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Let $Y_t= X_t - t X_1 -(1-t)X_0$ be a sample-continuous stochastic process defined on $[0,1]$. If $(Y_t)$ is Gaussian, does it imply that $(X_t)$ is Gaussian on $(0,1)$ ?

It feels wrong but I am unable to construct a counterexample.

Here are some preliminaries:

A time continuous stochastic process $\left\{X_{t} ; t \in T\right\}$ is Gaussian if and only if for every finite set of indices $t_{1}, \ldots, t_{k}$ in the index set $T$, $$\mathbf{X}_{t_{1}, \ldots, t_{k}}=\left(X_{t_{1}}, \ldots, X_{t_{k}}\right)$$

is a multivariate Gaussian random variable. That is the same as saying every linear combination of $\left(X_{t_{1}}, \ldots, X_{t_{k}}\right)$ has a univariate Gaussian distribution.

Using characteristic functions of random variables, the Gaussian property can be formulated as follows: $\left\{X_{t} ; t \in T\right\}$ is Gaussian if and only if, for every finite set of indices $t_{1}, \ldots, t_{k}$, there are real-valued $\sigma_{\ell j}, \mu_{\ell}$ with $\sigma_{j j}>0$ such that the following equality holds for all $s_{1}, s_{2}, \ldots, s_{k} \in \mathbb{R}$ \begin{equation*} \mathrm{E}\left(\exp \left(i \sum_{\ell=1}^{k} s_{\ell} \mathbf{X}_{t_{\ell}}\right)\right)=\exp \left(-\frac{1}{2} \sum_{\ell, j} \sigma_{\ell j} s_{\ell} s_{j}+i \sum_{\ell} \mu_{\ell} s_{\ell}\right) . \end{equation*} where $i$ denotes the imaginary unit such that $i^{2}=-1$. The numbers $\sigma_{\ell j}$ and $\mu_{\ell}$ can be shown to be the covariances and means of the variables in the process.

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1 Answer 1

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Here is a counterexample: Suppose, that $$X_t = W_t + tZ,$$ where $W_t$ is a Gaussian proces and $Z$ is any non-Gaussian random variable (independent of $W_t$), then $X_t$ is not Gaussian, however \begin{align*} Y_t &= X_t - tX_1 - (1-t)X_0 \\ &= (W_t + tZ) - t(W_1+Z) - (1-t)(W_0+0) \\ &= W_t - tW_1 - (1-t)W_0 \end{align*} is indeed a Gaussian proces.

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