I know that for two random variables X and Y, they are independent if they are jointly normal and uncorrelated. If the two random variables are the sample-mean $\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i$, and sample variance $s^2 = \frac{1}{n-1}\sum_{i=1}^{n} (x_i-\bar{x})^2$ of a normal sequence of iid normal random variables $x_i$, proving they are jointly normal can be done in a few different ways but I am trying to show that their linear combination is normal:
$\begin{aligned} a \bar{x}+b s^{2} &=\frac{a}{n} \sum_{i=1}^{n} x_{i}+\frac{b}{n-1} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} \\ &=\sum_{i=1}^{n}\left( \frac{a}{n} x_{i}+\frac{b}{n-1}\left(x_{i}-\bar{x}\right)^{2}\right) \\ &=\sum_{i=1}^{n}\left(\frac{a}{n} x_{i}+\frac{b}{n-1}\left(x_{i}^{2}-2x_{i}\bar{x}+\bar{x}^{2}\right)\right)\\ &=\sum_{i=1}^{n}\left(\frac{a}{n} x_{i}+\frac{b}{n-1}\left(x_i^2-2x_i\frac{1}{n}\sum_{i=1}^{n}x_i+\frac{1}{n^2}\sum_{i=1}^{n}x_i^2\right)\right) \end{aligned}$
But now that I have some chi-squared distributions, how can I simplify this and show it's normal?
