I know that $(n-1)s^2/\sigma^2$ is Chi-squared with $n-1$ degrees freedom. I'm currently working on a question where the population mean $\mu$ is known, i.e. $s^2=n^{-1}\sum_{i=1}^n(X_i-\mu)^2$.
My textbook states: "Clearly, $\frac{\sum_{i=1}^n(X_i-\mu)^2}{\sigma^2}$ is a Chi-squared random variable with $n$ degrees of freedom". I know that in this problem, it holds that $\frac{\sum_{i=1}^n(X_i-\mu)^2}{\sigma^2}$=$\frac{ns^2}{\sigma^2}$.
To me, this makes it seem like $\frac{qs^2}{\sigma^2}$ is chi squared with $q$ degrees of freedom, for any integer $q$, but I'm relatively positive that this is untrue. Can someone help me understand this discrepancy?
So let's be more explicit here. Let me "abuse" the notation a bit and do the following:
$$s_n^2=n^{-1}\sum_{i=1}^n(X_i-\mu)^2$$
Now, divide/ multiply by the variance $\sigma^2$ as $$s_n^2=n^{-1}\sigma^2\sum_{i=1}^n\frac{(X_i-\mu)^2}{\sigma^2}$$ The part $\frac{(X_i-\mu)^2}{\sigma^2}$ is Chi-square with $n$ d.o.f. (degrees of freedom) this is correct. Let's denote this Chi-square as $\chi_n^2$ with mean $n$ and variance $2n$. We have $$s_n^2=n^{-1}\sigma^2\chi_n^2$$ where $n^{-1}\sigma^2$ is a constant. To characterize it more we have \begin{align} E s_n^2 &= n^{-1}\sigma^2E\chi_n^2 = n^{-1}\sigma^2 (n) = \sigma^2\\ \text{var } s_n^2 &= n^{-1}\sigma^2\text{var } \chi_n^2 = n^{-1}\sigma^2 (2n) = 2\sigma^2 \end{align}
so addressing your question $\frac{qs_n^2}{\sigma^2}$ is Chi-square with $n$ d.o.f. with mean and variances as \begin{align} E \frac{qs_n^2}{\sigma^2} &= \frac{q}{\sigma^2}Es_n^2 = q\\ \text{var } \frac{qs_n^2}{\sigma^2} &=\frac{q}{\sigma^2}\text{var } \chi_n^2 = \frac{q}{\sigma^2}(2\sigma^2) =2q \end{align} In case you meant $\frac{qs_q^2}{\sigma^2}$, nothing changes except the d.o.f, i.e. we will have $q$ instead of $n$ d.o.f. That is why I insisted on $s_n$ instead of just $s$.