Proof: $T \sim \chi^2(N-1)$

Question

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,

$$T=\frac{\tilde{Z}\sqrt{N-1}}{\frac{\hat{\sigma}}{\sigma}\sqrt{N-1}}$$

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

Solution:

There's a clear proof of that here.

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. — Glen_b, Jul 13 '15 at 14:10
I think I have some issues in my handouts. Can you suggest me some good readings about this? Thanks — Phugo, Jul 13 '15 at 14:16
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 — Dilip Sarwate, Jul 13 '15 at 13:06
To get a $t_{n - 1}$ distribution, the definition of $\hat{\sigma}$ must be the earlier one. With $N$ as denominator, you can't get $t$ distribution. So what you can show is that $\left(\dfrac{\sqrt{N - 1}\hat{\sigma}}{\sigma}\right)^2 \sim \chi_{N - 1}^2$. — Zhanxiong, Jul 14 '15 at 22:22
I finally got it. There was a typo on my handouts which was driving me crazy. Thank you very very very much @Zhanxiong! — Phugo, Jul 14 '15 at 22:30
See proof provided in this thread: https://stats.stackexchange.com/questions/586626/how-to-prove-hatx-mu-hats-sqrtn-is-student-t-with-n-1-degrees — travelingbones, Aug 25 '22 at 18:34

score 0 · Accepted Answer · edited Aug 25 '22 at 03:16

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

$V=\sum_{i=1}^n(\frac{X_i-\mu}{\sigma})^2=\sum_{i=1}^n(\frac{(X_i-\bar{X})+(\bar{X}-\mu)}{\sigma})^2=\sum_{i=1}^{n}(\frac{X_i-\bar{X}}{\sigma})^2+2*\sum_{i=1}^n(X_i-\bar{X})(\bar{X}-\mu)/\sigma+(\frac{n(\bar{X}-\mu)}{\sigma})^2$

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.

Your question still has problems. You may post the original question, i think. — Deep North, Jul 13 '15 at 14:00
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: https://onlinecourses.science.psu.edu/stat414/node/174 This is a more rigorous proof following the scheme up here. — Phugo, Jul 14 '15 at 22:38
The proof writes =+, with following a 2() distribution and following a 2(1) distribution. How does one conclude that is indeed 2(-1)? Proof is incomplete. Once can write =-, but that looks like a 2(+1), except for dependence between , . — travelingbones, Aug 25 '22 at 00:53
see https://stats.stackexchange.com/questions/586626/how-to-prove-hatx-mu-hats-sqrtn-is-student-t-with-n-1-degrees which provides proof. — travelingbones, Aug 25 '22 at 18:35

Proof: $T \sim \chi^2(N-1)$

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