NIST Technical Note 1337: Characterization of Clocks and Oscillators

t ....chi -square. d. f. is the number of degrees ofAfL +.. "* .... freedom (possibly not an integer), and a 2 is the/f 4---~---4----~---~---"'t 1"""true" Allan Variance we're all interested ink- 1-I+- 2_1+- ~~ Jof ~ knowing--but can only estimate imperfectly.I+- ~ BUR: -+I+-~~-+'Chi-square is a random variable and its distri­r----------, bution has been studied extensively. For some:14-~~~ : reason, chi-square is defined so that d. f., the... _--------_ ....number of degrees of freedom. appears explicitlyFIGURE 5.2 in eq (5.1). Still, X 2 is a (implicit)first pair and calculate a sample Allan Variance.function of d. f., also.and we could calculate a second sample Allan The probability density for the chi-squareVariance from the second pair (i.e.• the third anddistribution is given by the relationfourth frequency measurements). The average ofthese two sample Allan Variances provides an d. f.-1 X1 2improved estimate of the "true" Allan Variance, p(x 2 ) = (x 2 T e (5.2)and ~ewould expect it to have a tighter confidenceinterval than in the previous example.This couldbution with TWO degrees of freedom.However, the re is another option. We cou1dalso consider the sample Allan Variance obtainedfrom the second and third frequency measurements.That is the middle sample variance.Now. however,Chi-square distributions are useful in deterwe'rein trouble because clearly this last sampleA11 an Vari ance is NOTtwo.independent of the otherIndeed, it is made up of parts of each ofthe other two.Thi s does NOT mean that we can'tuse it for improving our estimate of the "true"Allan Variance, but it does mean that wecan'tjust assume that the new average of three sampleAllan Variances is distributed as chi-square withcounter chi-square distributions with fractionaldegrees of freedom.And as one might expect. thenumber of degrees of freedom will depend upon theunderlying noise type, that is, white FM. flick.erFM, or whatever.Before going on with this, it is of value toreview somebution. Sample variances (like sample Allantion:2d. f. r(d/.)where r(d;/ ") is the gamma function, defined bybe expressed with the aid of the chi-square distritheintegralOIlt-l ·xr (t) = J0 X e dx (5.3)mining specified confidence intervals for variancesand standard deviations. Here is an example.Suppose we hav~a sample variance S2 = 3.0 and weknow that this variance has 10 degrees of fre~dom.(Just how we can k.now the degrees of freedom wi 11be discussed shortly.) Suppose a1so that we wantto know a range around our sample value of S2 = 3.0which "probably" contains the true value, a 2 .three degrees of freedom. Indeed, we will endesiredconfidence is, say, 90%.TheThat is, 10% ofthe time the true value will actually fall outsideof the stated bounds.to a 11 ocate S%The usual way to proceed isto the low end and 5% to the hi ghend for errors, leaving our 90% in the middle.This is arbitrary and a specific problem mightdictate a different allocation. We now resort toconcepts of the chi-square distritablesof the chi-square distribution and findthat for 10 degrees of freedom the 5% and 95%Variances) are distributed according to the equapointscorrespond to:x2(.05) :: 3.94x2. = (d. f. ) . 52(5.1)at for d.f. = 10 (5.4)where 52 is the sample Allan Variance, x2. is X 2 (.95) =18.314TN-27