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Jun 5, 2015 at 6:58 comment added Valentas Thank you for comments and the examples. I had actually realised that an arbitrary convergent subsequence was sufficient for what I wanted to do (apply a result for deterministic sequence to a random sequence).
Jun 5, 2015 at 6:44 vote accept Valentas
May 31, 2015 at 8:39 answer added Michael timeline score: 0
May 31, 2015 at 0:02 comment added Nate Eldredge @Michael: See Example 4 here.
May 31, 2015 at 0:01 comment added Michael @NateEldredge , what is the "standard typewriter sequence"?
Apr 30, 2015 at 15:58 comment added Nate Eldredge Is it even true for real-valued random variables $X_n$ that if $X_n \to X$ i.p. then you can find an a.s. convergent subsequences which consists of almost all members in your sense? Is it true for the standard "typewriter sequence" counterexample?
Apr 30, 2015 at 15:54 comment added Valentas By "consisting of almost all members" I mean is there, almost surely, a random set A such that $n^{-1} |A \cap \{1, .., n\}| \to 0$ and $\mu_{n, n\not \in A} \Rightarrow \mu$ (or something in this direction). Also I am interested in subsequences that converge to $\mu$ given in the assumption above.
Apr 30, 2015 at 15:38 comment added Nate Eldredge I am not really sure what you mean by "consisting of almost all members", can you be more explicit? You can certainly say that there is a subsequence $(\mu_{n_k})$ which converges weakly almost surely; this is a standard fact for real-valued random variables and the proof works for random variables taking values in any metrizable topological space, such as the weak topology on a bounded set of measures on a Polish space.
Apr 30, 2015 at 14:32 history edited Valentas CC BY-SA 3.0
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Apr 30, 2015 at 14:26 history edited Valentas CC BY-SA 3.0
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Apr 30, 2015 at 14:11 history asked Valentas CC BY-SA 3.0