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](https://mathoverflow.net/questions/43015/the-conditions-in-the-definition-of-brownian-motion). 
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. 



The conditions in the definition of Poisson process (and a Lévy process generalization)
You cannot define a Lévy process by the individual distributions of its increments, except in the trivial case of a deterministic process *X**t* − *X*0 = *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 *X*0 = 0 and *n* be any positive integer. Then, there is a cadlag process *Y* with a different distribution to *X*, but such that (*Y**t*1,*Y**t*2,…,*Y**t**n*) has the same distribution as (*X**t*1,*X**t*2,…,*X**t*n) for all times *t*1,*t*2,…,*t**n*.
> 
> 
> 


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](https://mathoverflow.net/questions/43015/the-conditions-in-the-definition-of-brownian-motion)) is to reduce it to the finite-time case. So, fix a set of times 0 = *t*0 < *t*1 < *t*2 < … < *t**m* for some *m* > 1.
We can look at the distribution of *X* conditioned on the ℝ*m*-valued random variable *U* ≡ (*X**t*1,*X**t*2,…,*X**t**m*). By the Markov property, it will consist of a set of independent processes on the intervals [*t**k*−1,*t**k*] and [*t**m*,∞), where the distribution of {*X**t* }*t* ∈[*t**k*−1,*t**k*] only depends on (*X**t**k*−1,*X**t**k*) and the distribution of {*X**t* }*t* ∈[*t**m*,∞) only depends on *X**t**m*. By the [disintegration theorem](http://en.wikipedia.org/wiki/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 *X**t**k*−1,*X**t**k*). The distribution of *X* at any set of *n* times depends on the values of at most 2*n* values of *U*.


Choosing *m* > 2*n*, the idea is to replace *U* by a differently distributed ℝ*m*-valued random variable for which any 2*n* 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 ℝ*m*-valued random variable with probability measure μ. Suppose that there exist (non-trival) measures μ1,μ2,…,μ*m* on the reals such that μ1(*A*1)μ2(*A*2)…μ*m*(*A**m*) ≤ μ(*A*1×*A*2×…×*A**m*) for all Borel subsets *A*1,*A*2,…,*A**m* ⊆ ℝ.
>  Then, there is an ℝ*m*-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 μ*k* 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 μ*k* are non-trivial, there will be measurable functions ƒ*k* on the reals, uniformly bounded by 1 and such that μ*k*(ƒ*k*) = 0 and μ*k*(|ƒ*k*|) > 0. Replacing μ*k* by the signed measure ƒ*k*·μ*k*, we can assume that μ*k*(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: ℝ*m* → ℝ+ independent of one of the dimensions. So, *V* has the same *m* − 1 dimensional marginals as *U*.


To apply (2) to *U* = (*X**t*1,*X**t*2,…,*X**t**m*), 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 μ*k* to be a multiple of the uniform measure on [0,1].
2. *X* is a Poisson process. In this case, we can take μ*k* to be a multiple of the (discrete) uniform distribution on {2*k*,2*k* + 1} and, as *X* can take any increasing nonnegative integer-valued path on the times *t**k*, 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 *N**t* be the Poisson process counting the number of jumps in intervals [0,*t*]. Also, let *Z**k* be the *k*'th jump of *X*. Then, *N* and the *Z**k* 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.


