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My question is simple. Suppose that $X_1,X_2,\ldots$ is an infinite sequence of Rademacher random variables (i.e. $\mathbb{P}(X_1 = -1) = \mathbb{P}(X_1 = 1) = 1/2$). Let $S_n$ denote the random walk $S_n = \sum_{i=1}^n X_i$. By the central limit theorem, $n^{-1/2}S_n$ has a limiting $N(0,1)$ distribution. My question is:

Is it true that for all $k \geq 1$, $n^{-k/2}\mathbb{E}(S_n^k) \rightarrow \mathbb{E} N(0,1)^k$?

Finally, is the above true, if the $X_i$'s are general i.i.d. mean $0$, variance $1$ random variables?

abcd
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  • You forgot independence? – user10354138 Jun 25 '19 at 01:56
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    For the first question, the answer is yes, since all the moments exist and the limiting distribution is normal. For the second question, not in general, since the higher moments may not exist. – herb steinberg Jun 25 '19 at 02:14
  • Hi user10354138, sorry, I assume i.i.d :) – abcd Jun 25 '19 at 02:30
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    Not in general, the distributions must have uniformly small tails in some sense. But it is true for the Bernoulli trials into which the Rademacher case can be rescaled. – Conifold Jun 25 '19 at 02:55
  • In general, Bernstein's extension of the central limit theorem states that this is true for all i.i.d $X_i$ with finite $k$th moment – Bananach May 25 '22 at 08:10

1 Answers1

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It is known that if $(Y_n)_{n\geqslant 1}$ is a uniformly integrable sequence that converges in distribution to $Y$, then $\mathbb E[Y_n]\to\mathbb E[Y]$. We want to apply this to $Y_n=(n^{-1/2}S_n)^k$. Uniform integrability follows from the fact that $\sup_{n\geqslant 1}\lVert Y_n\rvert_2$, which can be deduced by Hoeffding's inequality.

In the general case, that is, when the random variables are not necessarily Rademacher, one needs more assumptions, first of all in order to make the expectation of $(n^{-1/2}S_n)^k$ meaningful. Here again, a uniform integrability argument applies. Suppose that $\mathbb E[\lvert X_1\rvert^k]$ is finite for each $k$. Marcinkiewicz's inequality applied with $p=k+1$ gives that $$ \mathbb E\left[\left\lvert S_n\right\rvert^{k+1}\right]\leqslant C_{k+1}\mathbb E\left[\left(\sum_{i=1}^n X_i^2\right)^{\frac{k+1}2}\right] =C_{k+1}\left\lVert \sum_{i=1}^n X_i^2\right\rVert_{(k+1)/2}^{(k+1)/2} \leqslant C_{k+1}\left(\sum_{i=1}^n\lVert X_i\rVert_{k+1}^2\right)^{(k+1)/2} =C_{k+1}n^{k+1}\mathbb E[\lvert X_1\rvert^k] $$ hence $\sup_{n\geqslant 1}\lVert n^{-1/2}S_n\rVert_{k+1}<\infty$, which gives uniform integrability of $(Y_n)_{n\geqslant 1}$.

Davide Giraudo
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