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  1. Let $X_n$ and $Y_n$ be two sequences of random variables such that $X_n\stackrel{n}{\rightarrow}C$ almost sure and $Y_n\stackrel{n}{\rightarrow}C$ almost sure, $C\in \mathbb{R}$. Suppose that $X_n$ is uniformly integrable (UI). Can we say that $Y_n$ is uniformly integrable?

Remark: I know that almost sure convergence doesn't imply in U.I, otherwise almost sure convergence would imply in moment's convergence. However, the situation here looks like different to me, because both sequences have the same constant as limit.

  1. Suppose that $\sqrt{n}[Z_n-\mu]\stackrel{D}{\rightarrow}N(0,\sigma)$ and $\{\sqrt{n}[Z_n-\mu]\}_{n}$ is uniformly integrable. Is $\{\sqrt{n}[F^-(Z_n)-F⁻(\mu)]\}_{n}$ uniformly integrable, provide that $F$ is an absolutely continuous distribution?

Remark: In this case, $F⁻$ is the $F$'s inverse function. By Mean Value Theorem: $$\sqrt{n}[F^-(Z_n)-F⁻(\mu)] = \frac{\sqrt{n}}{f(F⁻(\xi))}[Z_n-\mu],$$ where $\min\{Z_n,\mu\}<\xi<\max\{Z_n,\mu\}$. But I'm note sure if this information is useful to answer this question.

Thanks in advance.

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Your first question seems oddly phrased as there is no connection between the $X_n$ and the $Y_n$ (except that they have the same almost sure limit - but this is very weak).

In any case, the answer is no. Work with $[0,1]$ with Borel sets and Lebesgue measure. Put $X_n:= n^2 \mathbb{1}_{[0,1/n]}$. Then $X_n \rightarrow 0$ a.s., but $EX_n = n$, so $(X_n)$ is not bounded in $L^1$. It consequently isn't UI. Put $Y_n := 0$ identically. Then $(Y_n)$ converges a.e. to $0$ and is UI.

Do you see what I mean by there being no connection?

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  • $\begingroup$ I've understood your answer. Indeed, it is clear to me now. The question 1 was my first doubt when the second question arose in my research. Some answer to the question 2 is my final goal. Thanks a lot. $\endgroup$ – S. W. M Apr 5 '15 at 4:22
  • $\begingroup$ Your second question is weird because there is no reason for $F$ to have an inverse. $\endgroup$ – Frank Apr 5 '15 at 10:41
  • $\begingroup$ Hi, Mr Frank. I've been reading about random variables based on F's inverse distribution. The conditions that imply the uniform integrability are clear to me when the random variable involves the distribution F, so I've been thinking how stablish the uniform integrability if the r.v above is based on F^{-}. The requirement "F is an absolutely continuous" looks very convenient because it can allow us to use (somehow) the (F^{-})'s regularity. Thanks for the feedback. $\endgroup$ – S. W. M Apr 7 '15 at 14:49

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