$X+Y \sim N(5,5)$ two random variables normal bi-variate The exercise is the following. Let $(X,Y)$ be a bi-variate normal vector with $\mu=(2,3)$ and covariance matrix $$\Sigma=\begin{pmatrix}1 & -2 \\ -2 & 4\end{pmatrix}.$$ What is the distribution of the random variable $X+Y$?
My try is: Consider the moment generating functions of $X$ and $Y$. Since $\sigma_X^2=1$, $\sigma_Y^2=4$, $\mu_X=2$ and $\mu_Y=3$, we have
$$\varphi_X(t)=\exp(2t+\frac{t^2}{2})$$
$$\varphi_Y(t)=\exp(3t+2t^2).$$
Thus, $$\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)=exp(5t+\frac{5t^2}{2}).$$
Hence, $X+Y\sim N(5,5)$ since the moment generating function of a normal r.v. is given by $\varphi_X(t)=\exp(\mu_X t + \frac{\sigma_X^2 t^2}{2})$.
I am asking this because my teacher put as a solution $X+Y\sim N(5,1)$. Any feedback?
EDIT:
I think that we can not use $\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)$ since $X$, $Y$ are not independent.

The variance of $X+Y$ is
$$
\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y)=1+4+2\times(-2)=1.
$$
So $N(5,1)$ is correct. Your problem is the assertion $\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)$ which doesn't work in this case because $X$ and $Y$ are dependent.
