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You cannot identify a Lévy process by the distribution of its increments, except in the trivial case of a deterministic process X_t − 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 (Y_t1,Y_t2,…,Y_tn) has the same distribution as (X_t1,X_t2,…,X_tn) for all times t₁,t₂,…,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) is to reduce it to the finite dimensional case. So, fix a set of times 0 = t₀ < t₁ < t₂ < … < t_m for some m > 1. We can look at the distribution of X conditioned on the ℝ^m-valued random variable U ≡ (X_t1,X_t2,…,X_tm). 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 ∈[tk−1,tk]} only depends on (X_tk−1,X_tk) and the distribution of {X_t }_{t ∈[tm,∞)} only depends on X_tm. 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 X_tk−1,X_tk). The distribution of X at any n times depends on the values of at most 2n values of U.

Choosing m > 2n, the idea is to replace U by a different distributed ℝ^m random variable for which any 2n 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^m-valued random variable with probability measure μ. Suppose that there exist finite (and non-trival) measures μ₁,μ₂,…,μ_m on the reals such that μ₁(A₁)μ₂(A₂)…μ_m(A_m) ≤ μ(A₁×A₂×…×A_m) for all Borel subsets A₁,A₂,…,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 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_k are non-trivial, there will by 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_kμ_k, we can assume that μ_k(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^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_t1,X_t2,…,X_tm), 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_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 {2k,2k + 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 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.