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$\overline{X}=\frac{1}{n}\sum_{i=1}^n X_i$ is the sample mean and $S^2=\frac{1}{n-1}\sum_{i=1}^n (X_i-\overline{X})^2$ is the sample variance. $X_1,...,X_n\sim N(\mu, \sigma^2)$ and independent.

I want to show that $\overline{X}$ and $S^2$ are independent.

I've already proven that $\overline{X}-X_i\sim N(0,\frac{n-1}{n}\sigma^2)$ and that $Cov(X_j-\overline{X},\overline{X}) = 0$.

But then I didn't know how to proceed. While searching for solutions, I found this document. Here it claims that I need to calculate the covariance matrix and then use the "multivariate normality" (I don't know what exactly they mean by that) so that $\overline{X}$ and $X=(X_1-\overline{X},...,X_n-\overline{X})^T$ are independent.

The rest is clear to me. I just want to understand the details to why $\overline{X}$ and $X$ are independent.

Quotenbanane
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  • $(X_j-\overline X,\overline X)$ is bivariate normal and whose components are uncorrelated for every $j$. Hence $\overline X$ and $X_j-\overline X$ are independent for every $j$. – StubbornAtom Jun 07 '21 at 11:54

1 Answers1

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Another way to prove the statemente is to use Basu's Theorem

First observe that

$$\overline{X}_n\sim N\left(\mu;\frac{\sigma^2}{n}\right)$$

$$(n-1)\frac{S_n^2}{\sigma^2}\sim \chi_{(n-1)}^2$$

As known, $\overline{X}_n$ is a complete and sufficient statistic (CSS) for $\mu$ as it is the Exponential family's Canonical Statistic, while $S_n^2$ is ancillary for $\mu$ as its distribution is free of $\mu$

Thus we can apply Basu's theorem concluding that $\overline{X}_n$ and $S_n^2$ are independent.

This proof is known also as Cochran's theorem

tommik
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