6

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.

Greenparker
  • 14,131
  • 3
  • 36
  • 80
  • 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
  • But the Cholesky decomposition is not a square root, although you do indeed have $Y^tY=T^2$. But you can do that with other choices of $\Sigma^{1/2}$ as well. Why this choice? – Sextus Empiricus Feb 28 '19 at 11:01
  • @MartijnWeterings I suppose you could define it via an eigen value decomposition. I found a paper by [Sepanski(1996)](https://www.sciencedirect.com/science/article/pii/0167715295002170) that provides the asympotic normality of $Y$. But it doesn't seem to say anything about the exact distributional form. – Greenparker Feb 28 '19 at 11:08
  • Sepanski writes *"The quantity we are normalizing by in the univaritate $T_n$ is $\sqrt{n}S_n$. The multivariate analogue of this is now the matrix $C_n^{1/2}$, where $C_n = \sum_{i=1}^n (X_i-\bar{X}_n)(X_i-\bar{X}_n)^t$. That $C_n$ has a unique nonnegative symmetric square root, denoted by $C_n^{1/2}$, follows from the fact that $ = \sum_{i=1}^n ^2$, so that $C_n$ is nonnegative."* This I find strange since $C_n$, which looks like a covariance matrix to me, can be negative. But maybe it is meant 'non-negative definite'? – Sextus Empiricus Feb 28 '19 at 11:39
  • So do you mean by $\Sigma^{1/2}$ the unique cholesky decomposition, or the unique positive definite quare root? – Sextus Empiricus Feb 28 '19 at 11:40
  • I suppose I mean the unique positive definite square root as defined by Sepanski. $C_n$ cannot be negative, since it is positive semi-definite. It can be singular, which is part of the theoretical work that Sepanski does. – Greenparker Feb 28 '19 at 11:58
  • What do you mean by '$C_n$ cannot be negative'? The matrix $C_n$ is not necessarily a nonnegative matrix. The definition by Sepanski is not so clear. There can be multiple nonnegative symmetric matrices that are square roots of the covariance matrix. – Sextus Empiricus Feb 28 '19 at 13:17
  • @MartijnWeterings $C_n$ by positive semi-definite by the argument that Sepanski gives. Are you confusing $C_n$ with $C_n^{1/2}$? – Greenparker Feb 28 '19 at 13:46
  • I am confusing 'nonnegative matrix' for 'positive semi-definite matrix'. – Sextus Empiricus Feb 28 '19 at 13:47
  • 1
    @MartijnWeterings Ah, I am certain when Sepanski says nonnegative, they mean positive semidefinite – Greenparker Feb 28 '19 at 13:47

0 Answers0