> "Smith, David W." wrote:
>
> I believe that the key assumption is stochastic ordering of the two
> distributions for the two samples. Under the alternative hypothesis
> this means that the two distribution functions do not cross. One of
> them has to be always to the right of the other. Lots of asymmetric
> distributions will do this. Under the null hypothesis of no
> difference in location (eg, median) the distributions are identical if
> they do not cross each other.
Do you mean that the *cumulative* distribution functions do not cross?
It would be an awfully strong assumption that the density functions did
not cross.
>
> If two distributions have the same location parameter (eg, mean) but
> different dispersion parameters (eg, variance), then their
> distributions would cross each other and the assumption of stochastic
> ordering would be violated. An example of this would be two normal
> distributions with equal means but different variances.
>
> I think this is somewhere in Hayek's little book on nonparametrics.
I hope not; if it is, Hayek is wrong. As Potthoff showed in '63,
the ranksum test is a test for the median between any two symmetric
distributions, not necessarily the same shape.
The proof is elementary (I do not recall if Potthoff used this
approach.) First, it may be shown easily that it is necessary and
sufficient that when one distribution is shifted to make the medians
equal, we then have
P(X+c < Y) = 1/2 .
To see this, set up the double integrals for P(X+c < Y) and P(Y < X+c)
and apply symmetry to show that they are equal.
There _has_ been some confusion on this topic in the literature; in
particular, H.A.David claimed (1963, The Method of Paired Comparisons,
p.19) that
"the paradox [of intransitivity] cannot occur
if y1,y2,y3 have continuous distributions,
since the locations of the medians then
determine whether a particular preference
probability is greater than, equal to, or
less than 1/2"
which is demonstrably wrong!
-Robert Dawson
.
.
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