Re: Some questions about centroid of a polytope
- From: mathsen <mohsen_303@xxxxxxxxx>
- Date: Tue, 02 Jun 2009 13:29:05 EDT
"mathsen" <mohsen_303@xxxxxxxxx> wrote in message
news:5389609.13556.1243876854089.JavaMail.jakarta@nitr
ogen.mathforum.org...
First of all, thank you very much for introducingthis wonderful
inequality to me.so the formula that you
Just to make something clear, I'm talking in R^{n}
gave for centroid might not be applicable here(because the number of
points in the polytope is infinite).expectation because what I
You know, I think we still need to keep the
want to show is arg_{y} min E{||x-y||^2} = centroidand if I want to use
the jensen's inequality then we will have somethinglike E(d(x)) >=
d(E{x}) which I don't think I can use to prove theabove statement.
After expanding the distance to the sqrt you can use
it to reverse the
inequality since sqrt is convex.
A simple outline of a proof might go along like this:
E[d(x,y)]' = E[d'(x,y)] = (E[x] - y)/d(x,y)
where ' is d/dy. y = E[X] clearly satisfies this. The
middle to last step
comes out by expanding d'(x,y) using the euclidean
norm.
Opps, I just noticed your problem is a bit different.
You want the square of
the distance.
In this case ,
E[d(x,y)^2]' = E[2*d(x,y)*d'(x,y)] = 2*(E[x] - y)
which has the same result of course(by jensons
inequality since the square
function is convex we could have used that directly
with the first result
too since E[d(x,y)]^2 <= E[d(x,y)^2] and argmin
E[d(x,y)]^2 = argmin
E[d(x,y)]).
I am making a few assumptions by using the derivative
to find the argmin as
that is only half the problem(need to check
boundaries and prove it is an
actual minimum, etc...).
That's great. Thank you very much.
.
- References:
- Re: Some questions about centroid of a polytope
- From: Jon Slaughter
- Re: Some questions about centroid of a polytope
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