Colloboration with George Judge: He has more general idea, My idea was more specialized:
IDEA: It is impossible to verify that a given observable sequence is i.i.d. -- even though it is possible to disprove this.
CONSEQUENCES: Fundamental indeterminacy about a basic assumption of regression models. Undermines inference. Basically we have a Popper type situation. Given a sequence of observations X1, X2, .... we have two possibilities:
A: Can PROVE it is NOT i.i.d.
B: Cannot prove that it is NOT i.i.d.
B is taken as equivalent to i.i.d. but it is not. There is a fundamental lack of determinacy.
OBJECT OF PAPER: CONFINE problem to testing heteroskedasticity. Show that this is impossible. Study the following THREE IMPOSSIBILITY results and distil common features and extract the lesson to be learned from this impossiblity
Suppose Xi/si are i.i.d. with common distribution F. The null hypothesis of homoskedasticity is that the s are all equal. The alternative is that they are not. If F is known than it is possible to test this hypothesis – that is, exactly in sense you use it, there is a test which has power greater than alpha at all alternatives [Pitmans test].In general, if F is unknown, then it is impossible to test – that is, for some heteroskedastic sequences, power will be less than alpha.
So to test heteroskedasticity, either one must restrict the
distribution, or else restrict the type of heteroskedasiticity by
putting some restriction on the alternative to equality of all scale
parameters.
A paper by myself and Jayasri Dutta proved this result (paper is attached)
Another paper by Freedman proves an impossiblity result which is more general than ours above but set in a different framework:
D.A. Freedman.
> “Diagnostics cannot have much power against general
> alternatives.”
> <A HREF="http://www.stat.berkeley.edu/~census/nopower.pdf">[PDF-Preprint]</A>
> To appear in <em>Journal of Forecasting</em>.
> </p>
A THIRD paper shows that testing for equality of variances is impossible without some restrictions is given in the paper below. Roughly speaking, consider several groups of observations and test that each group has the same variance
That is A1 A2, ...., An, B1, B2, ...., Bm, C1, C2, ... Cj D1, ... Dl
Under the assumption that all groups come from a common distribution (or that they have the same Kurtosis) one can test for equality of variances across groups. Without such an (implausible) commonalilty restriction one cant.
So again, effectively, it is impossible to test for i.i.d.
Rank Tests of Dispersion
Author(s): Lincoln E. Moses
Source: The Annals of Mathematical Statistics, Vol. 34, No. 3 (Sep., 1963), pp. 973-983