You cannot identify a Lévy process by the distribution of its increments, except in the trivial case of a deterministic process *X*<sub>*t*</sub> − *X*<sub>0</sub> = *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*<sub>0</sub> = 0 and *n* be any positive integer. Then, there is a cadlag process *Y* with a different distribution to *X*, but such that (*Y*<sub>*t*<sub>1</sub></sub>,*Y*<sub>*t*<sub>2</sub></sub>,…,*Y*<sub>*t*<sub>*n*</sub></sub>) has the same distribution as (*X*<sub>*t*<sub>1</sub></sub>,*X*<sub>*t*<sub>2</sub></sub>,…,*X*<sub>*t*<sub>n</sub></sub>) for all times *t*<sub>1</sub>,*t*<sub>2</sub>,…,*t*<sub>*n*</sub>.
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][1]) is to reduce it to the finite dimensional case. So, fix a set of times 0 = *t*<sub>0</sub> < *t*<sub>1</sub> < *t*<sub>2</sub> < … < *t*<sub>*m*</sub> for some *m* > 1.
We can look at the distribution of *X* conditioned on the ℝ<sup>*m*</sup>-valued random variable *U* ≡ (*X*<sub>*t*<sub>1</sub></sub>,*X*<sub>*t*<sub>2</sub></sub>,…,*X*<sub>*t*<sub>*m*</sub></sub>). By the Markov property, it will consist of a set of independent processes on the intervals [*t*<sub>*k*−1</sub>,*t*<sub>*k*</sub>] and [*t*<sub>*m*</sub>,∞), where the distribution of {*X*<sub>*t*</sub> }<sub>*t* ∈[*t*<sub>*k*−1</sub>,*t*<sub>*k*</sub>]</sub> only depends on (*X*<sub>*t*<sub>*k*−1</sub></sub>,*X*<sub>*t*<sub>*k*</sub></sub>) and the distribution of {*X*<sub>*t*</sub> }<sub>*t* ∈[*t*<sub>*m*</sub>,∞)</sub> only depends on *X*<sub>*t*<sub>*m*</sub></sub>. By the [disintegration theorem][2], 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*<sub>*t*<sub>*k*−1</sub></sub>,*X*<sub>*t*<sub>*k*</sub></sub>). The distribution of *X* at any *n* times depends on the values of at most 2<i>n</i> values of *U*.
Choosing *m* > 2<i>n</i>, the idea is to replace *U* by a different distributed ℝ<sup>*m*</sup> random variable for which any 2*n* elements still have the same distribution as *U*. The idea is to 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<sup>*m*</sup>-valued random variable with probability measure μ. Suppose that there exist finite (and non-trival) measures μ<sub>1</sub>,μ<sub>2</sub>,…,μ<sub>*m*</sub> on the reals such that μ<sub>1</sub>(*A*<sub>1</sub>)μ<sub>2</sub>(*A*<sub>2</sub>)…μ<sub>*m*</sub>(*A*<sub>*m*</sub>) ≤ μ(*A*<sub>1</sub>×*A*<sub>2</sub>×…×*A*<sub>*m*</sub>) for all Borel subsets *A*<sub>1</sub>,*A*<sub>2</sub>,…,*A*<sub>*m*</sub> ⊆ ℝ.
Then, there is an ℝ<sup>*m*</sup>-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<sub>*k</sub> is a on-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<sub>*k*</sub> are non-trivial, there will by measurable functions ƒ<sub>*k*</sub> on the reals, uniformly bounded by 1 and such that μ<sub>*k*</sub>(ƒ<sub>*k*</sub>) = 0 and μ<sub>*k*</sub>(|ƒ<sub>*k*</sub>|) > 0. Replacing μ<sub>*k*</sub> by the signed measure ƒ<sub>*k*</sub>μ<sub>*k*</sub>, we can assume that μ<sub>*k*</sub>(ℝ) = 0.
Then
$$
\mu_V = \mu + \mu_1\times\mu_2\times\cdots\times\mu_n
$$
is a probability. Choosing *V* with this distribution gives
$$
{\mathbb E}[f(V)]=\mu_V(f)=\mu(f)={\mathbb E}[f(U)]
$$
for any function ƒ: ℝ<sup>*m*</sup> → ℝ<sup>+</sup> independent of one of the dimensions. So, *V* has the same *m* − 1 dimensional marginals as *U*.
To apply (2), to *U* = (*X*<sub>*t*<sub>1</sub></sub>,*X*<sub>*t*<sub>2</sub></sub>,…,*X*<sub>*t*<sub>*m*</sub></sub>), 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 a joint-normal with nondegenerate covariance matrix. Its probability density is continuous and strictly positive so, in (2), we can take μ<sub>*k*</sub> to be a multiple of the uniform measure on [0,1].
2. *X* is a Poisson process. In this case, we can take μ<sub>*k*</sub> to be a multiple of the (discrete) uniform distribution on {2<i>k</i>,2<i>k</i> + 1} and, as *X* can take any increasing nonnegative integer-valued path on the times *t*<sub>*k*</sub>, 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*<sub>*t*</sub> be the Poisson process counting the number of jumps in intervals [0,*t*]. Also, let *Z*<sub>*k*</sub> be the *k*'th jump of *X*. Then, *N* and the *Z*<sub>*k*</sub> 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 distributions of *X* but, as the jump times of *X* no longer occur according to a Poisson process, *X* will no longer be a Lévy process.
[1]: http://mathoverflow.net/questions/43015/the-conditions-in-the-definition-of-brownian-motion
[2]: http://en.wikipedia.org/wiki/Disintegration_theorem