David
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Registered User
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Mar 22 |
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Liverani’s CLT (a question) deleted 171 characters in body |
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Mar 22 |
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Liverani’s CLT (a question) Very interesting. I will edit the answer in a few minutes. Thanks a lot for your help. |
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Mar 22 |
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Liverani’s CLT (a question) $E[|D_{1}(\lambda_{n})|]\to E[|D_{1}|]$ if and only if $(D_{1}(\lambda_{n}))$ are uniformly integrable. I don't know if this is provable here. Without the absolute values, by the way, I don't see a light using those results in Billingsley. |
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Mar 22 |
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Liverani’s CLT (a question) Yes you are right. Here are my current findings in that direction: apparently one can not say that the weak limit is $D_{1}$ just from the fact that there is a.e convergence to $D_{1}$ and the weak limit exists. Here is the reason: you can repeat the argument above taking absolute values to conclude that $|D_{1}(\lambda_{n})|$ has a weak limit $f$. In particular $E[|D_{1}(\lambda_{n})|]\to E[f]$ (take the constant function $1=\xi_{\Omega}$). Now, clearly $|D_{1}(\lambda_{n})|\to |D_{1}|$ but, according to Billingsley's "Probability and Measure", corollary to Theorem 16.14 |
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Mar 22 |
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Liverani’s CLT (a question) (I think I saw your point. I will think about it) |
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Mar 22 |
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Liverani’s CLT (a question) added 32 characters in body |
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Mar 22 |
answered | Liverani’s CLT (a question) |
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Mar 19 |
awarded | ● Critic |
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Mar 19 |
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Liverani’s CLT (a question) Yes, that is precisely the point. As you see this is the launching point for the Martingale approximation to run, but I don't see how to justify that statement. |
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Mar 19 |
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Liverani’s CLT (a question) Better explanation on the nature of $D_{k}(\lambda)$, $Y_{k}(\lambda)$. |
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Mar 19 |
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Liverani’s CLT (a question) added 63 characters in body |
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Mar 19 |
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Liverani’s CLT (a question) Fixed grammar in last paragraph. |
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Mar 19 |
asked | Liverani’s CLT (a question) |
