I have heard that a Poisson process "converges" to an Ito (diffusion) process in long term. However I do not see how the characteristic function of the form morphs into that of the latter. In what measure could this convergence be defined at all?

  • $\begingroup$ It is not clear what you mean "converges". Can you please provide a reference. $\endgroup$ – Gordon Feb 24 '16 at 13:52
  • $\begingroup$ Have you noticed I say "heard" and put quotation marks "..." on the word converges and write the last sentence in the question? What you said was precisely why I pose the question this way. I seek an answer to the same question we are both asking. $\endgroup$ – Hans Feb 24 '16 at 19:08

There are, as for random variables, different types of convergences for stochastic processes. Probably you mean convergence in the Skorokhod topology $J_1$. This is one convergence concept for $d$-dimensional cádlág processes.

Convergence of stochastic processes $X_n \xrightarrow {\mathscr L} X $ in this sense holds if and only if the laws $\mathscr{L}(X^n)$ converge in the space of probability measures on cádlág functions equipped with the Skorokhod topology.

Typically one has to show to things:

  1. $(X^n)$ is tight (i.e. relatively compact)
  2. $X_n \xrightarrow {\mathscr L(D)} X$ for some dense subset $D \in \mathbb{R}_{\ge 0}$ (for example, convergence of finite-dimensional distributions)

For a full treatment of this topic consider Jacod & Shirayev: Limit Theorems for Stochastic Processes, or the more classical reference Billingsley: Convergence of Probability Measures.

Your question for Poisson processes now can be answered with Theorem IX.4.8. in Jacod&Shiryaev. Consider a compound Poisson process $Y$, i.e. a Poisson process $N$ with intensity $\lambda$ and i.i.d. random variables $X_1,X_2, \dots$, s.t. $$ Y_t = \sum_{i=1}^{N_t} X_i, \quad t \ge 0. $$ We now expect if the jumps $X^1,X^2,...$ get smaller and the intensity $\lambda$ explodes, that a convergence to a Brownian motion can hold.

This is confirmed by Theorem IX.4.8: Consider $\lambda^n=n$, $X_1^n$ be normally distributed with mean zero and variance $n^{-(1/2)}$. Then the quadratic variation computes to $$ \int x^2 \lambda_n \phi\big(\frac{x}{a_n}\big) a_n^{-1}dx = \lambda_n (a_n)^2=1$$ where $a_n=n^{-(1/2)}$. This shows that $\tilde c^n \to 1$ in Theorem IX.4.8. Together with the fact that the limit process has no jumps ($K=0$ therein) and no drift ($b=0$ therein) this yields that the limit is a Brownian motion.

Theorem IX 4.8: enter image description here

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    $\begingroup$ Thank you very much for your answer. I have taken the liberty posting an image of Theorem IX 4.8 you referred to at the bottom of your answer. Is this the correct theorem? If so, do we need to identify $K^n$ and in particular check the convergence of the integral involving $K^n$ in line 4.11? $\endgroup$ – Hans Jun 12 '19 at 21:41
  • $\begingroup$ Yes, this is the right theorem ! Indeed, you have to check that the integrals over $K^n$ converge to zero (which guarantees that the limit is continuous). $\endgroup$ – Thorsten Jun 21 '19 at 16:22
  • $\begingroup$ In this regard, it is important to realize that $C_1(\mathbb{R}^d)$ is a not the class of once continuously differentiable functions. On page 395 in Jacod/Shirayev you find the precise definition. It is a countable subclass of non-negative continuous bounded functions, 0 around zero, containing functions of type $(a |x|-1)^+ \wedge 1$ and being convergence-determining for weak convergence induced by continuous bounded functions, vanishing around 0. $\endgroup$ – Thorsten Jul 9 '19 at 13:53
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    $\begingroup$ Thank you for pointing this out. I did misread $C_1$ as you are cautioning against. $\endgroup$ – Hans Jul 9 '19 at 19:35

If $X_t$ is a Poisson Counting Process with intensity $\lambda$ then the Martingale $M_t=X_t−\lambda t$ is called a Compensated Poisson Process. As $\lambda$ becomes large $M_t$ does converge to a Brownian motion with variance rate $\lambda$.

This can be seen by using the "heavy arrivals" approximation of the Poisson Distribution: when the arrival rate is large the number of events per second is approximately Normal with mean $\lambda$ and variance $\lambda$, therefore the increase in the compensated process per second is N(0,λ).

  • $\begingroup$ Your second paragraph is similar to user9403's answer below which is not sufficient to show "convergence" of two processes in any plausible sense. Please refer to my criticism below his answer. Could you please define what you mean by "convergence" of one stochastic process to another as I have asked in the question? Could you please provide a proof or a reference for the convergence you claim in the first paragraph? Otherwise, I do not see how your answer answers my question. Thank you. $\endgroup$ – Hans Feb 26 '16 at 20:03
  • $\begingroup$ According to you my answer does not show "convergence" yet you refuse to define "convergence". You have asked a question with no criteria leaving us to guess what you may want. Ask a precise question and you may get the answer you want. $\endgroup$ – user9403 Mar 27 '16 at 22:07
  • $\begingroup$ @user9403: In my question, the proper definition of "convergence" of a sequence of a stochastic process is part of what I am seeking. So I can not give an answer to what I am asking. Yes, it is an open ended question. I am trying to ask what justifies what I heard. It could just be something completely wrong. I would like to know that. It is one of the cases where one does not know the answer to the question but knows some claims could not be right. Also, I did make some suggestions as to what could constitutes a legitimate definition in my comment to your answer. $\endgroup$ – Hans Apr 30 '16 at 0:10
  • $\begingroup$ @user9403: Do you not agree that your answer does not look right in light of the counterexample I give below in my comment? $\endgroup$ – Hans Apr 30 '16 at 0:11
  • $\begingroup$ If you are asking for convergence in the sense of whether an increment of a Poisson process converges to an increment of Brownian Motion then no, they never converge. Poisson processes and Brownian Motions are Markov so the distribution over $T, T+dt$ is the same regardless of $T$. $\endgroup$ – user9403 Apr 30 '16 at 10:21

It is straightforward to show a weaker result: That a Poisson process becomes normally distributed as $T$ becomes large. A Poisson process has independent increments. Let $X_T-X_0$ be Poisson process. Take an arbitrary time step $\Delta t$. Then $X_T-X_0=\sum \left(X_{(i+1)\Delta t}-X_{i\Delta t}\right)$. By the Central Limit Theorem, the sum of IID random variables drawn from a distribution with finite variance converges to a normal random variable as the number of terms approaches infinity. Hence as $T \to \infty$, $X_T-X_0 \to \mathcal{N}(\cdot, \cdot)$.

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    $\begingroup$ I understand the Central Limit Theorem gives the distribution of the Poisson process converges to the Gaussian. However, the convergence of two processes surely can not just be convergence in distribution at any time. Is the convergence of two processes understood as all paths of one process convergent to those of the other almost surely? After resolving that, to show one process is Brownian motion, you need to prove the increment of the process is independent of each other and that the paths of the process are almost surely continuous. How do you show these for the limiting process? $\endgroup$ – Hans Feb 24 '16 at 19:00
  • $\begingroup$ A case in point to my previous comment is the process $A_t:=\sqrt{t}B_1$ where $B_t$ is the standard Brownian motion with respect to time $t$. $A_t$ and $B_t$ have exactly the same distribution at any time $t$ and $A_t$ has continuous paths but $A_t$ is no Brownian motion at all let alone $A_t$ being the same process as $B_t$. $\endgroup$ – Hans Feb 24 '16 at 23:32

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