MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 and was suggested to post it here :

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).

share|cite|improve this question
Here's how I would approach it. I would begin with a standard Brownian motion $W_t$, and an independent Poisson process $A_t$. I would try to "mix" the processes $W_t$ and $A_t$ together in a way that the resulting process, $N_t$, is still marginally a Poisson process. How should one try this mixing? Well, maybe convolutions with white noise, stochastic integrals, etc. I hope this helps. – Tom LaGatta Mar 8 '11 at 17:37
Any two Levy processes starting from zero and defined with respect to the same filtration are independent if and only if their quadratic covariation is (almost surely) zero. So, the quick answer to your question is no. On the other hand, there are constructions of a non-independent BM and Poisson process if they are not required to have independent increments wrt their joint filtration (As The Bridge's answer shows). – George Lowther Mar 8 '11 at 19:04
@George : do you have a reference for that fact ? – pgassiat Mar 8 '11 at 20:03
The OP should clarify what "$N_t$ is Poisson process with respect to $({\cal F}_t)$" means. The usual definition of Poisson process does not use a filtration. – Byron Schmuland Mar 8 '11 at 21:02
@Byron, @Ori: it means that it's adapted and that $N_t-N_s$ is independent of $\mathcal{F}_s$ for all $s < t$. – George Lowther Mar 9 '11 at 0:33
up vote 13 down vote accepted

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


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


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.

share|cite|improve this answer
This looks good, thanks ! – pgassiat Mar 8 '11 at 19:08
@kakuritsu Are the processes $(W_t)_t$ and $(N_t)_t$ independent in this example? (This was the OP original question.) – Did Mar 8 '11 at 20:51
@Didier Piau: seems like they are -- the same computation shows that all the finite dimensional laws are those of independent (Brownian, Poisson). – Alekk Mar 8 '11 at 22:01
@kakuritsu @Alekk Indeed the result is true and I know how to adapt the argument proving that $W_t$ and $N_t$ are independent for any given $t$, to prove the processes are independent. But this is not done in the post above, yet (the added paragraph merely acknowledges the problem, which is already something, but it does not provide a proof). Hence my first comment. – Did Mar 9 '11 at 9:18

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).

share|cite|improve this answer
Great answer !! – The Bridge Mar 9 '11 at 21:36
@ George. Interesting. Do you think this same theorem generalizes to semimartingales? I can see one could run through the proof by using the cumulant process instead of the Levy exponent. But then you somehow get stuck in getting useful information on the marginal processes, since you do'nt end up with a deterministic compensator and thus a characteristic function. Maybe you really need independence of increments for the proof. – user26695 Sep 21 '12 at 14:56

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.

share|cite|improve this answer
meditating on George Lowther and Kakuritsu, I am still wondering what is going wrong with the Markov process given by the above generator: is this simply not a real "Markov generator" so that this process simply does not exists (Hille-Yoshida does not apply)? Or is it more subtle that that ? – Alekk Mar 8 '11 at 19:34
@Alekk: No it's not a Markov generator, because of the $\partial\_xf(x,y+1)$ term. – George Lowther Mar 9 '11 at 1:47

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.


share|cite|improve this answer
Thanks for the suggestion and the reference. I am not familiar with copulas so it is not very clear to me how your construction works. In the reference, there is no construction of a bivariate stochastic process from two independent processe and a fixed copula, but only construction of one process from marginals and copulas for $t_1,\ldots,t_k$. In particular it is not obvious to me why the "new" $N$ and $W$ remain Poisson and Brownian. – pgassiat Mar 8 '11 at 13:47
What is clear though is that for every $t$ $W_t$ is gaussian and $N_t$ follows a Poisson law. Moreover by hypothesis $W_t$ is a Brownian Motion when forgetting about $N_t$ and "vice et versa" when forgetting about $W$, $N_t$ is a Poisson process. Isn't what you ware looking for ? So in my example, in a way what you are looking for is true only on average. – The Bridge Mar 8 '11 at 15:25
@Bridge : what I am looking for is $W$ and $N$ being BM and Poisson wrt to the same filtration. So the fact that $W$ is BM when forgetting about $N$ is not enough. Do you think that this goal is achievable with a copula construction like you described ? – pgassiat Mar 8 '11 at 15:54
Well that's a good question !! It is clear that it won't work for any copula as knowing the trajectory of N_t modifies the law of W_t (even if it is still gaussian). The problem is about independence of increments of W_t (and vice versa for the N_t process). – The Bridge Mar 8 '11 at 16:33
BTW could precise exactly what you mean by "wrt the same filtration". Regards – The Bridge Mar 8 '11 at 16:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.