On Fri, 10 Jun 2005 10:07:06 -0700 Eric Frazier <[EMAIL PROTECTED]> wrote: > > At 09:56 AM 6/10/2005, Fred Baker wrote: > > >you saw marshall's comment. If you're interested in a moving average, he's > >pretty close. > > > >If I understood your question, though, you simply wanted to quantify the > >jitter in a set of samples. I should think there are two obvious > >definitions there. > > > >A statistician would look, I should think, at the variance of the set. > >Reaching for my CRC book of standard math formulae and tables, it defines > >the variance as the square of the standard deviation of the set, which is > >to say > > That is one thing I have never understood, if you can pretty much just look > at a standard dev and see it is high, and yeah that means your numbers are > flopping all over the place, then what good is the square of it? Does it > just make graphing better in some way? >
Hello Eric; <statistics details> Because (under some broad assumptions, primarily that the underlying process is stationairy) estimates of the variance are distributed as a CHI**2 distribution. More exactly, summation( (x[i] - mean(x))^2) / true_variance is distributed as CHI**2(N), which means that as i increases, then summation( (x[i] - mean(x))^2) / true_variance is distributed as a normal distribution with a mean of N and a variance (of the variance estimate) of 2N, so that V = summation( (x[i] - mean(x))^2) / N is an efficient estimate of the true variance, with a sigma (of the variance estimate) of sqrt (2 / N) * V (Since you have to estimate the mean from the same data, you can show that the estimator is less biased if you use 1 / N - 1 rather than 1 / N in actual calculations.) Basically, if you want to perform the true rites in the Church of Linear Statistics, you worry about variances and CHI**2 distributions. If (like most of us in the real world) you are dealing with non-stationary processes and unknown distributions, you can ignore this, just calculate the standard deviation, see whether or not things differ by more than 3 standard deviations, and be done with it. </statistics details> Note (from the days when spacecraft had 4 kilobytes of memory) that if you estimate s[i] = s[i-1] + x[i] v[i] = v[i-1] + x[i]^2 then the mean estimate at any time is m[i] = s[i] / i the total variance is V[i] = v[i] / i and the standard deviation for i > 1 is sigma[i] = sqrt[(v[i] - (i * m[i]^2))/(i-1)] so, you can do this on the fly without storing all of the data. Regards Marshall > Thanks, > > Eric >