Re: The First Variation of a PDF
- From: William Elliot <marsh@xxxxxxxxxxxxxxxxxx>
- Date: Tue, 14 Oct 2008 04:36:13 -0700
On Mon, 13 Oct 2008, junoexpress wrote:
On Oct 13, 2:50 am, William Elliot <ma...@xxxxxxxxxxxxxxxxxx> wrote:
On Sun, 12 Oct 2008, junoexpress wrote:
Consider a pdf having compact support on [-X,+X] for some given
positive real X, that is a function of a real parameter c. When c=0,
the pdf is symmetric about x=0 and the pdf takes a very simple form.
When c is non-zero, the form of the pdf is non-symmetric about x=0
becomes quite complicated.
When c > 0, then for f_c(x,0) to be negative, assuming aI want to understand the effect c has when it is small but non-zero,
on the bias of x.
One way I considered was to construct an approximation to the pdf:
f(x,c) =~ f(x,0) + c.f_c(x,0).
smooth x curve >= 0, then f(x,c) > 0 so you've got lee way
to assume the approximation is not negative.
Similar in reverse with 0 < c.
My problem is that although the function integrates to one, theNow the RHS of (1) integrates to 1, but it is probably not a pdf
(since it is not obvious the first order approx is non-negative).
integral(-oo,oo) f(x,0) dx + integral(-oo,oo) c.f_c(x,c) dx
= 1 + c.integral(-oo,oo) f_c(x,0) dx = 1
function is not guaranteed to be a pdf because it has no guarantee of
being non-negative at all points in its domain.
Have you found examples (see my thoughts above) where it is negative?
Yes if |c| is large enought. However with small c, the approximaation
I'd think should remain non-negative.
So now, I'm trying to compute things like a bias with a function that is
not a pdf, and so how do I know what I am really getting is the bias? So
there's a point of confusion (on my part) as to what one is *gaining* by
making an approximation of this sort and OTOH, what one is *losing* by
not using a true pdf.
I don't know. Now were I to alter a metric to see what effect
it has and can only approximate the alteration, wouldn't I want
the approximation to be a metric? If it wasn't, then how could
I judge the effect it had on the space? It would be hard.
That probably is the best I can do at explaining my concern. Maybe I'mIf for all y, g(y) >= 0, then is there some c > 0
nitpicking here, but this bothers me, and I have to think that this is a
standard problem that people who know math well must have come across.
with g(0) + c.g'(c) >= 0?
That's the first simple step.
Now for each x, let g(y) = f(x,y).
Since the support for f(x,y) as a function of x is compact,
can we find a c > 0 with for all x in support
f(x,0) + c.f_y(x,c) >= 0. Outside the support,
isn't f(x,y) for each y, flat, ie zero? Thus
can we assume f_y(x,y) = 0 when x not in support?
Does this have the assumption that for each y, the
pdf of f(x,y) has the same compact support?
.
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- From: junoexpress
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- From: William Elliot
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