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vc ha scritto:
> pamelafluente_at_libero.it wrote:
> > vc ha scritto:
> > > Earlier, I said that the sample median regarded as an interval (50
> > > percentile) *is* invariant under monotone transformations.
> >
> > Ok let's assume that there exists such a thing like a "sample median
> > regarded as an interval " (it's not a correct expression but I
> > understand that you want to mean the interval of values which minimizes
> > the sum of absolute deviations), then you have to define what you mean
> > by "invariance", as the definition you provided:
> >
> > m(f(X)) = f(m(X)).
> >
> > does not apply to intervals.
>
> I am sorry, I did not realize that you did not understand the above
> elementary notation. As I said before let 'X' be a real-valued random
> variable, ' f' a transformation function and 'm' the function mapping
> the r.v. to a set possibly but not necessarily consisting of a single
> element and representing the median. Using the function compositionn
> notation, one could also write m o f o X = f o m o X (which is not
> much different). Does it help ?
You are doing it again.
I do understand the notation. That means
m(f(x)) = f(m(x)) for any x of the support of X
it's "for any x" . Does not apply to intervals. It applies to points.
For intervals for instance you could define invariance something that leave constant the probability measure of the interval or whatever. Or (another definition) you could say that the above must hold for each x in the interval. But if one takes this second definition, you have already proven that invariance does not hold.
So you still need to define invariance for intervals, and I have already warned you about the pointwise definition.
Further, we are not talking about median of rv's. But about median of a finite set of numerical values. We are not referring to probability measure.
>
>
> > Only after you have provided such a
> > definition of invariance, we can check if your statement is ok.
> >
> > Otherwise, we are left with a sentence that does not make sense : "
> > sample(?) median regarded as an interval (50 percentile(?) ) *is*
> > invariant under monotone transformations ".
>
> I hoped you'd realized that, up to now, we'd been talking about what
> is commonly called 'sample median' or the 'median of a data series' as
> opposed to the related 'median of probability distribution', but I
> guess I was wrong.
We are not within a sampling framework. There is no need to call for
sampling theory.
>Also, it may be useful to know that sample median
> *is* the 50th percentile. "Regarded as" can be replaced with "not
> reduced to a single value when the number of elements is even".
That I studied at elementary school. But it's not defined as an interval.
>
> It has as much to do with intervals as the median does because the
> median is the 50th percentile (or 1/2 -tile). In fact, any quantile
> can be an interval:
"any quantile can be an interval" is not a correct statement.
You are getting mixed up with the theory of countinuous probability
distribution.
We are referring to finite set. For a continuous cdf the median is a
point. For a finite set instead holds the definition I have provided
before.
>
> Let X be a r.v., P denotes probability, then any point 'q' is an
> R-tile iff it satisfies the two inequalities:
>
> P(X <= q) >= R,
> P(X >= q) >= 1-R
Same. We are dealing with finite sets. Leave alone the probability. Received on Tue Sep 26 2006 - 04:42:50 CDT