Re: Dirac-Delta function

From: Randy Poe (poespam-trap_at_yahoo.com)
Date: 10/03/04 

Date: 3 Oct 2004 15:16:07 -0700

"Adam" <addam@rogers.com> wrote in message news:<xOydnaVIz4Lbj_3cRVn-ig@rogers.com>...
> The only distribution I know of is the poisson and binomial. I'll do
> some reading.

Someone else pointed out that "distribution" here means something
different from "probability distribution". It's a generalization of
the concept of "function".

However, you reminded me of something from probability theory.

> > They have the properties which are important to physics. Most
> > importantly
> >
> > integral(x = -oo to oo) f(x) delta(x-x0) dx = f(x0)
> >
> > The Dirac delta "picks out" the value of f(x) at x0.
> >
> Yes, that is what is said over and over in my physics classes. It was
> introduced to take integrals over discrete charge distributions.

In a course in probability theory, the text used Riemann-Stieltjes
integrals, which are a generalization of Riemann integrals. In
the elementary form, an expectation value of f(x) where x has
probability density p(x) is

integral(x = -oo, oo) f(x) p(x) dx

If x is discrete over some or all of its range, then trying
to use densities to represent the distribution requires you
to use delta-functions, just as with charge distributions.
I think you see the connection here.

However, a more rigorous way to define expectation value
is integral(x = -oo, oo) f(x) dP(x)

where P(x) is the cumulative distribution, and the integral
with respect to dP(x) is a Riemann-Stieltjes integral, a
generalization of the basic Riemann integral which allows
integration with respect to functions with jump discontinuities.
P(x) for a discrete variable has such jumps.

I suspect that there may be an alternate, more rigorous
representation of the expressions you're worried about in
terms of Riemann-Stieltjes integrals rather than Riemann
integrals of delta functions. It's the integral that is
generalized rather than the concept of "function".

             - Randy

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