Solved – Fitting an exponential mixture model with interval constraints on the mixture weights

curve fittingexponential distributionmixture-distributionpredictive-modelsregression

What methods are there to fit a model of the form $y=A\mathrm e^{Bx}+C\mathrm e^{Dx}+E$?

Here is the actual scientific data to be fitted: http://dl.dropbox.com/u/39499990/Ben%2C%20real%20data.xlsx

B should be in the range of -1 to -100.

D should be in the range of -100 to -500.

E is a constant.

This specific model is of interest as it is an accepted one in the scientific community for describing the biological proccess in hand- Inactivation of a voltage dependent calcium channel. (for reference see for example: A novel molecular inactivation determinant of voltage-gated CaV1.2 L-type Ca2+ channel. A Livneh, R Cohen, and D Atlas; Neuroscience, Jan 2006; 139(4): 1275-87.
" The rate of inactivation was analyzed by a biexponential decay -A1exp(-t/Tao1)-A2exp(-t/Tao2)+C " )

Best would be a solution that I could implement in Excel, by the use of build-in functions or VBA code.

Best Answer

Best is not a solution in Excel -- spreadsheets are not a good environment for data analysis: http://www.burns-stat.com/pages/Tutor/spreadsheet_addiction.html

The 'nls' function in R would be one choice.

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Best Curve Fitting Function – How to Choose Between Linear, Exponential, and Logarithmic Models

You might want to check out the free software called Eureqa. It has the specific aim of automating the process of finding both the functional form and the parameters of a given functional relationship.
If you are comparing models, with different numbers of parameters, you will generally want to use a measure of fit that penalises models with more parameters. There is a rich literature on which fit measure is most appropriate for model comparison, and issues get more complicated when the models are not nested. I'd be interested to hear what others think is the most suitable model comparison index given your scenario (as a side point, there was recently a discussion on my blog about model comparison indices in the context of comparing models for curve fitting).
From my experience, non-linear regression models are used for reasons beyond pure statistical fit to the given data:
1. Non-linear models make more plausible predictions outside the range of the data
2. Non-linear models require fewer parameters for equivalent fit
3. Non-linear regression models are often applied in domains where there is substantial prior research and theory guiding model selection.

Solved – Exponential curve fitting with a constant

I don’t really get what you mean with AGWN, is this simply that $n(t_i)$ are independent with $n(t_i) \sim N(0,\sigma^2)$?

The least squares estimator (which, as usual with a normal model, coincide with maximum likelihood estimator, see eg this answer) is easy to find with numerical methods, here is a piece of R code:

a <- 2; b <- 0.2; c <- -1
t <- seq(-5,10, length=100)
y <- a*exp(-b*t)+c+rnorm(length(t), sd=2)
f <- function(par, t, y) { 
  a <- par[1]; b <- par[2]; c <- par[3]; 
  return(sum((a*exp(-b*t)+c-y)**2 ) ); }
nlm( function(par) f(par, t, y), c(0,0,0))

The result of the last call is

$minimum
[1] 430.8242

$estimate
[1]  2.0875336  0.1961210 -0.8672079

$gradient
[1]  1.301591e-05 -2.745537e-05  2.603429e-05

$code
[1] 1

$iterations
[1] 19

Our initial values (2, 0.2, -1) are estimated by (2.088, 0.196, -0.867). You can get a plot of the data points, the "true model" (red line) and the estimated model (dotted red line) as follows:

plot(t,y)
lines( t, a*exp(-b*t)+c, col="red")
nlm( function(par) f(par, t, y), c(0,0,0))$estimate -> r
lines( t, r[1]*exp(-r[2]*t)+r[3], col="red", lty=2)

Data points, with in red line the true model, in dotted red line the fitted model

Also, be sure not to miss this related question already quoted by whuber in the comments.

A few hints for numerical optimization

In the above code the numerical optimization is done using nlm, a function of R. Here are a few hints for a more elementary solution.

If $b\ne 0$ is fixed, it is easy to find $a$ and $c$ minimizing $f(a,b,c) = \sum_{i=1}^n (a \exp(-b\cdot t_i) +c -y_i)^2$: this a the classical least squares for the linear model. Namely, letting $x_i = \exp(-b\cdot t_i)$, we find $$ \begin{array}{rcccl} a &=& a(b) &=& { n \sum_i y_i x_i - \left( \sum_i y_i \right)\left(\sum_i x_i \right)\over n \sum x_i^2 - \left(\sum_i x_i \right)^2 }, \\ c &=& c(b) &=& { \sum_i y_i \sum_i x_i^2 - \left( \sum_i y_i x_i \right)\left(\sum_i x_i \right)\over n \sum x_i^2 - \left(\sum_i x_i \right)^2 }. \\ \end{array}$$

Then set $$ g(b) = f(a(b), b, c(b)).$$ The optimization problem is now reduced to find the minimum of $g(b)$. This can be done either using Newton method or a ternary search.

Best Answer

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