Linear combination of two independent normal random variables only normally distributed function? I have the intuition that it is, but I have no idea on how to show it.
More formally, let $X$, $Y$ be two independent random variables, with
$X \sim \mathcal{N}(\mu_1,\,\sigma_1^{2})$
$Y \sim \mathcal{N}(\mu_2,\,\sigma_2^{2})$
and in general $\mu_1 \neq \mu_2$, $\sigma_1 \neq \sigma_2$, and all of them are different from zero.
Let $Z$ be defined as $Z = f(X,Y)$, with $f:\,R^2 \rightarrow R$.
Is the linear combination (i.e. $f(X,Y) = aX + bY$, with $a,b \in R$) the only $f$ for which $Z \sim \mathcal{N}(\mu_3,\,\sigma_3^{2})$ ?
 A: The answer is Yes. 
Assume that $X$ and $Y$ are two independent and normal random variables that are not necessarily identically distributed. Then they are jointly normal (because they are independent), which is important here. 
Then the only function of $X$ and $Y$ leading to a normal random variable has to be linear; i.e has the form
$$ f(X,Y) : = aX + bY$$
for some constants $a$ and $b$.
This result is known as Cramér–Wold theorem (see for example: this lecture note). It is a consequence of the uniqueness of characteristic functions and the particular form of the characteristic function of normal distributions.
Another way to see this is to note that if $f (X,Y)$ is normal with positive variance,  you must be able to write it in terms of a standard normal random variable $N$ by whiting  (scaling and shift). But you can always do this also for the given random variables $X $ and $Y$.
For some counter examples when $X$ and $Y$ are not jointly normal, see the answer of this question: Are any linear combination of normal random variables, normally distributed? 
