1

A related post here. In which the author claimed, if $X_{i}\sim N(\mu,\sigma^2),i=1,\cdots,n$ are iid then

$$\frac{(n-1)s^2}{\sigma^2}\sim \chi^2(n-1)$$

in which $s^2$ is the unbiased sample variance. i.e. $s^2=\frac1{n-1}\sum(X_i-\bar X)^2$ where $\bar X=\frac1n\sum X_i$.

I want a proof of it. But I can't do one on my own. In fact, contrary to my initial suspicion that $X_i-X$ are uncorrelated, I actually got \begin{align} \mathsf{Cov}\,(X_i-\bar X,X_j-\bar X) &=\mathsf{Cov}\,(X_i,X_j)+\mathsf{Var}\,(\bar X,\bar X)-2\mathsf{Cov}\,(X_i,\bar X)\\ &=0+\frac1{n^2}n\sigma^2-2\frac1n\sigma^2\\ &=-\frac1n\sigma^2\ne 0 \end{align}

So where does the Chi square come from?

Vim
  • 227
  • 1
  • 9
  • By taking suitable linear combinations of the $X_i$, you can rewrite $s^2$ as the sum of squares of $n-1$ uncorrelated variables. Pick any orthogonal basis of $\mathbb{R}^n$ that includes the direction determined by the coefficients of $\bar X$; namely, $(1,1,\ldots,1)/n$. This is shown in many, many posts on this site: consider searching for questions related to variance. – whuber Jun 02 '17 at 15:06
  • @whuber this rings a bell. Yeah I recalled it can be done using an appropriate orthogonal transformation. I'll search around. Thanks! Please vote to close this question as dupe. – Vim Jun 02 '17 at 15:08
  • I would be grateful if, when you find a suitable answer, you could tell us which one it is. – whuber Jun 02 '17 at 15:09
  • @whuber sure. for example: https://stats.stackexchange.com/a/121676/109945 – Vim Jun 02 '17 at 15:11
  • 1
    Thank you! Even though that leads to closing your question, I am very happy to upvote it just for your effort to find the duplicate. – whuber Jun 02 '17 at 17:43

0 Answers0