Re: The First Variation of a PDF

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.
I 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:
(1)    f(x,c) \approx f(x,o) + c (df/dc)at c=0

f(x,c) =~ f(x,0) + c.f_c(x,0).

Now 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). This

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

bothers me somewhat, and has me wondering if this is the correct way
to approach this problem. Any suggestions?

Why should it bother you?  Doesn't it make you happy to have
an approximate pdf to work with?

Thank you for the notational changes: I like yours better.
My problem is that although the function integrates to one, the
function is not guaranteed to be a pdf because it has no guarantee of
being non-negative at all points in its domain. 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.
That probably is the best I can do at explaining my concern. Maybe I'm
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.

Matt
.

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