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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?

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  • $\begingroup$ 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 Mar 5 '17 at 1:05
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    $\begingroup$ 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 Mar 5 '17 at 1:09
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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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