Above is my question. $\overline X$ has distribution $N(0,1/n)$ - that's fine to work out. Similarly, $X_n / \sqrt{n}$ has distribution $N(0,1/n)$. These follow from the general relation $$ \sum_{m=1}^n a_m X_m ~\tilde ~~ N(\sum_m a_m \mu_m , \sum_m (a_m \sigma_m)^2)$$ (see Wikipedia for a derivation). However, I am stuck on determining the distribution of $\sum_{m=1}^n (X_m - \overline X)^2$ - or, rather, determining it in terms of $\sum_{m=1}^{n-1} X_m^2$.
Any help would be most appreciated.
Thanks, Sam.
A number of observations:
(i) As functions of independent random variables, $\dfrac{X_n}{\sqrt{n}}\,\,$ is independent of $\,\,\displaystyle \sum\limits_{m=1}^{n-1}{X_m^2}\,$.
(ii) $\,\,\dfrac{X_n}{\sqrt{n}}\sim\mathcal{N}\left(0, 1/n\right)\,\,$ and $\,\,\displaystyle \sum\limits_{m=1}^{n-1}{X_m^2}\sim\chi^2(n-1)\,$, since they are linear combinations of independent, standard normal and squared standard normal random variables, respectively.
By Cochran's Theorem,
(iii) $\,\,\bar{X}$ and $\displaystyle \sum\limits_{m=1}^{n}\left(X_m-\bar{X}\right)^2$ are independent;
(iv)$\,\,\,\displaystyle \sum\limits_{m=1}^{n}\left(X_m-\bar{X}\right)^2\sim\chi^2\left(n-1\right)$;
Finally,
(v) $\,\,\bar{X}\sim\mathcal{N}\left(0, 1/n\right)\,\,$, since it is a linear combination of independent standard normal random variables.
Together, (i) to (v) imply the result.