Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Last week, George Lowther provided a rather sophisticated counter-example of a continuous process $\{W(t):t \geq 0\}$ with $W(0)=0$ and $W(t)-W(s) \sim {\rm N}(0,t-s)$ for all $0 \leq s < t$, yet not a Brownian motion; see link text. Apparently, his approach relied heavily on special properties of BM.

Now, what about the analogue for a Poisson process: Can you find an example of a càdlàg (right-continuous with left limits) process $\{X(t):t \geq 0\}$ with $X(0)=0$ and $X(t)-X(s) \sim {\rm Poi}(t-s)$ for all $0 \leq s < t$, yet not being a Poisson process?

Bonus question: If the last question turns out to be too easy to answer (etc.), then what about the general Lévy process case? That is, given a Lévy process $X$ (defined below) with law $\mu_t$ at time $t>0$, does there exist a càdlàg process $\tilde X$ with $\tilde X(0) = 0$ and $\tilde X(t)-\tilde X(s) \sim \mu_{t-s}$ for all $0 \leq s < t$, which is yet not identical in law to $X$ (hence not a Lévy process)? [Here, assume that $X$ is non-deterministic, equivalently, $\mu_t$ is not a $\delta$-distribution.]

Definition: A stochastic process $X=\{X(t):t \geq 0\}$ is a Lévy process (say, real-valued) if the following conditions are satisfied: (1) $X(0)=0$ a.s.; (2) $X$ has independent increments; (3) $X$ has stationary increments; (4) $X$ is stochastically continuous; (5) Almost surely, the function $t \mapsto X(t)$ is right-continuous (for $t \geq 0$) and has left limits (for $t>0$). [In fact, condition (4) is implied by (1), (3), and (5).]

PS: you are still welcome to try and find a simpler counter-example for the Brownian motion case.

share|improve this question

2 Answers 2

up vote 7 down vote accepted

You cannot define a Lévy process by the individual distributions of its increments, except in the trivial case of a deterministic process Xt − X0 = bt with constant b. In fact, you can't identify it by the n-dimensional marginals for any n.

1) Let X be a nondeterministic Lévy process with X0 = 0 and n be any positive integer. Then, there is a cadlag process Y with a different distribution to X, but such that (Yt1,Yt2,…,Ytn) has the same distribution as (Xt1,Xt2,…,Xtn) for all times t1,t2,…,tn.

Taking n = 2 will give a process whose increments have the same distribution as for X.

The idea (as in my answer to this related question) is to reduce it to the finite-time case. So, fix a set of times 0 = t0 < t1 < t2 < … < tm for some m > 1. We can look at the distribution of X conditioned on the ℝm-valued random variable U ≡ (Xt1,Xt2,…,Xtm). By the Markov property, it will consist of a set of independent processes on the intervals [tk−1,tk] and [tm,∞), where the distribution of {Xt }t ∈[tk−1,tk] only depends on (Xtk−1,Xtk) and the distribution of {Xt }t ∈[tm,∞) only depends on Xtm. By the disintegration theorem, the process X can be built by first constructing the random variable U, then constructing X to have the correct probabilities conditional on U. Doing this, the distribution of X at any one time only depends on the values of at most two elements of U (corresponding to Xtk−1,Xtk). The distribution of X at any set of n times depends on the values of at most 2n values of U.

Choosing m > 2n, the idea is to replace U by a differently distributed ℝm-valued random variable for which any 2n elements still have the same distribution as for U. We can apply a small bump to the distribution of U in such a way that the m − 1 dimensional marginals are unchanged. To do this, we can use the following.

2) Let U be an ℝm-valued random variable with probability measure μ. Suppose that there exist (non-trival) measures μ12,…,μm on the reals such that μ1(A12(A2)…μm(Am) ≤ μ(A1×A2×…×Am) for all Borel subsets A1,A2,…,Am ⊆ ℝ. Then, there is an ℝm-valued random variable V with a different distribution to U, but with the same m − 1 dimensional marginal distributions.

By 'non-trivial' I mean that μk is a non-zero measure and does not consist of a single atom.

By changing the distribution of U in this way, we construct a new cadlag process with a different distribution to X, but with the same n dimensional marginals.

Proving (2) is easy enough. As μk are non-trivial, there will be measurable functions ƒk on the reals, uniformly bounded by 1 and such that μkk) = 0 and μk(|ƒk|) > 0. Replacing μk by the signed measure ƒk·μk, we can assume that μk(ℝ) = 0. Then $$ \mu_V = \mu + \mu_1\times\mu_2\times\cdots\times\mu_m $$ is a probability measure different from μ. Choosing V with this distribution gives $$ {\mathbb E}[f(V)]=\mu_V(f)=\mu(f)={\mathbb E}[f(U)] $$ for any function ƒ: ℝm → ℝ+ independent of one of the dimensions. So, V has the same m − 1 dimensional marginals as U.

To apply (2) to U = (Xt1,Xt2,…,Xtm), consider the following cases.

  1. X is continuous. In this case, X is just a Brownian motion (up to multiplication by a constant and addition of a constant drift). So, U is joint-normal with nondegenerate covariance matrix. Its probability density is continuous and strictly positive so, in (2), we can take μk to be a multiple of the uniform measure on [0,1].

