Let $S_n = X_1 + X_2 + \cdots + X_n$ where $X_1, x_2, \dots, X_n$ are iid random variables with mean $\mu$ and variance $\sigma^2$. Then Theorem 5.1 in Gut's Probability - A Graduate Course book, says that if $E[X_i^4] < \infty$ and $r \ge 2$, as $n \to \infty$, $$ E\bigg[\bigg|\frac{S_n - n \mu}{\sigma \sqrt{n}}\bigg|^r\bigg] \to E[N(0,1)]^r, $$ or equivalently $$ E[|S_n - n \mu|^r] \to \sigma^r n^{r/2} E[N(0,1)]^r. $$ Does this allow us to say anything about the magnitude of $E[|S_n|^r]$ as $n \to \infty$ when $\mu \neq 0$? It seems it should be something like $E[|S_n|^r] = O(n^r)$?
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Remember that $\|Y\|_{r}:=E[|Y|^r]^{1/r}$ is a semi-norm on the space of random variables with finite $r$-th moment.
The theorem in Gut rewrites as $\|\frac{S_n - n \mu}{\sigma \sqrt{n}}\|_r \to \|N(0,1)\|_r$. By the reverse triangle inequality, $$\|\frac{S_n }{\sigma \sqrt{n}}\|_r \leq \|\frac{S_n - n \mu}{\sigma \sqrt{n}}\|_r + \frac{\sqrt n \mu}{\sigma} = O(\sqrt n)$$ thus $$\|S_n\|_r = O(n)$$ and $$E[|S_n|^r] = O(n^r).$$
Gabriel Romon
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Interesting, thanks. Do you think it this could be tightened a bit more so that we can say that $E[|S_n|^r] \sim C n^r$ for $C \in \mathbb{R}$ at leading order? As it stands $E[|\frac{S_n }{\sigma \sqrt{n}}|^r] = O(n^r)$ leaves open the possibility that $E[|\frac{S_n }{\sigma \sqrt{n}}|^r] \sim C n^s$ where $s < r$. It would be really nice if we could firmly say that at leading-order $E[|S_n|^r] \sim C n^r$. – sonicboom Apr 08 '21 at 16:33