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I have seen this result stated countless times: assume the metric space $(\theta,d)$ is separable; then $(\theta,d)$ is complete if and only if the space $(\mathcal{P}(\Theta),\rho)$ (the space of probability measures, taken with the Prohorov metric--which is equivalent to the weak* topology) is complete. See, for instance,

  • Billingsley, Convergence of Probability Measures (1968), pg 240
  • W. Whitt, Weak Convergence of Probability Measures on the Function Space $C[0,\infty)$, Annals of Math. Stat 41 (1970), Corollary 2
  • http://www.math.leidenuniv.nl/~vangaans/jancol1.pdf , Theorem 9.2

But if this result holds (say, for $\Theta = \mathbb{R}$), then $(\mathcal{P}(\Theta),\rho)$ is weak* closed, which is false (see, e.g., milanmerkle.com/documents/radovi/WEACO2a.pdf , Section 5.4; a similar discussion has been had on these boards: Is the set of all probability measures weak*-closed?)

The proof of the first result typically relies on Prohorov's Theorem: take a Cauchy sequence $\{P_{n}\}$, show that the sequence is tight, therefore it is relatively sequentially compact. But in all the cases mentioned above, the authors use relative sequential compactness to conclude that the sequence must have a convergent subsequence in the space of probability measures (rather than the closure of that space). This seems to be the error, but the result is stated so ubiquitously that I feel I may be missing something....

UPDATE: I have located another related question: Tightness of a sequence of probability measures and weak convergence of a subsequence In Billingsley's proof (of my result), we take a Cauchy sequence $\{P_{n}\}$ and show that the sequence is tight, which is taken to prove the convergence of a subsequence. But this result seems incorrect, since tightness is only sufficient to prove the relative compactness of the set of measures in the sequence (by Prohorov's Theorem).

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Why "if the result holds, then $(\cal P(\Theta),\rho)$ is weak$^*$ closed"? –  Davide Giraudo Nov 29 '12 at 15:00
    
If $(\mathcal{P}(\Theta),\rho)$ is weak* complete, then it must be weak* closed –  Clay Nov 29 '12 at 15:41
    
You mean the weak star topology of which set? –  Davide Giraudo Nov 29 '12 at 20:37
    
I think this may be very closely related to the problem I am having... It is trivial that $(\mathcal{P}(\Theta),\rho)$ is weak* closed in $(\mathcal{P}(\Theta),\rho)$ (in the subspace topology). But perhaps $(\mathcal{P}(\Theta),\rho)$ is not weak* closed in the larger space of measures? It seems possible that this may explain the divergent results obtained above, but it is not clear (to me). –  Clay Nov 29 '12 at 21:02

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