- October 24th, 2017, 4:53 am
#297694
The estimation for those constants is mostly a regression line or polynomial. What most of the people don't understand is, that if one uses a regression line as an estimator for a constant in physics , the regression line functional has a confidence-interval dependent on the measure point which is dependent on the made measurment. This should be clear if a measurement has a dense point set in some regions and some scattered point in another region not only the slope or the rise of the regression line has a variance but every measure point then has for the corresponding line a functional estimator f(x)=a x_meas +b . But this estimator has a confidence in which is dependent on x_meas so $f(x_meas) \in [m-conf_low, m+conf_high]$ but this is dependent on the point. Some regions are densly measured others small. This is normally not taken into account of the corresponding considerations.
-- Updated October 24th, 2017, 4:55 am to add the following --
So it could be, that from a statistical viewpoint so called constants are for some parameter intervall in such big confidence intervalls, that those number are no constants at all.
-- Updated October 24th, 2017, 5:07 am to add the following --
Just assume a function wich is constant in the middle of it's measured intervall, but then rises on both edges of the intervall. If one takes many points for estimation of a regression line in the middle as measurements, an just two or three pertubed to the outside which still fit the regression model one has an extremely low variance for the regression line although the confidence interval at the edges of the intervall rises in it's extent very fast. Many people would claim that this is a constant . This is even more severe if this problem in non-linearly coupled into other equations.
-- Updated October 24th, 2017, 5:22 am to add the following --
So the socalled constant functional f(x)= 0*x +b is a good estimator for the middle points as a hypothesis but a bad estimator at the border of the intervall although the variance of the slope close to 0 is small (as well as the variance of b).
-- Updated October 24th, 2017, 8:01 am to add the following --
Even for statistical learning, neural networks and bayesian learning algorithms this is important. Another thing would be , that one doesn't use the variance but median as a measurment for calculation of the confidence intervals.