Independence of a combination of iid random variables I encountered 2 questions regarding independence of 2 random variables.
1) A and B are i.i.d normal and independent, Let $X = 3A + 2B$ and $Y = 2A - 3B$, prove that they are independent
2) Will they still be independent if the normality condition is dropped?

1) I assume I should notice that linear combination of i.i.d normal variables forms a jointly normally distributed variables X and Y, which means, that if the covariance is 0, then the independence is proven. Covariance ends up being 0.
2) I'm really unsure about how to approach this one. I have a hunch that the answer is negative but I'm simply bad at coming up with some simple examples. 
I'd appreciate and insight into the approaches for such questions.
 A: Your answer to question 1 is correct. For question 2, try the following $A$ and $B$ are i.i.d., take the values $0$ and $1$ with probability $1/2$. Then the vector $(X,Y)$ takes the values $(0,0)$, $(3,2)$, $(2,-3)$ and $(5,-1)$ with probability $1/4$. Let $E_1:= \{X=0\}$ and $E_2:= \{Y=2\}$. These events are not independent because their intersection have probability $0$ while both of them have probability $1/4$.
A: ad 1)
Your approach to 1) is valid, although as you correctly state it relies on the fact that $(X,Y)$ has a bivariate normal distribution, i.e. $aX+bY$ is normally distributed for all $a,b \in \mathbb{R}$. Note that in general two random variables can both be normally distributed, uncorrelated, but not independent.
ad 2) 
It can be shown that $\text{cov}(X,Y) = \mathbb{E}[XY] -\mathbb{E}[X]\mathbb{E}[Y] = 0$, simply by plugging in the definition of $X$ and $Y$ and using the iid assumption of $A$ and $B$. Hence $X$ and $Y$ are uncorrelated. Hence we are looking for a specific counterexample to the statement that no correlation implies independence.
One simple counterexample is taking the Rademacher distribution, i.e. $$P(A = -1) = P(A=1) = 0.5$$
Now we use the fact that $X$ and $Y$ are linked through the realizations of $A$ and $B$. 
$$
P\left(
\begin{bmatrix}
X\\
Y
\end{bmatrix}
= \begin{bmatrix}
5\\
1
\end{bmatrix}
\right) = 0 < P(X = 5) P(Y = 1)=\frac{1}{16}
$$
This shows that $X$ and $Y$ are dependent.
