Psychometric curve fitting using Levenberg–Marquardt algorithm

i'm trying to make (and understand) a psychometric curve fitting (that is used in a scientific paper) using a cumulative gaussian function between two vectors R_objective_distances and X_reel_distances :

 $g(m,n,RObjectiveDistances)=\frac{1}{\sqrt{2*\pi}}\int_{m+nR}^{+\propto 
}e^{-t^2/2} dt$

to do this, i used the toolbox Curve Fit of Matalab, and specified the library model 'Gaussians'. However, this Gaussian model is not as the one i want to perform. How can i use the model i want in Matlab? (i'm not familiarized with the Curve Fitting)

[Walter Roberson]

The equation is missing a } in the frac definition.

[Anass]

ok, sorry, I edited the equation

[Walter Roberson]

[Walter Roberson]

Could you check whether that should be an infinity instead of a propto ?

RObjectiveDistances appears on the left, but the right instead has R; does that R on the right have anything to do with RObjectiveDistances ?

[Anass]

no it's like this (see the image file i attached)

[Walter Roberson]

If the propto is in fact infinity then the right hand side works out as

1/(Pi*(1 - erf(sqrt(1/2)*(m+n*R))))

[Anass]

they are the same. I wanted to explicit it. RObjectiveDistances is the Same as R.

[Walter Roberson]

It looks to me that that is an infinity that is partly cut off.

What is the original paper being used?

[Anass]

this is the original paper that use this equation : http://www.gipsa-lab.grenoble-inp.fr/~kai.wang/papers/CG12.pdf (see section 4.1-Test and Comparisons)

(thanks for helping me)

Answer 1

[Walter Roberson]

If you zoom in to the equation the propto is an infinity.

The right hand side resolves to

1/2 * erfc(sqrt(1/2)*(m+n*R))

I think it unlikely that the Gaussian model happens to match exactly that, but you could try creating a custom model.

But you are looking for a curve fitting, presumably with parameters m, n, and R, and you have the difficulty that any combination of m+n*R that come out the same would match. But if you let R be a known parameter then you can solve for n in terms of m. If you let E0 be such that erfc(E0) = 0, then E0 = 9.150795341104318, and then

n = (sqrt(2)*E0-m)/R

Possibly you have a number of values to fit.

[Anass]

I tried to understand, but it's a little bit difficult for me. In the paper, they said that m and n are approximated with a least-square. I have read the documentation about the Gaussian model in the fitting tool of Matlab, and i saw how they approximate the 2 parameter of the model (in our case m and n). But here, i didn't understand how you can approximate them. You are no using a least square ?