Test of tbe Random Walk equally spaced intervals and consider the following estimators for the unknown parameters u and o: 1 2排 (X。-X-)= 1(X2 -X) (4a) 2ne-1 2 品= 2 (4b) 2nk-1 (X。-X-1-)2 2(&-,-2 (4c) 2n0=1 The estimators a and correspond to the maximum-likelihood estimators of the u and o parameters;is also an estimator of o but uses only the subset of n 1 observations Xo,X2,X,...,X2m and corre. sponds formally to times the variance estimator for increments of even-numbered observations.Under standard asymptotic theory,all three estimators are strongly consistent;that is,holding all other parameters constant,as the total number of observations 2n increases without bound the estimators converge almost surely to their population values.In addi- tion,it is well known that both and possess the following gaussian limiting distributions: V2n(2-σ)4N(0,2o) (5a) V2n(3-o)&N(0,4o) (5b) where indicates that the distributional equivalence is asymptotic.Of course,it is the limiting distribution of the difference of the variances that interests us.Although it may readily be shown that such a difference is also asymptotically gaussian with zero mean,the variance of the limiting distribution is not apparent since the two variance estimators are clearly not asymptotically uncorrelated.However,since the estimator is asymp- totically efficient under the null hypothesis H,we may apply Hausman's (1978)result,which shows that the asymptotic variance of the difference is simply the difference of the asymptotic variances.3 If we define=- then we have the result V2n Ja 4 N(0,200) (6) Using any consistent estimator of the asymptotic variance of /a standard Briefy,Hausman (1978)exploits the fact that any asymptotically efficient estimator of a parameter 0,say 8.,must possess the property that it is asymptotically uncorrelated with the difference .-0.where 8.is any other estimator of 0.If not,then there exists a linear combination of and .-8,that is more efficient than 8,contradicting the assumed efficiency of 6.The result follows directly,then,since avar (0)avar (0.0.)=avar (0.)+avar (0-0) -avar(6.-6)=avar()-avar(6.) where avar (denotes the asymptotic variance operator. 45