I'm a maths student, this is my first statics class and I'm studying Confidence Intervals. When it's time to estimate the CI for the mean of $\{X_i\}_{i=1..N}$ - gaussian random variables i.i.d. with unknown standard deviation - we use the estimator $\hat{\sigma}$ to compute the quantity

$$T:=\frac{\overline{X}_N - \mu}{\frac{\hat{\sigma}}{\sqrt{N}}}$$

where $\overline{X}_N$ is the sample mean, $\mu$ the mean and

$$\hat{\sigma}:=\sqrt{\frac{1}{N-1}\sum^N (X_i-\overline{X}_N)^2}$$

I would prove that $T \sim t(N-1)$, so:

$$T=\frac{\sigma}{\sigma} \frac{\overline{X}_N - \mu}{\frac{\hat{\sigma}}{\sqrt{N}}}=\frac{\frac{(\overline{X}_N - \mu)\sqrt{N}}{\sigma}}{\frac{\hat{\sigma}}{\sigma}}$$

Now the numerator is a standard gaussian and I'll call it $\tilde{Z}$. Hence,


In order to have a Student t-distribution I have to prove $(\frac{\hat{\sigma}}{\sigma}\sqrt{N-1})^2\sim\chi^2(N-1)$.


There's a clear proof of that here.

  • 5
    $\begingroup$ You can't. It isn't. $\endgroup$
    – Glen_b
    Jul 13, 2015 at 12:39
  • $\begingroup$ I think you'll find a lengthy explanation/proof here $\endgroup$
    – user82170
    Jul 13, 2015 at 12:39
  • $\begingroup$ You might want to read the question more closely and consider whether the "lengthy explanation/proof" that you link to really does answer the question asked. Moderator Glen_b's comment on the question is right on target $\endgroup$ Jul 13, 2015 at 13:06
  • $\begingroup$ Still can't, still isn't. Under certain conditions $(n-1)s^2/\sigma^2$ is $\chi^2$ -- and this is covered by a number of answers on site. $\endgroup$
    – Glen_b
    Jul 13, 2015 at 14:10
  • $\begingroup$ I think I have some issues in my handouts. Can you suggest me some good readings about this? Thanks $\endgroup$
    – Phugo
    Jul 13, 2015 at 14:16

1 Answer 1


Let $V=\sum_{i=1}^n(\frac{X_i-\mu}{\sigma})^2$, you know for each $\frac{X_i-\mu}{\sigma} $it has a $N(0,1)$ distribution because the $X_i$ are $N(\mu, \sigma)$, and you should know that square of a normal distribution has a $\chi^2(1)$ distribution.

Therefore, $V=\chi^2(1)+...+\chi^2(1)=\chi^2(n)$


Let $W:= \sum_{i=1}^{n}(\frac{X_i-\bar{X}}{\sigma})^2$, which is our quantity of interest.

The term $2*\sum_{i=1}^n(X_i-\bar{X})(\bar{X}-\mu)/\sigma$ is equal to 0.

Let $U:= (\frac{n(\bar{X}-\mu)}{\sigma})^2$. $U$ is is another $\chi^2(1)$, using that the square of a $N(0,1)$ is $\chi^2(1)$ and that $\frac{n(\bar{X}-\mu)}{\sigma}$ has a $N(0,1)$ distribution because $X_i$ are $N(\mu, \sigma)$.

Now you have $V=W+U$, with $V$ following a $\chi^2(n)$ distribution and $U$ following a $\chi^2(1)$ distribution.

Therefore, $\sum_{i=1}^{n}(\frac{X_i-\bar{X}}{\sigma})^2$ has a $\chi^2(n-1)$ distribution.

So $W = \frac{\sum_{i=1}^{n}(X_i-\bar{X})^2}{\sigma^2}$ has a $\chi^2(n-1)$

Usually people define sample variance as $s^2=\frac{\sum_{i=1}^{n}(X_i-\bar{X})^2}{n-1}$ which is an unbiased estimator of $\sigma^2$

Therefore, it should be $\frac{(n-1)s^2}{\sigma^2}$ has a $\chi^2(n-1)$ distribution.

Therefore your question has a problem there, you should be very careful.

  • $\begingroup$ Yeah. I just edited the question $\endgroup$
    – Phugo
    Jul 13, 2015 at 13:57
  • $\begingroup$ Your question still has problems. You may post the original question, i think. $\endgroup$
    – Deep North
    Jul 13, 2015 at 14:00
  • $\begingroup$ With the help of Zhanxiong I understood what was the right question and this is the answer for that. Thank you @Deep North! I would add a reference for future readers: onlinecourses.science.psu.edu/stat414/node/174 This is a more rigorous proof following the scheme up here. $\endgroup$
    – Phugo
    Jul 14, 2015 at 22:38
  • $\begingroup$ The proof writes 𝑉=π‘Š+π‘ˆ, with 𝑉 following a πœ’2(𝑛) distribution and π‘ˆ following a πœ’2(1) distribution. How does one conclude that π‘Š is indeed πœ’2(𝑛-1)? Proof is incomplete. Once can write π‘Š=𝑉-π‘ˆ, but that looks like a πœ’2(𝑛+1), except for dependence between 𝑉 , π‘ˆ. $\endgroup$ Aug 25, 2022 at 0:53
  • $\begingroup$ see stats.stackexchange.com/questions/586626/… which provides proof. $\endgroup$ Aug 25, 2022 at 18:35

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