3
$\begingroup$

Let us define the following cumulative distribution: \begin{align} \Pr (Y(t)\geq a)=\sum_{n=0}^\infty \frac{e^{-kt}(kt)^n }{n!}\int_a^\infty[\circledast_{f}\circ H_a ]^n\delta (x) dx \end{align} where the operator $[\circledast_{f_J}\circ H_a ]$ denotes the multiplication by a Heaviside function with a cut off at $a$ followed by the convolution by a pdf $f$. This is a cumulative distribution function in the sense that $\Pr (Y(t)> 0)=0$ and $\Pr (Y(t)\geq -\infty)=1$ (or we could say that $1-\Pr (Y(t)\geq a)$ is the CDF).

A global question would be: what can I know about this distribution, given the pdf $f$?

An intuitive property would be $\langle Y(t) \rangle\xrightarrow[t\rightarrow +\infty]{}-\infty \quad \forall f \;|\; \mu_f<0$

A more specific one is: how to compute $\langle Y(+\infty) \rangle$ given $f$?

My first idea was to notice that for $t\rightarrow + \infty$, we can just consider terms with a large $n$ in the series. Then I wanted to take inspiration from the central limit theorem. Given that the pdf $f$ has definite first and second moments: \begin{align} [\circledast_{f}]^n\delta (x)=f*f*...*f\sim\mathcal{N}\Big(\frac{x-\mu_f n}{\sigma_f \sqrt{n}}\Big)\quad \text{as }\quad n\rightarrow \infty \end{align} However we have here \begin{align} \int_a^\infty[\circledast_{f}\circ H_a]^n\delta (x)dx\sim \quad ? \quad <\int_a^\infty \mathcal{N}\Big(\frac{x-\mu_f n}{\sigma_f \sqrt{n}}\Big)dx\quad \text{as }\quad n\rightarrow \infty \end{align}

Here are other complementary informations:

  • For the special case $f(x)=p_-\delta(x+1)+p_+\delta(x-1)$, I have found, with a random walk approach, that $\langle Y(+\infty) \rangle = \frac{1}{1-\frac{p_+}{p_-}}$ for $p_+>p_-$.

  • following this kind of approximation along with the previous result, we end up with the following approximation: $\langle Y(+\infty) \rangle \approx \frac{1}{2}\Big(\sqrt{\mu_f^2+\sigma_f^2}-\frac{\mu_f^2+\sigma_f^2}{\mu_f}\Big)$ for $\mu_f>0$, which is not exact according to numerical simulations but which gives the order of magnitude.

  • This post is related to this one.

$\endgroup$
  • $\begingroup$ Same remark as your related question : the operator has maybe to be corrected, and the $ e^{-knt} $ changed into $ e^{-kt} $ ? In which case, you want the moments of $ Y_{N_t(k)} $ where $ Y_n = x + \sum_{m = 1}^n J_m $ and $ N_t(k) $ is an independent Poisson Process of parameter $k$ ? $\endgroup$ – Synia Aug 18 '17 at 0:30
  • $\begingroup$ Almost: I would like the moments of the \textit{infimum} of Y_{N_t} (or at least the mean at infinite time), which is at the origin of the Heaviside in the operator. Thank you for enlightening the meaning of all this, which was not clearly expressed. I will discuss the operator form in the thread of my the related question, you may be right. $\endgroup$ – Alexandre Aug 18 '17 at 13:58
  • $\begingroup$ Ouch ! You then look for the inf of the skeleton on a finite set $ \{ 1, \dots, [t] \} $ (with the integer part $[t]$ of $t$), which amounts to the running infinimum of a discrete random walk... This is a hard problem. For the Bernoulli increments case, you have results using the reflection principle. In general, I don't know... $\endgroup$ – Synia Aug 19 '17 at 0:13
  • 1
    $\begingroup$ You have a general type of process called "compound Poisson process" which is a simple type of Lévy process ; the study of running sup/inf is the study of its fluctuations identities. You may want to have a look at the following book : "Fluctuations of Lévy Processes with Applications: Introductory Lectures" by Andreas Kyprianou. In particular, the chapter 3. $\endgroup$ – Synia Aug 19 '17 at 0:17
  • 1
    $\begingroup$ There seems to be a general theorem, found on arxiv.org/pdf/1501.04542.pdf page 2 : define $ \tau_x := \inf\{t > 0/ X_t > x\} $ i.e. $ \tau_x = \overline{X}^{<-1>}_x $ where $ \overline{X}_t =\sup_{s \in [0, t]} X_s $ and $ <-1> $ is the inverse bijection (since the running sup is strictly increasing). Then, $\tau$ is a subordinator and its Laplace transform is known using the Laplace transform of the increment of the walk. You can certainly find it with the proof in the book loc. cit. and adapt from that to the Laplace transform of the running sup. $\endgroup$ – Synia Aug 19 '17 at 0:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.