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

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.

• You can't. It isn't. Jul 13, 2015 at 12:39
• I think you'll find a lengthy explanation/proof here
– user82170
Jul 13, 2015 at 12:39
• 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 Jul 13, 2015 at 13:06
• 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. Jul 13, 2015 at 14:10
• I think I have some issues in my handouts. Can you suggest me some good readings about this? Thanks Jul 13, 2015 at 14:16

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.

• Yeah. I just edited the question Jul 13, 2015 at 13:57
• Your question still has problems. You may post the original question, i think. Jul 13, 2015 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: onlinecourses.science.psu.edu/stat414/node/174 This is a more rigorous proof following the scheme up here. Jul 14, 2015 at 22:38
• The proof writes 𝑉=𝑊+𝑈, with 𝑉 following a 𝜒2(𝑛) distribution and 𝑈 following a 𝜒2(1) distribution. How does one conclude that 𝑊 is indeed 𝜒2(𝑛-1)? Proof is incomplete. Once can write 𝑊=𝑉-𝑈, but that looks like a 𝜒2(𝑛+1), except for dependence between 𝑉 , 𝑈. Aug 25, 2022 at 0:53
• see stats.stackexchange.com/questions/586626/… which provides proof. Aug 25, 2022 at 18:35