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Let $\left(X_n\right)_{n\geqslant 1}$ be a tight sequence of stochastic processes, so there exists a weakly convergent subsequence. Is it possible to show that the entire sequence converges weakly?

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  • $\begingroup$ Isn't tightness a description that applies to metric-space-valued RVs? I don't see how to think of a stochastic process as one such, unless you mean a discrete time process in which case this still seems awkward. $\endgroup$
    – Jeff
    Sep 30 '13 at 3:01
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In general no, even for real values sequences. For example, take two different probability distribution $\mu_1$ and $\mu_2$ and a sequence $\left(\nu_n\right)_{n\geqslant 1}$ such that $\nu_n=\mu_1$ for $n$ odd and $\nu_n=\mu_2$ for $n$ even. The sequence is tight but does not converge in distribution.

In order to show that the whole sequence converges weakly, we have to prove that each subsequence converge to the same limit. Usually (depending on the function space we work with), this can be done by considering the finite dimensional distributions.

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