Let $X_1, X_2, ..., X_n$ be normally distributed with mean $\mu$ and variance $\sigma ^2$, then $\frac{(n-1) S^2}{ \sigma^2}$ has a Chi-Square distribution with $n-1$ degrees of freedom.

How come it's chi-square distributed?


$S^2 = \frac{1}{n-1} \sum(X_i-\overline X)^2 = \frac{1}{n-1}(\sum X_i ^2 - (\sum X_i)^2)$.

Here, the first term $\sum X_i^2$ is chi-squared. But you also need to subtract the second term (which is normal distributed with mean $\mu$, and variance $\frac{\sigma^2}{n}$. How does it make it chi-square distributed?


The first proof people generally encounter involves MGFs. Note,

$$ \frac{(n-1)S^2}{\sigma^2} = \sum\limits_i^n\left(\frac{X_i-\bar{X}}{\sigma}\right)^2 = \sum\limits_i^n\left(\frac{X_i-\mu}{\sigma}\right)^2 - \left(\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\right)^2$$

Observe that the first term on the RHS is a chi-square RV with $n$ degrees of freedom. The second term is a chi-square with $1$ degree of freedom. Rearranging,

$$ \frac{(n-1)S^2}{\sigma^2} + \left(\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\right)^2 = \sum\limits_i^n\left(\frac{X_i-\mu}{\sigma}\right)^2 $$

Now we can find the MGFs both sides,

$$ (1-2t)^{-1/2}M(t) = (1-2t)^{-n/2} \Longrightarrow M(t) = (1-2t)^{-n/2+1/2} = (1-2t)^{-1/2(n-1)}$$

where $M(t)$ is the MGF of $\frac{(n-1)S^2}{\sigma^2}$. Thus, the MGF of $\frac{(n-1)S^2}{\sigma^2}$ is that of a chi-square RV with $n-1$ degrees of freedom.

Note: The above used the identity that $M_{X+Y}(t) = M_X(t)M_Y(t)$ where $X$ and $Y$ are independent. This detail was skipped, but one also needs to show that $S^2$ and $\bar{X}$ are independent which is true in the case of a Normally distributed sample.

| cite | improve this answer | |
  • $\begingroup$ thanks for your answer. May I ask how did you get from here to here? $$\sum\limits_i^n\left(\frac{X_i-\bar{X}}{\sigma}\right)^2 = \sum\limits_i^n\left(\frac{X_i-\mu}{\sigma}\right)^2 - \left(\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\right)^2$$ $\endgroup$ – kou Aug 6 '17 at 23:54
  • $\begingroup$ Ignoring the $\sigma^2$, note that $(X_i - \bar{X})^2 = (X_i - \mu + \mu - \bar{X})^2 = (X_i - \mu)^2 - 2(X_i - \mu)(\mu - \bar{X}) + (\bar{X}-\mu)^2$. Taking sums results in $\sum\limits_i^n(X_i - \bar{X})^2 = \sum\limits_i^n(X_i - \mu)^2 - 2n(\bar{X} -\mu)^2 + n(\bar{X}-\mu)^2$. $\endgroup$ – Flowsnake Aug 7 '17 at 0:09
  • $\begingroup$ $$ \frac{(n-1)S^2}{\sigma^2} + \left(\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\right)^2 $$ Your argument is at best incomplete if it does not mention that these two terms are independent. And their independence is not instantly obvious; that should be explained as well. A substantial step in showing that is to show that $\operatorname{cov}\big( \,\overline X, X_i-\overline X\,\big) = 0. \qquad$ $\endgroup$ – Michael Hardy Nov 29 '19 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.