# Silly physics...

Andrew Calkin calkina at ece.curtin.edu.au
Wed Feb 12 22:11:00 PST 2003

Hiya,
if you consider the measurement to be a stochastic quantitiy (random distribution), then if X is random, B is constant, then a measurement of error is the
mean squared error E[Y-\hat{Y}]:
MSE(Y)=bias(Y)^{2} + variance(Y)
and bias(Y)=B*bias(X)
variance(Y) =B^{2}*variance(X)
and E[Z] is the expectation operator.

Averaging reduces the variance of the given estimator (given it is an
ergodic random process), and if the estimator used is unbiased, then
MSE(Y)=B^{2}*variance(X). The question is, what is the variance of X?
If your answer _appeared_ to be reasonable, then I would say that probably
var[X] was also small. If you do a series of measurements, look at the
empirical distribution of Y, and calculate the variance of it. Then, you
can get an experimental estimate of the variance of X.

//Andrew

On Wed, Feb 12, 2003 at 11:32:28PM +0000, Ian Molton wrote:
> On Wed, 12 Feb 2003 23:20:20 +0000 (UTC)
> rbpark-NOSPAM at ualberta.ca (Rob Park) wrote:
>
> > > > dz=|k|dx
> > >
> > > bollocks.
> >
> > Oh?
>
> You dont apply error to constants, unless they are only experimentally
> derived, or expressed to a finite number of significant places.
>
> If you have an equation for measuring the area of a circle:
>
> area = PI * r^2
>
> The error is NOT
>
> darea = PI * dr^2
>
> it is:
>
> darea = dPI * dr^2
>
> Or are you suggesting that even if a near perfect measurement is made
> that there is still an error of over 3 times the measurement?
>
> PS. did you notice the error equations break down if any ONE measurement
> is assumed to be perfect? ;-)
> --
> Unsubscribe: send email to listar at linuxfromscratch.org
> and put 'unsubscribe lfs-chat' in the subject header of the message
--
Unsubscribe: send email to listar at linuxfromscratch.org
and put 'unsubscribe lfs-chat' in the subject header of the message