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Julian Newman's user avatar
Julian Newman's user avatar
Julian Newman's user avatar
Julian Newman
  • Member for 13 years, 6 months
  • Last seen this week
  • London, UK
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Square-integral involving Brownian bridge
Just to clarify, are you expecting that there is some normalising factor $g(x)$ such that as $x \downarrow 0$, the distribution of the random variable $g(x)\int_0^{1-x} (B(t+x)-B(t))^2 \, dt$ converges to a non-trivial probability distribution on $[0,\infty)$? ["trivial" would just be convergence to 0.] And are you expecting that this limiting distribution can be expressed in some "explicit" manner by some formula potentially involving other well-established distributions? If so, are you able to provide some explanation or intuition behind why you're expecting these things?
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Does the "random Krylov-Bogolyubov theorem" hold in a non-skew-product setting?
minor correction to wording; added time-series tag
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Does the Krylov-Bogolyubov construction preserve "ergodic statistics"?
In your last sentence (before "The claim follows"), are you not assuming that $\mu$ is ergodic, as opposed to just invariant? [Incidentally, I think the dominated convergence theorem should imply that the whole sequence $\left(\frac{1}{N} \sum_{i=0}^{N-1} f^i_\ast\nu\right)_{N \geq 1}$ - without having to take a subsequence - will converge weakly to $\mu$ as $N \to \infty$, and so $\mathbb{P}$ is just equal to $\mu$.]
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Does the Krylov-Bogolyubov construction preserve "ergodic statistics"?
Thanks. I'm a bit confused - I think I've just realised that any example of the Bowen-Mañé phenomenon (physical measures that are not ergodic) will immediately be a counterexample to my claim, and yet you have proved that the claim is true.
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Does the "random Krylov-Bogolyubov theorem" hold in a non-skew-product setting?
weakened convergence criterion on $\nu$, added "full support" statements, clarified some parts of motivation
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Does the "random Krylov-Bogolyubov theorem" hold in a non-skew-product setting?
changed full convergence to Cesaro convergence in part about $\nu$
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Heuristic interpretations of the PA-unprovability of Goodstein's Theorem
Sorry, I'm asking questions that I should probably find the answer to by just learning some basic formal logic; I see that what I was suggesting in my above two comments is rubbish, as reflected in mathoverflow.net/q/331897/15570 and mathoverflow.net/a/40826/15570.
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Heuristic interpretations of the PA-unprovability of Goodstein's Theorem
I do realise that the way I had formulated my second concluding thought seemed to imply that the people who answered the linked question about Kronecker's acceptance of a proof of Goodstein's Theorem had incomplete understanding of the issue - I don't know if that is connected with the downvotes, but I've now re-written that bit.
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