I am wondering how to prove that if $X,Y$ are normal with mean $0$ and variance $\sigma_X^2, \sigma_Y^2$ respectively, and I know $\mathbb{E}[XY]$, how do I prove that $X - Y$ is also normally distributed using characteristic functions.
I know that $$\Phi_{X-Y}(\theta) = \mathbb{E}[e^{i\theta(X-Y)}] = \mathbb{E}[e^{i\theta X} e^{-i \theta Y}]$$ but since $X,Y$ are not independent I can't split this into the product of the characteristic function of $X$ and of $Y$. Thank you
You won't be able to prove it because if you don't assume anything on the joint distribution of $(X,Y)$ (e.g. independence) then $X-Y$ may not be gaussian.
The well-known example is $Y=\varepsilon X$ for a random variable $\varepsilon$ independent of $X$, equal to $-1$ or $1$ both with probability $1/2$. Then $X-Y=(1-\varepsilon)X$ is equal to $0$ with probability $1/2$ so it cannot be gaussian.