[R-sig-Geo] LM tests
Munroe, Darla K
dkmunroe at email.uncc.edu
Fri Feb 27 21:39:05 CET 2004
yes, I'm sorry - you are right - I somehow temporarily forgot what
row-standardization exactly does!
But some of logic still holds - you preserve the ranking from low to high in
distance from all j to each individual i; the problem is you will lose the
differences among each of the is - i.e., you'll lose the effect of
observations that are relatively far away, and all the dijs are higher, for
example.
So does that make sense - row-standardizing a distance matrix will yield
something more like a k-nearest neighbors matrix? I.e., for each i, you
will just rank the neighbors from low to high, but you will lose the
variation to distinguish which observations are more fringe, vs. more
central.
sorry for the confusion - I should think more before I send email.
-----Original Message-----
From: Roger Bivand
To: Munroe, Darla K
Cc: 'Jill Caviglia-Harris '; 'r-sig-geo at stat.math.ethz.ch '
Sent: 2/27/04 3:33 PM
Subject: RE: [R-sig-Geo] LM tests
On Fri, 27 Feb 2004, Munroe, Darla K wrote:
> I was thinking about this issue, and correct me if I'm wrong -
>
> If you row-standardize the distance weights, you will in effect
rescale
> them, but you will not change the scale of the weights themselves,
correct?
> I.e., row standardization means dividing the weight for each
observation by
> the total # of non-zero elements for that row, correct? Well, each
> observation by definition in a distance matrix has the same number of
> "neighbors" (i.e., all n-1), correct? So 1/dij (or whatever your
distance
> matrix is) becomes 1/dij/n.
>
I'm not sure about that. What you are dividing by is the row sum:
\sum_j w_{ij}, and w_{ij} = 1/d_{ij}, so the sum will be smaller for
places a long way from others, and larger for places near most others,
won't it?
In the R spirit, try it out:
> set.seed(1)
> try <- 1/as.matrix(dist(cbind(rnorm(100), rnorm(100))))
> diag(try) <- 0
> summary(rowSums(try))
Min. 1st Qu. Median Mean 3rd Qu. Max.
45.07 68.97 90.05 91.78 113.40 153.10
So places with different "connectedness" will get "flattened", I think.
But then I'm not sure that full matrices are so very informative (there
is
a literature about reconstructing maps of relative position from
neighbour
graphs, I think, so the sparse binary weights actually carry quite a lot
of information - more parsimonious anyway).
Roger
> Is that going to affect your fundamental interpretation of the
structure of
> spatial dependence? Probably not - unless you're trying to interpret
rho or
> lambda in terms of the distance units (which I wouldn't presume to do,
> anyway...).
>
> Or am I off base?
>
> -----Original Message-----
> From: Roger Bivand
> To: Jill Caviglia-Harris
> Cc: r-sig-geo at stat.math.ethz.ch
> Sent: 2/27/04 2:40 PM
> Subject: Re: [R-sig-Geo] LM tests
>
> On Fri, 27 Feb 2004, Jill Caviglia-Harris wrote:
>
> > List members:
> >
> > I have been using the function lm.LMtests developed using the spdep
> > package to test for spatial lag and error. My problem is that these
> > tests assume that the weights matrix is row standardized, while I
have
> a
> > weights matrix that is set up as the inverse distance between
> neighbors.
>
> Certainly lm.LMtests() prints a warning, and the tradition it comes
from
>
> usually presupposes row standardisation. Curiously, quite a lot of the
> distribution results in Cliff and Ord actually assume symmetry, which
> can
> lead to fun with negative variance in Geary's C and join count
> statistics
> even with row standardised weights.
>
> > Converting it into a row standardized matrix would result in the
> loss
> > of important information. Have there been any functions developed
> that
> > any of you know about that are not dependent upon this assumption?
>
> Have you tried (probably yes) and does it make a difference? Are the
> results from a binary IDW and a row standardised IDW very different?
Is
> your IDW matrix full or sparse? Can Moran's I be applied instead
> (despite
> its covering lots of misspecification problems)? Are the IDW weights
> symmetric (probably, but not always)?
>
> I'm not sure why distances should be helpful if the data are observed
on
>
> areal units, so that measuring distances is between arbitrarily chosen
> points in those units, a change of support problem. That may be why
> there
> aren't methods too, though there's no reason not to try to develop
> things.
> But error correlation specified by distance does movbe rather close to
> geostatistics, doesn't it?
>
> Any other views, anyone?
>
> Roger
>
> > Thanks. -Jill
> >
> >
> > ***************************************************
> > Jill L. Caviglia-Harris, Ph.D.
> > Assistant Professor
> > Economics and Finance Department
> > Salisbury University
> > Salisbury, MD 21801-6860
> > phone: (410) 548-5591
> > fax: (410) 546-6208
> >
> > _______________________________________________
> > R-sig-Geo mailing list
> > R-sig-Geo at stat.math.ethz.ch
> > https://www.stat.math.ethz.ch/mailman/listinfo/r-sig-geo
> >
>
>
--
Roger Bivand
Economic Geography Section, Department of Economics, Norwegian School of
Economics and Business Administration, Breiviksveien 40, N-5045 Bergen,
Norway. voice: +47 55 95 93 55; fax +47 55 95 93 93
e-mail: Roger.Bivand at nhh.no
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