If two different linear combinations of two random variables are Gaussian, can we deduct both of them are Gaussian. If two different linear combinations of two random variables are Gaussian, can we deduct both of them are Gaussian. Mathematically, if we know that
$a_1X+b_1Y$ and $a_2X+b_2Y$ have Gaussian distributions and $a_1b_2 \neq a_2b_1$, can we prove that $X$ and $Y$ must be Gaussian? 
 A: Let $U$ be standard normal. Let $V=RU$, where $R$ is a Rademacher random variable that takes values $-1$ and $1$, each with probability $1/2$. Suppose $U$ and $R$ are independent. Let $X=\frac{U+V}{2}$ and $Y=\frac{U-V}{2}$. 
Then $X+Y$ and $X-Y$ are each normal. But $X$ is not normal, for it takes on value $0$ with probability $\frac{1}{2}$.
Remark: Suppose that $U=a_1X+b_1Y$ and $V=a_2X+B_2Y$ are independent normal, and  $a_1b_2-a_2b_1\ne 0$.  Then we  can almost  conclude that $X$ and $Y$ are normal. "Almost," because we have to declare constant random variables to be normal.  We can solve for $X$ and $Y$ in terms of $U$ and $V$, and use the fact that a linear combination of independent normals is normal. The condition of independence can be weakened, for example to $(U,V)$ being bivariate normal.
A: Yes. Write this in matrix form, so you have $\mathbf {AX} = \mathbf Z$ where $\mathbf Z$ is bivariate normal. Then $\mathbf X = \mathbf {A^{-1}Z}$. The RHS is a linear combination of Gaussians, therefore the LHS is Gaussian.
A: To be a lot simpler than the other answers, just use the fact that the sum of two Gaussian random variables is gaussian and the fact that the scalar multiple of a Gaussian random variable is also Gaussian. The first of these facts can be worked out by computing the convolution of two arbitrary Gaussian distributions. The second requires some thinking about how things scale.
That is, $a_2(a_1X+b_1Y)$ and $-a_1(a_2X+b_2Y)$ must both be Gaussian and so must their sum $(a_2b_1-a_1b_2)Y$. Here we use the fact that $a_2b_1\neq a_1b_2$ to divide out the front term to conclude that $Y$ must be Gaussian.
Then by using a similar procedure as above, we can work out that $X$ must be Gaussian too, and then we are done.
