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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.

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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.

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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.

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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

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