For proving the convergence of a certain sequence of stochastic processes (which take values on a compact set), I am taking the following approach (as taken in previous papers I am looking at):

1) First proving the sequence of stochastic processes is tight.

2) Having proven tightness, I try to identify a possible limit of the sequence of stochastic processes (e.g. using tools like Donsker's invariance principle).

I want to understand why, with the relevant additional conditions, these two steps are sufficient for proving that the sequence of stochastic processes does indeed converge to the possible limit (which is also a stochastic process) I find in step 2 above. My understanding so far:

By Prohorov's theorem on a family $\mathcal{M}$ of probability measures on a complete, separable metric space (S,d), tightness is equivalent to relative compactness. Here relative compactness is equivalent to saying every sequence in $\mathcal{M}$ has a subsequence which converges in $\mathcal{M}_1(S)$ (the complete space of probability measures in $(S,\mathcal{B}(S)))$. So taking the family of probability measures to be the laws of the stochastic processes in the sequence gives: if the sequence of stochastic processes is tight, then there is a subsequence which has a weak convergence limit.

However, at this point I am lost- how can I now show that the entire sequence (rather than one of its subsequences) has a weak convergence limit? Do we need some form of Cauchy criterion now to be satisfied by the sequence, so that the subsequence limit turns out to indeed be the sequence limit under completeness?



1 Answer 1


There is the following general statement

Let $(X,d)$ be a metric space and $(x_n)_{n \in \mathbb{N}} \subseteq X$. Then $x_n$ converges (in $X$) if, and only if, for any subequence of $(x_n)_{n \in \mathbb{N}}$ there exists a (further) subsequence which converges to a limit $x \in X$ and this limit does not depend on the chosen subsequence.

The proof is not difficult; the implication "$\Rightarrow$" is obvious (if the sequence converges, then any subsequence converges and the limit does not depend on the chosen subsequence) and "$\Leftarrow$" can be proved by contradiction.

Applying this statement in your framework, we can proceed as follows to prove the weak convergence of a sequence of probability measures, say $(\mu_n)_{n \in \mathbb{N}}$:

  1. Fix an arbitrary subsequence $(\mu_{n_k})_{k \in \mathbb{N}}$.
  2. Using compactness, show that this subsequence admits a convergent subsequence $(\mu_{n_{k_{\ell}}})_{\ell \in \mathbb{N}}$.
  3. Identify the possible limit of the sequence to conclude that the limit of the (convergent) subsequence does not depend on the chosen subsequence.

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