# David

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 Name David Member for 2 years Seen Apr 5 at 1:14 Website Location Age
 Mar22 revised Liverani’s CLT (a question)deleted 171 characters in body Mar22 comment Liverani’s CLT (a question)Very interesting. I will edit the answer in a few minutes. Thanks a lot for your help. Mar22 comment 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. Mar22 comment 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 Mar22 comment Liverani’s CLT (a question)(I think I saw your point. I will think about it) Mar22 revised Liverani’s CLT (a question)added 32 characters in body Mar22 answered Liverani’s CLT (a question) Mar19 awarded ● Critic Mar19 comment 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. Mar19 revised Liverani’s CLT (a question)Better explanation on the nature of $D_{k}(\lambda)$, $Y_{k}(\lambda)$. Mar19 revised Liverani’s CLT (a question)added 63 characters in body Mar19 revised Liverani’s CLT (a question)Fixed grammar in last paragraph. Mar19 asked Liverani’s CLT (a question)