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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 < … < 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 set of n times depends on the values of at most 2n values of U.

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 < … < 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 set of n times depends on the values of at most 2n values of U.

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 $$ \mu_V = \mu + \mu_1\times\mu_2\times\cdots\times\mu_n $$ is a probability measure. 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.

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.

You cannot characterize a Lévy process by the individual distributions 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.

You cannot define a Lévy process by the individual distributions 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.