# Goodness of fit : how to perform weighted Pearson test (or equivalent)?

I wish to perform goodness of fit. Currently I use NonLinearModelFit :

nlmSimple =
NonlinearModelFit[data, model, {{a, 20000}, {k1, 300}, {b, 20000}}, t, Weights -> 1/dataErr^2, VarianceEstimatorFunction -> (1 &)];


with $$data = \{\{x_1, y_1\},...,\{x_n,y_n\}\}$$ and $$dataErr = \{w_1,...,w_n\}$$. $$w_n$$ is the standard deviation follow by the $$n-th$$ distribution (asssume to be a normal distribution) in which $$y_n$$ is picked up. So each data point is pick up in distribution with different width. This is working fine but I would like to get a Chi Square test (or equivalent, I am "use" to Chi square but I know there are other goodness test, so I am open to any proposal) for the overall goodness of the fit.

I use function from here Performing a chi-square goodness of fit test. I added the degree of Freedom :

pearsonTest[obs_, exp_, dof_] /; Length[obs] == Length[exp] :=
Block[{t}, t = Total[(obs - exp)^2/exp] // N;
{t/(Length[exp] - dof),
SurvivalFunction[ChiSquareDistribution[Length[exp] - dof], t]}];


with $$obs = \{y_1,...,y_n\}$$ and $$exp= \{model[x_1],...,model[x_n]\}$$ But this does not take into account the standard deviation of $$y_n$$ and will output the same $$\chi^2$$ for any set of $$w_n$$. $$model$$ is a function than can be anything from simple exponential to "complicated" function with plenty of parameters.

So my question : does NonLinearMdelFit include some build-in tool for the overall fit's goodness I can use (I used the property of fitted model but this is only for parameters error)? And if no, how to add weighted data in a Pearson test (so this is more a mathematical problem).

• Using $(o-e)^2/e$ doesn't make any sense in a regression model. Maybe you're thinking about the following: en.wikipedia.org/wiki/Reduced_chi-squared_statistic. I suggest asking the question first on CrossValidated (stats.stackexchange.com) and then coming back here for implementation.
– JimB
Jun 18 '19 at 18:39
• OK so I think about it a bit, and you right it doesn't make sense since the underlying assuption is that the variable $o$ is following poisson disitrbution (which is true in my case) with mean = $e$ and variance=$e$ (which in not true). In my case the variance is $w_i^2$ so I should use $(o_i−e_i)^2/w_i^2$ . I will ask on CrossValidated as you suggest (I edit the first post for adding few details about the problem). Jun 19 '19 at 8:51

OK I solve the mathematical part of the problem. Since my data follow poisson normal distribution of mean $$\mu=y_i$$ and $$\sigma = w_i^2$$ I need to use $$\chi^2= \sum_{i=1}^n (f(x_i)- x_i)^2/w_i^2$$ instead of $$\chi^2= \sum_{i=1}^n (f(x_i)- x_i)^2/x_i$$ (assumption $$\sigma = x_i$$).
• That is a good book but definitely a bit dated and doesn't cover generalized linear models or generalized linear mixed models. This answer mentions that your data follows a Poisson distribution but a comment in your original post says that it does not. You might consider looking at Mathematica's GeneralizedLinearModelFit, get a more up-to-date textbook written by a statistician, and wean yourself off of thinking only in terms of $\chi^2$ statistics.
• Ok I will look at GeneralizedLinearModelFit. About the distribution : I edited my posts. At first this is poisson distribution (count data) but this can be approximazie by normal distribution (high count rate) and when I apply systematics correction and errors I assume normal distribution ( error propagation). So $y_i$ follow normal distribution of $w_i$. Jun 20 '19 at 8:46