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Re: Calculating Jitter

  • From: Marshall Eubanks
  • Date: Fri Jun 10 15:05:16 2005

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

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.


> Thanks,
> Eric