Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For the sake of example, suppose we are on the real line and that $H(x)$ is the Heaviside step function and $\delta(x)$ is the Dirac Delta function. According to the theory of distributional derivatives, one has

$\frac{dH}{dx}=\delta(x)$

in the sense of distribution, so that $\int \phi(x) \delta(x)dx = -\int \phi'(x) H(x)dx$ for all compactly supported nice functions $\phi$.

My question is whether the same definition carries over to Riemann-Stieltjes integration. In particular, for a cumulative distribution function $F$, I am trying to evaluate $\int \phi(x)H'(x)dF$ but, does this expression equal $\int \phi(x)\delta(x)dF=\phi(0)dF(0)$?.

My suspicion is that the distributional derivative here must take into account the $dF$ term, so integration by parts of $\int \phi(x)H'(x)dF$ does not give the usual expression. If $F$ has a density $f(x)$, then one gets $\int \phi(x)H'(x)f(x)dx=-\int (\phi f)' H(x)dx=\phi(0)f(0)$ which is as expected, so what happens when $F$ does NOT have a density function $f$? In particular how does this generalize to arbitrary distributions?

share|cite|improve this question
    
Try it for $F(x) = H(x)$. The square of the Dirac Delta does not make sense. – Robert Israel Jun 8 '12 at 21:12
    
@Robert Israel: I agree. But does that mean there is no way to define the distributional derivative under that particular Stieltjes integral – Alex R. Jun 8 '12 at 22:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.