Suppose that ${y_1}, ..., {y_n}$ is a random sample from an ${N(\mu,\sigma^2)}$ distribution. Then $$ {\sum_{i=1}^{n}{\frac {(y_{i}-\bar{y})^{2}}{\sigma^{2}}}}$$

has a $\chi^{2}_{n-1}$ distribution. Why is this the sum of ${\chi^{2}}$ distributions that sum to a chi-square distribution with ${n-1}$ degrees of freedom instead ${n}$ degree? Can anyone prove it for me?

  • $\begingroup$ There's some discussion here; Why is the sampling distribution of variance a chi-squared distribution?. One way to prove it is via Cochran's theorem ; the wikipedia page for it ((see the link in the answer) contains a proof of Cochran's theorem and has a section showing how the distribution of the variance follows as a consequence. Two other methods of proof are mentioned by Henry L in comments under the answer. $\endgroup$
    – Glen_b
    Commented Mar 5, 2017 at 1:05
  • 1
    $\begingroup$ I expect a proof (rather than a discussion and a link to one) has been given directly in an answer in CV before (if I locate one of them this should close as a duplicate). $\endgroup$
    – Glen_b
    Commented Mar 5, 2017 at 1:09

1 Answer 1


It is well-known that this is because the sample mean is used in place of $\mu$. Note that this is the sum of n chi-square random variables but they are dependent due to the use of the sample mean in place of $\mu$.


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