# What is the distribution of a chi-square minimizing function?

Suppose I have a set of $N$ experimental points of the form

$$\{x_i, y_i, d_i\},$$

where $i=1,...,N,$ and $d_i$ are errorbars for $y_i$. To fit the data, I minimize the reduced chi-square

$$\chi^2(p) = \sum_{i=1}^N \frac{[y_i - f(x_i,p)]^2}{d_i^2},$$

where $f(x,p)$ is a (generally non-linear) function parametrized by some parameter $p$ (there might be more than one parameter, but it doesn't really matter).

My question is: given the optimal parameter $p_0$, i.e. $\chi^2(p)$ is minimal at $p=p_0$, and assuming the $y_i$'s are independent and are Normally distributed, what can be said about the distribution of $f(x, p0)$?

## 1 Answer

Another term for your fitting procedure would be weighted non-linear least squares. The weights are a very minor complication. Fitting non-linear least squares is more tricky than ordinary least squares, but once the fitting is done the asymptotic ($N \to \infty$) distribution of the estimates is given by the same large-sample theory.

• Thanks for the links! From what my understanding is worth though, these links discuss the distribution of the fit parameters (in my clumsy notation, p), not the distribution of the fit function itself at a fixed value of the argument. Or do I miss something simple? – ev-br Nov 4 '10 at 15:22
• You're right, I didn't read your question properly. Once you've got an estimate and standard error for p from asympototic theory, however, an estimate and standard error for any function of p follows straightforwardly using the delta method. en.wikipedia.org/wiki/Delta_method – onestop Nov 4 '10 at 18:43
• Let me make sure I got it right. So, basically, once I have the best-fit p0 and its variance dp, I take f(x,p0+dp), Taylor expand it, and identify the deviation from f(x,p0) as a variance of f(x,p0) itself, is this right? – ev-br Nov 4 '10 at 19:40
• I think you're right though I'm having trouble understanding your notation - dp for variance looks odd and adding a variance to an estimate is wrong on purely dimensional grounds. – onestop Nov 5 '10 at 7:20
• Indeed, it's not variation, is it standard deviation? [apologies, I'm not accustomed to this terminology]. I mean, if p is gaussian, then the probability of having p within [p0-dp,p0+dp] is 66%. Does it make it right then? – ev-br Nov 5 '10 at 10:07