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.

  • $\begingroup$ 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? $\endgroup$ – Sextus Empiricus Feb 25 '19 at 15:10
  • $\begingroup$ @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. $\endgroup$ – Greenparker Feb 25 '19 at 15:15
  • $\begingroup$ How do you (uniquely) define $\Sigma^{1/2}$ ? $\endgroup$ – Sextus Empiricus Feb 26 '19 at 16:10
  • $\begingroup$ @MartijnWeterings Cholesky decomposition I suppose. $\endgroup$ – Greenparker Feb 26 '19 at 17:14
  • 1
    $\begingroup$ @MartijnWeterings Ah, I am certain when Sepanski says nonnegative, they mean positive semidefinite $\endgroup$ – Greenparker Feb 28 '19 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.