = A \times (r/t)^B + C$ is given, where $r$ are the value in the following table and $t$ is $-51$.

The input data are:

y       r
0.25    -41
0.5     -43
1       -49
2       -65
3       -58
4       -57
5       -67
6       -67
7       -77
8       -70
9       -69
10      -75
12      -72
14      -72
16      -78
18      -83
20      -81
25      -81
30      -75
40      -83

Now, the coefficients or constants $A, B, C$ was calculated with power regression and the result is:

Intercept: 0.1820634  
Multiplier: 0.8229884
Power: 6.6525179

Unfortunately, there is no hint how they result in this numbers.

I tried to do the power regression in excel but it also gives me only a formula $y = ax^b$.

Can anyone give some hints how to reproduce this?

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2 Answers 2

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Let us consider the model $$y=Ax^B+C$$ for which you want the best estimates of parameters $A,B,C$ in the least square sense based on $n$ data points$(x_i,y_i)$.

The model is nonlinear so, at a point, nonlinear regression will be required; but this implies to have good (or at least reasonable) starting values.

If there was no $C$, a logarithmic transform $$\log(y)=\alpha+B \log(x)$$ will allow to get $A=e^\alpha$ and $B$.

Here, the problem is more delicate but you can notice that, if $B$ is given a value, then parameters $A$ and $C$ are easily obtained from a linear regression $$y=A z +C$$ with $z_i=x_i^B$.

So, define you sum of squares as a function of $B$ $$SSQ(B)=\sum_{i=1}^n(A x_i^B+C-y_i)^2$$ and run the linear regression for different values of $B$ until you locate a minimum. For this temporaly "best" value of $B$, you know the corresponding $A$ and $C$; from here, you can start the nonlinear regression work.

Let us do it using your data. For the preliminary search step, the results are $$\left( \begin{array}{cc} B & SSQ \\ 1 & 955.644 \\ 2 & 851.165 \\ 3 & 770.371 \\ 4 & 714.426 \\ 5 & 680.846 \\ 6 & 665.380 \\ 7 & 663.513 \\ 8 & 671.277 \\ 9 & 685.527 \end{array} \right)$$

which shows a rather flat minimum around $B=7$; for this value of $B$, we have from the linear regression $A=0.902384$, $C=0.55644$.

Now, we can start the nonlinear regression using the above numbers as estimates. We end with $$y=1.07393 x^{6.65252}+0.182063$$ to which corresponds $SSQ=662.884$ which is effectively lower than any value in the above table. You should notice that my results do not coincide with your for the multiplier.

Looking at the confidence interval is very interesting $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ A & 1.07393 & 1.48505 & \{-2.07424,4.2221\} \\ B & 6.65252 & 2.73792 & \{0.84839,12.4567\} \\ C & 0.18206 & 3.87194 & \{-8.02609,8.39021\} \\ \end{array}$$

The above table shows really large standard errors on the parameters and huge confidence intervals.

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  • $\begingroup$ Very useful work as usual. Cheers. $\endgroup$
    – JJacquelin
    Feb 11, 2016 at 23:15
  • $\begingroup$ At calculating SSQ(B), I didn't understand how did you choose A and C? Obviously you are giving different values to B but I am not sure how SSQ(B) output is provided when A and C are still unknowns. $\endgroup$
    – revo
    Dec 8, 2023 at 16:34
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Before answering to the question I would like to make a prelimirary comment. The significance of the regression depends of several factors among them the scatter of the experimental data, the number of adjustable parameters of the model and others are important.

In the present example of data, the scatter appears rather large. Considering the data represented in logarithmic coordinates, at first sight a linear regression seems already not bad, as shown on the next figure :

enter image description here

The question is what are the avantage and the drawback to add a adjustable parameter $C$, that is to go from $\quad y=AX^B\quad$ to $\quad y=AX^B+C\quad$ ?

Certainely, if the scatter is not too large, one can expect a better fitting. But if the scatter is large, the improvement of fitting will be not evident : Almost equivalent results will be obtained with a lot of different triplets $(A,B,C)$ which will make doudbtful the significance of the previous parameters $(A,B)$.

This possible drawback must be taken into account. The previous valuable work by Claude Leibovici shows that the confidence intervals on the estimates are large and even more concerning the parameter $B$. This casts a shadow on the ability to obtain significant results.

Nevertheless, even if the available data is not favourable for showing a particular method of regression, the process is presented in full details on the joint page.

The theory can be found in the paper publish on Scribd : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales . Page 17 : The practical application to the three parameters power function. The notations are not the same as in the present question. This can be confusing. In order to make it consistent with the notations used here, the notations used in the published paper were changed to write the page below.

The advantage of this method is that it is direct (not iterative) and doesn't require guessed values to start. Of course, if it was really necessary, the estimates obtained could be used as initial values to more advanced methods of non-linear regression.

Note that the order of the data has been changed because it is necessary that the values of $X_k$ be ranked in increassing order.

enter image description here

The curve with the estimated parameters is drawn in blue.

As expected, the values $(A,B,C)$ are far from whose already given by Claude Leibovici, but with almost the same result : the coresponding green curve on the figure is quite the same as the blue curve. A difference is only clearly visible at low values of $X$.

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  • $\begingroup$ Interesting problem, isn't it ? I fully agree that, in this case, we could discard the $C$ parameter. But, admit that we add 10 to each of the $y$'s $\endgroup$ Feb 12, 2016 at 5:48
  • $\begingroup$ Of course, if 10 is added to each $y$'s , the parameter $C$ must not be discarded. But the main snag remains : the large scatter. With my method of solving, the computed value of $B$ remains the same, any value added to the $y$'s. $\endgroup$
    – JJacquelin
    Feb 12, 2016 at 6:39
  • $\begingroup$ For sure, I totally agree. When I first saw the data, I was sure that $C$ was not significant; ignoring it makes my first step useless and we have estimates for $A$ and $B$ by linear regression; from these, we can start the nonlinear regression without any problem. What I was commenting is that, not using your methodology, the problem is quite difficult if $C$ is significant. Once more, I am impressed by your method ! Cheers. $\endgroup$ Feb 12, 2016 at 6:45
  • $\begingroup$ By the way, I would really appreciate your opinion about this question math.stackexchange.com/questions/99299/… . Am I correct if I say that, in such a problem, there is no dependent or independent variables and that the problem is not to fit a line in a three dimension space ? $\endgroup$ Feb 12, 2016 at 6:49
  • $\begingroup$ hi Claude ! I am not sure to well understand the question. Nevertheless I gave an answer on :math.stackexchange.com/questions/99299/… $\endgroup$
    – JJacquelin
    Feb 14, 2016 at 9:46

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