# Distribution of multivariate “$Z$-score”?

Suppose $$\mathbf{X}_1, \dots, \mathbf{X}_n \sim N_p(\mathbf{\mu}, \Sigma)$$ where $$\mu \in \mathbb{R}^p$$ and $$\Sigma$$ is a $$p \times p$$ covariance matrix.

Suppose $$\hat{\Sigma}$$ is the sample covariance matrix, and $$\bar{\mathbf{X}}$$ is the sample mean, then we know that

$$n(\mathbf{\bar{X}} - \mu)^T \hat{\Sigma}^{-1}(\mathbf{\bar{X}} - \mu) \sim T^2_{p,n-1}\,,$$ where $$T^2_{p,n-1}$$ is the Hotelling T-squared distribution with dimensionality parameter $$p$$ and degrees of freedom $$n-1$$. Discussion on this can be found here. There is also an alternative $$F$$-distribution representation of the Hotelling $$T^2$$.

Q. Is there a known distributional form of $$Y = \sqrt{n}\hat{\Sigma}^{-1/2}(\bar{\mathbf{X}} - \mu)$$?

When $$p = 1$$, we know that $$Y \sim t_{n-1}$$ distribution. However, for $$p > 1$$, from the description of the multivariate $$t$$ distribution here, $$Y$$ is not distributed like a multivariate $$t$$ distribution.

• First thought: the distribution of $(\bar{X}- \mu)$ is a multivariate distribution. Second thought: but the distribution of $\sqrt{n}\hat{\Sigma}^{-1/2}(\bar{X}- \mu)$ is not so easy because of $\hat \Sigma$. Third thought: why do you consider this distribution? – Sextus Empiricus Feb 25 '19 at 15:10
• @MartijnWeterings 1) Yes, $(\bar{X} - \mu)$ is a multivariate normal distribution with variance $\Sigma/n$. 2) I don't think it's easy either, but I am wondering if there is already a known result somewhere. 3) It is a natural question to ask once you try to generalize from a 1-dimensional $t$ distribution to a general framework. – Greenparker Feb 25 '19 at 15:15
• How do you (uniquely) define $\Sigma^{1/2}$ ? – Sextus Empiricus Feb 26 '19 at 16:10
• @MartijnWeterings Cholesky decomposition I suppose. – Greenparker Feb 26 '19 at 17:14
• @MartijnWeterings Ah, I am certain when Sepanski says nonnegative, they mean positive semidefinite – Greenparker Feb 28 '19 at 13:47