Theorem 1.8(viii) Proof of Mathematical Statistics - Jun Shao

I have some questions about the proof of Theorem 1.8(viii) of Jun Shao's Mathematical Statistics.

The theorem is stating that If $X_n$ converges to $X$ in distribution, then for any $r > 0$, $lim_{n \rightarrow \infty} E\|X_n\|^r_r = E\|X\|^r_r < \infty$ if and only if ${\|X_n\|^r_r}$ is uniformly integrable.

My first question (highlighted in blue) is, why may we assume $X_n \rightarrow X$ almost surely by (iv)? The theorem guarantees only another set of random vectors $Y, Y_1, \cdots$ such that $Y_n \rightarrow Y$ almost surely.

My second question (highlighted in green) is, why is it that $\lim \sup_n E(\|X_n\|^r_r I_{B_n}) \leq \lim \sup_n E(\|X\|^r_r I_{B_n})$? There doesn't seem to be a statement that $X_n$ is monotone or ascending?

Thanks much in advance.

Here is the theorem and its corresponding proof:

1 Answer

For the blue question - this is a little subtle. Skorokhod's representation theorem indeed only gives a sequence $Y_n$ that converges a.s. to $Y$, and this may indeed be distinct from $X_n$ (roughly the point is that this representation doesn't need to preserve joint distributions, only the marginal for each $X_n$, so you may well have an $X_n$ that doesn't converge a.s. and only does in distro). However, in this case, all you care about are functionals on these marginal distributions, namely the $r$th moments of each of the $X_n$. So you might as well make the argument for this $Y_n$ and $Y$ - the $X_n$ will inherit the same since $P_{X_n} = P_{Y_n}$ for each $n$ and $P_X = P_Y$. This allows you to assume that $X_n \to X$ a.s. without loss of generality.

For the green question - no this doesn't imply monotonicity, and is just a property of sequences. Suppose for two sequences $a_n, b_n$ it holds that $a_n - b_n \to 0.$ Then for any $\varepsilon > 0,$ taking large enough $n$, $a_n \le b_n + \varepsilon$, and so $\limsup a_n \le \limsup b_n + \varepsilon$. But $\varepsilon$ is arbitrary, so this tells you that $\limsup a_n \le \limsup b_n$. In this case $a_n = \mathbb{E}[ \|X_n\|_r^r I_{B_n}], b_n = \mathbb{E}[\|X\|_r^r I_{B_n}].$