  2. X is a Poisson process. In this case, we can take μk to be a multiple of the (discrete) uniform distribution on {2k,2k + 1} and, as X can take any increasing nonnegative integer-valued path on the times tk, this satisfies the hypothesis of (2).

  3. If X is any non-continuous Lévy process, case 2 can be used to change the distribution of its jump times without affecting the n dimensional marginals: Let ν be its jump measure, and A be a Borel set such that ν(A) is finite and nonzero. Then, X decomposes as the sum of its jumps in A (which occur according to a Poisson process of rate ν(A)) and an independent Lévy process. In this way, we can reduce to the case where X is a Lévy process whose jumps occur at a finite rate, with arrival times given by a Poisson process. In that case, let Nt be the Poisson process counting the number of jumps in intervals [0,t]. Also, let Zk be the k'th jump of X. Then, N and the Zk are all independent and, $$ X_t=\sum_{k=1}^{N_t}Z_k. $$ As above, the Poisson process N can be replaced by a differently distributed cadlag process which has the same n dimensional marginals. This will not affect the n dimensional marginals of X but, as its jump times no longer occur according to a Poisson process, X will no longer be a Lévy process.

share|improve this answer
    
It will take me a while to go over such an answer. –  Shai Covo Oct 26 '10 at 0:21
    
Very elegant solution! [Note the typo before "is a probability measure" (you may wish to add here "different from $\mu$"), where $n$ should be $m$.] –  Shai Covo Oct 27 '10 at 14:23
    
@Shai: Thanks. I've fixed the typo. –  George Lowther Oct 27 '10 at 14:37

Based on the comments to this answer, I no longer believe what I initially wrote (still appearing at the bottom of the answer). It seems to me a construction should be possible. It is at least possible in the case of a Binomial point process.

Let $\{X_i\}_{i \in \mathbb{N},i\neq 4}$ be independent Bernoulli$(1/2)$ random variables. Let $X_4'$ be Bernoulli$(1/2)$ and independent of the $X_i$. Then define $X_4$ as follows:

  • $X_4 = 1$ if $(X_1,X_2,X_3)$ is either $(0,0,1)$ or $(1,1,0)$,
  • $X_3 = 0$ if $(X_1,X_2,X_3)$ is either $(0,1,0)$ or $(1,0,1)$,
  • $X_4=X_4'$ otherwise.

Then for all $j \geq 0$, $n \geq 1$, $X_{j+1}+\ldots+X_{j+n}$ has Binomial$(n,1/2)$ distribution but the family $(X_n)_{n \in \mathbb{N}}$ are not iid.


What appears below (where I suggested such a construction was impossible) is false.


For a standard Poisson process, this won't be possible. (See this question and its answer.)

Edit: Given the comments perhaps I should provide more detail.

With probability one, for every pair $0 < p < q$, $p,q$ rationals, $X(q)−X(p)$ is a non-negative integer. Since X is cadlag the same property must hold for every real pair $0 < s < t$, i.e. $X$ is increasing and integer-valued.

Let us also show that $X$ has no jumps of size more than one: with probability one, for all $x > 0$, $X(x^-) := \lim_{y \uparrow x} X(y) \geq X(x)-1$. If this failed to hold then there would be $\epsilon > 0$ and $t < \infty$ so that $$ \mathbb{P}(\exists x \in [0,t), X(x)-X(x^-) \geq 2) > \epsilon/2. $$ But since $X$ is increasing, for any positive integer $n$ we can bound this probability from above by $$ \sum_{1 \leq i < 2n} \mathbb{P}(X((i+1)t/2n)-X((i-1)t/2n) \geq 2) $$ the point being that these intervals are chosen to overlap so that a jump of size $\geq 2$ must fall in at least one of them. Each of the differences above is distributed as Poisson$(t/n)$, so the associated probability is $o(n^{-1})$ as $n \to \infty$ and thus the whole sum tends to zero as $n \to \infty$.

We then know that a process $X$ such as you describe must be increasing and integer valued, with all jumps of size $1$. In other words, $X$ is a point process on $[0,\infty)$. Now the answer from the other thread implies that $X$ must be a rate one Poisson process.

share|improve this answer
1  
In our situation, we only know $P(X(A)=0)$ for sets $A$ of the form $A=(s,t]$, which cannot characterize the law of $X$. Hence, the question is still unanswered. –  Shai Covo Oct 25 '10 at 15:48
    
My last comment is incorrect, but the situation is only more complicated, as the process $X$ is not even known to be a point process (for example, the function $t \mapsto X(t)$ might not be monotone). –  Shai Covo Oct 25 '10 at 16:34
    
With probability one, for every pair $0 < p < q$, $p,q$ rationals, $X(q)-X(p)$ is a non-negative integer. Since $X$ is cadlag the same property must hold for every real pair $0 < s < t$, i.e. $X$ is increasing and integer-valued, so it is a point process. –  Louigi Addario-Berry Oct 25 '10 at 17:42
    
Of course, a trivial mistake of mine. It would have been correct if $X$ was only assumed cadlag with $X(0)=0$ and $X(t) \sim Poi(t)$. However, the question remains unanswered. –  Shai Covo Oct 25 '10 at 18:04
    
I've elaborated on my answer. If I'm misunderstanding something please let me know. –  Louigi Addario-Berry Oct 25 '10 at 18:37

Your Answer

 
discard

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.