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Suppose I have a sequence of stochastic processes $X_{N}(t)$, $N=1,2,3,\ldots$ with mean zero and that I know for every fixed $t$, the random variable $X_{N}(t)$ converges in law to a Gaussian random variable with finite variance as $N \to \infty$. Suppose I also know that the covariance $\mathbb{E}(X_{N}(t_{1})X_{N}(t_{2}))$ converges to $\phi(t_{1},t_{2})$ for every fixed $t_{1}$ and $t_{2}$. Can I now claim that for any natural k, the random vector

$(X_{N}(t_{1}),X_{N}(t_{2}),\ldots,X_{N}(t_{k}))$

converges in law to the k-dimensional Gaussian random vector with covariance matrix $\Sigma_{i,j} = \phi(t_{i},t_{j})$? If not, what are some counterexamples to this statement?

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1 Answer 1

This has nothing to do with convergence, as there are counterexamples which do not depend on $N$. It also has nothing to do with time, as you didn't specify any condition that would restrict how the stochastic process depends on time. One can also work with just two times. The simplified question is then:

Suppose I have two variables $(X_1,X_2)$ that are individually Gaussian, and that have some variance-covariance matrix. Must they be the multivariate Gaussian distribution with that variance-covariance matrix?

Answer:

No. Let $X_1$ be Gaussian of mean $0$ and variance $1$, and let $X_2= \alpha X_1$, where $\alpha= \pm 1$ with equal probability and independent of $X_1$. Then $X_2$ is also Gaussian of mean $0$ and varaiance $1$, and they have covariance $0$, but they are not independent Gaussians.

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