Data

{2.5984*10^-56,1.2266*10^-25,2.77402*10^-19,1.14182*10^-15,3.43647*10^-13,2.46313*10^-11,7.13917*10^-10,1.11045*10^-8,1.09754*10^-7,7.68589*10^-7,4.10723*10^-6,0.0000176576,0.0000634956,0.000196694,0.000537035,0.00131592,0.00293618,0.00603686,0.0115497,0.0207306,0.035151,0.0566368,0.0871538,0.128646,0.18284,0.25105,0.333986,0.43162,0.543096,0.666724,0.800034,0.939904,1.08273,1.22464,1.36169,1.49013,1.60652,1.70795,1.79211,1.85739,1.90287,1.92831,1.93411,1.92122,1.89103,1.84532,1.78606,1.71543,1.6356,1.54874,1.45694,1.36212,1.26602,1.17018,1.07592,0.984329,0.896293,0.812488,0.733406,0.659374,0.590571,0.527055,0.468779,0.415615,0.36737,0.323804,0.284642,0.249587,0.218334,0.190572,0.165997,0.144312,0.125236,0.1085,0.0938559,0.0810734,0.0699408,0.0602653,0.0518725,0.0446053,0.038323,0.0329003,0.0282261,0.0242021,0.0207417,0.0177691,0.0152178,0.0130299,0.0111549,0.0095491,0.00817446,0.00699822,0.00599209,0.00513167,0.00439598,0.003767,0.00322925,0.00276948,0.00237632,0.00204006}

I'm having a little problem with the NonLinearModelFit function. I'm hoping someone can point me in the right direction please. I have this data points: https://www.dropbox.com/s/5hflg2nvko26hpz/question.nb

When I plot the data points I have this:

Then I use a Rice (Rician) Distribution Curve as model for fitting. For this, I used $$\frac{x \exp \left(-\frac{v^2+x^2}{2 \sigma ^2}\right) I_0\left(\frac{x v}{\sigma ^2}\right)}{\sigma ^2}$$

But when I plot both curves together I get something like this figure where the fit is the very tiny one below:

I checked the Residual fit and I got something really disturbing. Showing that my fit has a large error margin:

Honestly, I'm not sure what's wrong. I've spent some time on this and I think I need an outside eye to help look at it. Any idea will be much appreciated. Thanks friends.

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I think you will need an extra parameter (an amplitude) in order to match your data (which does not have an area of 1). –  chuy Jul 18 '13 at 16:38
When I added the 3rd parameter $m$ as @PatoCriollo suggested, it doesn't seem to work well with the data. I've provided the data above maybe you may want to play around with it. The plot of the data point itself [Fig 1.] is smooth so I'm not sure what's going on. Thanks –  Afloz Jul 19 '13 at 3:23
I think you're making some confusion. The plot of the data themselves does not have to match the plot of their distribution function. –  b.gatessucks Jul 19 '13 at 6:22
What do the 20…40 range and the values in the data represent? Is there any particular reason why you decided to use a Rice density? I'm getting reasonably good fits with a shifted gamma density, but if that's not allowed then there's little point in pursuing it. –  Ray Koopman Jul 21 '13 at 1:56
@RayKoopman, apology for late response. I considered Rice a good fit for the original curve however, if it's not too late, I'd love to see your idea on the shifted gamma function. –  Afloz Jul 26 '13 at 8:28

@Chuy makes a valid point in the comments section of the OP. Here's the fit with a "fudge" factor, using the equation for f in the notebook you reference and assigning your data to the variable data.

xfit = NonlinearModelFit[data, fudge x/s^2 Exp[-((x^2 + v^2)/(2 s^2))] BesselI[0, (x v)/
s^2] , {{fudge, 1}, {s, 0.5}, {v, 30}}, x]
Plot[xfit[x], {x, 0, 100}, Epilog -> {Red, PointSize[0.011],
Point@Transpose[{Range[Length[data]], data}]}]
xfit["BestFitParameters"]
(* {fudge -> 49.3597, s -> 10.3848, v -> 43.0833} *)


The result: a better but still not great fit.

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Yes, this is the best I came up with as well using Rician distr. but I had to normalize the data with data/Total[data] before I got this. Would you please explain the fudge parameter means? It appears this is what @Chuy was talking about. –  Afloz Jul 26 '13 at 8:31
Oh by "fudge" you mean something like an "enabler" or a"heuristic" a "weighting factor"? is that what you mean? –  Afloz Jul 26 '13 at 8:49
@Methyl yes. I chose the more 'technical' term :-). There are cases when you may not want to normalize the data and this is one option in such cases. –  bobthechemist Jul 26 '13 at 12:21
Ahaha. That makes sense. I'll call it the "Fudge Fit Factor". Thanks a lot B. You've done more that fit my data ;) –  Afloz Jul 26 '13 at 15:55
sample = RandomVariate[RiceDistribution[2.5, 3], 1000];

Your last line is not going to work; you can use this, straight from the docs : Histogram[data, Automatic, "PDF", Epilog -> First@Plot[PDF[mymodel, x], {x, 0, 2}, PlotStyle -> Red]]. –  b.gatessucks Jul 19 '13 at 6:23