by wij
Last Updated April 27, 2017 23:19 PM

Let $p(x)$ be an arbitrary distribution defined on $\mathbb{R}^d$. Define $\mu = \mathbb{E}[x]$. Given an i.i.d. sample $x_1, \ldots, x_n \sim p(x)$, consider the following $T^2$ statistic for testing the hypotheses $H_0: \mu=0$ vs $H_1: \mu\neq0$.

$$T^2 = n\bar{x}^\top S^{-1} \bar{x},$$

where $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ and $S = \frac{1}{n-1} \sum_{i=1}^n (x_i-\bar{x})(x_i-\bar{x})^\top$. It is known that if $\mu=0$ (i.e., under $H_0$) as $n\to\infty$ the statistic $T^2$ follows $\chi^2(d)$, a central chi-squared distribution with $d$ degrees of freedom.

I am interested in showing that the CDF of $T^2$ when $\mu\neq0$ roughly depends on only $\lambda_n = n\mu^\top \Sigma^{-1} \mu$ where $\Sigma = \mathbb{E}[(x-\mu)(x-\mu)^\top]$ for $n$ that is sufficiently large. How can I show this for an arbitrary $p(x)$? I am actually not interested in seeing the CDF, just the fact that it depends on only $\lambda_n$.

Here is what I know. If $p(x) = \mathcal{N}(x|\mu, \Sigma)$, then it is known that under $H_1$, $T^2 \sim \chi^2(d, \lambda_n)$, a non-central chi-squared distribution with $d$ degrees of freedom, and non-centrality parameter $\lambda_n$. If $p(x)$ is not normal, then intuitively for large $n$, $\bar{x}$ will roughly follow a normal distribution by the central limit theorem (CLT), and the resulting distribution of $T^2$ should get closer to $\chi^2(d, \lambda_n)$ as $n$ increases. The problem is to mathematically quantify the phrase "roughly follow a normal distribution". Technically, if $n$ is fixed, the CLT does not hold. On the other hand, if $n\to\infty$, then $\lambda_n\to\infty$.

To quantify the distribution of $\bar{x}$ to the target normal distribution, Berry-Esseen bound came to my mind. However, I am not sure how to bring it into this context. Any advice? Thanks.

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