Conditions for Poisson Jump Processes to Approach Brownian Motion

# Does a Poisson process converge to an Ito process in long term?


# Does a Poisson process converge to an Ito process in long term?












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?

## Answer by Thorsten (score 3, accepted)

https://quant.stackexchange.com/a/46036

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:

- $(X^n)$ is tight (i.e. relatively compact)

- $X_n \xrightarrow {\mathscr L(D)} X$ 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:

## Answer by Alex C (score 1)

https://quant.stackexchange.com/a/24489

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).

## Answer by user9403 (score 0)

https://quant.stackexchange.com/a/24492

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 $T \to \infty$, $X_T-X_0 \to \mathcal{N}(\cdot, \cdot)$.