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If $X = (X_1,...,X_n)$ is a random vector from $N(\mu, \Sigma)$, where $$ \Sigma = \begin{bmatrix} 1 & \rho & \dots & \rho \\ \rho & 1 & \dots & \rho \\ \vdots & \vdots & \ddots & \vdots \\ \rho & \rho & \dots & 1 \end{bmatrix}, $$ we can get that $S^2$ is independent with $\bar{X}$, since $\Sigma = \sigma^2[(1-\rho)I + \rho11^T]$. If we set $$ A = \begin{bmatrix} 1-1/n & 1/n & \dots & 1/n \\ -1/n & 1-1/n & \dots & -1/n \\ \vdots & \vdots & \ddots & \vdots \\ -1/n & -1/n & \dots & 1-1/n \end{bmatrix}, $$ then we can get that $AX \sim N(A\mu, A\Sigma A^T)$ and then we can verify $S^2$ and $\bar{X}$ are independent because of the independence between $\bar{X}$ and $X_i - \bar{X}$. However, do we know that $(n-1)S^2/\sigma^2$ still follows a $\chi^2$ distribution with $df = n-1$? Since from the covariance\variance matrix of $AX$ we know the $X_i - \bar{X}$ and $X_j - \bar{X}$ are not independent, where $i$ is not equal to $j$.

If yes, how can we prove it? If no, which distribution does it follow?

Michael Hardy
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StatsLearner
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  • $(n-1)S^2$ still follows $\chi^2_{n-1}$ when suitably scaled: https://math.stackexchange.com/q/3915321/321264. But I don't see the point of using this $A$. – StubbornAtom Nov 23 '20 at 08:07
  • But how can we scale it? Since if $X_i$ is independent, then $\frac{(n-1)S^2}{\sigma^2}$ follows chi-square distribution with df=n-1, but now it seems like the denominator is not $\sigma^2$ and rather other scale factor? – StatsLearner Nov 23 '20 at 09:14
  • @StubbornAtom : This matrix $A$ is used for defining $S^2,$ thus: $$ AX = A\left[ \begin{array}{c} X_1 \ \vdots \ X_n \end{array} \right] = \left[ \begin{array}{c} X_1-\overline X \ \vdots \ X_n-\overline X \end{array}\right] \quad \text{ where } \overline X = \frac{X_1+\cdots+X_n} n, $$ and then $$ (n-1)S^2 = |AX|^2. $$ – Michael Hardy Nov 24 '20 at 02:31

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Recall that $A$ is a symmetric idempotent matrix, i.e. $A = A^T = A^2.$ And $A1= 0$ if $1$ denotes the column vector whose every entry is the scalar $1.$

You have $\Sigma = (1-\rho)I + \rho 1 1^T$ but also $\Sigma = \sigma^2[(1-\rho)I + \rho 1 1^T].$ I will assume the latter was intended.

Then you have $$ \operatorname{var}(AX) = \sigma^2 A[(1-\rho)I + \rho 11^T] A^T = \sigma^2 (1 - \rho ) A. $$ Recall next that $(n-1)S^2 = \|AX\|^2.$

And recall that a suitable rotation of $\mathbb R^n$ transforms $X$ to a vector whose expected value is $(\mu\sqrt n,0,0,\ldots,0)^T$ and transforms $\sigma^2(1-\rho)A$ to $$ \sigma^2(1-\rho)\left[ \begin{array}{ccc} 0 \\ & 1 \\ & & 1 \\ & & & \ddots \\ & & & & 1 \end{array} \right]. $$ Hence we have $$ \frac{(n-1)S^2}{\sigma^2(1-\rho)} \sim \chi^2_{n-1}. $$

Michael Hardy
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