# Equivalent Definitions of Gaussian Process?

The Gaussian process $$\{X_t\}_{t \in T}$$ ($$T=[0,1]$$ for example) is usually defined using its finite-dimensional distribution. I came across this statement many times: linear operator (not necessarily bounded) transformation of Gaussian process is also Gaussian. [See1,2,3]Thus, obviously, any bounded linear operator on Gaussian process is Gaussian. I am wondering if there exists another equivalent definition: $$X_t$$ is Gaussian process iff for any bounded linear operator $$A: \mathcal{H} \to \mathbb{R}^d$$ ($$\mathcal{H}$$ is the Hilbert space where $$X_t$$ lies in), $$A X_t$$ is a $$d$$-dimensional Gaussian. Any proof or reference on the statement I read? Is my guess on equivalent definition correct?

• A discontinuous (unbounded) linear functional of a Gaussian process need not be Gaussian; indeed, typically it won't even be measurable, so saying it's "Gaussian" has no meaning. – Nate Eldredge Apr 15 at 2:00
• @NateEldredge Any reference on this? And if the operator is bounded and linear, does the equivalence hold? – jwyao Apr 15 at 2:01
• But if you start with bounded linear functionals, the equivalence is clear. One direction is trivial by taking $d=1$. For the other direction, it's an elementary exercise to show that a finite dimensional random vector is Gaussian iff every linear functional of it is Gaussian (easy with Fourier transforms, for instance); and note that if $f : \mathbb{R}^d \to \mathbb{R}$ is linear then $f \circ A$ is a bounded linear functional on $\mathcal{H}$. – Nate Eldredge Apr 15 at 2:04
• – Nate Eldredge Apr 15 at 2:06
• @NateEldredge Thanks. But I still don't understand why bounded linear operator on Gaussian process is Gaussian. Can you explain more? – jwyao Apr 15 at 2:32