Correlated Brownian motion and Poisson process

Question

Is there an (easy) way to construct, on the same filtered probability space,a Brownian motion $W$ and a Poisson process $N$, such that $W$ and $N$ are not independent ?

I first asked this question at math.stackexchange.com and was suggested to post it here : https://math.stackexchange.com/questions/25519/correlated-brownian-motion-and-poisson-process.

Note that I want $W$ and $N$ to be BM and Poisson for the same filtration $({\cal F}_t)$ (see Shai Covo's answer on math.stackexchange for a construction without this requirement).

Answer 1

You can not construct a Poisson process $N$, and a Brownian motion $W$ on the same filtration such that they are dependent. They are always independent on the given filtration.

Let $(W_t)_{t\geq 0}$ be a standard Brownian motion and $(N_t)_{t\geq 0}$ be a Poisson process with intensity $\lambda$, both defined on the same probability space $(\Omega,\mathcal{F},P )$ and with respect to same filtration $F=(\mathcal{F}_t,t\geq 0)$. Then define the process $L$ for $u_1,u_2\in \mathbb{R}$ by

$L_t:=\exp(u_1W_t+u_2N_t-\frac{1}{2}u_1^2t-\lambda(e^{u_2}-1)t)$.

Now you can show by using Ito`s lemma for semi-martingales (or jump-diffusions) that $L$ is a martingale and then taking expectation of both sides and using the martingale property, you have

$E[\exp(u_1W_t+u_2N_t)]=e^{\frac{1}{2}u_1^2t}e^{\lambda(e^{u_2}-1)t}$.

Hence you can conclude that $N_t$ and $W_t$ are independent random variables for each $t$, since the last expression involves the product of moment generating function of a normal distribution and a Poisson distribution.

Edit : @Did, more precise statement would be,

Let $(W_t)_{t\geq 0}$ be a standard Brownian motion and $(N_t)_{t\geq 0}$ be a Poisson process with intensity $\lambda$, both defined on the same probability space $(\Omega,\mathcal{F},P )$ and with respect to the same filtration $F=(\mathcal{F}_t,t\geq 0)$. Then the processes $W$ and $N$ are independent.

In the above, only first step is shown, that is, $W_t$ and $N_t$ are independent of each for fixed $t$.The next step is to show that for any finite set of times $(t_1,t_2,...,t_n)$, $(W_{t_1},...,W_{t_n})$ is independent of $(N_{t_1},...,N_{t_n})$. Hence the assertion follows from this.

Answer 2

To further elaborate on my comment, it is a theorem that if $X^1,X^2,\ldots,X^n$ are Lévy processes with respect to a common filtration, all starting from zero, then they are independent if and only if their quadratic covariations $[X^i,X^j]$ are all (almost surely) zero. This is stated as Theorem 11.43 of He, Wang & Yan, Semimartingale Theory and Stochastic Calculus. It's not difficult to prove with a bit of stochastic calculus, and I'll give a proof for two Lévy processes below.

In the situation described in the question, there are two Lévy processes $W$ and $N$, where $N$ is a pure jump process. So, the quadratic covariation is simply a sum over the jumps of the processes. $$ [W,N]\_t=\sum_{s\le t}\Delta W_s\Delta N_s. $$ But as Brownian motion is continuous, $\Delta W$ is zero, so the covariation $[W,N]$ is zero. Therefore, they are independent.

