What is the distribution of this random variable

Question

From an exam question regarding hypothesis testing and the t-distribution

For $X_1,\cdots,X_n\sim\mathcal{N}(\mu,\sigma^2)$ what is the distribution of

$ \sum\limits_{i=1}^{n}\frac{(X_i-\bar{X})^2}{\sigma} $

I'm struggling to solve this one, if the denominator were $\sigma^2$ then the under a hypothesis $H_0:\mu=0$ I believe that the variable would be distributed $\chi^2(n-1)$ however as it is I am at a loss, am I missing something blatantly obvious?

Answer 1

I'd do $$ \sum\limits_{i=1}^{n}\frac{(X_i-\bar{X})^2}{\sigma} = \sigma \sum\limits_{i=1}^{n}\frac{(X_i-\bar{X})^2}{\sigma^2}. $$

Working out the distribution of $\sum_{i=1}^{n}\frac{(X_i-\bar{X})^2}{\sigma^2}$:

$$ \sum_i \left(\frac{X_i-\bar{X}}{\sigma}\right)^2 = \sum_i \left(\frac{X_i - \mu}{\sigma} - \frac{\bar X - \mu}{\sigma}\right)^2 $$ $$ = \sum_i (Z_i - \bar Z)^2 $$ where the $Z_i$ are iid $\mathcal N(0,1)$. $$ \sum_i (Z_i - \bar Z)^2 = \sum_i Z_i^2 - n\bar Z^2 \sim \chi_{n-1}^2 $$ by Cochran's theorem.

This means $$ \sum\limits_{i=1}^{n}\frac{(X_i-\bar{X})^2}{\sigma} \sim \sigma \cdot \chi^2_{n-1}. $$ and I think that's the best we can say. There isn't an $n-1$ term because this is the sum of squares, not the variance estimate.

In a Gaussian sample $\sum_i (X_i - \bar X)^2$ is ancillary to $\mu$ so that's why this result doesn't actually depend on your hypothesis about $\mu$.