# distribution of $T^2$ statistic without Gaussian assumption

I would like to know if the Gaussian assumption is needed on $x_i$ in deriving the asymptotic distribution of the $T^2$ statistic. Here is the presentation sequence I got from the Wiki page on Hotelling's statistic.

Suppose $x_1,\ldots, x_n \sim \mathcal{N}(\mu, \Sigma)$ and $x_i \in \mathbb{R}^p$. Let $\bar{x} := \sum_{i=1}^n x_i/n$. According to the page, it can be shown that

$$y := n(\bar{x}-\mu)^\top \Sigma^{-1} (\bar{x}-\mu) \sim \mathcal{\chi}^2_p,$$

where $\mathcal{\chi}^2_p$ is the chi-squared distribution with $p$ degrees of freedom. This result is obtained by showing that the characteristic function of $y$, which is denoted by $\phi_y(\theta)$, is that of $\chi^2_p$. To do so, one starts with

$$\phi_y(\theta) =\mathbb{E}_{\mathbb{\bar{x}} \sim \mathcal{N}(\mu, \Sigma/n)} \exp\left( i\theta n(\bar{x}-\mu)^\top \Sigma^{-1} (\bar{x}-\mu) \right),$$ which requires only the fact that $\bar{x} \sim \mathcal{N}(\mu, \Sigma/n)$.

Here is my question. Instead of $x_1,\ldots, x_n \sim \mathcal{N}(\mu, \Sigma)$, let's assume that $x_1, \ldots, x_n \sim q(x)$ for some arbitrary distribution $q(x)$, such that $\mathbb{E}_{x\sim q}[x] = \mu$ and $\mathbb{V}[x]=\Sigma$. Assume that $n \to \infty$, and that the central limit theorem holds so that $\bar{x} \sim \mathcal{N}(\mu, \Sigma/n)$. I understand that I still have $y \sim \chi^2_p$.

Is this true? If not, what would I need for it to be true? I am sure this question was considered in the statistics literature some time ago. Could you also please give me some references?

• The fact that $\overline{x}\sim\mathcal{N}_{p}\left(\mu\,,\,\Sigma/n\right)$ is due to the normality of the $x_{i}$'s as far as I know. It is quite unlikely that $\overline{x}$ will be of this form if the $x_{i}$'s are not normal. Of course, asymptotically, you have the normality and then I believe the Hotelling test applies. I think that, in general setups (where we don't make the assumption of normality), we use things like rank statistics,... when there is not much assumptions about the distributions. See, for example, "Theory of rank tests", from Hajek and Spivak. – MoebiusCorzer Dec 3 '15 at 17:25
• " asymptotically, you have the normality and then I believe the Hotelling test applies.": Do you have a reference for this statement? – wij Dec 3 '15 at 18:50
• I checked some lecture notes in multivariate analysis after I made my statement and my professor quotes this result (when $\Sigma$ is not known) but does not prove it, so it does probably exist but I haven't found any reference yet. – MoebiusCorzer Dec 3 '15 at 19:39