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Suppose a stochastic process is both independent-increment and Gaussian. Are all its increments Gaussian distributed?

Thanks!

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Yes, they are. The independence of the increments is not even necessary.

Let $(X_t)_{t \geq 0}$ a Gaussian stochastic process. Suppose we want to prove that the random vector $Z:=(X_{t_1}-X_{t_{0}},X_{t_{2}}-X_{t_{1}},\ldots,X_{t_{n}}-X_{t_{n-1}})^T$ is Gaussian where $0 \leq t_0 < t_1 <\ldots <t_n$, $n \in \mathbb{N}$ arbritary.

Since $(X_t)_{t \geq 0}$ is Gaussian the random vector $(X_{t_0},\ldots,X_{t_n})$ is Gaussian. Moreover, linear transforms of Gaussian random vectors are still Gaussian. Since $$Z = \begin{pmatrix} X_{t_1}-X_{t_0} \\ \vdots \\ X_{t_n}-X_{t_{n-1}} \end{pmatrix}= A \cdot \begin{pmatrix} X_{t_0}\\ \vdots \\ X_{t_n} \end{pmatrix}$$ where $$A := \begin{pmatrix} -1 & 1 & 0 & 0 & \ldots 0 \\ 0 & -1 & 1 & 0 & \ldots 0 \\ & \ddots & \ddots & \ddots & \\ 0 & 0 & 0 & -1 & 1 \end{pmatrix}$$

we conclude that $Z$ is Gaussian.

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