Now, let's show that if $X$, $Y$ are Lévy processes w.r.t. the filtration $\{\mathcal{F}\_t\}\_{t\in\mathbb{R}^+}$ with $X_0=Y_0=0$ and $[X,Y]=0$ then they are independent. The characteristic functions of $X$ and $Y$ can be written as $$ \begin{align} &\mathbb{E}\left[e^{iaX_t}\right]=\exp(t\psi_X(a)),\\\\ &\mathbb{E}\left[e^{iaY_t}\right]=\exp(t\psi_Y(a)). \end{align} $$ Independence of the increments w.r.t. $\mathcal{F}\_{\cdot}$ implies that $M_t\equiv\exp(iaX_t-t\psi_X(a))$ and $N_t\equiv\exp(ibY_t-t\psi_Y(b))$ are martingales. As the jumps of the quadratic covariation equals the product of the jumps of the processes, $\Delta [X,Y]=\Delta X\Delta Y$, it follows that $X$ and $Y$ cannot jump simultaneously. So, $\Delta [M,N]=\Delta M\Delta N=0$. Also, the continuous part of the quadratic covariation $[M,N]^{c}$ is just an integral with respect to $[X,Y]^{c}$ (which follows from Ito's formula for non-continuous semimartingales). So, the covariation $[M,N]$ is zero. Using integration by parts, $$ d(M_tN_t)=M_{t-}dN_t + N_{t-}dM_t+d[M,N]\_t=M_{t-}dN_t + N_{t-}dM_t. $$ As a sum of integrals with respect to martingales, $MN$ is a local martingale. As it is also bounded at any time, this is a proper martingale. So, $\mathbb{E}[M_tN_t\mid\mathcal{F}\_s]=M_sN_s$ for $s < t$. Plugging in the definitions of $M$ and $N$, $$ \mathbb{E}\left[e^{iaX_t+bY_t}\;\Big\vert\;\mathcal{F}\_s\right]=\exp(iaX_s+(t-s)\psi_X(a))\exp(ibY_s+(t-s)\psi_Y(b)). $$ This determines the joint characteristic function of $(X_t,Y_t)$ conditional on $\mathcal{F}\_s$, showing that they are independent. As the distributions of $(X_t,Y_t)$ conditional on $\mathcal{F}\_s$ ($s < t$) determine all finite distributions, $X$ and $Y$ are independent. I'll leave the converse ($X,Y$ independent implies $[X,Y]=0$) as an exercise. It's not needed for the question anyway.

You can also compare this with the argument given by kakuritsu. It is essentially the same thing. Rather than working under the generality of Lévy processes, he (or she?) works directly with the Brownian motion and Poisson process, for which $\psi_W(a)=-\frac12a^2$ and $\psi_N(a)=\lambda(e^{ia}-1)$, and uses the moment generating rather than characteristic function (effectively, $a$ and $b$ above are imaginary).

Answer 3

At least you can do that on your computer: take a time discretization parameter $\Delta t$ and an iid sequence of random variables $(\xi_k, P_k)$ where $\delta_k$ is $Poisson(\Delta t)$ and $\xi$ is centred Gaussian with variance $\Delta t$ and define $W_{t} = \sum_{k \Delta t < t} \xi_k$ and $N_t = \sum_{k \Delta t < t} P_k$.

There are many non-trivial coupling $(\xi,P)$, and each one of them gives you a non-trivial approximation of what you are looking for. I would not be surprised if a limiting argument $\Delta t \to 0$ gives you a genuine example of a coupling of a Brownian motion and a Poisson process adapted to the same filtration: nevertheless, this does not seem to be a very tractable way of defining things -- this might be good for simulation purposes though, as I am guessing that there are some maths-finance questions related to this construction.

[Edit]: it seems like that we careful coupling $(\xi,P)$, you might be able to get a limiting process that has a generator equal to $$\mathcal{L}f(x,y) = f(x,y+1)-f(x,y) + \frac{1}{2} \partial^2_{xx}f(x,y) + K \cdot \Big(\partial_x f(x,y) - \partial_x f(x,y+1) \Big)$$ where $K \neq 0$ is a constant. I have not checked that $\mathcal{L}$ is actually a generator and that one can then define a Markov process from such a generator. If this happens to work, it seems like this is a good way to define such a coupling of a Brownian motion and a Poisson process.

Answer 4

Use any copula (fixed) (except the "independence copula" of course) over the marginals of the original independent bivariate process $(N_t,W_t)$ then you get two non-independent stochastic processes. On each coordinate you have respectively a BM and a Poisson Process but the joint law is determined by the copula you are using at each time.

here is a Phd thesis that explain all that in a rigourous way.

Regards