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Yes, without any additional assumptions — the relevant technical conditions are satisfied because your limit distribution is normal. For sufficiency in the univariate case, see any probability textbo …

More generally, given $p > 1$, take any bounded function on $\mathbb{R}$ which behaves like $1/|x|^p$ as $x\to \infty$, for example $1/(1+|x|^p)$. After renormalizing, this is will be the density of a …

You're right that things change for m>1; I was thinking sloppily. Assume $U=\{1,\ldots,n\}$ for concreteness. If $Y_1,\ldots,Y_m$ are chosen independently and uniformly from $U$, then for any $k_1,\ …

D'Aristotile, Diaconis, and Freedman called this "merging" of probability measures, and studied relationships among different notions of it in this paper.

As you suggest, the fact that $e^x$ is increasing is a useful property here. In particular, that lets you in some cases to apply Markov's/Chebyshev's inequality to the random variable $e^{tX}$ in ord …

I think this isn't hard if you don't care at all about covariance structure or regularity of $Z_t$. For any given $t$, your formula defines a valid cumulative distribution function, so such a random …

Any class of distributions which is closed under independent sums and almost surely nonzero will work here (and of course will also give an example for geometric means corresponding to the log-normal) …

The easiest condition would be a bound on $\sup_i \mathbb{E} X_i^6$, which would allow you to apply the Berry–Esseen theorem. More generally, if for some $0<\varepsilon < 2$ you have a uniform bound …

The answer to your first question is yes. See Chapter 19 of *Normal Approximation and Asymptotic Expansions by Bhattacharya and Rao. I believe the answer to your second question is no, but don't kno …

For the expected norm see equation (19) in this paper (or equation (12) in the published version of that paper), which, after some renormalizing, states that $$ \mathbb{E} x_1^{r_1} \cdots x_n^{r_n} = …

No. Let $X$ be, say, a standard normal random variable, $Z$ an independent random variable with $P[Z=1] = P[Z=-1] = 1/2$, and $Y=XZ$. Then $X$ and $Y$ are uncorrelated standard normal (in particular …

Stein's method doesn't give total variation approximation in one dimension, either, without some kind of additional assumptions. This has nothing to do with Stein's method; for an impossibility result …

Edit: The "second moment method" you've stated is false, as shown for example by $P(X=1)=p$ and $P(X=0)=1-p$ with $p>1/2$. See this Wikipedia article for a discussion of the more complicated inequalit …

Your question is part of what's called Littlewood-Offord theory, which has seen a lot of progress lately in work of Tao and Vu and of Rudelson and Vershynin. Take a look at Section 1.2 and especially …

There is a Hilbert space interpretation of independence, which follows from the interpretation of conditional expectation as an orthogonal projection, though it may be more complicated than you had in …