# Why are my fitted coefficients so well-determined? [closed]

I have a dataset of $$N$$ points each with some different value $$y$$ I try to fit the data to the form $$a\cos(2\pi t)+b\sin(2\pi t)+c\cos(4\pi t)+d\sin(4\pi t)+e$$. When I'm looking at the standard errors in the fitted parameters for $$a,b,c,d,e$$, they are really, really small. For $$a=-11$$, the standard error in $$a$$ is only around $$0.03$$. This well-determinedness is kind of unsettling and I doubt its correctness.

Should note that my $$N$$ is fairly large, around 400, and I'm using Mathematica to find standard errors in the fitted parameters (i.e. [ParameterTable]).

How can one compute the errors in the fitted parameters? Are the standard errors for the parameters it? They seem too small to be true.

• Your question is absolutely opaque. One might speak of fitting a function $y = a\cos(2\pi t)+b\sin(2\pi t)+c\cos(4\pi t)+d\sin(4\pi t)+e$ where your sample includes some observed values of $y$ and of $t,$, but you speak of $a_i,$ $i=1,\ldots,N$ without saying whether those are in the role of $y$ or of $t.$ Your question as now written cannot be understood. – Michael Hardy Nov 2 '19 at 4:19
• Removed the a's. I just want to say each point has some value. The main point of my question i suppose is how to find the errors in the fitted parameters for a NonlinearModelfit of the said function performed on a dataset? – Houndbobsaw Nov 2 '19 at 4:27
• Your question remains incomprehensible. You say "each with some different value $y$", but then nothing called $\text{“}y\text{''}$ appears in what follows. – Michael Hardy Nov 2 '19 at 4:46
• Suppose one has $$y = a\cos(2\pi t) + b \sin(2\pi t)+c\cos(4\pi t)+d\sin(4\pi t)+e.$$ Then data could consist of observed values of $y$ and of $t,$ and one might be able to estimate the coefficients. But if you say you have observed values of $y,$ but not of $t,$ and you write $$a\cos(2\pi t)+b\sin(2\pi t) + c\cos(4\pi t)+d\sin(4\pi t)+e$$ with no variable called $y,$ then at best we can guess what you mean, and guessing is what I am doing here. – Michael Hardy Nov 2 '19 at 4:55
• Could you tell us why you think that the confidence intervals of the fit parameters are too small? – Semoi Nov 2 '19 at 11:13

I don't know how Mathematica calculates the standard errors of a the fit. However, I reckon that the point estimator and its standard deviation is calculated via \begin{align} \hat{{\beta}} &= ({X}^T{X})^{-1} {X}^T {y} \\ %%% \hat{Sd}[\hat{\beta}_j] &=\sqrt{ \hat{\textrm{Cov}}[\hat{{\beta}}]_{jj}} = \hat{\sigma} \cdot \sqrt{\big[({X}^T {X})^{-1}\big]_{jj}} \end{align} where $${X}$$ is the design matrix! [Note, that $$y$$ and $$\beta$$ are vectors in the upper formulae and the hat indicates an estimate.]
• Well, you just use your formula and generate normally distributed random numbers for your $e$. – Semoi Nov 3 '19 at 9:00