1

In Resnick's Adventures of Stochastic Processes, I've seen the following convolution $\int^t_0 z(t-u) U(du)$, in which there's an abuse of notation, since $U(t)=\sum^{\infty}_{n=0}F^{n*}(t)$, where $F^{n*}$ is the n-fold convolution of the distribution $F$. It's similar to the same abuse of notation as in $\int g(u) F(du)$.

My question is what measure is $U$, i.e. $U(t)=U(-\infty,t)$ as we do for $F$?

And how can we prove that it determines a distribution?

If $F$ is a distribution, then by some of the properties of convolution we also have that $F^{n*}$ is a distribution. But, $U$ is an infinitely countable sum of distributions. So, how can we show that $U$ is a distribution?

An old man in the sea.
  • 5,241
  • 1
    It depends on $F$, for example $F(t) = \delta(t)$ means $F^{n \ast}(t) = \delta(t)$ and the series diverges in the sense of distributions. Here the convolution is well-defined because $F$ is a non-negative finite measure, so $\langle F ,\varphi \rangle$ is well-defined for $\varphi$ continuous, and $\langle F^{(n+1)\ast},\varphi \rangle=\langle F,F^{n\ast} \ast\varphi \rangle$ where $F^{n\ast}\ast\varphi$ is continuous. If $F$ is supported on $[a,\infty)$ for some $a>0$ then $F^{n\ast}$ is supported on $[na,\infty)$ so $\sum_{n=0}^\infty F^{n\ast}(t)$ converges in the sense of distributions. – reuns Dec 05 '17 at 23:23
  • @reuns yes you're right. I've just found that out. It may not be a distribution, but it's always a measure, since it's a countable sum of non-negative measures (distributions). Should I keep the question, or delete it? – An old man in the sea. Dec 05 '17 at 23:25
  • It doesn't have to converge in measure. Try $F(t) = - \delta(t)$ – reuns Dec 05 '17 at 23:27
  • For $F$ non-negative, it is a series of non-negative measures, thus strictly increasing in the sense of non-negative measures, thus it converges in the sense of non-locally finite measures : $\mu_U(A) = \lim_{N \to \infty} \sum_{n=0}^N \mu_{F^{n \ast}}(A)$. Is it what you meant ? – reuns Dec 05 '17 at 23:35
  • @reuns I meant this https://math.stackexchange.com/questions/805857/countable-sum-of-measures-is-a-measure

    Maybe I misunderstood something in the link.

     – An old man in the sea. Dec 05 '17 at 23:37
  • Yes, an increasing sequence of measures $\mu_n$ converges in the space of non-locally finite measures. Increasing means $\mu_{n+1}(A) \ge \mu_n(A)$ for any set $A$ in the $\sigma$-algebra. – reuns Dec 05 '17 at 23:42
  • @reuns ok. Thanks ;) – An old man in the sea. Dec 05 '17 at 23:48

0 Answers0