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Let $X_1, X_2, \dots, X_n$ be an iid sample from an unknown distribution finite mean $\mu$ and finite variance $\sigma^2$. Furthermore, let $R_1,R_2,\dots,R_n$. denote iid Rademacher random variables.

My goal is to compute the asymptotic distribution of $$T_n := \sqrt n\left(\frac 1n\sum_{i=1}^n R_i(X_i - \hat\mu)\right)$$ given the data $X_1, X_2,\dots, X_n$. Here $\hat\mu := n^{-1}\sum_{i=1}^nX_i$.

If the $R_i$'s are constant (e.g., $R_i = c$ for all $i$) and $\hat\mu$ is replaced by $\mu$, the CLT kicks in and I obtain that $T_n$ is asymptotically normal with mean $0$ and variance $\sigma^2$.

The mean of $T_n$ and variance is $\sigma^2$ conditional on the data equals $0$ and $\sigma^2$, respectively. But how do I obtain the asymptotic distribution of $T_n$ given the data?

StubbornAtom
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Syd Amerikaner
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1 Answers1

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$$ T_n=\sqrt n\left(\frac1n\sum_{i=1}^nR_i(X_i-\mu)\right)+\sqrt n\left(\frac1n\sum_{i=1}^nR_i(\mu-\hat\mu_n)\right), $$ where I denote $\hat\mu_n$ instead of $\hat\mu$ to highlight its dependence in $n$.

On the one hand, $(R_i(X_i-\mu))_{i\ge0}$ is a sequence of i.i.d. random variables with mean $0$ and variance $\sigma^2$, hence the central limit theorem yields $$ \sqrt n\left(\frac1n\sum_{i=1}^nR_i(X_i-\mu)\right)\overset{d}{\underset{n\to+\infty}{\longrightarrow}}\mathcal \sigma G, $$ where $G$ is a standard gaussian variable.

On the other hand, $$ \sqrt n\left(\frac1n\sum_{i=1}^nR_i(\mu-\hat\mu_n)\right)=\sqrt n(\mu-\hat\mu_n)\times\frac1n\sum_{i=1}^nR_i. $$

By the central limit theorem and the strong law of large numbers, $\sqrt n(\mu-\hat\mu_n)$ converges in distribution to $\sigma G$, and $\frac1n\sum_{i=1}^nR_i$ converges almost surely to $0$. By Slutsky's lemma, their product converges in distribution to $\sigma G\times0=0$.

By Slutsky's lemma again, $T_n$ converges in distribution to $\sigma G+0=\sigma G$. We deduce that $$ T_n\overset{d}{\underset{n\to+\infty}{\longrightarrow}}\mathcal \sigma G\sim\mathcal N(0,\sigma^2). $$

Will
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