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?
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1$\begingroup$ 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. $\endgroup$– Nate EldredgeCommented Apr 15, 2019 at 2:00
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$\begingroup$ @NateEldredge Any reference on this? And if the operator is bounded and linear, does the equivalence hold? $\endgroup$– jwyaoCommented Apr 15, 2019 at 2:01
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$\begingroup$ 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 $\mathcal{H}$. $\endgroup$– Nate EldredgeCommented Apr 15, 2019 at 2:04
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$\begingroup$ mathoverflow.net/questions/256541/…. $\endgroup$– Nate EldredgeCommented Apr 15, 2019 at 2:06
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$\begingroup$ @NateEldredge Thanks. But I still don't understand why bounded linear operator on Gaussian process is Gaussian. Can you explain more? $\endgroup$– jwyaoCommented Apr 15, 2019 at 2:32
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