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Proof of $\frac{(n-1)S^2}{\sigma^2} \backsim \chi^2_{n-1}$

If $X_1,...,X_n$ is a random sample from a normal distribution and $S^2=\sum_{i=1}^n\frac{(X_i-\bar{X})^2}{n-1}$

Could anyone help me show that $\frac{(n-1)S^2}{\sigma^2}\sim \chi_{n-1}^2$ and that it's independent to $\bar{X}$

Many thanks!


marked as duplicate by cardinal, Freeman, user940, Mike Spivey, Sasha Mar 14 '12 at 17:04

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  • $\begingroup$ Also related: math.stackexchange.com/questions/68555 $\endgroup$ – cardinal Mar 14 '12 at 16:41
  • $\begingroup$ A reference: Section 7.7.2 in Sheldon Ross' A First Course in Probability. $\endgroup$ – David Mitra Mar 14 '12 at 16:43
  • $\begingroup$ A second reference: Any textbook on the statistical theory of linear regression. $\endgroup$ – cardinal Mar 14 '12 at 16:44
  • $\begingroup$ Thank you for that reference. Ah ok.. it's a duplicate, sorry about that $\endgroup$ – Freeman Mar 14 '12 at 16:45

$X=(X_1,\dots, X_n)^\prime$ has a multivariate normal distribution with $\mu_X=\mu {\bf 1}$ and $\Sigma_X=\sigma^2 I$. Here ${\bf 1}$ is the column vector of all $1$s, while $I$ is the $n\times n$ identity matrix.

Let $A$ be the matrix of an orthogonal transformation that takes the vector $\bf 1$ into the vector $\sqrt{n}\, {\bf e}_1$. Then the vector $U=AX$ is multivariate normal with $\mu_U=\mu \sqrt{n}\, {\bf e}_1 $ and $\Sigma_U=\sigma^2 I$.

The first coordinate of the random vector $U$ is $$U_1=(AX)^\prime{\bf e}_1=X^\prime A^\prime {\bf e}_1= {1\over \sqrt{n}}\, X^\prime A^\prime A{\bf 1} = {1\over \sqrt{n}}\, X^\prime {\bf 1}=\sqrt{n}\,\bar X.$$

Also, $$\sum_{i=1}^n X^2_i=X^\prime X= X^\prime A^\prime A X=U^\prime U =n\bar X^2+\sum_{i=2}^n U_i^2,$$ and hence $${(n-1)S^2 \over \sigma^2} = \sum_{i=2}^n (U_i/\sigma)^2\sim \chi^2_{n-1}.$$

  • $\begingroup$ Thank you, I assume the end of the last line is a typo? $\endgroup$ – Freeman Mar 14 '12 at 16:44
  • $\begingroup$ @LHS Well, it is my notation for a chi-squared distribution with $n-1$ degrees of freedom. I will change it to match your notation. $\endgroup$ – user940 Mar 14 '12 at 16:46

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