Method for determining distributions of sum of Normal distribution unknown mean and variance I've been trying to complete this question but have been struggling to see how to approach it. 

Any help would be greatly appreciated. 
Is there a standard way of approaching and answering questions like this? I've thought about Chi square and T- distributions but their definitions don't seem to be helping me. I know the first one follows a Chi (m+n-2) square distribution, but how do I solve the second one?
Thank you for your help!
 A: For finding the distribution of the first one
$$(m+n-2)\frac{S^2}{\sigma^2}$$
let
$S^2_1=\frac{1}{m-1}\sum_{i=1}^m(X_i-\overline X)^2$ and $S^2_2=\frac{1}{n-1}\sum_{j=1}^n(X_i-\overline X)^2$. Then
$$(m+n-2)\frac{S^2}{\sigma^2}=(m-1)\frac{S_1^2}{\sigma^2}\ +\ (n-1)\frac{S_2^2}{\sigma^2}$$
But as you correctly guessed, each summand of the last equation follows a $\chi_{m-1}^2$ and a $\chi_{n-1}^2$ respectively. And since $X_1\dots X_m$ and $Y_1\dots Y_n$ are independent random samples, we can add up those two distributions to get that
$$(m+n-2)\frac{S^2}{\sigma^2}(m+n-2)\sim \chi_{m+n-2}^2$$
For the last one, remember that if a sample of size $n$ distributes as a $N(\mu,\sigma^2)$, then $\overline X\sim N(\mu,\frac{\sigma^2}{\sqrt n})$. And that the difference of sample means $\overline X -\overline Y$, where $(X_1\dots X_m)$ and $(Y_1,\dots Y_n)$ are independent samples with distribution $N(\mu_1,\sigma^2)$ and $N(\mu_2,\sigma^2$) respectively, follows a $N(\mu_1-\mu_2,\sigma^2\sqrt{\frac{1}{m}+\frac{1}{n}})$. Using this and the previous distribution, you should be able to get the distribution of the last expression.
